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AMUSEMENTS IN MATHEMATICS 



AMUSEMENTS IN 
MATHEMATICS 



BY 

HENRY ERNEST DUDENEY '~^ 

AUTHOR OF "the CANTERBURY PUZZLES: AND 
OTHER CURIOUS PROBLEMS," ETC. 



In Mathematicks he was greater 
Than Tycho Brahe or Erra Pater : 
For he, by geometrick scale, 
Could take the size of pots of ale ; 
Resolve, by sines and tangents, straight, 
If bread or butter wanted weight ; 
And wisely tell what hour o' th' day 
The clock does strike, by algebra. 

Butler's Hudibras, 



BOBTOH roT,i.i?OE LIBBABIT 
npv^v ^ .ILL, MASt^. 



THOMAS NELSON AND SONS, LTD. 
LONDON, EDINBURGH, AND NEW YORK 



150395 



PREFACE. 

In issuing this new volume of my Mathematical Puzzles, of which some have 
appeared in the periodical press and others are given here for the first time, 
I must acknowledge the encouragement that I have received from many 
unknown correspondents, at home and abroad, who have expressed a desire 
to have the problems in a collected form, with some of the solutions given 
at greater length than is possible in magazines and newspapers. Though 
I have included a few old puzzles that have interested the world for 
generations, where I felt that there was something new to be said about 
them, the problems are in the main original. It is true that some of these 
have become widely known through the press, and it is possible that the 
reader may be glad to know their source. 

On the question of Mathematical Puzzles in general there is, perhaps, 
little more to be said than I have written elsewhere. The history of the 
subject entails nothing short of the actual story of the beginnings and 
development of exact thinking in man. The historian must start from the 
time when man first succeeded in counting his ten fingers and in dividing 
an apple into two approximately equal parts. Every puzzle that is worthy 
of consideration can be referred to mathematics and logic. Every man, 
woman, and child who tries to " reason out " the answer to the simplest 
puzzle is working, though not of necessity consciously, on mathematical 
lines. Even those puzzles that we have no way of attacking except by 
haphazard attempts can be brought under a method of what has been 
called " glorified trial" — a system of shortening our labours by avoiding or 
eliminating what our reason tells us is useless. It is, in fact, not easy to 
say sometimes where the " empirical " begins and where it ends. 

When a man says, ** I have never solved a puzzle in my life," it is difficult 
to know exactly what he means, for every intelligent individual is doing it 
every day. The unfortunate inmates of our lunatic asylums are sent there 
expressly because they cannot solve puzzles — ^because they have lost their 
powers of reason. If there were no puzzles to solve, there would be no 
questions to ask ; and if there were no questions to be asked, what a world 
it would be ! We should all be equally omniscient, and conversation would 
be useless and idle. 

It is possible that some few exceedingly sober-minded mathematicians, 
who are impatient of any terminology in their favourite science but the 
academic, and who object to the elusive % and y appearing under any other 
names, will have wished that various problems had been presented in a 
less popular dress and introduced with a less flippant phraseology. I can 



vi PREFACE. 

only refer them to the first word of my title and remind them that we are 
primarily out to be amused — not, it is true, without some hope of picking 
up morsels of knowledge by the way. If the manner is light, I can only say, 
in the words of Touchstone, that it is " an ill-favoured thing, sir, but my 
own ; a poor humour of mine, sir." 

As for the question of difficulty, some of the puzzles, especially in 
the Arithmetical and Algebraical category, are quite easy. Yet some of 
those examples that look the simplest should not be passed over without a 
little consideration, for now and again it will be found that there is some 
more or less subtle pitfall or trap into which the reader may be apt to fall. 
It is good exercise to cultivate the habit of being very wary over the exact 
wording of a puzzle. It teaches exactitude and caution. But some of the 
problems are very hard nuts indeed, and not unworthy of the attention of 
the advanced mathematician. Readers will doubtless select according to 
their individual tastes. 

In many cases only the mere answers are given. This leaves the be- 
ginner something to do on his own behalf in working out the method of 
solution, and saves space that would be wasted from the point of view of 
the advanced student. On the other hand, in particular cases where it 
seemed likely to interest, I have given rather extensive solutions and treated 
problems in a general manner. It will often be found that the notes on one 
problem will serve to elucidate a good many others in the book ; so that 
the reader's difiiculties will sometimes be found cleared up as he advances. 
Where it is possible to say a thing in a manner that may be " understanded 
of the people " generally, I prefer to use this simple phraseology, and so 
engage the attention and interest of a larger public. The mathematician will 
in such cases have no difiiculty in expressing the matter under consideration 
in terms of his familiar symbols. 

I have taken the greatest care in reading the proofs, and trust that any 
errors that may have crept in are very few. If any such should occur, I 
can only plead, in the words of Horace, that " good Homer sometimes nods,'* 
or, as the bishop put it, " Not even the youngest curate in my diocese is 
infallible." 

I have to express my thanks in particular to the proprietors of The Strand 
Magazine, Cassell's Magazine, The Queen, Tit-Bits, and The Weekly Dispatch 
for their courtesy in allowing me to reprint some of the puzzles that have 
appeared in their pages. 

The Authors* Club, 
March 25, 1917- 



CONTENTS. 



Preface 

Arithmetical and Algebraical Problems — 

Money Puzzles 

Age and Kinship Puzzles 

Clock Puzzles 

Locomotion and Speed Puzzles 

Digital Puzzles . 

Various Arithmetical and Algebraical Problems 
Geometrical Problems — 

Dissection Puzzles 

Greek Cross Puzzles . 

Various Dissection Puzzles . 

Patchwork Puzzles 

Various Geometrical Puzzles 
Points and Lines Problems 
Moving Counter Problems . 
Unicursal and Route Problems 
Combination and Group Problems 
Chessboard Problems — 

The Chessboard . 

Statical Chess Puzzles 

The Guarded Chessboard 

Dynamical Chess Puzzles 

Various Chess Puzzles . 
Measuring, Weighing, and Packing Puzzles 
Crossing River Problems . 
Problems concerning Games 
Puzzle Games .... 
Magic Square Problems 

Subtracting, Multiplying, and Dividing Magics 

Magic Squares of Primes 
Mazes and how to thread Them 
The Paradox Party . 
Unclassified Problem? 

Solutions 

Index . , . » ^ • 



vii 



AMUSEMENTS IN MATHEMATICS. 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



" And what was he ? 
Forsooth, a great arithmetician." 

Oihello, I. i. 

The puzzles in this department are roughly thrown together in classes for the 
convenience of the reader. Some are very easy, others quite difficult. But 
they are not arranged in any order of difficulty — and this is intentional, for 
it is well that the solver should not be warned that a puzzle is just what 
it seems to be. It may, therefore, prove to be quite as simple as it looks, 
or it may contain some pitfall into which, through want of care or over- 
confidence, we may stumble. 

Also, the arithmetical and algebraical puzzles are not separated in the 
manner adopted by some authors, who arbitrarily require certain problems 
to be solved by one method or the other. The reader is left to make his own 
choice and determine which puzzles are capable of being solved by him on 
purely arithmetical lines. 



MONEY PUZZLES. 

" Put not your trust in money, but put your 
money in trust." 

Oliver Wendell Holmes. 



I.— A POST-OFFICE PERPLEXITY. 

In every business of life we are occasion- 
ally perplexed by some chance question that 
for the moment staggers us. I quite pitied a 
young lady in a branch post-office when a gen- 
tleman entered and deposited a crown on the 
counter with this request : " Please give me 
some twopenny stamps, six times as many 
penny stamps, and make up the rest of the 
money in twopence-halfpenny stamps." For 
a moment she seemed bewildered, then her 
brain cleared, and with a smile she handed over 
stamps in exact fulfilment of the order. How 
long would it have taken you to think it out ? 

2.— YOUTHFUL PRECOCITY. 

The precocity of some youths is surprising. 
One is disposed to say on occasion, " That boy 
of yours is a genius, and he is certain to do 
great things when he grows up ; " but past ex- 
perience has taught us that he invariably be- 
comes quite an ordinary citizen. It is so often 
the case, on the contrary, that the dull boy 
(1,926) 



becomes a great man. You never can tell. 
Nature loves to present to us these queer para- 
doxes. It is well known that those wonderful 
" lightning calculators," who now and again 
surprise the world by their feats, lose all their 
mysterious powers directly they are taught the 
elementary rules of arithmetic. 

A boy who was demolishing a choice banana 
was approached by a young friend, who, regard- 
ing him with envious eyes, asked, "How much did 
you pay for that banana, Fred ? " The prompt 
answer was quite remarkable in its way : " The 
man what I bought it of receives just half as 
many sixpences for sixteen dozen dozen bana- 
nas as he gives bananas for a fiver." 

Now, how long will it take the reader to say 
correctly just how much Fred paid for his rare 
and refreshing fruit ? 

3.— AT A CATTLE MARKET. 

Three countrjnnen met at a cattle market. 
'- Look here," said Hodge to Jakes, " I'U give 
you six of my pigs for one of yovu: horses, and 
then you'U have twice as many animals here 
as I've got." " If that's your way of doing 
business," said Durrant to Hodge, " I'U give 
you fourteen of my sheep for a horse, and then 
you'll have three times as many animals as I." 
" Well, I'll go better than that," said Jakes to 
Durrant; "I'll give you four cows for a horse, 



AMUSEMENTS IN MATHEMATICS. 



and then you'll have six times as many animals 
as I've got here." 

No doubt this was a very primitive way of 
bartering animals, but it is an interesting little 
puzzle to discover just how many animals Jakes, 
Hodge, and Durrant must have taken to the 
cattle market. 

4.— THE BEANFEAST PUZZLE. 

A NUMBER of men went out together on a bean- 
feast. There were four parties invited — namely, 
25 cobblers, 20 tailors, 18 hatters, and 12 glovers. 
They spent altogether £6, 13s. It was found 
that five cobblers spent as much as four tailors ; 
that twelve tailors spent as much as nine hatters ; 
and that six hatters spent as much as eight 
glovers. The puzzle is to find out how much 
each of the four parties spent. 

5.— A QUEER COINCIDENCE. 

Seven men, whose names were Adams, Baker, 
Carter, Dobson, Edwards, Francis, and Gud- 
geon, were recently engaged in play. The 
name of the particular game is of no conse- 
quence. They had agreed that whenever a 
player won a game he should double the money 
of each of the other players — that is, he was to 
give the players just as much money as they 
had already in their pockets. They played 
seven games, and, strange to say, each won a 
game in turn, in the order in which their names 
are given. But a more curious coincidence is 
this — that when they had finished play each 
of the seven men had exactly the same amount 
— two shillings and eightpence — in his pocket. 
The puzzle is to find out how much money each 
man had with him before he sat down to play. 

6.— A CHARITABLE BEQUEST. 

A MAN left instructions to his executors to dis- 
tribute once a year exactly fifty-five shillings 
among the poor of his parish; but they were 
only to continue the gift so long as they could 
make it in different ways, always giving eigh- 
teenpence each to a number of women and half 
a crown each to men. During how many years 
could the charity be administered ? Of course, 
by " different ways " is meant a different num- 
ber of men and women every time. 

7.— THE WIDOW'S LEGACY. 

A GENTLEMAN who recently died left the sum of 
£8,000 to be divided among his widow, five sons, 
and four daughters. He directed that every 
son should receive three times as much as a 
daughter, and that every daughter should have 
twice as much as their mother. What was the 
widow's share ? 

8.— INDISCRIMINATE CHARITY. 

A CHARITABLE gentleman, on his way home one 
night, was appealed to by three needy persons 
in succession for assistance. To the first person 
he gave one penny more than half the money 



he had in his pocket ; to the second person he 
gave twopence more than half the money he 
then had in his pocket ; and to the third person 
he handed over threepence more than half of 
what he had left. On entering his house he 
had only one penny in his pocket. Now, can 
you say exactly how much money that gentle- 
man had on him when he started for home ? 

9.— THE TWO AEROPLANES. 

A MAN recently bought two aeroplanes, but 
afterwards found that they would not answer 
the purpose for which he wanted them. So he 
sold them for £600 each, making a loss of 20 per 
cent, on one machine and a profit of 20 per cent, 
on the other. Did he make a profit on the whole 
transaction, or a loss ? And how much ? 

ID.— BUYING PRESENTS. 

*• Whom do you think I met in town last week, 
Brother William ? " said Uncle Benjamin. 
" That old skinflint Jorkins. His family had 
been taking him around bu)ring Christmas pres- 
ents. He said to me, ' Why caimot the gov- 
ernment abolish Christmas, and make the giving 
of presents pimishable by law ? I came out 
this morning with a certain amount of money 
in my pocket, and I find I have spent just half 
of it. In fact, if you will believe me, I take 
home just as many shillings as I had pounds, 
and half as many pounds as I had shillings. It 
is monstrous ! * " Can you say exactly how much 
money Jorkins had spent on those presents ? 

II.— THE CYCLISTS' FEAST. 

'TwAS last Bank Holiday, so I've been told, 

Some cyclists rode abroad in glorious weather. 
Resting at noon within a tavern old, 

They all agreed to have a feast together. 
" Put it all in one bill, mine host," they said, 

" For every man an equal share will pay." 
The bill was promptly on the table laid, 

And four pounds was the reckoning that day. 
But, sad to state, when they prepared to square, 

'Twas found that two had sneaked outside 
and fled. 
So, for two shillings more than his due share 

Each honest man who had remained was bled. 
They settled later with those rogues, no doubt. 
How many were they when they first set out ? 

12.— A QUEER THING IN MONEY. 

It will be found that £66, 6s. 6d. equals 15,918 
pence. Now, the four 6's added together make 
24, and the figures in 15,918 also add to 24. It 
is a curious fact that there is only one other sum 
of money, in pounds, shillings, and pence (all 
similarly repetitions of one figure), of which the 
digits shall add up the same as the digits of the 
amount in pence. What is the other sum of 
money ? 

13.— A NEW MONEY PUZZLE. 

The largest sum of money that can be written 
in pounds, shillings, pence, and farthings, using 
each of the nine digits once and only once, is 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



£98,765, 4S. 34d. Now, try to discover the 
smallest sum of money that can be written 
down imder precisely the same conditions. 
There must be some value given for each de- 
nomination — ^pounds, shillings, pence, and far- 
things — and the nought may not be used. It 
requires just a little judgment and thought. 

14.— SQUARE MONEY. 

" This is queer," said McCrank to his friend. 
" Twopence added to twopence is fourpence, and 
twopence multiplied by twopence is also four- 
pence." Of course, he was wrong in thinking 
you can multiply money by money. The multi- 
plier must be regarded as an abstract number. 
It is true that two feet multiplied by two feet 
will make four square feet. Similarly, two pence 
multiplied by two pence will produce four square 
pence ! And it will perplex the reader to say 
what a " square penny " is. But we will assume 
for the purposes of our puzzle that twopence 
multiplied by twopence is fourpence. Now, 
what two amounts of money will produce the 
next smallest possible result, the same in both 
cases, when added or multiplied in this manner ? 
The two amounts need not be alike, but they 
must be those that can be paid in current coins 
of the realm. 

15.— POCKET MONEY. 

What is the largest sum of money — -all in cur- 
rent silver coins and no four-shilling piece — 
that I could have in my pocket without being 
able to give change for a half-sovereign ? 

16.— THE MILLIONAIRE'S PERPLEXITY. 

Mr. Morgan G. Bloomgarten, the millionaire, 
known in the States as the Clam King, had, for 
his sins, more money than he knew what to do 
with. It bored him. So he determined to per- 
secute some of his poor but happy friends with 
it. They had never done him any harm, but he 
resolved to inoculate them with the " source of 
all evil." He therefore proposed to distribute 
a million dollars among them and watch them 
go rapidly to the bad. But he was a man of 
strange fancies and superstitions, and it was an 
inviolable rule with him never to make a gift 
that was not either one dollar or some power 
of seven — such as 7, 49, 343, 2,401, which num- 
bers of dollars are produced by simply multi- 
plying sevens together. Another rule of his was 
that he would never give more than six persons 
exactly the same sum. Now, how was he to 
distribute the 1,000,000 dollars ? You may 
distribute the money among as many people 
as you like, under the conditions given. 

17.— THE PUZZLING MONEY-BOXES. 

Four brothers — named John, William, Charles, 
and Thomas — had each a money-box. The 
boxes were all given to them on the same day, | 
and they at once put what money they had into j 
them : only, as the boxes were not very large, | 
they first changed the money into as few coins ; 
as possible. After they had done this, they told | 



one another how much money they had saved, 
and it was found that if John had had 2S. more 
in his box than at present, if William had had 2S. 
less, if Charles had had twice as much, and if 
Thomas had had half as much, they would all 
have had exactly the same amoimt. 

Now, when I add that all four boxes together 
contained 45s., and that there were only six 
coins in all in them, it becomes an entertaining 
puzzle to discover just what coins were in each 
box. 

18.— THE MARKET WOMEN. 

A NUMBER of market women sold their various 
products at a certain price per pound (different 
in every case), and each received the same 
amount — 2s. zjd. What is the greatest number 
of women there could have been ? The price 

Eer pound in every case must be such as could 
e paid in current money. 

19.— THE NEW YEAR'S EVE SUPPERS. 

The proprietor of a small London caf6 has given 
me some interesting figures. He says that the 
ladies who come alone to his place for refresh- 
ment spend each on an average eighteenpence, 
that the unaccompanied men spend half a crown 
each, and that when a gentleman brings in a 
lady he spends half a guinea. On New Year's 
Eve he supplied suppers to twenty-five persons, 
and took five pounds in all. Now, assuming 
his averages to have held good in every case, 
how was his company made up on that occa- 
sion ? Of course, only single gentlemen, single 
ladies, and pairs (a lady and gentleman) can be 
supposed to have been present, as we are not 
considering larger parties. 

20.— BEEF AND SAUSAGES. 

" A NEIGHBOUR of mine," said Aunt Jane, 
" bought a certain quantity of beef at two shil- 
lings a pound, and the same quantity of sausages 
at eighteenpence a potmd. I pointed out to her 
that if she had divided the same money equally 
between beef and sausages she would have 
gained two pounds in the total weight. Can 
you tell me exactly how much she spent ? " 

" Of course, it is no business of mine," said 
Mrs. Sunniborne ; " but a lady who could pay 
such prices must be somewhat inexperienced in 
domestic economy." 

" I quite agree, my dear," Aunt Jane replied, 
" but you see that is not the precise point under 
discussion, any more than the name and morals 
of the tradesman." 

21.— A DEAL IN APPLES. 

I PAID a man a shilling for some apples, but they 
were so small that I made him throw in two 
extra apples. I find that made them cost just 
a penny a dozen less than the first price he asked. 
How many apples did I get for my shilling ? 

22.~A DEAL IN EGGS. 

A MAN went recently into a dairyman's shop tc 
buy eggs. He wanted them of various qualities. 



AMUSEMENTS IN MATHEMATICS. 



The salesman had new-laid eggs at the high price 
of fivepence each, fresh eggs at one penny each, 
eggs at a halfpenny each, and eggs for election- 
eering purposes at a greatly reduced figure, but 
as there was no election on at the time the buyer 
had no use for the last. However, he bought 
some of each of the three other kinds and ob- 
tamed exactly one hundred eggs for eight and 
foiUT)ence. Now, as he brought away exactly 
the same number of eggs of two of the three 
quaUties, it is an interesting puzzle to determine 
just how many he bought at each price. 

23.— THE CHRISTMAS-BOXES. 

Some years ago a man told me he had spent one 
hundred English silver coins in Christmas-boxes, 
giving every person the same amount, and it 
cost him exactly £1, los. id. Can you tell just 
how many persons received the present, and 
how he could have managed the distribution ? 
That odd penny looks queer, but it is aU right. 

24.— A SHOPPING PERPLEXITY. 

Two ladies went into a shop where, through 
some curious eccentricity, no change was given, 
and made purchases amounting together to less 
than five shillings. " Do you know," said one 
lady, "I find I shall require no fewer than six 
current coins of the realm to pay for what I 
have bought." The other lady considered a 
moment, and then exclaimed : " By a peculiar 
coincidence, I am exactly in the same dilemma." 
" Then we will pay the two bills together." 
But, to their astonishment, they still required 
six coins. What is the smallest possible amount 
of their purchases — ^both different ? 

25.— CHINESE MONEY. 

The Chinese are a curious people, and have 
strange inverted ways of doing things. It is 
said that they use a saw with an upward pressure 
instead of a downward one, that they plane a 
deal board by pulling the tool toward them in- 
stead of pushing it, and that in building a house 
they first construct the roof and, having raised 
that into position, proceed to work downwards. 
In money the currency of the country consists 
of taels of fluctuating value. The tael became 
thinner and thinner imtil 2,000 of them piled 
together made less than three inches in height. 
The common cash consists of brass coins of vary- 
ing thicknesses, with a round, square, or tri- 
angular hole in the centre, as in our illustration. 




These are strung on wires like buttons. Sup- 
posing that eleven coins with round holes are 
worth fifteen ching-changs, that eleven with 
square holes are worth sixteen ching-changs, and 



that eleven with triangular holes are worth 
seventeen ching-changs, how can a Chinaman 
give me change for half a crown, using no coins 
other than the three mentioned ? A ching- 
chang is worth exactly twopence and four- 
fifteenths of a ching-chang. 

26.— THE JUNIOR CLERK'S PUZZLE. 

Two youths, bearing the pleasant names of 
Moggs and Snoggs, were employed as junior 
clerks by a merchant in Mincing Lane. They 
were both engaged at the same salary — that is, 
commencing at the rate of £50 a year, payable 
half-yearly. Moggs had a yearly rise of £10, 
and Snoggs was offered the same, only he asked, 
for reasons that do not concern our puzzle, that 
he might take his rise at £2, los. half-yearly, to 
which his employer (not, perhaps, unnaturally !) 
had no objection. 

Now we come to the real point of |he puzzle. 
Moggs put regidarly into the Post Office Savings 
Bank a certain proportion of his salary, while 
Snoggs saved twice as great a proportion of his, 
and at the end of five years they had together 
saved £268, 15s. How much had each saved ? 
The question of interest can be ignored. 

27.— GIVING CHANGE. 

Every one is familiar with the difficulties that 
frequently arise over the giving of change, and 
how the assistance of a third person with a few 
coins in his pocket will sometimes help us to set 
the matter right. Here is an example. An 
Englishman went into a shop in New York and 
bought goods at a cost of thirty-four cents. 
The only money he had was a dollar, a three-cent 
piece, and a two-cent piece. The tradesman had 
only a half-dollar and a quarter-dollar. But 
another customer happened to be present, and 
when asked to help produced two dimes, a five- 
cent piece, a two-cent piece, and a one-cent 
piece. How did the tradesman manage to give 
change ? For the benefit of those readers who 
are not familiar with the American coinage, it is 
only necessary to say that a dollar is a hundred 
cents and a dime ten cents. A puzzle of this 
kind should rarely cause any difficulty if at- 
tacked in a proper manner. 

28.— DEFECTIVE OBSERVATION. 

Our observation of Uttle things is frequently 
defective, and our memories very liable to lapse. 
A certain judge recently remarked in a case that 
he had no recollection whatever of putting the 
wedding-ring on his wife's finger. Can you 
correctly answer these questions without having 
the coins in sight ? On which side of a penny 
is the date given ? Some people are so unob- 
servant that, although they are handling the 
coin nearly every day of their Uves, they are at 
a loss to answer this simple question. If I lay 
a penny flat on the table, how many other pennies 
can I place around it, every one also lying flat 
on the table, so that they all touch the first one ? 
The geometrician will, of course, give the answer 
at once, and not need to make any experiment. 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



He will also know that, since all circles are 
similar, the same answer will necessarily apply 
to any coin. The next question is a most in- 
teresting one to ask a company, each person 
writing down his answer on a slip of paper, so 
that no one shall be helped by the answers of 
others. What is the greatest number of three- 
penny-pieces that may be laid flat on the surface 
of a half-crown, so that no piece lies on another 
or overlaps the surface of the half-crown ? It 
is amazing what a variety of different answers 
one gets to this question. Very few people will 
be found to give the correct number. Of course 
the answer must be given without looking at 
the coins. 

29.— THE BROKEN COINS. 

A MAN had three coins — a sovereign, a shilling, 
and a penny — and he found that exactly the 
same fraction of each coin had been broken 
away. Now, assuming that the original in- 
trinsic value of these coins was the same as 
their nominal value — that is, that the sovereign 
was worth a pound, the shilling worth a shilling, 
and the penny worth a penny — what proportion 
of each coin has been lost if the value of the 
three remaining fragments is exactly one pound ? 



30.— TWO QUESTIONS I^ 
BILITIES. 



'•"xtOB '■'I 



There is perhaps no class of puzzle over which 
people so frequently blunder as that which in- 
volves what is called the theory of probabilities. 
I will give two simple examples of the sort of 
puzzle I mean. They are really quite easy, and 
yet many persons are tripped up by them. A 
friend recently produced five pennies and said 
to me : " In throwing these five pennies at the 
same time, what are the chances that at least 
four of the coins will turn up either all heads or 
all tails ? " His own solution was quite wrong, 
but the correct answer ought not to be hard to 
discover. Another person got a wrong answer 
to the following little puzzle which I heard him 
propound : "A man placed three sovereigns 
and one shilling in a bag. How much should 
be paid for permission to draw one coin from 
it ? " It is, of course, understood that you are 
as likely to draw any one of the four coins as 
another. 

31.— DOMESTIC ECONOMY. 

Young Mrs. Perkins, of Putney, writes to me as 
follows : "I should be very glad if you could 
give me the answer to a little sum that has been 
worrjdng me a good deal lately. Here it is : 
We have only been married a short time, and 
now, at the end of two years from the time when 
we set up housekeeping, my husband tells me 
that he finds we have spent a third of his yearly 
income in rent, rates, and taxes, one-half in 
domestic expenses, and one-ninth in other ways. 
He has a balance of £190 remaining in the 
bank. I know this last, because he acciden- 
tally left out his pass-book the other day, and I 
peeped into it. Don't you think that a husband 
ought to give his wife his entire confidence in 



his money matters ? Well, I do ; and — will 
you believe it ? — ^he has never told me what 
his income really is, and I want, very naturally, 
to find out. Can you tell me what it is from 
the figures I have given you? " 

Yes ; the answer can certainly be given from 
the figures contained in Mrs. Perkins's letter. 
And my readers, if not warned, will be practi- 
cally unanimous in declaring the income to be 
— something absurdly in excess of the correct 
answer ! 

32.— THE EXCURSION TICKET PUZZLE. 

When the big flaming placards were exhibited 
at the little provincial railway station, announ- 
cing that the Great Company would run 

cheap excursion trains to London for the Christ- 
mas holidays, the inhabitants of Mudley-cum- 
Turmits were in quite a flutter of excitement. 
Half an hour before the train came in the little 
booking ofi&ce was crowded with country pas- 
sengers, all bent on visiting their friends in the 
great Metropolis. The booking clerk was un- 
accustomed to dealing with crowds of such a 
dimension, and he told me afterwards, while 
wiping his manly brow, that what caused him 
so much trouble was the fact that these rustics 
paid their fares in such a lot of small money. 

He said that he had enough farthings to 
supply a West End draper with change for a 
week, and a sufficient number of threepenny 
pieces for the congregations of three parish 
churches. '* That excursion fare," said he, " is 
nineteen shillings and ninepence, and I should 
like to know in just how many different ways 
it is possible for such an amount to be paid in 
the current coin of this realm." 

Here, then, is a puzzle : In how many differ- 
ent ways may nineteen shillings and ninepence 
be paid in our current coin ? Remember that 
the fourpenny-piece is not now current. 

33._A PUZZLE IN REVERSALS. 

Most people know that if you take any sum of 
money in pounds, shillings, and pence, in which 
the number of pounds (less than £12) exceeds 
that of the pence, reverse it (calling the poimds 
pence and the pence pounds), find the difference, 
then reverse and add this difference, the result 
is always £12, i8s. ^iid. But if we omit the 
condition, " less than £12," and allow nought to 
represent shillings or pence — (i) What is the 
lowest amount to which the rule will not apply ? 
(2) What is the highest amount to which it will 
apply ? Of course, when reversing such a sxun 
as £14, 15s. 3d. it may be written £3, i6s. 2d., 
which is the same as £3, 15s. i4d. 

34.— THE GROCER AND DRAPER. 

A COUNTRY " grocer and draper " had two rival 
assistants, who prided themselves on their ra- 
pidity in serving customers. The young man 
on the grocery side could weigh up two one- 
pound parcels of sugar per minute, while the 
drapery assistant could cut three one-yard 
lengths of cloth in the same time. Their em- 
ployer, one slack day, set them a race, giving 



AMUSEMENTS IN MATHEMATICS. 



the grocer a barrel of sugar and telling him to 
weigh up forty-eight one-pound parcels of sugar 
while the draper divided a roll of forty-eight 
yards of cloth into yard pieces. The two men 
were interrupted together by customers for nine 
minutes, but the draper was disturbed seventeen 
times as long as the grocer. What was the re- 
sult of the race ? 

35.— JUDKINS'S CATTLE. 

Hiram B, Judkins, a cattle-dealer of Texas, 
had five droves of animals, consisting of oxen, 
pigs, and sheep, with the same number of ani- 
mals in each drove. One morning he sold aU 
that he had to eight dealers. Each dealer 
bought the same number of animals, paying 
seventeen dollars for each ox, four dollars for 
each pig, and two doUars for each sheep ; and 
Hiram received in all three himdred and one 
dollars. What is the greatest number of ani- 
mals he could have had ? And how many 
would there be of each kind ? 

36.— BUYING APPLES. 

As the purchase of apples in small quantities 
has always presented considerable difficulties, I 
think it well to offer a few remarks on this sub- 
ject. We all know the story of the smart boy 
who, on being told by the old woman that she 
was selling her apples at four for threepence, 
said : *' Let me see ! Four for threepence ; 
that's three for twopence, two for a penny, one 
for nothing — I'll take one I " 

There are similar cases of perplexity. For 
example, a boy once picked up a penny apple 
from a stall, but when he learnt that the wo- 
man's pears were the same price he exchanged 
it, and was about to walk off. " Stop ! " said 
the woman. " You haven't paid me for the 

?ear ! " " No," said the boy, " of course not. 
gave you the apple for it." " But you didn't 
pay for the apple ! " " Bless the woman ! You 
don't expect me to pay for the apple and the 
pear too ! " And before the poor creature could 
get out of the tangle the boy had disappeared. 

Then, again, we have the case of the man who 
gave a boy sixpence and promised to repeat 
the gift as soon as the yoimgster had made it 
into ninepence. Five minutes later the boy 
returned. " I have made it into ninepence," 
he said, at the same time handing his benefactor 
threepence. '* How do you make that out ? " 
he was asked. " I bought threepenny worth 
of apples." " But that does not make it into 
ninepence ! " "I should rather think it did," 
was the boy's reply. " The apple woman has 
threepence, hasn't she ? Very well, I have 
threepennyworth of apples, and 1 have just 
given you the other threepence. What's that 
but ninepence ? " 

I cite these cases just to show that the smaU 
boy really stands in need of a little instruction 
in the art of buying apples. So I will give a 
simple poser dealing with this branch of com- 
merce. 

An old woman had apples of three sizes for 
sale — one a penny, two a penny, and three a 



penny. Of course two of the second size and 
three of the third size were respectively equal 
to one apple of the largest size. Now, a gen- 
tleman who had an equal number of boys and 
girls gave his children sevenpence to be spent 
amongst them all on these apples. The puzzle is 
to give each child an equal distribution of apples. 
How was the sevenpence spent, and how many 
children were there ? 

37-— BUYING CHESTNUTS. 

Though the following little puzzle deals with 
the purchase of chestnuts, it is not itself of the 
" chestnut " type. It is quite new. At first 
sight it has certainly the appearance of being of 
the " nonsense puzzle " character, but it is all 
right when properly considered. 

A man went to a shop to buy chestnuts. He 
said he wanted a pennyworth, and was given five 
chestnuts. " It is not enough ; I ought to have 
a sixth," he remarked. " But if I give you one 
chestnut more," the shopman replied, " you 
will have five too many." Now, strange to say, 
they were both right. How many chestnuts 
should the buyer receive for half a crown ? 

38.— THE BICYCLE THIEF. 

Here is a little tangle that is perpetually crop- 
ping up in various guises. A cyclist bought a 
bicycle for £15 and gave in payment a cheque 
for £25. The seller went to a neighbouring 
shoplceeper and got him to change the cheque 
for him, and the cyclist, having received his £10 
change, mounted the machine and disappeared. 
The cheque proved to be valueless, and the sales- 
man was requested by his neighbour to refund 
the amoimt he had received. To do this, he 
was compelled to borrow the £25 from a friend, 
as the cyclist forgot to leave his address, and 
could not be foimd. Now, as the bicycle cost 
the salesman £11, how much money did he lose 
altogether ? 

39.~-THE COSTERMONGER'S PUZZLE. 

" How much did yer pay for them oranges, 
Bill ? " 

" I ain't a-goin' to tell yer, Jim. But I beat 
the old cove down fourpence a hundred." 

" What good did that do yer ? " 

" Well, it meant five more oranges on every 
ten shillin's-worth." 

Now, what price did Bill actually pay for the 
oranges ? There is only one rate that will fit 
in with his statements. 



AGE AND KINSHIP PUZZLES. 

" The days of our years are threescore years 
and ten." — Psalm xc. 10. 

For centuries it has been a favourite method 
of propounding arithmetical puzzles to pose 
them in the form of questions as to the age of 
an individual. They generally lend themselves 
to very easy solution by the use of algebra, 
though often the difficulty lies in stating them 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



correctly. They may be made very complex 
and may demand considerable ingenuity, but 
no general laws can well be laid down for their 
solution. The solver must use his own sagacity. 
As for puzzles in relationship or kinship, it 
is quite curious how bewildering many people 
find these things. Even in ordinary conversa- 
tion, some statement as to relationship, which 
is quite clear in the mind of the speaker, will 
immediately tie the brains of other people into 
knots. Such expressions as "He is my uncle's 
son-in-law's sister " convey absolutely nothing 
to some people without a detailed and laboured 
explanation. In such cases the best course is 
to sketch a brief genealogical table, when the 
eye comes immediately to the assistance of the 
brain. In these days, when we have a growing 
lack of respect for pedigrees, most people have 
got out of the habit of rapidly drawing such 
tables, which is to be regretted, as they would 
save a lot of time and brain racking on occasions. 

40.— MAMMA'S AGE. 

Tommy : " How old are you, mamma ? " 

Mamma : " Let me think. Tommy. Well, 
our three ages add up to exactly seventy years." 

Tommy : " That's a lot, isn't it ? And how 
old are you, papa ? " 

Papa : *' Just six times as old as you, my 
son.'*^ 

Tommy : " Shall I ever be half as old as you, 
papa ? " 

Papa : " Yes, Tommy ; and when that hap- 
pens our three ages will add up to exactly twice 
as much as to-day." 

Tommy : " And supposing I was bom before 
you, papa ; and supposing mamma had forgot 
all about it, and hadn't been at home when I 
came ; and supposing " 

Mamma : " Supposing, Tommy, we talk 
about bed. Come along, darling. You'll have 
a headache." 

Now, if Tommy had been some years older 
be might have calculated the exact ages of his 
parents from the information they had given 
him. Can you find out the exact age of 
mamma ? 

41.— THEIR AGES. 

" My husband's age," remarked a lady the other 
day, " is represented by the figures of my own 
age reversed. He is my senior, and the difEer- 
ence between our ages is one-eleventh of their 
sum." 

42.— THE FAMILY AGES. 

When the Smileys recently received a visit from 
the favourite uncle, the fond parents had all the 
five children brought into his presence. First 
came Billie and little Gertrude, and the uncle 
was informed that the boy was exactly twice as 
old as the girl. Then Henrietta arrived, and it 
was pointed out that the combined ages of her- 
self and Gertrude equalled twice the age of 
Billie. Then Charlie came running in, and 
somebody remarked that now the combined 
ages of the two boys were exactly twice the com- 
bined ages of the two girls. The uncle was ex- 



pressing his astonishment at these coincidences 
when Janet came in. *' Ah ! uncle," she ex- 
claimed, " you have actually arrived on my 
twenty-first birthday ! " To this Mr. Smiley 
added the final staggerer : " Yes, and now the 
combined ages of the three girls are exactly 
equal to twice the combined ages of the two 
boys." Can you give the age of each child ? 

43.— MRS. TIMPKINS'S AGE. 

Edwin : " Do you know, when the Timpkinses 
married eighteen years ago Timpkins was three 
times as old as his wife, and to-day he is just 
twice as old as she ? " 

Angelina : " Then how old was Mrs. Timpkins 
on the wedding day ? " 

Can you answer Angelina's question ? 

44.— A CENSUS PUZZLE. 

Mr. and Mrs. Jorkins have fifteen children, all 
bom at intervals of one year and a half. Miss 
Ada Jorkins, the eldest, had an objection to 
state her age to the census man, but she ad- 
mitted that she was just seven times older than 
little Johnnie, the yoimgest of all. What was 
Ada's age ? Do not too hastily assume that 
you have solved this little poser. You may 
find that you have made a bad blunder ! 

45.— MOTHER AND DAUGHTER. 

" Mother, I wish you would give me a bicycle," 
said a girl of twelve the other day. 

" I do not think you are old enough yet, my 
dear," was the reply. " When I am only three 
times as old as you are you shall have one." 

Now, the mother's age is forty-five years. 
When may the young lady expect to receive 
her present ? 

46.— MARY AND MARMADUKE. 

Marmaduke : " Do you know, dear, that in 
seven years' time our combined ages will be 
sixty-three years ? " 

Mary : "Is that really so ? And yet it is a 
fact that when you were my present age you 
were twice as old as I was then. I worked it 
out last night." 

Now, what are the ages of Mary and Marma- 
duke ? 

47.— ROVER'S AGE. 

" Now, then, Tommy, how old is Rover ? " 
Mildred's young man asked her brother. 

" Well, five years ago," was the youngster's 
reply, " sister was four times older than the dog, 
but now she is only three times as old." 

Can you tell Rover's age ? 

48.— CONCERNING TOMMY'S AGE. 

Tommy Smart was recently sent to a new school. 
On the first day of his arrival the teacher asked 
him his age, and this was his curious reply : 
" Well, you see, it is like this. At the time I 
was born — I forget the year — ^my only sister, 
Ann, happened to be just one-quarter the age 



8 



AMUSEMENTS IN MATHEMATICS. 



of mother, and she is now one-third the age 
of father." "That's aU very weU," said the 
teacher, " but what I want is not the age of 
your sister Ann, but your own age." " I was 
just coming to that," Tommy answered ; *' I 
am just a quarter of mother's present age, and 
in four years' time I shall be a quarter the age 
of father. Isn't that funny ? " 

This was all the information that the teacher 
could get out of Tommy Smart. Could you 
have told, from these facts, what was his precise 
age ? It is certainly a little puzzling. 

49.— NEXT-DOOR NEIGHBOURS. 

There were two families living next door to 
one another at Tooting Bee — the Jupps and the 
Simkins. The united ages of the four Jupps 
amounted to one hundred years, and the united 
ages of the Simkins also amounted to the same. 
It was foimd in the case of each family that the 
sum obtained .by adding the squares of each of 
the children's ages to the square of the mother's 
age equalled the square of the father's age. In 
the case of the Jupps, however, Julia was one 
year older than her brother Joe, whereas Sophy 
Simkin was two years older than her brother 
Sammy. What was the age of each of the eight 
individuals ? 

50.— THE BAG OF NUTS. 

Three boys were given a bag of nuts as a Christ- 
mas present, and it was agreed that they should 
be divided in proportion to their ages, which to- 
gether amounted to 17 J years. Now the bag 
contained 770 nuts, and as often as Herbert took 
four Robert took three, and as often as Herbert 
took six Christopher took seven. The puzzle 
is to find out how many nuts each had, and 
what were the boys' respective ages. 

51.— HOW OLD WAS MARY ? 

Here is a funny little age problem, by the late 
Sam Loyd, which has been very popular in the 
United States. Can you unravel the mystery ? 
The combined ages of Mary and Ann are forty- 
four years, and Mary is twice as old as Ann was 
when Mary was half as old as Ann will be when 
Ann is three times as old as Mary was when 
Mary was three times as old as Ann. How old 
is Mary ? That is all, but can you work it out ? 
If not, ask your friends to help you, and watch 
the shadow of bewilderment creep over their 
faces as they attempt to grip the intricacies of 
the question. 

52.— QUEER RELATIONSHIPS. 

*' Speaking of relationships," said the Parson, 
at a certain dinner-party, " our legislators are 
getting the marriage law into a frightful tangle. 
Here, for example, is a puzzling case that has 
come under my notice. Two brothers married 
two sisters. One man died and the other man's 
wife also died. Then the survivors married." 

'* The man married his deceased wife's sister, 
under the recent Act ? " put in the Lawyer. 

" Exactly. And therefore, under the civil 



law, he is legally married and his child is legiti- 
mate. But, you see, the man is the woman's 
deceased husband's brother, and therefore, also 
under the civil law, she is not married to him 
and her child is illegitimate." 

" He is married to her and she is not married 
to him ! " said the Doctor. 

" Quite so. And the child is the legitimate 
son of his father, but the illegitimate son of his 
mother." 

" Undoubtedly ' the law is a hass,' " the 
Artist exclaimed, " if I may be permitted to say 
so," he added, with a bow to the Lawyer. 

" Certainly," was the reply. " We lawyers 
try our best to break in the beast to the service 
of man. Our legislators are responsible for the 
breed." 

" And this reminds me," went on the Parson, 
" of a man in my parish who married the sister 
of his widow. This man " 

" Stop a moment, sir," said the Professor. 
" Married the sister of his widow ? Do you 
marry dead men in your parish ? " 

" No ; but I will explain that later. Well, 
this man has a sister of his own. Their names 
are Stephen Brown and Jane Brown. Last 
week a young fellow turned up whom Stephen 
introduced to me as his nephew. Naturally, I 
spoke of Jane as his aunt, but, to my astonish- 
ment, the youth corrected me, assiiring me that, 
.though he was the nephew of Stephen, he was 
not the nephew of Jane, the sister of Stephen. 
This perplexed me a good deal, but it is quite 
correct." 

The Lawyer was the first to get at the heart 
of the mystery. What was his solution ? 

53— HEARD ON THE TUBE RAILWAY. 

First Lady : " And was he related to you, 
dear ? " 

Second Lady : " Oh, yes. You see, that 
gentleman's mother was my mother's mother- 
in-law, but he is not on speaking terms with my 
papa." 

Furst Lady : " Oh, indeed ! " (But you could 
see that she was not much wiser.) 

How was the gentleman related to the Second 
Lady? 

54.— A FAMILY PARTY. 

A certain family party consisted of i grand- 
father, I grandmother, 2 fathers, 2 mothers, 4 
children, 3 grandchildren, i brother, 2 sisters, 
2 sons, 2 daughters, i father-in-law, i mother- 
in-law, and I daughter-in-law. Twenty-three 
people, you will say. No ; there were only 
seven persons present. Can you show how this 
might be ? 

55.— A MIXED PEDIGREE. 

Joseph Bloggs : "I can't follow it, my dear 
boy. It makes me dizzy ! " 

John Snoggs : " It's very simple. Listen 
again ! You happen to be my father's brother- 
in-law, my brother's father-in-law, and also my 
father-in-law's brother. You see, my father 
was " 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



But Mr. Bloggs refused to hear any more. 
Can the reader show how this extraordinary 
triple relationship might have come about ? 

56._WILSON'S POSER. 



" Speaking of perplexities- 



" said Mr. Wil- 
son, throwing down a magazine on the table in 
the commercial room of the Railway Hotel. 

" Who was speaking of perplexities ? " in- 
quired Mr. Stubbs. 

" WeU, then, reading about them, if you want 
to be exact — it just occurred to me that perhaps 
you three men may be interested in a little 
matter connected with myself." 

It was Christmas Eve, and the four commer- 
cial travellers were spending the holiday at 
Grassminster. Probably each suspected that 
the others had no homes, and perhaps each was 
conscious of the fact that he was in that pre- 
dicament himself. In any case they seemed 
to be perfectly comfortable, and as they drew 
round the cheerful fire the conversation became 

" What is the difficulty ? " asked Mr. Pack- 
hurst. 

" There's no difficulty in the matter, when 
you rightly understand it. It is like this. A 
man named Parker had a fl3^ing-machine that 
would carry two. He was a venturesome sort 
of chap — ^reckless, I should call him — and he 
had some bother in finding a man willing to risk 
his life in making an ascent with him. How- 
ever, an imcle of mine thought he would chance 
it, and one fine morning he took his seat in the 
machine and she started off well. When they 
were up about a thousand feet, my nephew 
suddenly " 

" Here, stop, Wilson ! What was your 
nephew doing there ? You said your uncle," 
interrupted Mr. Stubbs. 

'• Did I ? Well, it does not matter. My 
nephew suddenly turned to Parker and said 
that the engine wasn't running well, so Parker 
called out to my imcle " 

" Look here," broke in Mr. Waterson, " we 
are getting mixed. Was it your uncle or your 
nephew ? Lefs have it one way or the other." 

" What I said is quite right. Parker called 
out to my uncle to do something or other, when 
my nephew " 

" There you are again, Wilson," cried Mr. 
Stubbs ; " once for sdl, are we to understand 
that both yom: imcle and your nephew were on 
the machine ? " 

" Certainly. I thought I made that clear. 
Where was I ? Well, my nephew shouted back 
to Parker " 

" Phew ! I'm sorry to interrupt you again, 
Wilson, but we can't get on like this. Is it true 
that the machine would only carry two ? " 

" Of course. I said at the start that it only 
carried two." 

" Then what in the name of aerostation do you 
mean by saying that there were three persons on 
board ? " shouted Mr. Stubbs. 

" Who said there were three ? " 

" You have told us that Parker, your uncle, 



and your nephew went up on this blessed flying- 
machine." 

" That's right." 

" And the thing would only carry two ! " 

" Right again." 

" Wilson, I have known you for some time 
as a truthful man and a temperate man," said 
Mr. Stubbs, solemnly. " But I am afraid since 
you took up that new line of goods you have 
overworked yourself." 

" Half a minute, Stubbs," interposed Mr. 
Waterson. " I see clearly where we all slipped a 
cog. Of course, Wilson, you meant us to under- 
stand that Parker is either your uncle or your 
nephew. Now we shaU be all right if you will 
just tell us whether Parker is your uncle or 
nephew." 

" He is no relation to me whatever." 

The three men sighed and looked anxiously 
at one another. Mt. Stubbs got up from his 
chair to reach the matches, Mr. Packhurst pro- 
ceeded to wind up his watch, and Mr. Waterson 
took up the poker to attend to the fire. It was 
an awkward moment, for at the season of good- 
will nobody wished to teU Mr. Wilson exactly 
what was in his mind. 

" It's curious," said Mr. Wilson, very de- 
liberately, " and it's rather sad, how thick- 
headed some people are. You don't seem to 
grip the facts. It never seems to have occurred 
to either of you that my uncle and my nephew 
are one and the same man." 

" What ! " exclaimed all three together. 

" Yes ; David George Linklater is my uncle, 
and he is also my nephew. Consequently, I am 
both his uncle and nephew. Queer, isn't it ? 
I'll explain how it comes about." 

Mr. Wilson put the case so very simply that 
the three men saw how it might happen without 
any marriage within the prohibited degrees. 
Perhaps the reader can work it out for himself. 



CLOCK PUZZLES. 

" Look at the clock ! " 

Ingoldsby Legends. 

In considering a few puzzles concerning clocks 
and watches, and the times recorded by their 
hands under given conditions, it is well that a 
particular convention should always be kept in 
mind. It is frequently the case that a solution 
requires the assumption that the hands can 
actually record a time involving a minute frac- 
tion of a second. Such a time, of course, cannot 
be really indicated. Is the puzzle, therefore, im- 
possible of solution ? The conclusion deduced 
from a logical syllogism depends for its truth on 
the two premises assumed, and it is the same in 
mathematics. Certain things are antecedently 
assumed, and the answer depends entirely on 
the truth of those assumptions. 

" If two horses," says Lagrange, " can pull a 
load of a certain weight, it is natural to suppose 
that four horses could pull a load of double that 
weight, six horses a load of three times that 
weight. Yet, strictly speaking, such is not the 



lO 



AMUSEMENTS IN MATHEMATICS. 



case. For the inference is based on the assump- 
tion that the four horses pull alike in amount 
and direction, which in practice can scarcely ever 
be the case. It so happens that we are fre- 
quently led in our reckonings to results which 
diverge widely from reality. But the fault is 
not the fault of mathematics ; for mathematics 
always gives back to us exactly what we have 
put into it. The ratio was constant according 
to that supposition. The result is founded 
upon that supposition. If the supposition is 
false the result is necessarily false." 

If one man can reap a field in six days, we say 
two men will reap it in three days, and three 
men will do the work in two days. We here 
assume, as in the case of Lagrange's horses, that 
aU the men are exactly equally capable of work. 
But we assume even more than this. For when 
three men get together they may waste time in 
gossip or play ; or, on the other hand, a spirit of 
rivalry may spur them on to greater dihgence. 
We may assume any conditions we like in a 
problem, provided they be clearly expressed 
and understood, and the answer will be in ac- 
cordance with those conditions. 

57.— WHAT WAS THE TIME ? 

" I SAY, Rackbrane, what is the time ? " an 
acquaintance asked our friend the professor the 
other day. The answer was certainly curious. 

" If you add one quarter of the time from 
noon till now to half the time from now tiU noon 
to-morrow, you will get the time exactly." 

What was the time of day when the professor 
spoke ? 

58.— A TIME PUZZLE. 

How many minutes is it until six o'clock if fifty 
minutes ago it was four times as many minutes 
past three o'clock ? 

59.— A PUZZLING WATCH. 

A FRIEND puUed out his watch and said, " This 
watch of mine does not keep perfect time ; I 
must have it seen to. I have noticed that the 
minute hand and the hour hand are exactly 
t<^!gether every sixty-five minutes." Does that 
watch gain or lose, and how much per hour ? 

60.— THE WAPSHAW'S WHARF 
MYSTERY. 

There was a great commotion in Lower Thames 
Street on the morning of January 12, 1887. 
When the early members of the staff arrived at 
Wapshaw's Wharf they found that the safe had 
been broken open, a considerable sum of money 
removed, and the offices left in great disorder. 
The night watchman was nowhere to be found, 
but nobody who had been acquainted with him 
for one moment suspected him to be guilty of 
the robbery. In this belief the proprietors were 
confirmed when, later in the day, they were in- 
formed that the poor fellow's body had been 
picked up by the River Police. Certain marks 
of violence pointed to the fact that he had been 
brutally attacked and thrown into the river. 
A watch found in his pocket had stopped, as is 



invariably the case in such circumstances, and 
this was a valuable clue to the time of the out- 
rage. But a very stupid officer (and we in- 
variably find one or two stupid individuals in 
the most intelligent bodies of men) had actually 
amused himself by turning the hands round and 
round, trying to set the watch going again. 
After he had been severely reprimanded for this 
serious indiscretion, he was asked whether he 
could remember the time that was indicated by 
the watch when found. He replied that he 
could not, but he recollected that the hour hand 
and minute hand were exactly together, one 
above the other, and the second hand had just 
passed the forty-ninth second. More than this 
he could not remember. 

What was the exact time at which the watch- 
man's watch stopped ? The watch is, of course, 
I assumed to have been an accurate one. 



61.— CHANGING PLACES. 




The above clock face indicates a little before 
42 minutes past 4. The hands will again point 
at exactly the same spots a little after 23 minutes 
past 8. In fact, the hands will have changed 
places. How many times do the hands of a 
clock change places between three o'clock p.m. 
and midnight ? And out of all the pairs of 
times indicated by these changes, what is the 
exact time when the minute hand will be nearest 
to the point IX ? 

62.— THE CLUB CLOCK. 

One of the big clocks in the Cogitators' Club 
was found the other night to have stopped just 
when, as will be seen in the illustration, the 
second hand was exactly midway between the 
other two hands. One of the members proposed 
to some of his friends that they should tell him 
the exact time when (if the clock had not 



ARITHMETICAI. AND ALGEBRAICAL PROBLEMS. 



II 




stopped) the second hand would next again 
have been midway between the minute hand 
and the hour hand. Can you find the correct 
time that it would happen ? 



63.— THE STOP-WATCH. 




We have here a stop-watch with three hands. 
The second hand, which travels once round the 
face in a minute, is the one with the little ring 
at its end near the centre. Our dial indicates 
the exact time when its owner stopped the 
watch. You will notice that the three hands 
are nearly equidistant. The hour and minute 
hands point to spots that are exactly a third of 
the circumference apart, but the second hand 
is a little too advanced. An exact equidistance 
for the three hands is not possible. Now, we 
want to know what the time will be when the 
three hands are next at exactly the same dis- 
tances as shown from one another. Can you 
state the time ? 



64.— THE THREE CLOCKS. 

On Friday, April i, 1898, three new clocks were 
all set going precisely at the same time — twelve 
noon. At noon on the following day it was 
found that clock A had kept perfect time, that 
clock B had gained exactly one minute, and that 
clock C had lost exactly one minute. Now, sup- 
posing that the clicks B and C had not been 
regulated, but all three allowed to go on as they 
had begun, and that they maintained the same 
rates of progress without stopping, on what date 
and at what time of day would all three pairs of 
hands again point at the same moment at twelve 
o'clock ? 

65.— THE RAILWAY STATION CLOCK. 

A CLOCK hangs on the wall of a railway station, 
71 ft. 9 in. long and 12 ft. 5 in. high. Those are 
the dimensions of the wall, hot o! the clock ! 
While waiting for a train we noticed that the 
hands of the clock were pointing in opposite 
directions, and were parallel to one of the diag- 
onals of the wall. What was the exact time ? 

66.— THE VILLAGE SIMPLETON. 

A FACETIOUS individual who was taking a long 
walk in the country came upon a yokel sitting 
on a stile. As the gentleman was not quite 
sure of his road, he thought he would make 
inquiries of the local inhabitant; but at the 
first glance he jumped too hastily to the con- 
clusion that he had dropped on the village idiot. 
He therefore decided to test the fellow's intelli- 
gence by first putting to him the simplest ques- 
tion he could think of, which was, " What day 
of the week is this, my good man ? " The fol- 
lowing is the smart answer that he received : — 

*' When the day after to-morrow is yesterday, 
to-day will be as far from Sunday as to-day was 
from Simday when the day before yesterday 
was to-morrow." 

Can the reader say what day of the week it 
was ? It is pretty evident that the countryman 
was not such a fool as he looked. The gentle- 
man went on his road a puzzled but a wiser man. 



LOCOMOTION AND SPEED 
PUZZLES. 

"The race is not to the swift." — Ecclesiastes 
ix. II. 

67.— AVERAGE SPEED. 

In a recent motor ride it was found that we had 
gone at the rate of ten miles an hour, but we did 
the return journey over the same route, owing 
to the roads being more clear of traffic, at fifteen 
miles an hour. What was our average speed ? 
Do not be too hasty in your answer to this 
simple little question, or it is pretty certain 
that you will be wrong. 

68.— THE TWO TRAINS. 

I PUT this little question to a stationmaster, 
and his correct answer was so prompt that I am 



12 



AMUSEMENTS IN MATHEMATICS. 



convinced there is no necessity to seek talented 
railway officials in America or elsewhere. 

Two trains start at the same time, one from 
London to Liverpool, the other from Liverpool 
to London. If they arrive at their destinations 
one hour and four hours respectively after pass- 
ing one another, how much faster is one train 
running than the other ? 

69.— THE THREE VILLAGES. 

I SET out the other day to ride in a motor-car 
from Acrefield to Butterford, but by mistake I 
took the road going via Cheesebury, which is 
nearer Acrefield than Butterford, and is twelve 
miles to the left of the direct road I should have 
travelled. After arriving at Butterford I found 
that I had gone thirty-five miles. What are the 
three distances between these villages, each being 
a whole number of miles ? I may mention that 
the three roads are quite straight. 



half miles an hour, so that it takes her just six 
hours to make the double journey. Can any of 
you tell me how far it is from the bottom of the 
hiU to the top ? " 

71.— SIR EDWYN DE TUDOR. 

In the illustration we have a sketch of Sir Edw3ni 
de Tudor going to rescue his lady-love, the fair 
Isabella, who was held a captive by a neighbour- 
ing wicked baron. Sir Edwyn calculated that if 
he rode fifteen miles an hour he would arrive at 
the castle an hour too soon, while if he rode ten 
miles an hour he would get there just an hour 
too late. Now, it was of the first importance 
that he should arrive at the exact time ap- 
pointed, in order that the rescue that he had 
planned should be a success, and the time of the 
tryst was five o'clock, when the captive lady 
would be taking her afternoon tea. The puzzle 
is to discover exactly how far Sir Edwyn de 
Tudor had to ride, 







... ^^■- 






'^c 



70. -DRAWING HER PENSION. 

'* Speaking of odd figures," said a gentleman 
who occupies some post in a Government office, 
" one of the queerest characters I know is an 
old lame widow who climbs up a hill every week 
to draw her pension at the village post office. 
She crawls up at the rate of a mile and a half an 
hour and comes down at the rate of four and a 



72.— THE HYDROPLANE QUESTION. 

The inhabitants of Slocomb-on-Sea were greatly 
excited over the visit of a certain flying man. 
All the town turned out to see the flight of the 
wonderful hydroplane, and, of course, Dobson 
and his family were there. Master Tommy was 
in good form, and informed his father that Eng- 
lishmen made better airmen than Scotsmen 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



13 



and Irishmen because they are not so heavy. 
" How do you make that out ? " asked Mr. 
Dobson. " Well, you see," Tommy replied, " it 
is true that in Ireland there are men of Cork and 
in Scotland men of A)nr, which is better still, but 
in England there are lightermen." Unfortu- 
nately it had to be explained to Mrs. Dobson, 
and this took the edge ofi the thing. The hydro- 
plane flight was from Slocomb to the neighbour- 
ing watering-place Poodleville — five miles dis- 
tant. But there was a strong wind, which so 
helped the airman that he made the outward 
journey in the short time of ten minutes, though 
it took him an hour to get back to the starting 
point at Slocomb, with the wind dead against 
him. Now, how long would the ten miles have, 
taken him if there had been a perfect calm ? 
Of course, the hydroplane's engine worked uni- 
formly throughout. 

73.— DONKEY RIDING. 

During a visit to the seaside Tommy and Evan- 
geline insisted on having a donkey race over 
the mile course on the sands. Mr. Dobson and 
some of his friends whom he had met on the 
beach acted as judges, but, as the donkeys were 
familiar acquaintances and declined to part 
company the whole way, a dead heat was un- 
avoidable. However, the judges, being sta- 
tioned at difl'erent points on the course, which 
was marked off in quarter-miles, noted the fol- 
lowing results : — ^The first three-quarters were 
run in six and three-quarter minutes, the first 
half-mile took the same time as the second half, 
and the third quarter was run in exactly the 
same time as the last quarter. From these re- 
sults Mr. Dobson amused himself in discovering 
just how long it took those two donkeys to run 
the whole mile. Can you give the answer ? 



74.--THE BASKET OF POTATOES. 

A MAN had a basket containing fifty potatoes. 
He proposed to his son, as a little recreation, 
that he should place these potatoes on the 
ground in a straight line. The distance between 
the first and second potatoes was to be one yard, 
between the second and third three yards, be- 
tween the third and fourth five yards, between 
the fourth and fifth seven yards, and so on — an 
increase of two yards for every successive potato 
laid down. Then the boy was to pick them up 
and put them in the basket one at a time, the 
basket being placed beside the first potato. 
How far woiild the boy have to travel to accom- 
plish the feat of picking them all up ? We will 
not consider the journey involved in placing 
the potatoes, so that he starts from the basket 
with them aU laid out. 



75.— THE PASSENGER'S FARE. 

At first sight you would hardly think there was 
matter for dispute in the question involved in 
the following little incident, yet it took the two 
persons concerned some little time to come to 



an agreement. Mr. Smithers hired a motor-car 
to take him from Addleford to Clinkerville and 
back again for £3. At Bakenham, just midway, 
he picked up an acquaintance, Mr. Tompkins, 
and agreed to take him on to Clinkerville and 
bring him back to Bakenham on the return 
journey. How much should he have charged 
the passenger ? That is the question. What 
was a reasonable fare for Mr. Tompkins ? 



DIGITAL PUZZLES. 

" Nine worthies were they called." 

Dryden : The Flower and the Leaf. 

I GIVE these puzzles, dealing with the nine digits, 
a class to themselves, because I have always 
thought that they deserve more consideration 
than they usually receive. Beyond the mere 
trick of " casting out nines," very little seems 
to be generally known of the laws involved in 
these problems, and yet an acquaintance with 
the properties of the digits often supplies, among 
other uses, a certain number of arithmetical 
checks that are of real value in the saving of 
labour. Let me give just one example — the 
first that occurs to me. 

If the reader were required to determine 
whether or not 15,763,530,163,289 is a square 
number, how would he proceed ? If the number 
had ended with a 2, 3, 7, or 8 in the digits place, 
of course he would know that it could not be a 
square, but there is nothing in its apparent form 
to prevent its being one. I suspect that in such 
a case he would set to work, with a sigh or a 
groan, at the laborious task of extracting the 
square root. Yet if he had given a little atten- 
tion to the study of the digital properties of 
numbers, he woidd settle the question in this 
simple way. The sum of the digits is 59, the 
sum of which is 14, the sum of which is 5 (which 
I call the " digital root "), and therefore I know 
that the number cannot be a square, and for 
this reason. The digital root of successive 
square numbers from i upwards is always i, 4, 
7, or 9, and can never be anything else. In fact, 
the series, 1, 4, 9, 7, 7, 9, 4, i, 9, is repeated into 
infinity. The analogous series for triangular 
numbers is i, 3, 6, i, 6, 3, i, 9, 9. So here we 
have a similar negative check, for a number 

cannot be triangular (that is, ) if its digital 

2 

root be 2, 4, 5, 7, or 8. 



76.— THE BARREL OF BEER. 

A man bought an odd lot of wine in barrels and 
one barrel containing beer. These are shown in 
the illustration, marked with the number of 
gallons that each barrel contained. He sold a 
quantity of the wine to one man and twice the 
quantity to another, but kept the beer to him- 
self. The puzzle is to point out which barrel 
contains beer. Can you say which one it is ? 
Of course, the man sold the barrels just as h« 



14 



AMUSEMENTS IN MATHEMATICS. 




bought them, without manipulating in any way 
the contents. 

77.— DIGITS AND SQUARES. 

It will be seen in the diagram that we have so 
arranged the nine digits in a square that the 
number in the second row is twice that in the 



1 


9 


2 


3 


8 


4 


5 


7 


6 



first row, and the number in the bottom row 
three times that in the top row. There are 
three other ways of arranging the digits so 
as to produce the same result. Can you find 
them ? 

78.— ODD AND EVEN DIGITS. 

The odd digits, i, 3, 5, 7, and 9, add up 25, 
while the even figures, 2, 4, 6, and 8, only add up 
20. Arrange these figures so that the odd ones 
and the even ones add up alike. Complex and 
improper fractions and recurring decimals are 
not allowed. 

79— THE LOCKERS PUZZLE. 



gram. He told his clerk to place a different 
one-figure number on each locker of cupboard 
A, and to do the same in the case of B, and of C. 
As we are here allowed to call nought a digit, 
and he was not prohibited from usmg nought 
as a nmnber, he clearly had the option of 
omitting any one of ten digits from each cup- 
board. 

Now, the employer did not say the lockers 
were to be numbered in any numerical order, 
and he was surprised to find, when the work was 
done, that the figures had apparently been mixed 
up indiscriminately. Calling upon his clerk for 
an explanation, the eccentric lad stated that the 
notion had occurred to him so to arrange the 
figures that in each case they formed a simple 
addition sum, the two upper rows of figures pro- 
ducing the sum in the lowest row. But the 
most surprising point was this : that he had so 
arranged them that the addition in A gave the 
smallest possible sum, that the addition in C 
gave the largest possible sum, and that all the 
nine digits in the three totals were different. 
The puzzle is to show how this could be done. 
No decimals are allowed and the nought may 
not appear in the hundreds place. 

80.— THE THREE GROUPS. 

There appeared in " Nouvelles Annales de 
Mathematiques " the following puzzle as a modi- 
fication of one of my " Canterbury Puzzles." 
Arrange the nine digits in three groups of two, 
three, and four digits, so that the first two 
numbers when multiplied together make the 
third. Thus, 12 X 483=5,796. I now also pro- 
pose to include the cases where there are one, 
four, and four digits, such as 4x1,738=6,952. 
Can you find all the possible solutions in both 
cases ? 

81.— THE NINE COUNTERS. 










DDD 



DDD 
DDD 



DDD 




A MAN had in his office three cupboards, each 
containing nine lockers, as shown in the dia- 



I HAVE nine counters, each bearing one of the 
niae digits, i, 2, 3, 4, 5, 6, 7, 8 and 9. I arranged 
them on the table in two groups, as shown in 
the illustration, so as to form two multiplica- 
tion sums, and found that both sums gave 
the same product. You will find that 
158 multiplied by 23 is 3,634, and that 
79 multiplied by 46 is also 3,634. Now, 
the puzzle I propose is to rearrange the 
counters so as to get as large a product 
as possible. What is the best way of 
placing them? Remember both groups 
must multiply to the same amount, and 
there must be three counters multiplied 
by two in one case, and two multiplied 
by two counters in the other, just as at 
present. 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



15 



82.— THE TEN COUNTERS. 

In this case we use the nought in addition to the 
i> 2, 3, 4, 5, 6, 7, 8, 9. The puzzle is, as in the 
last case, so to arrange the ten counters that 
the products of the two multiplications shall be 
the same, and you may here have one or more 
figures in the multiplier, as you choose. The 
above is a very easy feat ; but it is also required 
to find the two arrangements giving pairs of the 
highest and lowest products possible. Of course 
every counter must be used, and the cipher may 
not be placed to the left of a row of figures where 
it would have no effect. Vulgar fractions or 
decimals are not allowed. 

83.— DIGITAL MULTIPLICATION. 

Here is another entertaining problem with the 
nine digits, the nought being excluded. Using 
each figure once, and only once, we can form 
two multiplication sums that have the same 
product, and this may be done in many ways. 
For example, 7x658 and 14x329 contain all 
the digits once, and the product in each case is 
the same — 4,606. Now, it will be seen that the 
sum of the digits in the product is 16, which is 
neither the highest nor the lowest sum so ob- 
tainable. Can you find the solution of the 
problem that gives the lowest possible sum of 
digits in the common product ? Also that 
which gives the highest possible sum ? 

84.— THE PIERROT'S PUZZLE. 



Pierrot as in the example shown, or to place a 
single figure on one side and three figures on 
the other. If we only used three digits instead 
of four, the only possible ways are these : 
3 multipUed by 51 equals 153, and 6 multiplied 
by 21 equals 126. 

85.— THE CAB NUMBERS. 

A London policeman one night saw two cabs 
drive off in opposite directions under suspicious 
circumstances. This officer was a particularly 
careful and wide-awake man, and he took out 
his pocket-book to make an entry of the num- 
bers of the cabs, but discovered that he had 
lost his pencil. Luckily, however, he fotmd a 
small piece of chalk, with which he marked 
the two numbers on the gateway of a wharf 
close by. When he returned to the same spot 
on his beat he stood and looked again at the 
numbers, and noticed this peculiarity, that all 
the nine digits (no nought) were used and that 
no figure was repeated, but that if he multiplied 
the two numbers together they again produced 
the nine digits, aU once, and once only. When 
one of the clerks arrived at the wharf in the 
early morning, he observed the chalk marks 
and carefully rubbed them out. As the police- 
man could not remember them, certain mathe- 
maticians were then consulted as to whether 
there was any known method for discovering 
all the pairs of numbers that have the peculi- 
arity that the officer had noticed ; but they 
knew of none. The investigation, however, was 




The Pierrot in the illustration is standing in a 
posture that represents the sign of multiplica- 
tion. He is indicating the peculiar fact that 
15 multiplied by 93 produces exactly the same 
figures (1,395). differently arranged. The puzzle 
is to take any four digits you Hke (all different) 
and similarly arrange them so that the number 
formed on one side of the Pierrot when multi- 
plied by the number on the other side shaU 
produce the same figures. There are very few 
ways of doing it, and I shaU give all the cases 
possible. Can you find them all ? You are 
allowed to put two figures on each side of the 



interesting, and the following question out of 
many was proposed : What two numbers, con- 
taining together all the nine digits, will, when 
multiplied together, produce another number 
(the highest possible) containing also aU the nine 
digits ? The nought is not allowed an3rwhere. 

86.— QUEER MULTIPLICATION. 

If I multiply 51,249,876 by 3 (thus using all the 
nine digits once, and once only), I get 153,749,628 
(which again contains sdl the nine digits once). 
Similarly, if I multiply 16,583,742 by 9 the 



i6 



AMUSEMENTS IN MATHEMATICS. 



result is 149,253,678, where in each case all the 
nine digits are used. Now, take 6 as your multi- 
plier and try to arrange the remaining eight 
digits so as to produce by multiplication a 
number containing all nine once, and once 
only. You will find it far from easy, but it 
can be done. 



■THE NUMBER-CHECKS PUZZLE. 




Where a large number of workmen are em- 
ployed on a building it is customary to provide 
every man with a little disc bearing his number. 
These are hung on a board by the men as they 
arrive, and serve as a check on punctuaUty. 
Now, I once noticed a foreman remove a number 
of these checks from his board and place them 
on a split-ring which he carried in his pocket. 
This at once gave me the idea for a good puzzle. 
In fact, I win confide to my readers that this is 
just how ideas for puzzles arise. You cannot 
really create an idea : it happens — and you 
have to be on the alert to seize it when it does 
so happen. 

It wiU be seen from the illustration that there 
are ten of these checks on a ring, numbered 
I to 9 and o. The puzzle is to divide them into 
three groups without taking any ofi the ring, 
so that the first group multiplied by the second 
makes the third group. For example, we can 
divide them into the three groups, 2 — 8 9 o 7 — 
I 5 4 6 3, by bringing the 6 and the 3 round to 
the 4, but unfortunately the first two when multi- 
pUed together do not make the third. Can you 
separate them correctly ? Of course you may 
have as many of the checks as you Uke in any 
group. The puzzle calls for some ingenuity, 
unless you have the luck to hit on the answer 
by chance. 

88.— DIGITAL DIVISION. 

It is another good puzzle so to arrange the nine 
digits (the nought excluded) into two groups so 
that one group when divided by the other pro- 
duces a given number without remainder. For 
example, 13458 divided by 6 7 2 9 gives 2. Can 
the reader find similar arrangements produc- 
ing 3, 4, 5, 6, 7, 8, and 9 respectively ? Also, 



can he find the pairs of smallest possible num- 
bers in each case ? Thus, 14658 divided by 
7 3 2 9 is just as correct for 2 as the other 
example we have given, but the numbers are 
higher. 

89.— ADDING THE DIGITS. 

If I write the sum of money, £987, 5s. 4id., and 
add up the digits, they sum to 36. No digit 
has thus been used a second time in the amount 
or addition. This is the largest amount possible 
under the conditions. Now find the smallest 
possible amount, pounds, shillings, pence, and 
farthings being all represented. You need not 
use more of the nine digits than you choose, but 
no digit may be repeated throughout. The 
nought is not allowed. 

90.— THE CENTURY PUZZLE. 

Can you write 100 in the form of a mixed num- 
ber, using all the nine digits once, and only once ? 
The late distinguished French mathematician, 
Edouard Lucas, found seven difierent ways of 
doing it, and expressed his doubts as to there 
being any other ways. As a matter of fact there 
are just eleven ways and no more. Here is one 
of them, 9iV^- Nine of the other ways have 
similarly two figures in the integral part of the 
number, but the eleventh expression has only 
one figure there. Can the reader find this last 
form ? 

91.— MORE MIXED FRACTIONS. 

When I first published my solution to the last 
puzzle, I was led to attempt the expression of 
all numbers in turn up to 100 by a mixed frac- 
tion containing aU the nine digits. Here are 
twelve numbers for the reader to try his hand 
at : 13, 14, 15, 16, 18, 20, 27, 36, 40, 69, 72, 94. 
Use every one of the nine digits once, and only 
once, in every case. 

92.— DIGITAL SQUARE NUMBERS. 

Here are the nine digits so arranged that they 
form four square numbers : 9, 81, 324, 576. 
Now, can you put them all together so as to 
form a single square number — (I) the smallest 
possible, and (II) the largest possible ? 

93.— THE MYSTIC ELEVEN. , 

Can you find the largest possible number con- 
taining any nine of the ten digits (calling nought 
a digit) that can be divided by 11 without a re- 
mainder ? Can you also find the smallest pos- 
sible number produced in the same way that is 
divisible by 11 ? Here is an example, where 
the digit 5 has been omitted : 896743012. This 
number contains nine of the digits and is divis- 
ible by II, but it is neither the largest nor the 
smallest number that wiU work. 

94.— THE DIGITAL CENTURY. 

12 3 4 5 6 7 8 9=100. 

It is required to place arithmetical signs be- 
tween the nine figures so that they shall equal 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



17 



100. Of course, you must not alter the present 
numerical arrangement of the figures. Can you 
give a correct solution that employs (i) the 
fewest possible signs, and (2) the fewest possible 
separate strokes or dots of the pen ? That is, 
it is necessary to use as few signs as possible, 
and those signs should be of the simplest form. 
The signs of addition and multiplication (+ and 
X) will thus count as two strokes, the sign of 
sulatraction ( — ) as one stroke, the sign of 
division (t-) as three, and so on, 

95.— THE FOUR SEVENS. 



(5+5)x(5>5) 








In the illustration Professor Rackbrane is seen 
demonstrating one of the little posers with 
which he is accustomed to entertain his class. 
He believes that by taking his pupils ofE the 
beaten tracks he is the better able to secure 
their attention, and to induce original and in- 
genious methods of thought. He has, it will be 
seen, just shown how four 5's may be written 
with simple arithmetical signs so as to represent 
100. Every juvenUe reader will see at a glance 
that his example is quite correct. Now, what 
he wants you to do is this : Arrange four 7's 
(neither more nor less) with arithmetical signs 
so that they shall represent 100. If he had 
said we were to use four 9's we might at once 
have written 99 9-9, but the four 7's call for 
rather more ingenuity. Can you discover the 
little trick ? 

96.— THE DICE NUMBERS. 

I HAVE a set of four dice, not marked with spots 
in the ordinary way, but with Arabic figures, as 
shown in the illustration. Each die, of course, 
bears the numbers i to 6. When put together 
(1,926) 



they will form a good many different numbers. 
As represented they make the number 1246. 
Now, if I make all the different four-figure 



y^^:^-7-g^Ts^, 



1 12 4 6 



numbers that are possible with these dice (never 
putting the same figure more than once in any 
number), what will they all add up to ? You 
are allowed to turn the 6 upside down, so as to 
represent a 9. I do not ask, or expect, the 
reader to go to all the labour of writing out the 
fxill list of numbers and then adding them up. 
Life is not long enough for such wasted energy. 
Can you get at the answer in any other way ? 



VARIOUS ARITHMETICAL AND 
ALGEBRAICAL PROBLEMS. 

" Variety's the very spice of life, 
That gives it all its flavour." 

CowpER : The Task. 

97.— THE SPOT ON THE TABLE. 

A BOY, recently home from school, wished to 
give his father an exhibition of his precocity. 
He pushed a large circular table into the comer 
of the room, as shown in the illustration, so that 
it touched both walls, and he then pointed to 
a spot of ink on the extreme edge. 




" Here is a little puzzle for you, pater," said 
the youth. " That spot is exactly eight inches 
from one wall and nine inches from the Other. 
Can you tell me the diameter of the table with- 
out measuring it ? " 

The boy was overheard to tell a friend, " It 



i8 



AMUSEMENTS IN MATHEMATICS. 



fairly beat the guv'nor ; " but his father is known 
to have remarked to a City acquamtance that 
he solved the thing in his head in a minute. I 
often wonder which spoke the truth. 

98.— ACADEMIC COURTESIES. 

In a certain mixed school, where a special fea- 
ture was made of the inculcation of good man- 
ners, they had a curious rule on assembling 
every morning. There were twice as many girls 
as boys. Every girl made a bow to every other 
girl, to every boy, and to the teacher. Every 
boy made a bow to every other boy, to every 
girl, and to the teacher. In all there were nine 
hundred bows made in that model academy 
every morning. Now, can you say exactly how 
many boys there were in the school ? If you 
are not very careful, you are likely to get a good 
deal out in your calculation. 

99.— THE THIRTY-THREE PEARLS. 




" A MAN I know," said Teddy Nicholson at a 
certain family party, " possesses a string of 
thirty-three pearls. The middle pearl is the 
largest and best of all, and the others are so 
selected and arranged that, starting from one 
end, each successive pearl is worth £100 
more than the preceding one, right up to the 
big pearl. From the other end the pearls in- 
crease in value by £150 up to the large pearl. 
The whole string is worth £65,000. What is the 
value of that large pearl ? " 

" Pearls and other articles of clothing," said 
Uncle Walter, when the price of the precious 
gem had been discovered, remind me of Adam 



and Eve. Authorities, you may not know, 
differ as to the number of apples that were eaten 
by Adam and Eve. It is the opinion of some 
that Eve 8 (ate) and Adam 2 (too), a total of 10 
only. But certain mathematicians have figured 
it out differently, and hold that Eve 8 and Adam 
8, a total of 16. Yet the most recent investi- 
gators think the above figures entirely wrong, 
for if Eve 8 and Adam 8 2, the total must be 90." 

" Well," said Harry, " it seems to me that if 
there were giants in those days, probably Eve 
8 I and Adam 8 2, which would give a total of 
163." 

" I am not at all satisfied," said Maud. " It 
seems to me that if Eve 8 i and Adam 812, 
they together consumed 893." 

" I am sure you are aU wrong," insisted Mr. 
Wilson, " for I consider that Eve 814 Adam, 
and Adam 8124 Eve, so we get a total of 
8,938." 

"But, look here," broke in Herbert. "If 
Eve 814 Adam and Adam 81242 oblige Eve, 
surely the total must have been 82,056 ! " 

At this point Uncle Walter suggested that 
they might let the matter rest. He declared 
it to be clearly what mathematicians call an 
indeterminate problem. 

100.— THE LABOURER'S PUZZLE. 

Professor Rackbrane, during one of his 
rambles, chanced to come upon a man digging 
a deep hole. 

" Good morning," he said. " How deep is 
that hole ? " 

" Guess," replied the labourer. " My height 
is exactly five feet ten inches." 

" How much deeper are you going ? " said 
the professor. 

" I am going twice as deep," was the answer, 
" and then my head will be twice as far below 
ground as it is now above ground." 

Rackbrane now asks if you could tell how 
deep that hole would be when finished. 

loi.— THE TRUSSES OF HAY. 

Farmer Tompkins had five trusses of hay, 
which he told his man Hodge to weigh before 
delivering them to a customer. The stupid fel- 
low weighed them two at a time in all possible 
ways, and informed his master that the weights 
in pounds were no, 112, 113, 114, 115, 116, 1x7, 
118, 120, and 121. Now, how was Farmer 
Tompkins to find out from these figures how 
much every one of the five trusses weighed 
singly ? The reader may at first think that he 
ought to be told " which pair is which pair," 
or something of that sort, but it is quite un- 
necessary. Can you give the five correct 
weights ? 

102.— MR. GUBBINS IN A FOG. 

Mr. Gubbins, a diligent man of business, was 
much inconvenienced by a London fog. The 
electric light happened to be out of order and 
he had to manage as best he could with two 
candles. His clerk assured him that though 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



19 



both were of the same length one candle would 
burn for four hours and the other for five hours. 
After he had been working some time he put 
the candles out as the fog had lifted, and he 
then noticed that what remained of one candle 
was exactly four times the length of what was 
left of the other. 

When he got home that night Mr. Gubbins, who 
liked a good puzzle, said to himself, " Of course 
it is possible to work out just how long those 
two candles were burning to-day. I'U have a 
shot at it." But he soon found himself in a 
worse fog than the atmospheric one. Could 
you have assisted him in his dilemma ? How 
long were the candles burning ? 

103.— PAINTING THE LAMP-POSTS. 

Tim Murphy and Pat Donovan were engaged 
by the local authorities to paint the lamp-posts 
in a certain street. Tim, who was an early 
riser, arrived first on the job, and had painted 
three on the south side when Pat turned up and 
pointed out that Tim's contract was for the 
north side. So Tim started afresh on the north 
side and Pat continued on the south. When 
Pat had finished his side he went across the 
street and painted six posts for Tim, and then 
the job was finished. As there was an equal 
number of lamp-posts on each side of the street, 
the simple question is : Which man painted the 
more lamp-posts, and just how many more ? 

104.— CATCHING THE THIEF. 

" Now, constable," said the defendant's counsel 
in cross-examination, " you say that the prisoner 
was exactly twenty-seven steps ahead of you 
when you started to rim after him ? " 

" Yes, sir." 

" And you swear that he takes eight steps to 
your five ? " 

" That is so." 

" Then I ask you, constable, as an intelligent 
man, to explain how you ever caught him, if 
that is the case ? " 

" Well, you see, I have got a longer stride. 
In fact, two of my steps are equal in length to 
five of the prisoner's. If you work it out, you 
will find that the number of steps I required 
would bring me exactly to the spot where I 
captured him." 

Here the foreman of the jury asked for a few 
minutes to figure out the number of steps the 
constable must have taken. Can you also say 
how many steps the officer needed to catch the 
thief ? 

105.— THE PARISH COUNCIL ELECTION, 

Here is an easy problem for the novice. At 
the last election of the parish council of Tittle- 
bury-in-the-Marsh there were twenty-three can- 
didates for nine seats. Each voter was qualified 
to vote for nine of these candidates or for any 
less number. One of the electors-wants to know 
in just how many different ways it was possible 
for him to vote. 



106.— THE MUDDLETOWN ELECTION. 

At the last Parliamentary election at Muddle- 
town 5,473 votes were polled. The Liberal was 
elected by a majority of 18 over the Conserva- 
tive, by 146 over the Independent, and by 575 
over the Sociahst. Can you give a simple rule 
for figuring out how many votes were polled for 
each candidate ? 

107.— THE SUFFRAGISTS' MEETING. 

At a recent secret meeting of Suffragists a serious 
difEerence of opinion arose. This led to a split, 
and a certain number left the meeting, " I 
had half a mind to go myself," said the chair- 
woman, "and if I had done so, two-thirds of 
us would have retired." " True," said another 
member; "but if I had persuaded my friends 
Mrs. Wild and Christine Armstrong to remain 
we should only have lost half our number." 
Can you tell how many were present at the 
meeting at the start ? 

108,— THE LEAP-YEAR LADIES. 

Last leap-year ladies lost no time in exercising 
the privilege of making proposals of marriage. 
If the figures that reached me from an occult 
source are correct, the following represents the 
state of affairs in this country. 

A number of women proposed once each, of 
whom one-eighth were widows. In conse- 
quence, a number of men were to be married 
of whom one-eleventh were widowers. Of the 
proposals made to widowers, one-fifth were de- 
clined. AU the widows were accepted. Thirty- 
five forty-fourths of the widows married bache- 
lors. One thousand two hundred and twenty- 
one spinsters were declined by bachelors. The 
number of spinsters accepted by bachelors was 
seven times the number of widows accepted by 
bachelors. Those are aU the particulars that I 
was able to obtain. Now, how many women 
proposed ? 

109.— THE GREAT SCRAMBLE. 

After dinner, the five boys of a household 
happened to find a parcel of sugar-plums. It 
was quite unexpected loot, and an exciting 
scramble ensued, the full details of which I wiU 
recount with accuracy, as it forms an interesting 
puzzle. 

You see, Andrew managed to get possession 
of just two-thirds of the parcel of sugar-plums. 
Bob at once grabbed three-eighths of these, and 
Charlie managed to seize three-tenths also. 
Then young David dashed upon the scene, and 
captured aU that Andrew had left, except one- 
seventh, which Edgar artfully secured for him- 
self by a cimning trick. Now the fun began in 
real earnest, for Andrew and Charlie jointly set 
upon Bob, who stumbled against the fender and 
dropped half of all that he had, which were 
equally picked up by David and Edgar, who 
had crawled imder a table and were waiting. 
Next, Bob sprang on CharUe from a chair, and 
upset all the latter's collection on to the floor. 
Of this prize Andrew got just a quarter, Bob 



20 



AMUSEMENTS IN MATHEMATICS. 



gathered up one-third, David got two-sevenths, 
while Charlie and Edgar divided equally what 
was left of that stock. 

They were just thinking the fray was over 




when David suddenly struck out in two direc- 
tions at once, upsetting three-quarters of what 
Bob and Andrew had last acquired. The two 
latter, with the greatest difficulty, recovered 
five-eighths of it in equal shares, but the three 
others each carried ofE one-fifth of the same. 
Every sugar-plum was now accounted for, and 
they called a truce, and divided equally amongst 
them the remainder of the parcel. What is the 
smallest number of sugar-plums there could 
have been at the start, and what proportion did 
each boy obtain ? 

no.— THE ABBOT'S PUZZLE. 

The first English puzzUst whose name has come 
down to us was a Yorkshireman — ^no other than 
Alciiin, Abbot of Canterbury (a.d. 735-804). 
Here is a little puzzle from his works, which is 
at least interesting on account of its antiquity. 
"If 100 bushels of corn were distributed among 
100 people in such a manner that each man re- 
ceived three bushels, each woman two, and each 
chUd half a bushel, how many men, women, and 
children were there ? " 

Now, there are six different correct answers, 
if we exclude a case where there would be no 
women. But let us say that there were just 
five times as many women as men, then what 
is the correct solution ? 

III.— REAPING THE CORN. 

A FARMER had a square cornfield. The corn was 
all ripe for reaping, and, as he was short of men, 
it was arranged that he and his son should share 
the work between them. The farmer first cut 
one rod wide all round the square, thus leaving 
a smaller square of standing corn in the middle 



of the field. " Now," he said to his son, " I 
have cut my half of the field, and you can do 
your share." The son was not quite satisfied 
as to the proposed division of labour, and as the 
village schoolmaster happened to be passing, he 
appealed to that person to decide the matter. 
He fovmd the farmer was quite correct, provided 
there was no dispute as to the size of the field, 
and on this point they were agreed. Can you 
tell the area of the field, as that ingenious school- 
master succeeded in doing ? 

112.— A PUZZLING LEGACY. 

A MAN left a hundred acres of land to be divided 
among his three sons — Alfred, Benjamin, and 
Charles — in the proportion of one-third, one- 
fourth, and one-fifth respectively. But Charles 
died. How was the land to be divided fairly 
between Alfred and Benjamin ? 

113.— THE TORN NUMBER. 

I HAD the other day in my possession a label 
bearing the number 3 o 2 5 in large figures. This 
got accidentally torn in half, so that 3 o was on 
one piece and 2 5 on the other, as shown on the 
illustration. On looking at these pieces I began 




to make a calculation, scarcely conscious of what 
I was doing, when I discovered this little peculi- 
arity. If we add the 3 o and the 2 5 together 
and square the sum we get as the result the com- 
plete original number on the label! Thus, 30 
added to 2 5 is 5 5, and 5 5 multiplied by 5 5 is 
3025. Curious, is it not ? Now, the puzzle is 
to find another number, composed of four fig- 
ures, all different, which may be divided in the 
middle and produce the same result. 

114.— CURIOUS NUMBERS. 

The number 48 has this peculiarity, that if you 
add I to it the result is a square number (49, the 
square of 7), and if you add i to its hali, you 
also get a square nimiber (25, the square of 5). 
Now, there is no Umit to the numbers that have 
this peculiarity, and it is an interesting puzzle 
to find three more of them — the smallest possible 
numbers. What are they ? 

ii5._A PRINTER'S ERROR. 

In a certain article a printer had to set up the 
figures 5^.2'*, which, of course, means that the 
fourth power of 5 {625) is to be multiplied by the 
cube of 2 (8), the product of which is 5,000. But 
he printed 5^.2^ as 5 4 2 3, which is not correct. 
Can you place four digits in the manner shown, 
so that it will be equally correct if the printer 
sets it up aright or makes the same blunder ? 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



21 



ii6.— THE CONVERTED MISER. 

Mr. Jasper Bullyon was one of the very few 
misers who have ever been converted to a sense 
of their duty towards their less fortunate fellow- 
men. One eventful night he counted out his 
accumulated wealth, and resolved to distribute 
it amongst the deserving poor. 

He foimd that if he gave away the same 
number of pounds every day in the year, he 
could exactly spread it over a twelvemonth 
without there being anything left over ; but if 
he rested on the Sundays, and only gave away 
a fixed number of pounds every week-day, there 
would be one sovereign left over on New Year's 
Eve. Now, putting it at the lowest possible, 
what was the exact number of pounds that he 
had to distribute ? 

Could any question be simpler ? A sum of 
pounds divided by one number of days leaves 
no remainder, but divided by another number 
of days leaves a sovereign over. That is all ; 
and yet, when you come to tackle this little 
question, you will be surprised that it can be- 
come so puzzling. 

117.— A FENCE PROBLEM. 



<y 



%i 




The practical usefulness of puzzles is a point 
that we are liable to overlook. Yet, as a matter 
of fact, I have from time to time received quite 
a large number of letters from individuals who 
have found that the mastering of some little 
principle upon which a puzzle was built has 
proved of considerable value to them in a most 
unexpected way. Indeed, it may be accepted 
as a good maxim that a puzzle is of little real 
value unless, as well as being amusing and per- 
plexing, it conceals some instructive and pos- 
sibly useful featiure. It is, however, very curi- 
ous how these little bits of acquired knowledge 
dovetail into the occasional requirements of 
everyday life, and equally curious to what 
strange and mysterious uses some of our readers 
seem to apply them. What, for example, can 
be the object of Mr. Wm. Oxley, who writes to 
me all the way from Iowa, in wishing to ascer- 
tain the dimensions of a field that he proposes 
to enclose, containing just as many acres as 
there shall be raUs in the fence ? 



The man wishes to fence in a perfectly square 
field which is to contain just as many acres as 
there are rails in the required fence. Each 
hurdle, or portion of fence, is seven rails high, 
and two lengths would extend one pole (16^ ft.) : 
that is to say, there are fourteen rails to the pole, 
lineal measure. Now, what must be the size 
of the field ? 

118.— CIRCLING THE SQUARES. 




The puzzle is to place a different number in each 
of the ten squares so that the sum of the squares 
of any two adjacent numbers shall be equal to 
the sum of the squares of the two numbers dia- 
metrically opposite to them. The four numbers 
placed, as examples, must stand as they are. 
The square of 16 is 256, and the square of 2 is 4. 
Add these together, and the result is 260. Also 
— the square of 14 is 196, and the square of 8 is 
64. These together also make 260. Now, in 
precisely the same way, B and C should be equal 
to G and H (the sum will not necessarily be 260), 
A and K to F and E, H and I to C and D, and 
so on, with any two adjoining squares in the 
circle. 

All you have to do is to fill in the remaining 
six numbers. Fractions are not allowed, and 
I shall show that no number need contain 
more than two figures. 

119.— RACKBRANE'S LITTLE LOSS. 

Professor Rackbrane was spending an even- 
ing with his old friends, Mr. and Mrs. Potts, and 
they engaged in some game (he does not say 
what game) of cards. The professor lost_ the 
first game, which resulted in doubling the money 
that both Mr. and Mrs. Potts had laid on the 
table. The second game was lost by Mrs. Potts, 
which doubled the money then held by her hus- 
band and the professor. Curiously enough, the 
third game was lost by Mr. Potts, and had the 



22 



AMUSEMENTS IN MATHEMATICS. 



effect of doubling the money then held by his 
wife and the professor. It was then found that 
each person had exactly the same money, but 
the professor had lost five shillings in the course 
of play. Now, the professor asks, what was the 
sum of money with which he sat down at the 
table ? Can you tell him ? 

120.— THE FARMER AND HIS SHEEP. 

Farmer Longmore had a curious aptitude for 
arithmetic, and was known in his district as the 
" mathematical farmer." The new vicar was 
not aware of this fact when, meeting his worthy 
parishioner one day in the lane, he asked him 



will know exactly how many sheep Farmer 
Longmore owned. 

121.— HEADS OR TAILS. 

Crooks, an inveterate gambler, at Goodwood 
recently said to a friend, " I'll bet you half the 
money in my pocket on the toss of a coin — 
heads I win, tails I lose." The coin was tossed 
and the money handed over. He repeated the 
offer again and again, each time betting half the 
money then in his possession. We are not told 
how long the game went on, or how many times 
the coin was tossed, but this we know, that the 
number of times that Crooks lost was exactly 




in the course of a short conversation, " Now, 
how many sheep have you altogether ? " He 
was therefore rather surprised at Longmore's 
answer, which was as follows : " You can divide 
my sheep into two different parts, so that the 
difference between the two numbers is the same 
as the difference between their squares. Maybe, 
Mr. Parson, you will like to work out the little 
sum for yourself." 

Can the reader say just how many sheep the 
farmer had ? Supposing he had possessed only 
twenty sheep, and he divided them into the two 
parts 12 and 8. Now, the difference between 
12 and 8 is 4, but the difference between their 
squares, 144 and 64, is 80. So that will not do, 
for 4 and 80 are certainly not the same. If you 
can find numbers that work out correctly, you 



equal to the number of times that he won. Now, 
did he gain or lose by this little ventmre ? 

122.— THE SEE-SAW PUZZLE. 

Necessity is, indeed, the mother of invention. 
I was amused the other day in watching a boy 
who wanted to play see-saw and, in his failure 
to find another child to share the sport with him, 
had been driven back upon the ingenious resort 
of tying a number of bricks to one end of the 
plank to balance his weight at the other. 

As a matter of fact, he just balanced against 
sixteen bricks, when these were fixed to the short 
end of plank, but if he fixed them to the long 
end of plank he only needed eleven as balance. 

Now, what was that boy's weight, if a brick 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



23 



weighs equal to a three-quarter brick and 
three-quarters of a pound ? 

123.— A LEGAL DIFFICULTY. 

" A CLIENT of mine," said a lawyer, " was on 
the point of death when his wife was about to 
present him with a child. I drew up his will, 
in which he settled two-thirds of his estate upon 
his son (if it should happen to be a boy) and one- 
third on the mother. But if the child should be 
a girl, then two-thirds of the estate should go 
to the mother and one-third to the daughter. 
As a matter of fact, after his death twins were 
born — a boy and a girl. A very nice point then 
arose. How was the estate to be equitably 
divided among the three in the closest possible 
accordance with the spirit of the dead man's 
will ? " 

124.— A QUESTION OF DEFINITION. 

" My property is exactly a mile square," said 
one landowner to another. 

" Curiously enough, mine is a square mile," 
was the reply. 

" Then there is no difference ? " 

Is this last statement correct ? 

125.— THE MINERS' HOLIDAY. 

Seven coal-miners took a holiday at the sea- 
side during a big strike. Six of the party 
spent exactly half a sovereign each, but Bill 
Harris was more extravagant. BUI spent three 
shillings more than the average of the party. 
What was the actual amount of Bill's expen- 
diture ? 

126.— SIMPLE MULTIPLICATION. 

If we number six cards i, 2, 4, 5, 7, and 8, and 
arrange them on the table in this order : — 

142857 
we can demonstrate that in order to multiply 
by 3 all that is necessary is to remove the i to 
the other end of the row, and the thing is done. 
The answer is 428571. Can you find a number 
that, when multiplied by 3 and divided by 2, 
the answer will be the same as if we removed 
the first card (which in this case is to be a 3) 
from the beginning of the row to the end ? 

127.— SIMPLE DIVISION. 

Sometimes a very simple question in elementary 
arithmetic will cause a good deal of perplexity. 
For example, I want to divide the four numbers, 
701, 1,059, i>4i7, and 2,312, by the largest 
number possible that will leave the same re- 
mainder in every case. How am I to set to 
work ? Of course, by a laborious system of 
trial one can in time discover the answer, but 
there is quite a simple method of doing it if you 
can only find it. 

128.— A PROBLEM IN SQUARES. 

We possess three square boards. The surface 
of the first contains five square feet more than 
the second, and the second contains five square 
feet more than the third. Can you give exact 



measurements for the sides of the boards ? If 
you can solve this little puzzle, then try to find 
three squares in arithmetical progression, with 
a common difference of 7 and also of 13. 

129.— THE BATTLE OF HASTINGS. 

All historians know that there is a great deal 
of mystery and uncertainty concerning the de- 
tails of the ever-memorable battle on that fatal 
day, October 14, 1066. My puzzle deals with 
a curious passage in an ancient monldsh chron- 
icle that may never receive the attention that 
it deserves, and if I am unable to vouch for the 
authenticity of the document it will none the 
less serve to furnish us with a problem that can 
hardly fail to interest those of my readers who 
have arithmetical predilections. Here is the 
passage in question. 

" The men of Harold stood well together, as 
their wont was, and formed sixty and one 
squares, with a like number of men in every 
square thereof, and woe to the hardy Norman 
who ventured to enter their redoubts ; for a 
single blow of a Saxon war-hatchet would break 
his lance and cut through his coat of mail. . . . 
When Harold threw himself into the fray the 
Saxons were one mighty square of men, shouting 
the battle-cries, ' Ut ! ' ' Olicrosse ! ' ' Gode- 
mit6 ! ' " 

Now, I find that aU the contemporary author- 
ities agree that the Saxons did actually fight 
in this solid order. For example, in the " Car- 
men de Bello Hastingensi," a poem attributed 
to Guy, Bishop of Amiens, living at the time of 
the battle, we are told that " the Saxons stood 
fixed in a dense mass," and Henry of Hunting- 
don records that " they were like unto a castle, 
impenetrable to the Normans ; " while Robert 
Wace, a century after, tells us the same thing. 
So in this respect my newly-discovered chronicle 
may not be greatly in error. But I have reason 
to believe that there is something wrong with 
the actual figures. Let the reader see what he 
can make of them. 

The number of men would be sixty-one times 
a square number ; but when Harold himself 
joined in the fray they were then able to form 
one large square. What is the smallest possible 
number of men there could have been ? 

In order to make clear to the reader the sim- . 
plicity of the question, I will give the lowest 
solutions in the case of 60 and 62, the numbers 
immediately preceding and following 61. They 
are 60x42+1 = 312, and 62x82+1 = 632. That 
is, 60 squares of 16 men each would be 960 
men, and when Harold joined them they would 
be 961 in number, and so form a square with 
31 men on every side. Similarly in the case 
of the figures I have given for 62. Now, find 
the lowest answer for 61. 

130.— THE SCULPTOR'S PROBLEM. 

An ancient sculptor was commissioned to supply 
two statues, each on a cubical pedestal. It is 
with these pedestals that we are concerned. 
They were of unequal sizes, as will be seen in 
the illustration, and when the time arrived for 



24 



AMUSEMENTS IN MATHEMATICS. 



payment a dispute arose as to whether the agree- 
ment was based on lineal or cubical measure- 
ment. But as soon as they came to measure the 
two pedestals the matter was at once settled, be- 
cause, curiously enough, the number of lineal 
feet was exactly the same as the number of 




cubical feet. The puzzle is to find the dimen- 
sions for two pedestals having this peculiarity, 
in the smallest possible figures. You see, if the 
two pedestals, for example, measure respectively 
3 ft. and i ft. on every side, then the Uneed 
measurement would be 4 ft. and the cubical 
contents 28 ft., which are not the same, so 
these measurements will not do. 

131.— THE SPANISH MISER. 

There once lived in a small town in New Castile 
a noted miser named Don Manuel Rodriguez. 
His love of money was only equalled by a strong 
passion for arithmetical problems. These puzzles 
usually dealt in some way or other with his accu- 
mulated treasure, and were propounded by him 
solely in order that he might have the pleasure 
of solving them himself. Unfortunately very 
few of them have survived, and when travelling 
through Spain, collecting material for a proposed 
work on " The Spanish Onion as a Cause of 
National Decadence," I only discovered a very 
few. One of these concerns the three boxes 
that appear in the accompanying authentic 
portrait. 

Each box contained a different number of 
golden doubloons. The difference between the 
number of doubloons in the upper box and the 
number in the middle box was the same as the 
difference between the number in the middle 
box and the number ia the bottom box. And 
if the contents of any two of the boxes were 
united they would form a square number. 
What is the smallest number of doubloons that 
there could have been in any one of the boxes ? 



132.— THE NINE TREASURE BOXES. 

The following puzzle will illustrate the impor- 
tance on occasions of being able to fix the mini- 
mum and maximum limits of a required number. 
This can very frequently be done. For ex- 
ample, it has not yet been ascertained in how 
many different ways the knight's tour can be 
performed on the chess board ; but we know 
that it is fewer than the number of combinations 
of 168 things taken 63 at a time and is greater 
than 31,054,144 — for the latter is the number 
of routes of a particular type. Or, to take a 
more familiar case, if you ask a man how many 
coins he has in his pocket, he may tell you that 
he has not the slightest idea. But on further 
questioning you will get out of him some such 
statement as the following : " Yes, I am posi- 
tive that I have more than three coins, and 
equally certain that there are not so many as 
twenty-five." Now, the knowledge that a cer- 
tain number lies between 2 and 12 in my puzzle 
will enable the solver to find the exact answer ; 
without that information there would be an 
infinite number of answers, from which it would 
be impossible to select the correct one. 

This is another puzzle received from my friend 
Don Manuel Rodriguez, the cranky miser of 
New Castile. On New Year's Eve in 1879 he 
showed me nine treasure boxes, and after in- 
forming me that every box contained a square 
number of golden doubloons, and that the differ- 
ence between the contents of A and B was the 
same as between B and C, D and E, E and F, 
G and H, or H and I, he requested me to tell 
him the number of coins in every one of the 
boxes. At first I thought this was impossible, 
as there would be an infinite number of different 
answers, but on consideration I found that this 




was not the case. I discovered that while every 
box contained coins, the contents of A, B, C in- 



ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 



25 



creased in weight in alphabetical order ; so did 
D, E, F ; and so did G, H, I ; but D or E need 
not be heavier than C, nor G or H heavier than 

F. It was also perfectly certain that box A 
could not contain more than a dozen coins at 
the outside ; there might not be half that 
number, but I was positive that there were not 
more than twelve. With this knowledge I was 
able to arrive at the correct answer. 

In short, we have to discover nine square 
numbers such that A, B, C ; and D, E, F ; and 

G, H, I are three groups in arithmetical pro- 
gression, the common difference being the same 
in each group, and A being less than 12, How 
many doubloons were there in every one of the 
nine boxes ? 

133.— THE FIVE BRIGANDS. 

The five Spanish brigands, Alfonso, Benito, 
Carlos, Diego, and Esteban, were counting their 
spoils after a raid, when it was found that they 
had captured altogether exactly 200 doubloons. 
One of the band pointed out that if Alfonso had 
twelve times as much, Benito three times as 
much, Carlos the same amount, Diego half as 
much, and Esteban one-third as much, they 
would still have altogether just 200 doubloons. 
How many doubloons had each ? 

There are a good many equally correct an- 
swers to this question. Here is one of them : 



A. . 


. . 6 X 


12 = 72 


B. . 


12 X 


3 = 36 


C. . 


. 17 X 


I = 17 


D. . 


. 120 X 


i = 60 


E . . 


. . 45 X 


i = 15 



200 



200 



The puzzle is to discover exactly how many 
different answers there are, it being understood 
that every man had something and that there 
is to be no fractional money — only doubloons 
in every case. 

This problem, worded somewhat differently, 
was propounded by Tartaglia (died 1559), and 
he flattered himself that he had found one solu- 
tion ; but a French mathematician of note 
(M. A. Labosne), in a recent work, says that his 
readers will be astonished when he assures them 
that there are 6,639 different correct answers 
to the question. Is this so ? How many 
answers are there ? 

'134.— THE BANKER'S PUZZLE. 

A BANKER had a sporting customer who was 
always anxious to wager on anything. Hoping 
to cure him of his bad habit, he proposed as a 
wager that the customer would not be able to 
divide up the contents of a box containing only 
sixpences into an exact number of equal piles 
of sixpences. The banker was first to put in 
one or more sixpences (as many as he Uked) ; 
then the customer was to put in one or more 
(but in his case not more than a poimd in value), 
neither knowing what the other put in. Lastly, 
the customer was to transfer from the banker's 



counter to the box as many sixpences as the 
banker desired him to put in. The puzzle is to 
find how many sixpences the banker should 
first put in and how many he should ask the 
customer to transfer, so that he may have the 
best chance of winning. 

135.— THE STONEMASON'S PROBLEM. 

A STONEMASON oHce had a large number of 
cubic blocks of stone in his yard, all of exactly 
the same size. He had some very fanciful 
little ways, and one of his queer notions was to 
keep these blocks piled in cubical heaps, no two 
heaps containing the same number of blocks. 
He had discovered for himself (a fact that is 
well known to mathematicians) that if he took 
all the blocks contained in any number of heaps 
in regular order, beginning with the single cube, 
he could always arrange those on the ground so 
as to form a perfect square. This will be clear 
to the reader, because one block is a square, 
1+8 = 9 is a square, 1 + 8 + 27 = 36 is a square, 
I + 8 + 27+ 64 = 100 is a square, and so on. In 
fact, the sum of any number of consecutive 
cubes, beginning always with i, is in every case 
a square number. 

One day a gentleman entered the mason's 
yard and offered him a certain price if he would 
supply him with a consecutive number of these 
cubical heaps which should contain altogether 
a number of blocks that could be laid out to 
form a square, but the buyer insisted on more 
than three heaps and declined to take the single 
block because it contained a flaw. What was 
the smallest possible number of blocks of stone 
that the mason had to supply ? 

136.— THE SULTAN'S ARMY. 

A CERTAIN Sultan wished to send into battle an 
army that could be formed into two perfect 
squares in twelve different ways. What is the 
smallest number of men of which that army 
could be composed ? To make it clear to the 
novice, I will explain that if there were 130 men, 
they could be formed into two squares in only 
two different ways — 81 and 49, or 121 and 9. 
Of course, all the men must be used on every 
occasion. 

I37-— A STUDY IN THRIFT. 

Certain numbers are called triangular, because 
if they are taken to represent counters or coins 
they may be laid out on the table so as to form 
triangles. The number i is always regarded as 
triangular, just as i is a square and a cube 
number. Place one counter on the table — that 
is, the first triangular number. Now place two 
more counters beneath it, and you have a tri- 
angle of three counters ; therefore 3 is tri- 
angular. Next place a row of three more 
counters, and you have a triangle of six coun- 
ters ; therefore 6 is triangular. We see that 
every row of counters that we add, containing 
just one more counter than the row above it, 
makes a larger triangle. 



26 



AMUSEMENTS IN MATHEMATICS. 



Now, half the sum of any number and its 
square is always a triangular number. Thus 
half of 2 + 22 = 3; half of 3 + 32 = 6; half of 
4 + 4^=10 ; half of 5 + 52 = 15 ; and so on. So 
if we want to form a triangle with 8 counters 
on each side we shall require half of 8 + 82, or 
36 coimters. This is a pretty little property of 
numbers. Before going further, I will here say 
that if the reader refers to the " Stonemason's 
Problem " (No. 135) he will remember that the 
sum of any number of consecutive cubes be- 
ginning with I is always a square, and these 
form the series 12, 32, 52, io2, etc. It will now 
be understood when I say that one of the keys 
to the puzzle was the fact that these are always 
the squares of trieuigular numbers — that is, the 
squares of i, 3, 6, 10, 15, 21, 28, etc., any of 
which numbers we have seen will form a tri- 
angle. 

Every whole number is either triangular, or 
the sum of two triangular numbers or the 
sum of three triangular nimabers. That is, if 
we take any number we choose we can always 
form one, two, or three triangles with them. 
The number i will obviously, and imiquely, 
only form one triangle ; some numbers wiU 
only form two triangles (as 2, 4, 11, etc.) ; some 
numbers will only form three triangles (as 5, 8, 
14, etc.). Then, again, some numbers will form 
both one and two triangles (as 6), others both 
one and three triangles (as 3 and 10), others 
both two and three triangles (as 7 and 9), while 
some nimibers (like 21) will form one, two, or 
three triangles, as we desire. Now for a little 
puzzle in triangular numbers. 

Sandy McAllister, of Aberdeen, practised 
strict domestic economy, and was anxious to 
train his good wife in his own habits of thrift. 
He told her last New Year's Eve that when she 
had saved so many sovereigns that she could lay 
them all out on the table so as to form a perfect 
square, or a perfect triangle, or two triangles, 
or three triangles, just as he might choose to 
ask, he would add five pounds to her treasure. 
Soon she went to her husband with a little bag 
of £36 in sovereigns and claimed her reward. 
It will be found that the thirty-six coins wiU 
form a square (with side 6), that they will form 
a single triangle (with side 8), that they will 
form two triangles (with sides 5 and 6), and that 
they win form three triangles (with sides 3, 5, 
and 5). In each of the four cases all the thirty- 
six coins are used, as required, and Sandy there- 
fore made his wife the promised present like an 
honest man. 

The Scotsman then vmdertook to extend his 
promise for five more years, so that if next year 
the increased number of sovereigns that she has 
saved can be laid out in the same four different 
ways she will receive a second present ; if she 
succeeds in the following year she will get a 
third present, and so on until she has earned 
six presents in all. Now, how many sovereigns 
must she put together before she can win the 
sixth present ? 

What you have to do is to find five numbers, 
the smallest possible, higher than 36, that can 
be displayed in the four ways — to form a square. 



to form a triangle, to form two triangles, and to 
form three triangles. The highest of your five 
numbers will be your answer. 

138.— THE ARTILLERYMEN'S DILEMMA. 




" All cannon-balls are to be piled in square 
pyramids," was the order issued to the regiment. 
This was done. Then came the further order, 
" All p3rramids are to contain a square number 
of balls." Whereupon the trouble arose. " It 
can't be done," said the major. " Look at this 
pyramid, for example ; there are sixteen balls 
at the base, then nine, then four, then one at 
the top, making thirty balls in aU. But there 
must be six more balls, or five fewer, to make a 
square number." " It must be done," insisted 
the general. " All you have to do is to put the 
right number of balls in your pjTamids." " I've 
got it ! " said a lieutenant, the mathematical 
genius of the regiment. " Lay the balls out 
singly." " Bosh ! " exclaimed the general. 
" You can't pile one ball into a pjnramid ! " 1 
Is it really possible to obey both orders ? 

139.— THE DUTCHMEN'S WIVES. 

I WONDER how many of my readers are ac- 
quainted with the puzzle of the " Dutchmen's 
Wives " — in which you have to determine the 
names of three men's wives, or, rather, which 
wife belongs to each husband. Some thirty 
years ago it was " going the rounds," as some- 
thing quite new, but I recently discovered it in 
the Ladies' Diary for 1739-40, so it was clearly 
familiar to the fair sex over one hundred and 
seventy years ago. How many of our mothers, 
wives, sisters, daughters, and aunts could solve 
the puzzle to-day ? A far greater proportion 
than then, let us hope. 

Three Dutchmen, named Hendrick, Elas, and 
Cornelius, and their wives, Gurtriin, Katriin, 
and Anna, purchase hogs. Each buys as 
many as he (or she) gives shillings for one. Each 
husband pays altogether three guineas more 
than his wife. Hendrick buys twenty-three 
more hogs than Katriin, and Elas eleven more 



GEOMETRICAL PROBLEMS. 



27 




than Gurtriin. Now, what was the name of 
each man's wife ? 

140.— FIND ADA'S SURNAME. 

This puzzle closely resembles the last one, my 
remarks on the solution of which the reader may 
like to apply in another case. It was recently 
submitted to a Sydney evening newspaper that 
indulges in " intellect sharpeners," but was re- 
jected with the remark that it is childish and 
that they only published problems capable of 
solution ! Five ladies, accompanied by their 



daughters, bought cloth at the same shop. 
Each of the ten paid as many farthings per 
foot as she bought feet, and each mother 
spent 8s. 5}d. more than her daughter. 
Mrs. Robinson spent 6s. more than Mrs. 
Evans, who spent about a quarter as 
much as Mrs. Jones. Mrs. Smith spent 
most of aU. Mrs. Brown bought 21 yards 
more than Bessie — one of the girls. 
Annie bought 16 yards more than Mary 
and spent £3, os. 8d. more than Emily. 
The Christian name of the other girl was 
Ada. Now, what was her surname ? 

141.— SATURDAY MARKETING. 

Here is an amusing little case of mar- 
keting which, although it deals with a good 
many items of money, leads up to a ques- 
tion of a totally different character. Four 
married couples went into their village on 
a recent Saturday night to do a little 
marketing. They had to be very econo- 
mical, for among them they only pos- 
sessed forty shilling coins. The fact is, Ann 
spent rs., Mary spent 2S., Jane spent 3s., and 
Kate spent 4s. The men were rather more ex- 
travagant than their wives, for Ned Smith 
spent as much as his wife, Tom Brown twice as 
much as his wife, Bill Jones three times as much 
as his wife, and Jack Robinson four times as 
much as his wife. On the way home somebody 
suggested that they should divide what coin 
they had left equally among them. This was 
done, and the puzzling question is simply this : 
What was the surname of each woman ? Can 
you pair off the four couples ? 



GEOMETRICAL PROBLEMS. 



" God geometrizes continually." 

Plato. 

" There is no study," said Augustus de Morgan, 
" which presents so simple a beginning as that 
of geometry ; there is none in which difficulties 
grow more rapidly as we proceed." This will 
be found when the reader comes to consider the 
following puzzles, though they are not arranged 
in strict order of difficulty. And the fact that 
they have interested and given pleasure to man 
for imtold ages is no doubt due in some measure 
to the appeal they make to the eye as well as 
to the brain. Sometimes an algebraical for- 
mula or theorem seems to give pleasmre to the 
mathematician's eye, but it is probably only 
an intellectual pleasmre. But there can be no 
doubt that in the case of certain geometrical 
problems, notably dissection or superposition 
puzzles, the aesthetic faculty in man contributes 
to the delight. For example, there are probably 
few readers who will examine the various cut- 
tings of the Greek cross in the following pages 
without being in some degree stirred by a sense 
of beauty. Law and order in Nature are always 
pleasing to contemplate, but when they come 



under the very eye they seem to make a spe- 
cially strong appeal. Even the person with 
no geometrical knowledge whatever is induced 
after the inspection of such things to exclaim, 
"How very pret ty ! " In fact , I have known more 
than one person led on to a study of geometry 
by the fascination of cutting-out puzzles. I 
have, therefore, thought it well to keep these 
dissection puzzles distinct from the geometrical 
problems on more general lines. 



DISSECTION PUZZLES. 

" Take him and cut him out in little stars." 
Romeo and Juliet, iii. 2. 

Puzzles have infinite variety, but perhaps there 
is no class more ancient than dissection, cutting- 
out, or superposition puzzles. They were cer- 
tainly known to the Chinese several thousand 
years before the Christian era. And they are 
just as fascinating to-day as they can have been 
at any period of their history. It is supposed 
by those who have investigated the matter that 
the ancient Chinese philosophers used these 



28 



AMUSEMENTS IN MATHEMATICS. 



puzzles as a sort of kindergarten method of im- 
parting the principles of geometry. Whether 
this was so or not, it is certain that all good 
dissection puzzles (for the nursery type of jig- 
saw puzzle, which merely consists in cutting 
up a picture into pieces to be put together 
again, is not worthy of serious consideration) 
are reaUy based on geometrical laws. This 
statement need not, however, frighten off the 
novice, for it means little more than this, that 
geometry will give us the " reason why," if we 
are interested in knowing it, though the solu- 
tions may often be discovered by any intelligent 
person after the exercise of patience, ingenuity, 
and common sagacity. 

If we want to cut one plane figure into parts 
that by readjustment will form another figure, 
the first thing is to find a way of doing it at all, 
and then to discover how to do it in the fewest 
possible pieces. Often a dissection problem is 
quite easy apart from this limitation of pieces. 
At the time of the publication in the Weekly 
Dispatch, in 1902, of a method of cutting an 
equilateral triangle into four parts that wiU 
form a square (see No. 26, " Canterbury 
Puzzles "), no geometrician would have had any 
difficulty in doing what is required in five 
pieces : the whole point of the discovery lay 
in performing the little feat in fo\ir pieces only. 

Mere approximations in the case of these 
problems are valueless ; the solution must be 
geometrically exact, or it is not a solution at 
aU. Fallacies are cropping up now and again, 
and I shaU have occasion to refer to one or two 
of these. They are interesting merely as fal- 
lacies. But I want to say something on two 
little points that are always arising in cutting- 
out puzzles — the questions of " hanging by a 
thread " and " turning over." These points 
can best be illustrated by a puzzle that is fre- 
quently to be found in the old books, but in- 
variably with a false solution. The puzzle is 
to cut the figure shown in Fig. i into three 




Fig. I. 



Fig. 2. 



pieces that will fit together and form a half- 
square triangle. The answer that is invariably 
given is that shown in Figs, i and 2. Now, it 
is claimed that the four pieces marked C are 
really only one piece, because they may be so 
cut that they are left " hanging together by a 
mere thread." But no serious puzzle lover will 
ever admit this. If the cut is made so as to 
leave the four pieces joined in one, then it can- 
not result in a perfectly exact solution. If, on 
the other hand, the solution is to be exact, then 



there will be four pieces — or six pieces in all. 
It is, therefore, not a solution in three pieces. 

If, however, the reader will look at the solu- 
tion in Figs. 3 and 4, he will see that no such 



__ — - 

[D_ 

2£ 




Fig. 3- 



Fig. 4. 



fault can be found with it. There is no question 
whatever that there are three pieces, and the 
solution is in this respect quite satisfactory. 
But another question arises. It will be foimd 
on inspection that the piece marked F, in Fig. 3, 
is turned over in Fig. 4 — that is to say, a different 
side has necessarily to be presented. If the 
puzzle were merely to be cut out of cardboard 
or wood, there might be no objection to this 
reversal, but it is quite possible that the ma- 
terial would not admit of being reversed. There 
might be a pattern, a polish, a difference of 
texture, that prevents it. But it is generally 
understood that in dissection puzzles you are 
allowed to turn pieces over unless it is dis- 
tinctly stated that you may not do so. And 
very often a puzzle is greatly improved by the 
added condition, " no piece may be turned 
over." I have often made puzzles, too, in which 
the diagram has a small repeated pattern, and 
the pieces have then so to be cut that not only 
is there no turning over, but the pattern has to 
be matched, which cannot be done if the pieces 
are turned round, even with the proper side 
uppermost. 

Before presenting a varied series of cutting- 
out puzzles, some very easy and others difi&cult, 
I propose to consider one family alone — those 
problems involving what is known as the Greek 
cross with the square. This will exhibit a 
great variety of curious transpositions, and, by 
having the solutions as we go along, the reader 
will be saved the trouble of perpetually turning 
to another part of the book, and will have every- 
thing under his eye. It is hoped that in this 
way the article may prove somewhat instructive 
to the novice and interesting to others. 



GREEK CROSS PUZZLES. 

" To fret thy soul with crosses." 

Spenser. 
" But, for my part, it was Greek to me." 
Julius CcBsar, i. 2. 

Many people are accustomed to consider the 
cross as a whoUy Christian symbol. This is 
erroneous : it is of very great antiquity. The 
ancient Egyptians employed it as a sacred 



GEOMETRICAL PROBLEMS. 



29 



s3nnbol, and on Greek sculptures we find repre- 
sentations of a cake (the supposed real origin 
of our hot cross buns) bearing a cross. Two 
such cakes were discovered at Herculaneum. 
Cecrops offered to Jupiter Olympus a sacred 
cake or boun of this kind. The cross and ball, 
so frequently found on Egyptian figures, is a 
circle and the tau cross. The circle signified 
the eternal preserver of the world, and the T, 
named from the Greek letter tau, is the mono- 
gram of Thoth, the Egyptian Mercury, meaning 
wisdom. This tau cross is also called by Chris- 
tians the cross of St. Anthony, and is borne on 
a badge in the bishop's palace at Exeter. As 
for the Greek or mundane cross, the cross with 
four equal arms, we are told by competent 
antiquaries that it was regarded by ancient 
occultists for thousands of years as a sign of the 
dual forces of Nature — the male and female 
spirit of everything that was everlasting. 




Fig. 5. 

The Greek cross, as shown in Fig. 5, is formed 
by the assembling together of five equal squares. 
We will start with what is known as the Hindu 
problem, supposed to be upwards of three thou- 
sand years old. It appears in the seal of 
Harvard College, and is often given in old works 
as symbolical of mathematical science and ex- 




FiG. 6. 



Fig. 7. 



is done. It was not until the middle of the 
nineteenth century that we found that the cross 
might be transformed into a square in only 
four pieces. Figs. 8 and 9 will show how to do 





Fig. 8. 



Fig. 9. 



actitude. Cut the cross into five pieces to 
form a square. Figs. 6 and 7 show how this 



it, if we further require the four pieces to be 
all of the same size and shape. This Fig. 9 
is remarkable because, according to Dr. Le 
Plongeon and others, as expounded in a 
work by Professor Wilson of the Smithsonian 
Institute, here we have the great Swastika, or 
sign, of " good luck to you " — the most ancient 
symbol of the human race of which there is any 
record. Professor Wilson's work gives some 
four hundred illustrations of this curious sign as 
found in the Aztec mounds of Mexico, the 
pyramids of Egypt, the ruins of Troy, and the 
ancient lore of India and China. One might 
almost say there is a curious af&nity between 
the Greek cross and Swastika ! If, however, 
we require that the four pieces shall be produced 
by only two clips of the scissors (assuming the 
puzzle is in paper form), then we must cut as in 
Fig. 10 to form Fig. 11, the first clip of the 
scissors being from a to b. Of course folding 
the paper, or holding the pieces together after 
the first cut, would not in this case be allowed. 
But there is an infinite number of different ways 
of making the cuts to solve the puzzle in four 
pieces. To this point I propose to return. 





b 






X 




^ 


c 


V 






A 




Fig. 10. 



Fig. II. 



It will be seen that every one of these puzzles 
has its reverse puzzle — to cut a square into 
pieces to form a Greek cross. But as a square 
has not so many angles as the cross, it is not 
always equally easy to discover the true direc- 
tions of the cuts. Yet in the case of the ex- 
amples given, I will leave the reader to deter- 
mine their direction for himself, as they are 
rather obvious from the diagrams. 



30 



AMUSEMENTS IN MATHEMATICS. 



Cut a square into five pieces that will form 
two separate Greek crosses of different sizes. 
This is quite an easy puzzle. As will be seen in 
Fig. 12, we have only to divide our square into 
25 little squares and then cut as shown. The 
cross A is cut out entire, and the pieces B, C, D, 
and E form the larger cross in Fig. 13, The 
reader may here like to cut the single piece, B, 
into four pieces all similar in shape to itself, 
and form a cross with them in the manner 
shown in Fig. 13. I hardly need give the 
solution. 



B: 



;c 



Sir 



D 





B: 












1 






i 
















i e: 





Fig. 12. 



Fig. 13. 



Cut a square into five pieces that will form 
two separate Greek crosses of exactly the same 
size. This is more difi&cult. We make the cuts 
as in Fig. 14, where the cross A comes out entire 
and the other four pieces form the cross in 
Fig. 15. The direction of the cuts is pretty 
obvious. It wiU be seen that the sides of the 
square in Fig. 14 are marked off into six equal 
parts. The sides of the cross are found by 
ruling lines from certain of these points to 
others. 





Fig. 14. 



Fig. 15. 



I wiU now explain, as I promised, why a 
Greek cross may be cut into four pieces in an 
infinite number of different ways to make a 
square. Draw a cross, as in Fig. 16, Then 
draw on transparent paper the square shown in 
Fig. 17, taking care that the distance c \.o d is 
exactly the same as the distance a to 6 in the 
cross. Now place the transparent paper over 
the cross and slide it about into different posi- 
tions, only be very careful always to keep the 
square at the same angle to the cross as shown, 
where a h is parallel to c d. If you place the 
point c exactly over a the lines will indicate the 
solution (Figs. 10 and 11). If you place c in 
the very centre of the dotted square, it will give 
the solution in Figs. 8 and 9. You will now see 
that by sliding the square about so that the 
point c is always within the dotted square you 
may get as many different solutions as you like ; 



because, since an infinite number of different 
points may theoretically be placed within this 
square, there must be an infinite number of 




Fig. 16. 



Fig. 17. 



different solutions. But the point c need not 
necessarily be placed within the dotted square. 
It may be placed, for example, at point e to 
give a solution in four pieces. Here the joins 
at a and / may be as slender as you like. Yet if 
you once get over the edge at a or / you no longer 
have a solution in four pieces. This proof will 
be found both entertaining and instructive. 
If you do not happen to have any transparent 
paper at hand, any thin paper will of course do 
if you hold the two sheets against a pane of 
glass in the window. 

It may have been noticed from the solutions of 
the puzzles that I have given that the side of the 
square formed from the cross is always equal to 
the distance a to 6 in Fig. 16. This must neces- 
sarily be so, and I will presently try to make the 
point quite clear. 

We wiU now go one step further. I have al- 
ready said that the ideal solution to a cutting- 
out puzzle is always that which requires the 
fewest possible pieces. We have just seen that 
two crosses of the same size may be cut out of 
a square in five pieces. The reader who suc- 
ceeded in solving this perhaps asked himself: 
" Can it be done in fewer pieces ? " This is 
just the sort of question that the true puzzle 
lover is always asking, and it is the right atti- 
tude for him to adopt. The answer to the 
question is that the puzzle may be solved in 
four pieces — the fewest possible. This, then, 
is a new puzzle. Cut a square into four pieces 
that will form two Greek crosses of the same size. 






/ 




c 


/ 


x> 




/ 





Fig. 18. 



Fig. 19. 



Fig. 20. 



The solution is very beautiful. If you divide 
by points the sides of the square into three equal 
parts, the directions of the lines in Fig. 18 will 
be quite obvious. If you cut along these lines. 



GEOMETRICAL PROBLEMS. 



31 



the pieces A and B will form the cross in Fig. 19 
and the pieces C and D the similar cross in 
Fig. 20, In this square we have another form 
of Swastika. 

The reader will here appreciate the truth of 
my remark to the effect that it is easier to find 
the directions of the cuts when transforming a 
cross to a square than when converting a square 
into a cross. Thus, in Figs. 6, 8, and 10 the 
directions of the cuts are more obvious than in 
Fig. 14, where we had first to divide the sides 
of the square into six equal parts, and in Fig. 18, 
where we divide them into three equal parts. 
Then, supposing you were required to cut two 
equal Greek crosses, each into two pieces, to 
form a square, a glance at Figs. 19 and 20 will 
show how absurdly more easy this is than the 
reverse puzzle of cutting the square to make 
two crosses. 

Referring to my remarks on " fallacies," I 
will now give a little example of these " solu- 
tions " that are not solutions. Some years ago 
a young correspondent sent me what he evi- 
dently thought was a brilliant new discovery — 
the transforming of a square into a Greek cross 
in four pieces by cuts all parallel to the sides 
of the square. I give his attempt in Figs. 21 
and 22, where it wiU be seen that the four pieces 
do not form a symmetrical Greek cross, because 
the foiu: arms are not really squares but oblongs. 
To make it a true Greek cross we should require 
the additions that I have indicated with dotted 
lines. Of course his solution produces a cross, 
but it is not the symmetrical Greek variety re- 
quired by the conditions of the puzzle. My 
young friend thought his attempt was " near 
enough " to be correct ; but if he bought a penny 
apple with a sixpence he probably would not 
have thought it " near enough " if he had been 
given only fourpence change. As the reader 
advances he will realize the importance of this 
question of exactitude. 







i 










A 


' — 


B 






C 












A 






B 


"7 

I 


C 








D 








j 










2 


















Fig. 21. 



Fig. 22. 



In these cutting-out puzzles it is necessary 
not only to get the directions of the cutting lines 
as correct as possible, but to remember that 
these lines have no width. If after cutting up 
one of the crosses in a manner indicated in these 
articles you find that the pieces do not exactly 
fit to form a square, you may be certain that the 
fault is entirely your own. Either your cross 
was not exactly drawn, or your cuts were not 
made quite in the right directions, or (if you 
used wood and a fretsaw) your saw was not 
sufficiently fine. If you cut out the puzzles in 



paper with scissors, or in cardboard with a pen- 
knife, no material is lost ; but with a saw, how- 
ever fine, there is a certain loss. In the case of 
most puzzles this slight loss is not sufficient to 
be appreciable, if the puzzle is cut out on a large 
scale, but there have been instances where I have 
found it desirable to draw and cut out each part 
separately — not from one diagram — in order to 
produce a perfect result. 





Fig. 23. 



Fig. 24. 



Now for another puzzle. If you have cut out 
the five pieces indicated in Fig. 14, you will find 
that these can be put together so as to form the 
curious cross shown in Fig. 23. So if I asked 
you to cut Fig. 24 into five pieces to form either 
a square or two equal Greek crosses you would 
know how to do it. You would make the cuts 
as in Fig. 23, and place them together as in 
Figs. 14 and 15. But I want something better 
than that, and it is this. Cut Fig. 24 into only 
four pieces that will fit together and form a 
square. 





Fig. 25. 



Fig. 26. 



The solution to the puzzle is shown in Figs. 
25 and 26. The direction of the cut dividmg 
A and C in the first diagram is very obvious, and 
the second cut is made at right angles to it. 
That the four pieces should fit together and form 
a square will surprise the novice, who will do 
well to study the puzzle with some care, as it 
is most instructive. 

I will now explain the beautiful rule by which 
we determine the size of a square that shall have 
the same area as a Greek cross, for it is appli- 
cable, and necessary, to the solution of almost 
every dissection puzzle that we meet with. It 
was first discovered by the philosopher Pytha- 
goras, who died 500 b.c, and is the 47th 
proposition of Euclid. The young reader who 
knows nothing of the elements of geometry wiU 
get some idea of the fascinating character of 
that science. The triangle A B C in Fig. 27 is 
what we call a right-angled triangle, because the 
side B C is at right angles to the side A B. Now 
if we build up a square on each side of the tri- 



32 



AMUSEMENTS IN MATHEMATICS. 



angle, the squares on A B and B C will together 
be exactly equal to the square on the long side 
A C, which we call the hypotenuse. This is 
proved in the case I have given by subdividing 
the three squares into cells of equal dimensions. 





y 






























B 










- 




- 





D 



F 



Fig. 27. 



Fig. 28. 



It will be seen that 9 added to 16 equals 25, the 
number of cells in the large square. If you make 
triangles with the sides 5,12 and 13, or with 8, 15 
and 17, you will get similar arithmetical proofs, 
for these are all " rational " right-angled tri- 
angles, but the law is equally true for aU cases. 
Supposing we cut off the lower arm of a Greek 
cross and place it to the left of the upper arm, 
as in Fig. 28, then the square on E F added to 
the square on D E exactly equals a square on 
D F. Therefore we know that the square of 
D F will contain the same area as the cross. 
This fact we have proved practically by the 
solutions of the earlier puzzles of this series. 
But whatever length we give to D E and E F, 
we can never give the exact length of D F 
in numbers, because the triangle is not a 
" rational " one. But the law is none the less 
geometrically true. 




Fig. 29. 



Fig. 30. 



Now look at Fig. 29, and you will see an ele- 
gant method for cutting a piece of wood of the 
shape of two squares (of any relative dimen- 
sions) into three pieces that will fit together 
and form a single square. If you mark off the 
distance a b equal to the side c d the directions 
of the cuts are very evident. From what we 
have just been considering, you will at once see 
why b c must be the length of the side of the 
new square. Make the experiment as often as 
you like, taking different relative proportions 
for the two squares, and you will find the rule 
always come true. If you make the two squares 
of exactly the same size, you will see that the 
diagonal of any square is always the side of a 
square that is twice the size. All this, which is 
so simple that anybody can understand it, is 



very essential to the solving of cutting-out 
puzzles. It is in fact the key to most of them. 
And it is aU so beautiful that it seems a pity 
that it should not be familiar to everybody. 

We will now go one step further and deal with 
the half-square. Take a square and cut it in 
half diagonally. Now try to discover how to 
cut this triangle into four pieces that will form 
a Greek cross. The solution is shown in Figs. 
31 and 32. In this case it will be seen that we 
divide two of the sides of the triangle into three 
equal parts and the long side into four equal 
parts. Then the direction of the cuts will be 
easily found. It is a pretty puzzle, and a little 
more difficult than some of the others that I 
have given. It should be noted again that it 
would have been much easier to locate the cuts 
in the reverse puzzle of cutting the cross to 
form a half-square triangle. 




Fig. 31. 



Fig. 32. 



Another ideal that the puzzle maker always 
keeps in mind is to contrive that there shall, if 
possible, be only one correct solution. Thus, 
in the case of the first puzzle, if we only require 
that a Greek cross shall be cut into four pieces 
to form a square, there is, as I have shown, an 
infinite number of different solutions. It makes 
a better puzzle to add the condition that aU the 
four pieces shall be of the same size and shape, 
because it can then be solved in only one way, 
as in Figs. 8 and 9. In this way, too, a puzzle 
that is too easy to be interesting may be im- 
proved by such an addition. Let us take an 
example. We have seen in Fig. 28 that Fig. 33 
can be cut into two pieces to form a Greek cross. 
I suppose an intelligent child would do it in five 



A 

§1 I— 1 

' 3 C 











A 






C 




3 





Fig. 33. 



Fig. 34. 



minutes. But suppose we say that the puzzle 
has to be solved with a piece of wood that has 



GEOMETRICAL PROBLEMS. 



33 



a bad knot in the position shown in Fig. 33 — a 
knot that we must not attempt to cut through 
— then a solution in two pieces is barred out, 
and it iDecomes a more interesting puzzle to 
solve it in three pieces. I have shown in Figs. 
33 and 34 one way of doing this, and it will be 
found entertaining to discover other ways of 
doing it. Of course I could bar out all these 
other ways by introducing more knots, and so 
reduce the puzzle to a single solution, but it 
would then be overloaded with conditions. 

And this brings us to another point in seeking 
the ideal. Do not overload your conditions, or 
you will make your puzzle too complex to be 
interesting. The simpler the conditions of a 
puzzle are, the better. The solution may be as 
complex and difficult as you like, or as happens, 
but the conditions ought to be easily imderstood, 
or people will not attempt a solution. 

If the reader were now asked " to cut a half- 
square into as few pieces as possible to form a 
Greek cross," he would probably produce our 
solution, Figs, 31-32, and confidently claim that 
he had solved the puzzle correctly. In this way 
he would be wrong, because it is not now stated 
that the square is to be divided diagonally. 
Although we should always observe the exact 
conditions of a puzzle we must not read into it 
conditions that are not there. Many puzzles 
are based entirely on the tendency that people 
have to do this. 

The very first essential in solving a puzzle is 
to be sure that you understand the exact condi- 
tions. Now, if you divided your square in half 
so as to produce Fig. 35 it is possible to cut it 
into as few as three pieces to form a Greek cross. 
We thus save a piece. 

I give another puzzle in Fig. 36. The dotted 
lines are added merely to show the correct 
proportions of the figure — a square of 25 cells 
with the four corner cells cut out. The puzzle is 
to cut this figure into five pieces that will form 
a Greek cross (entire) and a square. 



example, that if we continue the cut that divides 
B and C in the cross, we get Fig. 15. 



Fig. 35. 



Fig. 36. 



The solution to the first of the two puzzles 
last given — to cut a rectangle of the shape of 
a half-square into three pieces that will form a 
Greek cross — ^is shown in Figs. 37 and 38. It 
will be seen that we divide the long sides of the 
oblong into six equal parts and the short sides 
into three equal parts, in order to get the points 
that will indicate the direction of the cuts. The 
reader should compare this solution with some 
of the previous illustrations. He will see, for 
(1,926) 






Fig. 37. 



Fig. 38. 



The other puzzle, like the one illustrated in 
Figs. 12 and 1 3, will show how useful a little arith- 
metic may sometimes prove to be in the solution 
of dissection puzzles. There are twenty-one of 
those little square cells into which our figure is 
subdivided, from which we have to form both 
a square and a Greek cross. Now, as the cross 
is built up of five squares, and 5 from 21 leaves 
16— a square number — ^we ought easily to be 
led to the solution shown in Fig. 39. It will be 





B 




c 




A 




V 




a 







B 








E. 








C 








D 





Fig. 39. 



Fig. 40. 



seen that the cross is cut out entire, while the 
four remaining pieces form the square in Fig. 40. 
Of course a heilf-square rectangle is the same 
as a double square, or two equal squares joined 
together. Therefore, if you want to solve the 
puzzle of cutting a Greek cross into four pieces 
to form two separate squares of the same size, 
all you have to do is to continue the short cut 
in Fig. 38 right across the cross, and you will 
have four pieces of the same size and shape. Now 
divide Fig. 37 into two equal squares by a hori- 




zontal cut midway and you will see the four 
pieces forming the two squares. 



34 



AMUSEMENTS IN MATHEMATICS. 



Cut a Greek cross into five pieces that will 
form two separate squares, one of which shall 
contain half the area of one of the arms of the 
cross. In further illustration of what I have 





. C 
/aN 


1 "i 


V 


A 








Fig. 42. 



Fig. 43- 



already written, if the two squares of the same 
size A B D and b c f e, in Fig. 41, are cut in the 
manner indicated by the dotted lines, the four 
pieces will form the large square a g e c. We 
thus see that the diagonal a c is the side of a 
square twice the size of a b c d. It is also clear 
that half the diagonal of any square is equal 



pieces B, C, D, and E. After what I have written, 
the reader will have no difficulty in seeing that 
the square A is half the size of one of the arms of 
the cross, because the length of the diagonal of 
the former is clearly the same as the side of the 
latter. The thing is now self-evident. I have 
thus tried to show that some of these puzzles 
that many people are apt to regard as quite 
wonderful and bewildering, are really not diflS- 
cult if only we use a little thought and judgment. 
In conclusion of this particular subject I will 
give four Greek cross puzzles, with detached 
solutions. 

142.— THE SILK PATCHWORK. 

The lady members of the Wilkinson family had 
made a simple patchwork quilt, as a smaU 
Christmas present, all composed of square pieces 
of the same size, as shown in the illustration. 
It only lacked the four comer pieces to make it 
complete. Somebody pointed out to them that 
if you unpicked the Greek cross in the middle 
and then cut the stitches along the dark joins, 
the four pieces aU of the same size and shape 
would fit together and form a square. This 
the reader knows, from the solution in Fig, 39, 




to the side of a square of half the area. There- 
fore, if the large square in the diagram is one of 
the arms of your cross, the small square is the 
size of one of the squares required in the puzzle. 
The solution is shown in Figs. 42 and 43. It 
will be seen that the small square is cut out 
whole and the large square composed of the four 



is quite easily done. But George Wilkinson 
suddenly suggested to them this poser. He 
said, " Instead of picking out the cross entire, 
and forming the square from four equal pieces, 
can you cut out a square entire and four equal 
pieces that will form a perfect Greek cross ? " 
The puzzle is, of course, now quite easy. 



GEOMETRICAL PROBLEMS. 



35 



143.— TWO CROSSES FROM ONE. 

Cut a Greek cross into five pieces that will form 
two such crosses, both of the same size. The 
solution of this puzzle is very beautiful. 

144.— THE CROSS AND THE TRIANGLE. 

Cut a Greek cross into six pieces that will 
fornai an equilateral triangle. This is another 
hard problem, and I will state here that a solu- 
tion is practically impossible without a previ- 
ous knowledge of my method of transforming 
an equilateral triangle into a square (see No. 26, 
" Canterbury Puzzles "). 

145.— THE FOLDED CROSS. 

Cut out of paper a Greek cross ; then so fold 
it that with a single straight cut of the scissors 
the four pieces produced will form a square. 

VARIOUS DISSECTION PUZZLES. 

We will now consider a small miscellaneous 
selection of cutting-out puzzles, varying in 
degrees of difl&culty. 

146.—AN EASY DISSECTION PUZZLE. 



First, cut out a piece of paper or cardboard of 
the shape shown in the illustration. It will be 
seen at once that the proportions are simply 
those of a square attached to half of another 
similar square, divided diagonally. The puzzle 
is to cut it into four pieces all of precisely the 
same size and shape. 

147.— AN EASY SQUARE PUZZLE. 



diagonally, you will get two of the pieces shown 
in the illustration. The puzzle is with five such 
pieces of equal size to form a square. One of 
the pieces may be cut in two, but the others 
must be used intact. 



148.— THE BUN PUZZLE. 




The three circles represent three buns, and it 
is simply required to show how these may be 
equally divided among four boys. The buns 
must be regarded as of equal thickness through- 
out and of equal thickness to each other. Of 
course, they must be cut into as few pieces as 
possible. To simplify it I will state the rather 
surprising fact that only five pieces are neces- 
Sciry, from which it will be seen that one boy 
gets his share in two pieces and the other three 
receive theirs in a single piece. I am aware 
that this statement " gives away " the puzzle, 
but it should not destroy its interest to those 
who like to discover the " reason why." 

149— THE CHOCOLATE SQUARES. 








































' ' I * 






..T...J » 

















If you take a rectangular piece of cardboard, 
twice as long as it is broad, and cut it in half 



Here is a slab of chocolate, indented at the 
dotted lines so that the twenty squares can 
be easily separated. Make a copy of the slab 
in paper or cardboard and then try to cut it 
into nine pieces so that they will form four 
perfect squares all of exactly the same size. 

150.— DISSECTING A MITRE. 

The figure that is perplexing the carpenter in 
the illustration represents a mitre. It will be 
seen that its proportions are those of a square 
with one quarter removed. The puzzle is to 
cut it into five pieces that wiU fit together and 
form a perfect square. I show an attempt, 
published in America, to perform the feat in 



36 



AMUSEMENTS IN MATHEMATICS. 




four pieces, based on what is known as the 
" step principle," but it is a fallacy. 

We are told first to cut off the pieces i and 2 
and pack them into the triangular space marked 
off by the dotted line, and so form a rectangle. 



l\ 


A. 


3 


/ 




4 



So far, so good. Now, we are directed to apply 
the old step principle, as shown, and, by moving 
down the piece 4 one step, form the required 
square. But, unfortimately, it does not produce 
a square : only an oblong. CaU the three long 



sides of the mitre 84 in. each. Then, before 
cutting the steps, our rectangle in three pieces 
will be 84 X 63. The steps must be io| in. in 
height and 12 in. in breadth. Therefore, by 
moving down a step we reduce by 12 in. the 
side 84 in. and increase by 10^ in. the side 63 in. 
Hence our final rectangle must be 72 in. x 
73^ in., which certainly is not a square ! The 
fact is, the step principle can only be applied 
to rectangles with sides of particular relative 
lengths. For example, if the shorter side in 
this case were 6if (instead of 63), then the step 
method would apply. For the steps would 
then be io| in. in height and 12 in. in breadth. 
Note that 61^ X 84= the square of 72. At pres- 
ent no solution has been found in four pieces, 
and I do not believe one possible. 

151.— THE JOINER'S PROBLEM. 

I HAVE often had occasion to remark on the 
practical utility of puzzles, arising out of an 
application to the ordinary affairs of life of the 
little tricks and " wrinkles " that we learn 
while solving recreation problems. 

The joiner, in the illustration, wants to cut 
the piece of wood into as few pieces as possible 
to form a square table-top, without any waste 



GEOMETRICAL PROBLEMS. 



37 




by cutting it into five pieces, the parts fit to- 
gether and form a square, as shown in the illus- 
tration. Now, it is quite an interesting puzzle 



of material. How should he go to work ? How 
many pieces would you require ? 

152.— ANOTHER JOINER'S PROBLEM. 



A JOINER had two pieces of wood of the shapes 
and relative proportions shown in the diagram. 
He wished to cut them into as few pieces as 
possible so that they could be fitted together, 
without waste, to form a perfectly square table- 
top. How should he have done it ? There is 
no necessity to give measurements, for if the 
smaller piece (which is half a square) be made a 
little too large or a little too small it will not 
affect the method of solution. 

153— A CUTTING-OUT PUZZLE. 

Here is a little cutting-out poser. I take a strip 
of paper, measuring five inches by one inch, and, 



y< 


>< 


6- 




to discover how we can do this in only four 
pieces. 

154.— MRS. HOBSON'S HEARTHRUG. 
30 6 





Mrs. Hobson's boy had an accident when play- 
ing with the fire, and burnt two of the corners 
of a pretty hearthrug. The damaged comers 
have been cut away, and it now has the appear- 
ance and proportions shown in my diagram. 
How is Mrs. Hobson to cut the rug into 
the fewest possible pieces that wiU fit to- 
gether and form a perfectly square rug ? 

It will be seen that the rug is in the pro- 
portions 36 X 27 (it does not matter 
whether we say inches or yards), and each 
piece cut away measured 12 and 6 on the 
outside. 

155 —THE PENTAGON AND SQUARE. 

I WONDER how many of my readers, amongst 
those who have not given any close attention 
to the elements of geometry, could draw a regu- 
lar pentagon, or five-sided figure, if they sud- 
denly required to do so. A regular hexagon, or 
six-sided figure, is easy enough, for everybody 
knows that all you have to do is to describe a 
circle and then, taking the radius as the length 
of one of the sides, mark off the six points round 
the circumference. But a pentagon is quite 
another matter. So, as my puzzle has to do 
with the cutting up of a regular pentagon, it 
will perhaps be well if I first show my less ex- 
perienced readers how this figure is to be cor- 
rectly drawn. Describe a circle and draw the 
two lines H B and D G, in the diagram, 
through the centre at right angles. Now find 
the point A, midway between C and B. Next 
place the point of your compasses at A and with 



38 



AMUSEMENTS IN MATHEMATICS. 



the distance A D describe the arc, cutting H B 
at E. Then place the point of your compasses 
at D and with the distance D E describe the 
arc cutting the circumference at F. Now, D F 




is one of the sides of your pentagon, and you 
have simply to mark off the other sides round 
the circle. Quite simple when you know how, 
but otherwise somewhat of a poser. 

Having formed your pentagon, the puzzle is 
to cut it into the fewest possible pieces that will 
fit together and form a perfect square. 



has simply cut out of paper an equilateral tri- 
angle — that is, a triangle with all its three sides 
of the same length. He proposes that it shall 
be cut into five pieces in such a way that they 
wiU fit together and form either two or three 
smaller equilateral triangles, using all the ma- 
terial in each case. Can you discover how the 
cuts should be made ? 

Remember that when you have made yom: 
five pieces, you must be able, as desired, to 
put them together to form either the single 
original triangle or to form two triangles or to 
form three triangles — all equilateral. 

157.— THE TABLE-TOP AND STOOLS. 

I HAVE frequently had occasion to show that 
the published answers to a great many of the 
oldest and most widely known puzzles are either 
quite incorrect or capable of improvement. I 
propose to consider the old poser of the table- 
top and stools that most of my readers have 
probably seen in some form or another in 
books compiled for the recreation of child- 
hood. 

The story is told that an economical and in- 
genious schoolmaster once wished to convert a 
circular table-top, for which he had no use, into 
seats for two oval stools, each with a hand-hole 
in the centre. He instructed the carpenter to 
make the cuts as in the illustration and then 
join the eight pieces together in the manner 
shown. So impressed was he with the in- 
genuity of his performance that he set the 
puzzle to his geometry class as a little study in 
dissection. But the remainder of the story has 







156.— THE DISSECTED TRIANGLE. 

A GOOD puzzle is that which the gentleman in 
the illustration is showing to his friends. He 



never been published, because, so it is said, it 
was a characteristic of the principals of acad- 
emies that they would never admit that they 
could err. I get my information from a de- 



GEOMETRICAL PROBLEMS. 



39 



scendant of the original boy who had most reason 
to be interested in the matter. 

The clever youth suggested modestly to the 
master that the hand-holes were too big, and 
that a small boy might perhaps fall through 
them. He therefore proposed another way of 
making the cuts that woiild get over this objec- 
tion. For his impertinence he received such 
severe chastisement that he became convinced 
that the larger the hand-hole in the stools the 
more comfortable might they be. 



Korean ensign and merchant flag, and has been 
adopted as A trade sign by the Northern Pacific 
Railroad Company, though probably few are 
aware that it is the Great Monad, as shown in 
the sketch below. This sign is to the Chinaman 
what the cross is to the Christian. It is the 
sign of Deity and eternity, while the two parts 
into which the circle is divided are called the 
Yin and the Yan — the male and female forces 
of nature. A writer on the subject more than 
three thousand years ago is reported to have 




Now what was the method the boy proposed ? 

Can you show how the circular table-top may 
be cut into eight pieces that will fit together and 
form two oval seats for stools (each of exactly 
the same size and shape) and each having similar 
hand-holes of smaller dimensions than in the 
case shown above ? Of course, all the wood 
must be used. 

158.— THE GREAT MONAD. 




Here is a symbol of tremendous antiquity 
which is worthy of notice. It is borne on the 



said in reference to it : "The illimitable produces 
the great extreme. The great extreme pro- 
duces the two principles. The two principles 
produce the four quarters, and from the four 
quarters we develop the quadrature of the eight 
diagrams of Feuh-hi." I hope readers will 
not ask me to explain this, for I have not the 
slightest idea what it means. Yet I am per- 
suaded that for ages the symbol has had occult 
and probably mathematical meanings for the 
esoteric student. 

I will introduce the Monad in its elementary 
form. Here are three easy questions respect- 
ing this great S5nnbol : — 

(I.) Which has the greater area, the inner 
circle containing the Yin and the Yan, or the 
outer ring ? 

(II.) Divide the Yin and the Yan into four 
pieces of the same size and shape by one cut. 

(III.) Divide the Yin and the Yan into four 
pieces of the same size, but different shape, by 
one straight cut. 

159.— THE SQUARE OF VENEER. 

The following represents a piece of wood in my 
possession, 5 in. square. By markings on the 
surface it is divided into twenty-five square 
inches. I want to discover a way of cutting 
this piece of wood into the fewest possible pieces 
that will fit together and form two perfect 
squares of different sizes and of known dimen- 
sions. But, unfortunately, at every one of the 
sixteen intersections of the cross lines a smaU 
nail has been driven in at some time or other, 
and my fret-saw will be injured if it comes in 



40 



AMUSEMENTS IN MATHEMATICS. 



contact with any of these. I have therefore 
to find a method of doing the work that will not 
necessitate my cutting through any of those 






















































sixteen points. How is it to be done ? Re- 
member, the exact dimensions of the two squares 
must be given. 

i6o.— THE TWO HORSESHOES. 




Why horseshoes should be considered " lucky " 
is one of those things which no man can under- 
stand. It is a very old superstition, and John 
Aubrey (1626-1700) says, " Most houses at the 
West End of London have a horseshoe on the 
threshold." In Monmouth Street there were 
seventeen in 181 3 and seven so late as 1855. 
Even Lord Nelson had one nailed to the mast 
of the ship Victory. To-day we find it more 
conducive to " good luck " to see that they are 
securely nailed on the feet of the horse we are 
about to drive. 

Nevertheless, so far as the horseshoe, like the 
Swastika and other emblems that I have had 
occasion at times to deal with, has served to 
symbolize health, prosperity, and goodwill to- 
wards men, we may well treat it with a certain 
amount of respectful interest. May there not, 
moreover, be some esoteric or lost mathematical 
mystery concealed in the form of a horseshoe ? 
I have been looking into this matter, and I wish 



to draw my readers' attention to the very re- 
markable fact that the pair of horseshoes shown 
in my illustration are related in a striking and 
beautiful manner to the circle, which is the 
symbol of eternity. I present this fact in the 
form of a simple problem, so that it may be 
seen how subtly this relation has been concealed 
for ages and ages. My readers will, I know, be 
pleased when they find the key to the mystery. 
Cut out the two horseshoes carefully round 
the outline and then cut them into four pieces, 
all different in shape, that will fit together and 
form a perfect circle. Each shoe must be cut 
into two pieces and all the part of the horse's 
hoof contained within the outline is to be used 
and regarded as part of the area. 

161.— THE BETSY ROSS PUZZLE. 

A CORRESPONDENT askcd me to supply him with 
the solution to an old puzzle that is attributed 
to a certain Betsy Ross, of Philadelphia, who 
showed it to George Washington. It consists 
in so folding a piece of paper that with one clip 
of the scissors a five-pointed star of Freedom 
may be produced. Whether the story of the 
puzzle's origin is a true one or not I cannot say, 
but I have a print of the old house in Phila- 
delphia where the lady is said to have lived, and 
I believe it still stands there. But my readers 
will doubtless be interested in the little poser. 

Take a circular piece of paper and so fold it 
that with one cut of the scissors you can pro- 
duce a perfect five-pointed star. 

162.— THE CARDBOARD CHAIN. 




Can you cut this chain out of a piece of card- 
board without any join whatever ? Every link 
is solid, without its having been split and after- 
wards joined at any place. It is an interesting 
old puzzle that I learnt as a child, but I have no 
knowledge as to its inventor. 

163.— THE PAPER BOX. 

It may be interesting to introduce here, though 
it is not strictly a puzzle, an ingenious method 
for making a paper box. 

Take a square of stout paper and by succes- 
sive foldings make all the creases indicated by 
the dotted lines in the illustration. Thenxut 
away the eight little triangular pieces that are 
shaded, and cut through the paper along the 
dark lines. The second illustration shows the 
box half folded up, and the reader will have no 
difficulty in effecting its completion. Before 
folding up, the reader might cut out the circular 
piece indicated in the diagram, for a purpose 
I wiU now explain. 

This box will be found to serve excellently 
for the production of vortex rings. These rings, 



GEOMETRICAL PROBLEMS. 



45 



which were discussed by Von Helmholtz in 
1858, are most interesting, and the box (with 
the hole cut out) will produce them to perfection. 
Fill the box with tobacco smoke by blowing it 







/ Vr(iilllliiii.h..iii 



gently through the hole. Now, if you hold it 
horizontally, and softly tap the side that is 
opposite to the hole, an immense number of 
perfect rings can be produced from one mouth- 
ful of smoke. It is best that there should be 
no currents of air in the room. People often 
do not realise that these rings are formed in the 
air when no smoke is used. The smoke only 
makes them visible. Now, one of these rings, 
if properly directed on its course, wiU travel 
across the room and put out the flame of a 
candle, and this feat is much more striking if 
you can manage to do it without the smoke. 
Of course, with a little practice, the rings may 
be blown from the mouth, but the box produces 
them in much greater perfection, and no skill 
whatever is required. Lord Kelvin propounded 
the theory that matter may consist of vortex 



rings in a fluid that fills all space, and by a de- 
velopment of the hypothesis he was able to 
explain chemical combination. 

164.— THE POTATO PUZZLE. 

Take a circular slice of potato, place it on the 
table, and see into how large a number of pieces 
you can divide it with six cuts of a knife. Of 
course you must not readjust the pieces or pile 
them after a cut. What is the greatest number 
of pieces you can make ? 




The illustration shows how to make sixteen 
pieces. This can, of course, be easily beaten. 



THE SEVEN PIGS. 





Here is a little puzzle that was put to one of 
the sons of Erin the other day and perplexed 
him unduly, for it is really quite easy. It will be 
seen from the illustration that he was shown a 
sketch of a square pen containing seven pigs. 
He was asked how he would intersect the pen 
with three straight fences so as to enclose 
every pig in a separate sty. In other words, 
aU you have to do is to take your pencil 
and, with three straight strokes across the 
square, enclose each pig separately. Noth- 
ing could be simpler. 

The Irishman complained that the pigs would 
not keep still while he was putting up the fences. 
He said that they would all flock together, or 
one obstinate beast would go into a comer and 
flock all by himself. It was pointed out to him 
that for the purposes of the puzzle the pigs were 
stationary. He answered that Irish pigs are 
not stationery — they are pork. Being per- 
suaded to make the attempt, he drew three lines, 
one of which cut through a pig. When it was 
explained that this is not allowed, he protested 



42 



AMUSEMENTS IN MATHEMATICS. 













that a pig was no use until you cut its throat. 
" Begorra, if it's bacon ye want without cutting 
your pig, it will be all gammon." We wiU not 
do the Irishman the injustice of suggesting that 
the miserable pun was intentional. However, 
he failed to solve the puzzle. Can you do it ? 

i66.— THE LANDOWNER'S FENCES. 

The landowner in the illustration is consulting 
with his bailifi over a rather puzzling Uttle ques- 
tion. He has a large plan of one of his fields, in 
which there are eleven trees. Now, he wants to 
divide the field into just eleven enclosures by 
means of straight fences, so that every enclosure 
shall contain one tree as a shelter for his cattle. 
How is he to do it with as few fences as possible ? 
Take your pencil and draw straight lines across 
the field until you have marked off the eleven 
enclosures (and no more), and then see how 
many fences you require. Of course the fences 
may cross one another. 

167.— THE WIZARD'S CATS. 

A WIZARD placed ten cats inside a magic circle 
as shown in our illustration, and hypnotized 
them so that they should remain stationary 



during his pleasure. He then proposed to draw 
three circles inside the large one, so that no cat 
could approach another cat without crossing a 




GEOMETRICAL PROBLEMS. 



43 



magic circle. Try to draw the three circles 
so that every cat has its own enclosure and can- 
not reach another cat without crossing a line. 

1 68.— THE CHRISTMAS PUDDING. 




" Speaking of Christmas puddings," said the 
host, as he glanced at the imposing delicacy 
at the other end of the table, " I am reminded 
of the fact that a friend gave me a new puzzle 
the other day respecting one. Here it is," he 
added, diving into his breast pocket. 

" ' Problem : To find the contents,' I sup- 
pose," said the Eton boy. 

" No ; the proof of that is in the eating. I 
win read you the conditions." 

" ' Cut the pudding into two parts, each of 
exactly the same size and shape, without touch- 
ing any of the plums. The pudding is to be 
regarded as a flat disc, not as a sphere.' " 

" Why should you regard a Christmas pud- 
ding as a disc ? And why should any reason- 
able person ever wish to make such an accurate 
division ? " asked the cjmic. 

" It is just a puzzle — a problem in dissection." 

All in turn had a look at the puzzle, but no- 
body succeeded in solving it. It is a little diffi- 
cult unless you are acquainted with the prin- 
ciple involved in the making of such puddings, 
but easy enough when you know how it is done. 

169.— A TANGRAM PARADOX. 

Many pastimes of great antiquity, such as chess, 
have so developed and changed down the cen- 
turies that their original inventors would 
scarcely recognize them. This is not the case 
with Tangrams, a recreation that appears to be 
at least four thousand years old, that has appar- 
ently never been dormant, and that has not 
been altered or " improved upon " since the 
legendary Chinaman Tan first cut out the seven 



pieces shown in Diagram i. If you mark the 
point B, midway between A and C, on one side 
of a square of any size, and D, midway between 
C and E, on an adjoining side, the direction of 




the cuts is too obvious to need further explana- 
tion. Every design in this article is built up 
from the seven pieces of blackened cardboard. 
It will at once be understood that the possible 
combinations are infinite. 

The late Mr. Sam Loyd, of New York, who 
published a small book of very ingenious de- 
signs, possessed the manuscripts of the late Mr. 
Challenor, who made a long and close study of 
Tangrams. This gentleman, it is said, records 
that there were originally seven books of Tan- 
grams, compiled in China two thousand years 
before the Christian era. These books are so 
rare that, after forty years' residence in the 
coimtry, he only succeeded in seeing perfect 
copies of the first and seventh volumes with 
fragments of the second. Portions of one of 
the books, printed in gold leaf upon parchment, 
were found in Peking by an English soldier and 
sold for three hundred pounds. 

A few years ago a little book came into my 
possession, from the library of the late Lewis 
Carroll, entitled The Fashionable Chinese Puzzle. 
It contains three hundred and twenty-three 
Tangram designs, mostly nondescript geomet- 
rical figures, to be constructed from the seven 
pieces. It was " Published by J. and E. WaUis, 
42 Skinner Street, and J. Wallis, Jun., Marine 
Library, Sidmouth " (South Devon). There is 
no date, but the following note fixes the time 
of publication pretty closely : " This ingenious 
contrivance has for some time past been the 
favourite amusement of the ex-Emperor Na- 
poleon, who, being now in a debilitated state 
and living very retired, passes many hours a 
day in thus exercising his patience and in- 
genuity." The reader will find, as did the 
great exile, that much amusement, not whoUy 
uninstructive, may be derived from forming the 
designs of others. He will find many of the 
illustrations to this article quite easy to build 
up, and some rather difficult. Every picture 
may thus be regarded as a puzzle. 

But it is another pastime altogether to create 
new and original designs of a pictorial character, 
and it is surprising what extraordinary scope 
the Tangrams afford for producing pictures of 
real life — angular and often grotesque, it is true, 
but fuU of character. I give an example of a 
recumbent figure (2) that is particularly grace- 
ful, and only needs some slight reduction of 
its angularities to produce an entirely satis- 
factory outline. 

As I have referred to the author of Alice in 
Wonderland, I give also my designs of the March 



44 



AMUSEMENTS IN MATHEMATICS. 




Hare (3) and the Hatter {4). I also give an 
attempt at Napoleon {5), and a very excellent 
Red Indian with his Squaw by Mr. Loyd (6 and 
7). A large number of other designs will be 
foimd in an article by me in The Strand Maga- 
zine for November, 1908. 





On the appearance of this magazine article, 
the late Sir James Murray, the eminent phil- 
ologist, tried, with that amazing industry that 
characterized all his work, to trace the word 
" tangram " to its source. At length he wrote 
as follows : — " One of my sons is a professor 
in the Anglo-Chinese college at Tientsin. 
Through him, his colleagues, and his students, 
I was able to make inquiries as to the alleged 
Tan among Chinese scholars. Our Chinese 
pfofessor here (Oxford) also took an interest 
in the matter and obtained information from 
the secretary of the Chinese Legation in Lon- 




don, who is a very eminent representative of 
the Chinese literati. 

" The result has been to show that the man 
Tan, the god Tan, and the ' Book of Tan ' are 
entirely unknown to Chinese literature, history, 
or tradition. By most of the learned men the 
name, or allegation of the existence, of these 
had never been heard of. The puzzle is, of 
coiurse, well known. It is called in Chinese 
chH ch'iao t'u ; literally, ' seven-ingenious-plan ' 



or ' ingenious-puzzle figure of seven pieces.* 
No name approaching * tangram,* or even ' tan,' 
occurs in Chinese, and the only suggestions for 
the latter were the Chinese fan, ' to extend ' ; 
or Vang, Cantonese dialect for ' Chinese.' It 
was suggested that probably some American 




or Englishman who knew a little Chinese or 
Cantonese, wanting a name for the puzzle, 
might concoct one out of one of these words 
and the European ending ' gram.' I should 
say the name ' tangram ' was probably in- 
vented by an American some little time before 
1:864 and after 1847, but I cannot find it in 
print before the 1864 edition of Webster. I 
have therefore had to deal very shortly with 
the word in the dictionary, telling what it is 
applied to and what conjectures or guesses 
have been made at the name, and giving a 
few quotations, one from your own article, 
which has enabled me to make more of the 
subject than I could otherwise have done." 

Several correspondents have informed me 
that they possess, or had possessed, specimens 
of the old Chinese books. An American gentle- 
man writes to me as follows : — " I have in my 
possession a book made of tissue paper, printed 
in black (with a Chinese inscription on the 
front page), containing over three hundred de- 
signs, which belongs to the box of * tangrams,' 
which I also own. The blocks are seven in 
number, made of mother-of-pearl, highly pol- 
ished and finely engraved on either side. These 
are contained in a rosewood box 2\ in. square. 
My great imcle, , was one of the first mis- 
sionaries to visit China. This box and book, 
along with quite a collection of other relics, 
were sent to my grandfather and descended to 
myself." 

My correspondent kindly supplied me with 
rubbings of the Tangrams, from which it is clear 
that they are cut in the exact proportions that 
I have indicated. I reproduce the Chinese in- 
scription (8) for this reason. The owner of the 
book informs me that he has submitted it to 
a number of Chinamen in the United States 
and offered as much as a dollar for a translation. 
But they all steadfastly refused to read the 
words, offering the lame excuse that the in- 
scription is Japanese. Natives of Japan, how- 
ever, insist that it is Chinese. Is there some- 
thing occult and esoteric about Tangrams, that 



GEOMETRICAL PROBLEMS. 



45 



it is so difficult to lift the veil ? Perhaps this 
page will come under the eye of some reader 
acquainted with the Chinese language, who will 




supply the required translation, which may, or 
may not, throw a little light on this curious 
question. 

By using several sets of Tangrams at the 



iTWi 



same time we may construct more ambitious 
pictures. I was advised by a friend not to 
send my picture, " A Game of Billiards " (9), 
to the Academy. He assured me that it woiild 
not be accepted because the " judges are so 
hide-bound by convention." Perhaps he was 
right, and it will be more appreciated by Post- 
impressionists and Cubists. The players are con- 
sidering a very delicate stroke at the top of the 



My second picture is named " The Orchestra " 
(10), and it was designed for the decoration of a 
large haU of music. Here we have the conductor, 
the pianist, the fat little cornet -player, the left- 
handed player of the double-bass, whose atti- 
tude is life-like, though he does stand at an un- 
usual distance from his instrument, and the 
drummer-boy, with his imposing music-stand. 
The dog at the back of the pianoforte is not 
howling : he is an appreciative listener. 

One remarkable thing about these Tangram 
pictures is that they suggest to the imagination 
such a lot that is not really there. Who, for 
example, can look for a few minutes at Lady 
Belinda (ii) and the Dutch girl (12) without 





soon feeling the haughty expression in the one 
case and the arch look in the other ? Then 
look again at the stork (13), and see how it is 
suggested to the mind that the leg is actually 
much more slender than any one of the pieces 
employed. It is really an optical illusion. 
Again, notic« in the case of the yacht (14) how, 
by leaving that little angular point at the top, 
a complete mast is suggested. If you place 
your Tangrams together on white paper so that 
they do not quite touch one another, in some 
cases the effect is improved by the white lines ; 
in other cases it is almost destroyed. 

Finally, I give an example from the many 




table. Of course, the two men, the table, and the I ciirious paradoxes that one happens upon in 
clock are formed from four sets of Tangrams. 1 manipulating Tangrams. I show designs of 



46 



AMUSEMENTS IN MATHEMATICS. 





two dignified individuals (15 and 16) who appear 
to be exactly alike, except for the fact that one 
has a foot and the other has not. Now, both 




of these figures are made from the same seven 
Tangrams. Where does the second man get 
his foot from ? 

PATCHWORK PUZZLES. 

" Of shreds and patches." — Hamlet, iii. 4. 
170.— THE CUSHION COVERS. 






000 






0^0 
0^ 




000 




A 

'^'e 



o o Ct 

000 




Q O Q 




000 

<^ 




Q O Q 

000 



^^^ 

^ 



000 

© *S5 o 
000 



^ 



^ « o 
000 




O <» o 
0^0 



pieces will form one perfectly square cushion 
top, and the remaining two pieces another 
square cushion top. How is she to do it ? Of 
course, she can only cut along the lines that 
divide the twenty-five squares, and the pat- 
tern must "match" properly without any 
irregularity whatever in the design of the 
material. There is only one way of doing it. 
Can you find it ? 

171.— THE BANNER PUZZLE. 




^§S 

^ 



o « 
ft 



A LADY had a square piece of bimting with two 
lions on it, of which the illustration is an ex- 
actly reproduced reduction. She wished to cut 
the stuff into pieces that would fit together and 
form two square banners with a lion on each 
banner. She discovered that this could be done 
in as few as foiu: pieces. How did she manage 
it ? Of course, to cut the British Lion would be 
an impardonable offence, so you must be care- 
ful that no cut passes through any portion of 
either of them. Ladies are informed that no 
allowance whatever has to be made for " turn- 
ings," and no part of the material may 
be wasted. It is quite a simple little dis- 
section puzzle if rightly attacked. Remem- 
ber that the banners have to be perfect 
squares, though they need not be both of 
the same size. 




o o « 
009 




172.— MRS, SMILEY'S CHRISTMAS 
PRESENT. 



The above represents a square of brocade. A I 
lady wishes to cut it in four pieces so that two I 



Mrs. Smiley's expression of pleasure was 
sincere when her six granddaughters sent 
to her, as a Christmas present, a very pretty 
patchwork quilt, which they had made 
with their own hands. It was constructed 
of square pieces of silk material, all of one 
size, and as they made a large quilt with 
fourteen of these little squares on each side, 
it is obvious that just 196 pieces had been 
stitched into it. Now, the six granddaughters 
each contributed a part of the work in the 
form of a perfect square (all six portions being 
different in size), but in order to join them up 
to form the square quilt it was necessary 
that the work of one girl should be unpicked 
into three separate pieces. Can you show how 

the joins might have been made ? Of course, 

po portion can be turned over. 



GEOMETRICAL PROBLEMS. 



47 







173— MRS. PERKINS'S QUILT. 




It will be seen that in this case the square 
patchwork quilt is built up of 169 pieces. The 



puzzle is to find the smallest possible number 
of square portions of which the quilt could be 
composed and show how they might be joined 
together. Or, to put it the reverse way, divide 
the quilt into as few square portions as possible 
by merely cutting the stitches. 

174.— THE SQUARES OF BROCADE. 




I HAPPENED to be paying a call at the house of 
a lady, when I took up from a table two lovely 
squares of brocade. They were beautiful speci- 
mens of Eastern workmanship — both of the 
same design, a delicate chequered pattern. 

" Are they not exquisite ? " said my friend. 
" They were brought to me by a cousin who has 
just returned from India. Now, I want you 



48 



AMUSEMENTS IN MATHEMATICS. 




to give me a little assistance. You see, I have 
decided to join them together so as to make one 
large square cushion-cover. How should I do 
this so as to mutilate the ma- 
terial as little as possible ? Of 
course I propose to make my 
cuts only along the lines that 
divide the little chequers." 

I cut the two squares in the 
manner desired into four pieces 
that would fit together and form 
another larger square, taking 
care that the pattern should 
match properly, and when I had 
finished I noticed that two of 
the pieces were of exactly the 
same area ; that is, each of the 
two contained the same num- 
ber of chequers. Can you show 
how the cuts were made in ac- 
cordance with these conditions? 

175.— ANOTHER PATCHWORK PUZZLE. 

A LADY was presented, by two of her girl friends, 
with the pretty pieces of silk patchwork shown 
in our illustration. It will be seen that both 



pieces are made up of squares all of the same 
size — one 12 x 12 and the other 5x5. She 
proposes to join them together and make one 
square patchwork quilt, 13 X 13, but, of course, 
she will not cut any of the material — ^merely cut 
the stitches where necessary and join together 
again. What perplexes her is this. A friend 
assures her that there need be no more than 
four pieces in aU to join up for the new quilt. 
Could you show her how this little needlework 
puzzle is to be solved in so few pieces ? 

176.— LINOLEUM CUTTING. 

The diagram herewith represents two separate 
pieces of linoleum. The chequered pattern is 




not repeated at the back, so that the pieces 
cannot be turned over. The puzzle is to cut 
the two squares into four pieces so that they 
shall fit together and form one perfect square 
10 X 10, so that the pattern shall properly 



[?.v5?^' 





GEOMETRICAL PROBLEMS. 



49 



match, and so that the larger piece shall have 
as small a portion as possible cut from it. 





177. 


— 


-ANOTHER 


LINOLEUM PUZZLE. 


" 


















-.r.. 
























- 


_ 


- 


- 




















































































































































































































MU 












































^ 














































*^ 


^ 












































„ 


^ 


_ 






^, 


„, 


„ 


„ 


„ 




„ 




_ 


_ 


_j- 




_i 


tZD 



Can you cut this piece of linoleum into four 
pieces that will fit together and form a perfect 
square ? Of course the cuts may only be made 
along the lines. 



VARIOUS GEOMETRICAL 
PUZZLES. 

" So various are the tastes of men." 

Mark Akenside. 

178.— THE CARDBOARD BOX. 

This puzzle is not difl&cult, but it will be found 
entertaining to discover the simple rule for its 
solution. I have a rectangular cardboard box. 
The top has an area of 120 square inches, the 
side 96 square inches, and the end 80 square 
inches. What are the exact dimensions of the 
box ? 

179.— STEALING THE BELL-ROPES. 

Two men broke into a church tower one night 
to steal the bell-ropes. The two ropes passed 
through holes in the wooden ceiling high above 
them, and they lost no time in climbing to the 
top. Then one man drew his knife and cut the 
rope above his head, in consequence of which 
he fell to the floor and was badly injured. His 
fellow-thief called out that it served him right 
for being such a fool. He said that he should 
have done as he was doing, upon which he 
cut the rope below the place at which he held 
on. Then, to his dismay, he found that he was 
in no better plight, for, after hanging on as long 
as his strength lasted, he was compelled to let 
go and faU beside his comrade. Here they were 
both found the next morning with their limbs 
broken. How far did they fall ? One of the 
ropes when they found it was just touching the 
floor, and when you pulled the end to the wall, 
keeping the rope taut, it touched a point just 
three inches above the floor, and the waU was 
four feet from the rope when it hung at rest. 
How long was the rope from floor to ceiling ? 

180.— THE FOUR SONS. 

Readers will recognize the diagram as a fa- 
miliar friend of their youth. A man possessed 
a square-shaped estate. He bequeathed to his 
(1,926) 



widow the quarter of it that is shaded off. The 
remainder was to be divided equitably amongst 
his four sons, so that each should receive land 
of exactly the same area and exactly similar in 
shape. We are shown how this was done. But 
the remainder of the story is not so generally 
known. In the centre of the estate was a weU, 
indicated by the dark spot, and Benjamin, 
Charles, and David complained that the division 
was not " equitable," since Alfred had access 
to this well, while they could not reach it with- 
out trespassing on somebody else's land. The 
puzzle is to show how the estate is to be appor- 




tioned so that each son shall have land of the 
same shape and area, and each have access to 
the well without going off his own land. 

181.— THE THREE RAILWAY STATIONS. 

As I sat in a railway carriage I noticed at the 
other end of the compartment a worthy squire, 
whom I knew by sight, engaged in conversation 
with another passenger, who was evidently a 
friend of his. 

" How far have you to drive to your place from 
the railway station ? " asked the stranger. 

" Well," replied the squire, " if I get out at 
Appleford, it is just the same distance as if I 
go to Bridgefield, another fifteen miles farther 
on; and if I changed at Appleford and went 
thirteen miles from there to Carterton, it would 
still be the same distance. You see, I am equi- 
distant from the three stations, so I get a good 
choice of trains." 

Now I happened to know that Bridgefield is 
just fourteen miles from Carterton, so I amused 
myself in working out the exact distance that 
the squire had to drive home whichever station 
he got out at. What was the distance ? 

182.— THE GARDEN PUZZLE. 

Professor Rackbrane tells me that he was 
recently smoking a friendly pipe under a tree 
in the garden of a country acquaintance. The 
garden was enclosed by four straight walls, and 
his friend informed him that he had measured 
these and found the lengths to be 80, 45, 100, 
and 63 yards respectively. " Then," said the 
professor, " we can calculate the exact area of 
the garden." " Impossible," his host replied. 



50 



AMUSEMENTS IN MATHEMATICS. 



"because you can get an infinite number of 
different shapes with those four sides." " But 
you forget," Rackbrane said, with a twinkle in 
his eye, " that you told me once you had planted 
this tree equidistant from all the four corners 
of the garden." Can you work out the garden's 
area ? 

183.— DRAWING A SPIRAL. 

If you hold the page horizontally and give it a 
quick rotary motion while looking at the centre 
of the spiral, it wiU appear to revolve. Perhaps 
a good many readers are acquainted with this 
little optical illusion. But the puzzle is to show 
how I was able to draw this spiral with so much 
exactitude without using anything but a pair 




of compasses and the sheet of paper on which 
the diagram was made. How would you pro- 
ceed in such circumstances ? 

184.— HOW TO DRAW AN OVAL. 

Can you draw a perfect oval on a sheet of paper 
with one sweep of the compasses ? It is one of 
the easiest things in the world when you know 
how. 

185.— ST. GEORGE'S BANNER. 

At a celebration of the national festival of St. 
George's Day I was contemplating the familiar 
banner of the patron saint of our country. We 
all know the red cross on a white ground, shown 
in oxur illustration. This is the banner of St. 
George. The bannei of St. Andrew (Scotland) 
is a white "St. Andrew's Cross " on a blue 
ground. That of St. Patrick (Ireland) is a 
similar cross in red on a white ground. These 
three are united in one to form our Union 
Jack. 

Now on looking at St. George's banner it oc- 
curred to me that the following question would 
make a simple but pretty little puzzle. Sup- 
posing the flag measures four feet by three feet, 
how wide must the arm of the cross be if it is 




required that there shall be used just the same 
quantity of red and of white bunting ? 

186.— THE CLOTHES LINE PUZZLE. 

A BOY tied a clothes line from the top of each 
of two poles to the base of the other. He then 
proposed to his father the following question. 
As one pole was exactly seven feet above the 
ground and the other exactly five feet, what 
was the height from the ground where the two 
cords crossed one another ? 

187.— THE MILKMAID PUZZLE. 




Here is a little pastoral puzzle that the reader 
may, at first sight, be led into supposing is very 
profound, involving deep calculations. He may 
even say that it is quite impossible to give any 
answer unless we are told something definite as 
to the distances. And yet it is really quite 
" childlike and bland." 



GEOMETRICAL PROBLEMS. 



51 



In the comer of a field is seen a milkmaid 
milking a cow, and on the other side of the field 
is the dairy where the extract has to be deposited. 
But it has been noticed that the young woman 
always goes down to the river with her pail 
before returning to the dairy. Here the sus- 
picious reader will perhaps ask why she pays 
these visits to the river. I can only reply 
that it is no business of ours. The alleged milk 
is entirely for local consumption. 

" Where are you going to, my pretty maid ? " 
" Down to the river, sir," she said. 
" I'll not choose your dairy, my pretty maid." .. 
" Nobody axed you, sir," she said. 

If one had any curiosity in the matter, such 
an independent spirit would entirely disarm one. 
So we will pass from the point of commercial 
morality to the subject of the puzzle. 

Draw a line from the milking-stool down to 
the river and thence to the door of the dairy, 
which shall indicate the shortest possible route 
for the milkmaid. That is all. It is quite easy 
to indicate the exact spot on the bank of the 
river to which she should direct her steps if she 
wants as short a walk as possible. Can you 
find that spot ? 

188.— THE BALL PROBLEM. 



THE YORKSHIRE ESTATES. 



A STONEMASON was engaged the other day in 
cutting out a round ball for the purpose of some 
architectural decoration, when a smart school- 
boy came upon the scene. 

Look here," said the mason, '* you seem to 
be a sharp youngster, can you tell me this ? 
If I placed this ball on the level ground, how 
many other balls of the same size could I lay 
around it (also on the ground) so that every 
ball should touch this one ? " 

The boy at once gave the correct answer, and 
then put this little question to the mason : — 

" If the surface of that ball contained just as 
many square feet as its volume contained cubic 
feet, what would be the length of its dia- 
meter ? " 

The stonemason could not give an answer. 
Could you have replied correctly to the mason's 
and the boy's questions ? 





I WAS on a visit to one of the large towns of 
Yorkshire. While walking to the railway station 
on the day of my departure a man thrust a hand- 
bill upon me, and I took this into the railway 
carriage and read it at my leisure. It informed 
me that three Yorkshire neighbouring estates 
were to be offered for sale. Each estate was 
square in shape, and they joined one another at 
their corners, just as shown in the diagram. 
Estate A contains exactly 370 acres, B contains 
116 acres, and C 74 acres. 

Now, the little triangular bit of land enclosed 
by the three square estates was not offered for 
sale, and, for no reason in particular, I became 
curious as to the area of that piece. How many 
acres did it contain ? 

190.— FARMER WURZEL'S ESTATE. 




I WILL now present another land problem. The 
demonstration of the answer that I shall give 
will, I think, be found both interesting and easy 
of comprehension. 

Farmer Wurzel owned the three square fields 
shown in the annexed plan, containing respec- 
tively 18, 20, and 26 acres. In order to get a 
ring-fence round his property he bought the 



52 



AMUSEMENTS IN MATHEMATICS. 



four intervening triangular fields. The puzzle 
is to discover what was then the whole area of 
his estate. 

191.— THE CRESCENT PUZZLE. 




Here is an easy geometrical puzzle. The cres- 
cent is formed by two circles, and C is the centre 
of the larger circle. The width of the crescent 
between B and D is 9 inches, and between E and 
F 5 inches. What are the diameters of the two 
circles ? 

192.— THE PUZZLE WALL. 




There was a small lake, around which four poor 
men built their cottages. Four rich men after- 
wards buUt their mansions, as shown in the 
illustration, and they wished to have the lake 
to themselves, so they instructed a builder to 
put up the shortest possible wall that would 
exclude the cottagers, but give themselves free 
access to the lake. How was the wall to be built ? 



193.— THE SHEEPFOLD. 

It is a curious fact that the answers always 
given to some of the best-known puzzles that 
appear in every little book of fireside recreations 
that has been published for the last fifty or a 
hundred years are either quite unsatisfactory 
or clearly wrong. Yet nobody ever seems to 
detect their faults. Here is an example : — ^A 
farmer had a pen made of fifty hurdles, capable 
of holding a hundred sheep only. Supposing 
he wanted to make it sufficiently large to hold 
double that number, how many additional 
hurdles must he have ? 

194— THE GARDEN WALLS. 




A SPECULATIVE country builder has a circular 
field, on which he has erected four cottages, as 
shown in the illustration. The field is sur- 
roiuided by a brick wall, and the owner under- 
took to put up three other brick walls, so that 
the neighbours should not be overlooked by 
each other, but the foxu: tenants insist that there 
shall be no favouritism, and that each shaU have 
exactly the same length of wall space for his 
wall fruit trees. The puzzle is to show how the 
three walls may be built so that each tenant 
shall have the same area of groimd, and pre- 
cisely the same length of wall. 

Of course, each garden must be entirely en- 
closed by its walls, and it must be possible to 
prove that each garden has exactly the same 
length of wall. If the puzzle is properly solved 
no figures are necessary. 

195.— LADY BELINDA'S GARDEN. 

Lady Belinda is an enthusiastic gardener. In 
the illustration she is depicted in the act of 
worrying out a pleasant little problem which I 
will relate. One of her gardens is oblong in 
shape, enclosed by a high holly hedge, and sbe is 
turning it into 9. rosary for Idbiie ci^tivatioi^ ci 



GEOMETRICAL PROBLEMS. 



53 



some of her choicest roses. She wants to devote 
exactly half of the area of the garden to the 
flowers, in one large bed, and the other half to 
be a path going all round it of equal breadth 
throughout. Such a garden is shown in the 
diagram at the foot of the picture. How is she 




to mark out the garden under these simple con- 
ditions ? She has only a tape, the length of the 
garden, to do it with, and, as the holly hedge 
is so thick and dense, she must make all her 
measurements inside. Lady Belinda did not 
know the exact dimensions of the garden, and, 
as it was not necessary for her to know, I also 
give no dimensions. It is quite a simple task 
no matter what the size or proportions of the 
garden may be. Yet how many lady gardeners 
would know just how to proceed ? The tape 
may be quite plain — that is, it need not be a 
graduated measure. 

196.— THE TETHERED GOAT. 




Here is a little problem that everybody should 
know how to solve. The goat is placed in a 



half-acre meadow, that is in shape an equilateral 
triangle. It is tethered to a post at one corner 
of the field. What should be the length of the 
tether (to the nearest inch) in order that the 
goat shall be able to eat just half the grass in 
the field ? It is assumed that the goat can 
feed to the end of the tether. 

197.— THE COMPASSES PUZZLE. 

It is curious how an added condition or restric- 
tion wiU sometimes convert an absurdly easy 
puzzle into an interesting and perhaps difficult 
one. I remember buying in the street many 
years ago a little mechanical puzzle that had a 
tremendous sale at the time. It consisted of a 
medal with holes in it, and the puzzle was to 
work a ring with a gap in it from hole to hole 
until it was finally detached. As I was walking 
along the street I very soon acquired the trick 
of taking off the ring with one hand while hold- 
ing the puzzle in my pocket. A friend to whom 
I showed the little feat set about accomplishing 
it himself, and when I met him some days after- 
wards he exhibited his proficiency in the art. 
But he was a little taken aback when I then took 
the puzzle from him and, while simply holding 
the medal between the finger and thumb of one 
hand, by a series of little shakes and jerks 
caused the ring, without my even touching it, 
to fall oflf upon the floor. The following little 
poser will probably prove a rather tough nut 
for a great many readers, simply on account of 
the restricted conditions : — 

Show how to find exactly the middle of any 
straight line by means of the compasses only. 
You are not allowed to use any ruler, pencil, or 
other article — only the compasses ; and no trick 
or dodge, such as folding the paper, will be per- 
mitted. You must simply use the compasses 
in the ordinary legitimate way. 

198.— THE EIGHT STICKS. 

I HAVE eight sticks, four of them being exactly 
half the length of the others. I lay every one 
of these on the table, so that they enclose three 
squares, all of the same size. How do I do it ? 
There must be no loose ends hanging over. 

199.— PAPA'S PUZZLE. 

Here is a puzzle by Pappus, who lived at Alex- 
andria about the end of the third century. It 
is the fifth proposition in the eighth book of his 
Mathematical Collections. I give it in the form 
that I presented it some years ago imder the title 
" Papa's Puzzle," just to see how many readers 
would discover that it was by Pappus himself. 
" The little maid's papa has taken two different- 
sized rectangular pieces of cardboard, and has 
cUpped off a triangular piece from one of them, 
so that when it is suspended by a thread from 
the point A it hangs with the long side perfectly 
horizontal, as shown in the illustration. He has 
perplexed the child by asking her to find the 
point A on the other card, so as to produce a 
similar result when cut and suspended by a 
thread." Of course, the point must not be 



54 



AMUSEMENTS IN MATHEMATICS. 




found by trial clippings. A curious and pretty 
point is involved in this setting of the puzzle. 
Can the reader discover it ? 



a little calculation that ought to interest my 
readers. The Professor was paying out the 
wire to which his kite was attached from a winch 
on which it had been roUed into a perfectly 
spherical form. This ball of wire was just 
two feet in diameter, and the wire had a dia- 
meter of one-hvmdredth of an inch. What was 
the length of the wire ? 

Now, a simple little question like this that 
everybody can perfectly understand will puzzle 
many people to answer in any way. Let us see 
whether, without going into any profound ma- 
thematical calculations, we can get the answer 
roughly — say, within a mile of what is correct ! 
We will assume that when the wire is all woimd 
up the baU is perfectly solid throughout, and 
that no allowance has to be made for the axle 
that passes through it. With that simplifi- 
I cation, I wonder how many readers can state 
within even a mile of the correct answer the 
length of that wire. 

201.— HOW TO MAKE CISTERNS. 

Our friend in the illustration has a large sheet 
of zinc, measuring (before cutting) eight feet by 
three feet, and he has cut out square pieces (aU 
of the same size) from the four comers and now 
proposes to fold up the sides, solder the edges, 




^^^, 



200.— A KITE-FLYING PUZZLE. 

While accompanying my friend Professor High- 
flite during a scientific kite-flying competition 
on the South Downs of Sussex I was led into 



and make a cistern. But the point that puzz]«-.s 
him is this : Has he cut out those square pieces 
of the correct size in order that the cistern may 
hold the greatest possible quantity of water ? 
You see, if you cut them very small you get a 



GEOMETRICAL PROBLEMS. 



55 



very shallow cistern ; if you cut them large 
you get a tall and slender one. It is all a ques- 
tion of finding a way of cutting out these four 
square pieces exactly the right size. How are 
we to avoid making them too small or too large ? 

202.— THE CONE PUZZLE. 




I HAVE a wooden cone, as shown in Fig. i. How 
am I to cut out of it the greatest possible cyl- 
inder ? It will be seen that I can cut out one 
that is long and slender, like Fig. 2, or short and 
thick, like Fig. 3. But neither is the largest 
possible. A child could tell you where to cut, 
if he knew the rule. Can you find this simple 
rule ? 

203.— CONCERNING WHEELS. 




There are some curious facts concerning the 
movements of wheels that are apt to perplex 
the novice. For example : when a railway 
train is travelling from London to Crewe certain 
parts of the train at any given moment are 
actually moving from Crewe towards London. 
Can you indicate those parts ? It seems absurd 



that parts of the same train can at any time 
travel in opposite directions, but such is the 
case. 

In the accompanying illustration we have 
two wheels. The lower one is supposed to be 
fixed and the upper one running round it in the 
direction of the arrows. Now, how many times 
does the upper wheel turn on its own axis in 
making a complete revolution of the other 
wheel ? Do not be in a hurry with your answer, 
or you are almost certain to be wrong. Experi- 
ment with two pennies on the table and the 
correct answer will surprise you, when you 
succeed in seeing it. 

204.— A NEW MATCH PUZZLE. 



In the illustration eighteen matches are shown 
arranged so that they enclose two spaces, one 
just twice as large as the other. Can you re- 
arrange them (i) so as to enclose two four-sided 
spaces, one exactly three times as large as the 
other, and (2) so as to enclose two five-sided 
spaces, one exactly three times as large as the 
other ? All the eighteen matches must be 
fairly used in each case ; the two spaces must 
be quite detached, and there must be no loose 
ends or duplicated matches. 

205.— THE SIX SHEEP-PENS. 



A ' ' ' ■ k.' 



a»t=;= 



=1 



Here is a new little puzzle with matches. It 
wiU be seen in the illustration that thirteen 
matches, representing a farmer's hurdles, have 
been so placed that they enclose six sheep-pens 
all of the same size. Now, one of these hiurdles 
was stolen, and the farmer wanted still to en- 
close six pens of equal size with the remaining 
twelve. How was he to do it ? All the twelve 
matches must be fairly used, and there must 
be no duplicated matches or loose ends. 



56 



AMUSEMENTS IN MATHEMATICS. 



POINTS AND LINES PROBLEMS. 



" Line upon line, line upon line ; here a little 
and there a little." — Isa. xxviii. lo. 

What are known as " Points and Lines " puzzles 
are found very interesting by many people. 
The most familiar example, here given, to plant 
nine trees so that they shall form ten straight 
rows with three trees in every row, is attributed 
to Sir Isaac Newton, but the earliest collection 
of such puzzles is, I believe, in a rare little book 
that I possess — published in 1821 — Rational 
Amusement for Winter Evenings, by John Jack- 
son. The author gives ten examples of " Trees 
planted in Rows." 

These tree-planting puzzles have always been 
a matter of great perplexity. They are real 
" puzzles," in the truest sense of the word, be- 
cause nobody has yet succeeded in finding a 
direct and certaiu way of solving them. They 
demand the exercise of sagacity, ingenuity, and 
patience, and what we call " luck " is also some- 
times of service. Perhaps some day a genius 
will discover the key to the whole mystery. 
Remember that the trees must be regarded as 



mere points, for if we were allowed to make our 
trees big enough we might easily " fudge " our 
diagrams and get in a few extra straight rows 
that were more apparent than real. 

206.— THE KING AND THE CASTLES. 

There was once, in ancient times, a powerful 
king, who had eccentric ideas on the subject of 
military architecture. He held that there was 
great strength and economy in symmetrical 
forms, and always cited the example of the bees, 
who construct their combs in perfect hexagonal 
cells, to prove that he had nature to support 
him. He resolved to build ten new castles in his 
country, aU to be connected by fortified walls, 
which should form five lines with four castles 
in every Une. The royal architect presented 
his preliminary plan in the form I have shown. 
But the monarch pointed out that every castle 
could be approached from the outside, and com- 
manded that the plan should be so modified 
that as many castles as possible should be free 
from attack from the outside, and could only 
be reached by crossing the fortified walls. The 
arcliitect replied that he thought it impossible 
so to arrange them that even one castle, which 



the king proposed to use as a royal residence, 
could be so protected, but his majesty soon 
enlightened him by pointing out how it might 




be done. How would you have built the ten 
castles and fortifications so as best to fulfil the 
king's requirements ? Remember that they 
must form five straight lines with tova castles 
in every line, 

207.— CHERRIES AND PLUMS. 

The illustration is a plan of a cottage as it 
stands surrounded by an orchard of fifty-five 
trees. Ten of these trees are cherries, ten are 




plums, and the remainder apples. The cherries 
are so planted as to form five straight lines, with 
four cherry trees in every line. The plum trees 



POINTS AND LINES PROBLEMS. 



57 



are also planted so as to form five straight lines 
with four plum trees in every line. The puzzle 
is to show which are the ten cherry trees and 
which are the ten plums. In order that the 
cherries and plums should have the most fav- 
ourable aspect, as few as possible (under the 
conditions) are planted on the north and east 
sides of the orchard. Of course in picking out 
a group of ten trees (cherry or plum, as the case 
may be) you ignore all intervening trees. That 
is to say, four trees may be in a straight line 
irrespective of other trees (or the house) being 
in between. After the last puzzle this will be 
quite easy. 

208.— A PLANTATION PUZZLE. 



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A MAN had a square plantation of forty-nine 
trees, but, as will be seen by the omissions in the 
illustration, four trees were blown down and 
removed. He now wants to cut down all the 
remainder except ten trees, which are to be so 
left that they shall form five straight rows with 
four trees in every row. Which are the ten 
trees that he must leave ? 

209.— THE TWENTY-ONE TREES. 

A GENTLEMAN wishcd to plant twenty-one trees 
in his park so that they should form twelve 
straight rows with five trees in every row. 
Could you have supplied him with a pretty 
symmetrical arrangement that would satisfy 
these conditions ? 

210.— THE TEN COINS. 

Place ten pennies on a large sheet of paper or 
cardboard, as shown in the diagram, five on 
each edge. Now remove four of the coins, with- 
out disturbing the others, and replace them on 
the paper so that the ten shall form five straight 
lines with four coins in every line. This in 
itself is not difficult, but you should try to dis- 



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cover in how many different ways the puzzle 
may be solved, assuming that in every case the 
two rows at starting are exactly the same. 

311.— THE TWELVE MINCE-PIES. 

It will be seen in our illustration how twelve 
mince-pies may be placed on the table so as to 
form six straight rows with four pies in every 
row. The puzzle is to remove only four of them 
to new positions so that there shall be seven 



straight rows with four in every row. Which 
four would you remove, and where would you 
replace them ? 



58 



AMUSEMENTS IN MATHEMATICS. 



212.— THE BURMESE PLANTATION. 



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ft 


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A SHORT time ago I received an interesting 
communication from the British chaplain at 
Meiktila, Upper Burma, in which my corre- 
spondent informed me that he had found some 



amusement on board ship on his way out in 
trying to solve this little poser. 

If he has a plantation of forty-nine trees, 
planted in the form of a square as shown in 
the accompanying illustration, he wishes to 
know how he may cut down twenty-seven of 
the trees so that the twenty-two left standing 
shall form as many rows as possible with four 
trees in every row. 

Of course there may not be more than four 
trees in any row. 

213.— TURKS AND RUSSIANS. 

This puzzle is on the lines of the Afridi problem 
published by me in Tit-Bits some years ago. 

On an open level tract of country a party of 
Russian infantry, no two of whom were stationed 
at the same spot, were suddenly surprised by 
thirty-two Turks, who opened fire on the Rus- 
sians from all directions. Each of the Turks 
simultaneously fired a bullet, and each bullet 
passed immediately over the heads of three 
Russian soldiers. As each of these bullets when 
fired killed a different man, the puzzle is to dis- 
cover what is the smallest possible munber of 
soldiers of which the Russian party could have 
consisted and what were the casualties on each 
side. 



MOVING COUNTER PROBLEMS. 



" I cannot do't without counters." 

Winter's Tale, iv. 3. 

Puzzles of this class, except so far as they occur 
in connection with actual games, such as chess, 
seem to be a comparatively modern introduc- 
tion. Mathematicians in recent times, notably 
Vandermonde and Reiss, have devoted some 
attention to them, but they do not appear to 
have been considered by the old writers. So 
far as games with counters are concerned, per- 
haps the most ancient and widely known in old 
times is " Nine Men's Morris " (known also, as 
I shall show, tmder a great many other names), 
unless the simpler game, distinctly mentioned in 
the works of Ovid (No. no, " Ovid's Game," in 
The Canterbury Puzzles), from which " Noughts 
and Crosses " seems to be derived, is still more 
ancient. 

In France the game is called Marelle, in 
Poland Siegen Wulf MyU (She-goat Wolf Mill, 
or Fight), in Germany and Austria it is called 
Muhle (the Mill), in Iceland it goes by the name 
of Mylla, while the Bogas (or native bargees) of 
South America are said to play it, and on the 
Amazon it is called Trique, and held to be of 
Indian origin. In our own country it has 
different names in different districts, such as 
Meg Merrylegs, Peg Meryll, Nine Peg o'Merryal, 
Nine- Pin Miracle, Merry Peg, and Merry Hole. 
Shakespeare refers to it in " Midsummer Night's 
Dream " (Act ii., scene i) : — 

" The nine-men's morris is filled up with mud ; 
And the quaint mazes in the wanton green, 
For lack of tread, are undistinguishable." 



It was played by the shepherds with stones in 
holes cut in the turf. John Clare, the peasant 
poet of Northamptonshire, in " The Shepherd 
Boy " (1835) says : — " Oft we track his haunts 
. . . . By nine-peg-morris nicked upon the 
green." It is also mentioned by Drayton in 
his " Polyolbion." 

It was found on an old Roman tile discovered 
during the excavations at Silchester, and cut 
upon the steps of the Acropolis at Athens. 
When visiting the Christiania Museum a few 
years ago I was shown the great Viking ship that 
was discovered at Gokstad in 1880. On the 
oak planks forming the deck of the vessel were 
found holes and lines marking out the game, 
the holes being made to receive pegs. While 
inspecting the ancient oak furniture in the Rijks 
Museum at Amsterdam I became interested in 
an old catechiunen's settle, and was surprised 
to find the game diagram cut in the centre of 
the seat — quite conveniently for surreptitious 
play. It has been discovered cut in the choir 
stalls of several of our English cathedrals. In 
the early eighties it was found scratched upon 
a stone built into a waU (probably about the 
date 1200), during the restoration of Hargrave 
church in Northamptonshire. This stone is 
now in the Northampton Museum. A similar 
stone has since been found at Sempringham, 
Lincolnshire. It is to be seen on an ancient 
tombstone in the Isle of Man, and painted on 
old Dutch tiles. And in 1901 a stone was dug 
out of a gravel pit near Oswestry bearing an 
undoubted diagram of the game. 

The game has been played with different 



MOVING COUNTER PROBLEMS. 



59 



rules at dififetent periods and places. I give a 
copy of the board. Sometimes the diagonal 




lines are omitted, but this evidently was not 
intended to affect the play : it simply meant 
that the angles alone were thought sufficient 
to indicate the points. This is how Strutt, in 
Sports and Pastimes, describes the game, and it 
agrees with the way I played it as a boy : — 
" Two persons, having each of them nine pieces, 
or men, lay them down alternately, one by one, 
upon the spots ; and the business of either party 
is to prevent his antagonist from placing three 
of his pieces so as to form a row of three, with- 
out the intervention of an opponent piece. If 
a row be formed, he that made it is at liberty 
to take up one of his competitor's pieces from 
any part he thinks most to his advantage ; 
excepting he has made a row, which must not 
be touched if he have another piece upon the 
board that is not a component part of that row. 
When all the pieces are laid down, they are 
played backwards and forwards, in any direc- 
tion that the lines run, but only can move 
from one spot to another (next to it) at one 
time. He that takes off all his antagonist's 
pieces is the conqueror." 



THE SIX FROGS. 




The six educated frogs in the illustration are 
trained to reverse their order, so that their 
numbers shall read 6, 5, 4, 3, 2, i, with the blank 
square in its present position. They can jump 
to the next square (if vacant) or leap over one 
frog to the next square beyond (if vacant), just 
as we move in the game of draughts, and can 
go backwards or forwards at pleasure. Can 
you show how they perform their feat in the 
fewest possible moves ? It is quite easy, so 
when you have done it add a seventh frog to 
the right and try again. Then add more frogs 
until you are able tc give the shortest solution 
for any number. Foi it can always be done, 
with that single vacant square, no matter how 
many frogs there are. 



215.— THE GRASSHOPPER PUZZLE. 

It has been suggested that this puzzle was a 
great favourite among the yoimg apprentices of 
the City of London in the sixteenth and seven- 
teenth centuries. Readers will have noticed 
the curious brass grasshopper on the Royal 
Exchange. This long-lived creature escaped 
the fires of 1666 and 1838. The grasshopper, 
after his kind, was the crest of Sir Thomas 
Gresham, merchant grocer, who died in 1579, 
and from this cause it has been used as a sign 
by grocers in general. Unfortunately for the 
legend as to its origin, the puzzle was only pro- 
duced by myself so late as the year 1900. On 




twelve of the thirteen black discs are placed 
numbered coimters or grasshoppers. The puzzle 
is to reverse their order, so that they shall read, 
i> 2, 3, 4, etc., in the opposite direction, with 
the vacant disc left in the same position as at 
present. Move one at a time in any order, 
either to the adjoining vacant disc or by jump- 
ing over one grasshopper, like the moves in 
draughts. The moves or leaps may be made 
in either direction that is at any time possible. 
What are the fewest possible moves in which 
it can be done ? 

216.— THE EDUCATED FROGS. 

Our six educated frogs have learnt a new and 
pretty feat. When placed on glass tumblers, 
as shown in the illustration, they change sides 
so that the three black ones are to the left 
and the white frogs to the right, with the un- 
occupied tumbler at the opposite end — No. 7. 
They can jump to the next tumbler (if imoccu- 
pied), or over one, or two, frogs to an unoccupied 




tumbler. The jumps can be made in either 
direction, and a frog may jump over his own or 
the opposite colour, or both colours. Four sue- 



6o 



AMUSEMENTS IN MATHEMATICS. 



cessive specimen jumps will make everything 
quite plain : 4 to i, 5 to 4, 3 to 5, 6 to 3. Can 
you show how they do it in ten jumps ? 

217.— THE TWICKENHAM PUZZLE. 




In the illustration we have eleven discs in a 
circle. On five of the discs we place white 
coimters with black letters — as shown — and on 
five other discs the black counters with white 
letters. The bottom disc is left vacant. Start- 
ing thus, it is required to get the coimters into 
order so that they spell the word " Twicken- 
ham " in a clockwise direction, leaving the 
vacant disc in the original position. The black 
coimters move Ln the direction that a clock- 
hand revolves, and the white counters go the 
opposite way. A coimter may jump over one 
of the opposite colour if the vacant disc is next 
beyond. Thus, if your first move is with K, 
then C can jump over K. If then K moves 
towards E, you may next jump W over C, and 
so on. The puzzle may be solved in twenty- 
six moves. Remember a counter cannot jmnp 
over one of its own colour. 

218.— THE VICTORIA CROSS PUZZLE. 




often built up with the slenderest material*. 
Trivialities that might entirely escape the ob- 
servation of others, or, if they were observed, 
would be regarded as of no possible moment, 
often supply the man who is in quest of posers 
with a pretty theme or an idea that he thinks 
possesses some " basal value." 

When seated opposite to a lady in a railway 
carriage at the time of Queen Victoria's Dia- 
mond Jubilee, my attention was attracted to a 
brooch that she was wearing. It was in the 
form of a Maltese or Victoria Cross, and bore the 
letters of the word VICTORIA. The number 
and arrangement of the letters immediately 
gave me the suggestion for the puzzle which I 
now present. 

The diagram, it will be seen, is composed of 
nine divisions. The puzzle is to place eight 
counters, bearing the letters of the word VIC- 
TORIA, exactly in the manner shown, and then 
slide one letter at a time from black to white and 
white to black alternately, until the word reads 
round in the same direction, only with the 
initial letter V on one of the black arms of the 
cross. At no time may two letters be in the 
same division. It is required to find the shortest 
method. 

Leaping moves are, of course, not permitted. 
The first move must obviously be made with 
A, I, T, or R, Supposing you move T to the 
centre, the next counter played will be O or C, 
since I or R cannot be moved. There is some- 
thing a little remarkable in the solution of this 
puzzle which I will explain. 



219.— THE LETTER BLOCK PUZZLE. 




'' ^^m^^^mw^^^^^^m^^^ 



Here is a little reminiscence of our old friend 
the Fifteen Block Puzzle. Eight wooden blocks 
are lettered, and are placed in a box, as shown 
in the illustration. It will be seen that you can 
only move one block at a time to the place 
vacant for the time being, as no block may be 
The puzzle-maker is pecxiliarly a " snapper-up lifted out of the box. The puzzle is to shift 
of unconsidered trifles," and his productions are them about until you get them in the order — 



MOVING COUNTER PROBLEkS. 



6i 



A 


B 


C 


D 


E 


F 


G 


H 





This you will find by no means difficult if you i 
are allowed as many moves as you like. But | 
the puzzle is to do it in the fewest possible 
moves. I will not say what this smallest num- 
ber of moves is, because the reader may like 
to discover it for himself. In writing down 
your moves you will find it necessary to record 
no more than the letters in the order that they 
are shifted. Thus, your first five moves might 
be C, H, G, E, F ; and this notation can have 
no possible ambiguity. In practice you only 
need eight counters and a simple diagram on a 
sheet of paper. 

220.— A LODGING-HOUSE DIFFICULTY. 







CABINd 




Piano 


1 2 


3 


bF WRDfioei 




BWKCASE 


DRAVvERS 1 






4 1 5^ 


' 1 



The Dobsons secured apartments at Slocomb- 
on-Sea. There were six rooms on the same floor, 
all communicating, as shown in the diagram. 
The rooms they took were numbers 4, 5, and 6, 
aU facing the sea. But a little difficulty arose. 
Mr. Dobson insisted that the piano and the book- 
case should change rooms. This was wily, for 
the Dobsons were not musical, but they wanted 
to prevent any one else playing the instrument. 
Now, the rooms were very small and the pieces 
of furniture indicated were very big, so that no 
two of these articles could be got into any room 
at the same time. How was the exchange to 
be made with the least possible labour ? Sup- 
pose, for example, you first move the wardrobe 
into No. 2 ; then you can move the bookcase to 
No. 5 and the piano to No. 6, and so on. It 
is a fascinating puzzle, but the landlady had 
reasons for not appreciating it. Try to solre 
her difficulty in the fewest possible removals 
with counters on a sheet of paper. 

221.— THE EIGHT ENGINES. 

The diagram represents the engine-yard of a 
railway company under eccentric management. 
The engines are allowed to be stationary only 
at the nine points indicated, one of which is at 
present vacant. It is required to move the 
engines, one at a time, from point to point, in. 
seventeen moves, so that their niimbers shall be 
in numerical order round the circle, with the 




central point left vacant. But one of the 
engines has had its fire drawn, and therefore 
cannot move. How is the thing to be done ? 
And which engine remains stationary through- 
out ? 

222.— A RAILWAY PUZZLE. 




Make a diagram, on a large sheet of paper, like 
the illustration, and have three counters marked 
A, three marked B, and three marked C. It 
will be seen that at the intersection of lines there 
are nine stopping-places, and a tenth stopping- 



62 



AMUSEMENTS IN MATHEMATICS. 



place is attached to the outer circle like the tail 
of a Q. Place the three counters or engines 
marked A, the three marked B, and the three 
marked C at the places indicated. The puzzle 
is to move the engines, one at a time, along 
the lines, from stopping-place to stopping- 
place, until you succeed in getting an A, a B, 
and a C on each circle, and also A, B, and C 
on each straight line. You are required to do 
this in as few moves as possible. How many 
moves do you need ? 

223.— A RAILWAY MUDDLE. 

The plan represents a portion of the line of the 
London, Clodville, and Mudford Railway Com- 
pany. It is a single line with a loop. There is 
only room for eight wagons, or seven wagons 
and an engine, between B and C on either the 
left line or the right line of the loop. It hap- 
pened that two goods trains (each consisting of 
an engine and sixteen wagons) got into the 
position shown in the illustration. It looked 
Uke a hopeless deadlock, and each engine-driver 
wanted the other to go back to the next station 
and take off nine wagons. But an ingenious 
stoker undertook to pass the trains and send 
them on their respective journeys with their 
engines properly in front. He also contrived 
to reverse the engines the fewest times possible. 
Could you have performed the feat ? And how 
many times would you require to reverse the 




engines ? A " reversal " means a change of 
direction, backward or forward. No rope- 
shunting, fly-shunting, or other trick is al- 



lowed. All the work must be done legitimately 
by the two engines. It is a simple but interest- 
ing puzzle if attempted with counters. 

224.— THE MOTOR-GARAGE PUZZLE. 





The difficulties of the proprietor of a motor 
garage are converted into a little pastime of a 
kind that has a peculiar fascination. All you 
need is to make a simple plan or diagram on a 
sheet of paper or cardboard and number eight 
counters, i to 8. Then a whole family can enter 
into an amusing competition to find the best 
possible solution of the difficulty. 

The illustration represents the plan of a motor 
garage, with accommodation for twelve cars. 
But the premises are so inconveniently restricted 
that the proprietor is often caused considerable 
perplexity. Suppose, for example, that the 
eight cars numbered i to 8 are in the positions 
shown, how are they to be shifted in the quickest 
possible way so that i, 2, 3, and 4 shall change 
places with 5, 6, 7, and 8 — that is, with the 
numbers still running from left to right, as at 
present, but the top row exchanged with the 
bottom row ? What are the fewest possible 
moves ? 

One car moves at a time, and any distance 
counts as one move. To prevent misunder- 
standing, the stopping-places are marked in 
squares, and only one car can be in a square at 
the same time. 

225.— THE TEN PRISONERS. 

If prisons had no other use, they might stiU 
be preserved for the special benefit of puzzle- 
makers. They appear to be an inexhaustible 
mine of perplexing ideas. Here is a little poser 
that will perhaps interest the reader for a short 
period. We have in the illustration a prison 
of sixteen cells. The locations of the ten pris- 
oners will be seen. The jailer has queer super- 
stitions about odd and even numbers, and he 



MOVING COUNTER PROBLEMS. 



63 



wants to rearrange the ten prisoners so that 
there shall be as many even rows of men, verti- 
cally, horizontally, and diagonally, as possible. 
At present it wiU be seen, as indicated by the 
arrows, that there are only twelve such rows of 




5 and 4. I will state at once that the greatest 
lumber of such rows that is possible is sixteen. 
3ut the jailer only allows four men to be re- 
noved to other cells, and informs me that, as 
he man who is seated in the bottom right-hand 
iorner is infirm, he must not be moved. Now, 
low are we to get those sixteen rows of even 
lumbers under such conditions ? 

226.— ROUND THE COAST. 




Iere is a puzzle that will, I think, be found as 
musing as instructive. We are given a ring of 
ight circles. Leaving circle 8 blank, we are 
squired to write in the name of a seven-lettered 
ort in the United Kingdom in this manner. 



Touch a blank circle with your pencil, then 
jump over two circles in either direction round 
the ring, and write down the first letter. Then 
touch another vacant circle, jump over two 
circles, and write down your second letter. Pro- 
ceed similarly with the other letters in their 
proper order until you have completed the word. 
Thus, suppose we select " Glasgow," and pro- 
ceed as follows : 6 — i, 7 — 2, 8 — 3, 7 — 4, 8 — 5, 
which means that we touch 6, jump over 7 and 
8, and write down " G " on i ; then touch 7, 
jump over 8 and i, and write down " 1 " on 2 ; 
and so on. It will be found that after we have 
written down the first five letters — " Glasg " — 
as above, we cannot go any further. Either 
there is something wrong with " Glasgow," or 
we have not managed our jumps properly. Can 
you get to the bottom of the mystery ? 

227.— CENTRAL SOLITAIRE. 




This ancient puzzle was a great favourite with 
our grandmothers, and most of us, I imagine, 
have on occasions come across a " Solitaire " 
board — a round polished board with holes cut 
in it in a geometrical pattern, and a glass marble 
in every hole. Sometimes I have noticed one 
on a side table in a suburban front parlour, or 
found one on a shelf in a country cottage, or 
had one brought under my notice at a wayside 
inn. Sometimes they are of the form shown 
above, but it is equally common for the board 
to have four more holes, at the points indicated 
by dots. I select the simpler form. 

Though " Solitaire " boards are still sold at 
the toy shops, it will be suf&cient if the reader 
will make an enlarged copy of the above on 
a sheet of cardboard or paper, number the 
" holes," and provide himself with 33 counters, 
buttons, or beans. Now place a coimter in 
every hole except the central one. No. 17, and 
the puzzle is to take off all the coimters in a 
series of jumps, except the last counter, which 
must be left in that central hole. You are 



64 



AJVIUSEMENTS IN MATHEMATICS. 



allowed to jump one counter over the next one 
to a vacant hole beyond, just as in the game of 
draughts, and the counter jumped over is im- 
mediately taken off the board. Only remember 
every move must be a jump ; consequently you 
will take off a counter at each move, and thirty- 
one single jumps will of course remove all the 
thirty-one counters. But compound moves are 
allowed (as in draughts, again), for so long as 
one counter continues to jump, the jumps aU 
count as one move. 

Here is the beginning of an imaginary solu- 
tion which will serve to make the manner of 
moving perfectly plain, and show how the 
solver should write out his attempts : 5-1 7) 
12-10, 26-12, 24-26 (13-11, 11-25), 9-11 (26-24, 
24-10, 10-12), etc., etc. The jumps contained 
within brackets count as one move, because they 
are made with the same counter. Find the 
fewest possible moves. Of course, no diagonal 
jumps are permitted ; you can only jump in 
the direction of the lines. 



any diagonal moves — only moves parallel to 
the sides of the square. It is obvious that as 
the apples stand no move can be made, but you 
are permitted to transfer any single apple you 
like to a vacant plate before starting. Then 
the moves must be all leaps, taking off the 
apples leaped over. 

229.— THE NINE ALMONDS. 

" Here is a little puzzle," said a Parson, 
" that I have found peculiarly fascinating. It 
is so simple, and yet it keeps you interested 
indefinitely." 

The reverend gentleman took a sheet of paper 
and divided it off into twenty-five squares, like 
a square portion of a chessboard. Then he 
placed nine almonds on the central squares, as 
shown in the illustration, where we have repre- 
sented numbered coimters for convenience in 
giving the solution. 

" Now, the puzzle is," continued the Parson, 




228.— THE TEN APPLES. 

The family represented in the illustration are 
amusing themselves with this little puzzle, which 
is not very difficult but quite interesting. They 
have, it will be seen, placed sixteen plates on 
the table in the form of a square, and put an 
apple in each of ten plates. They want to 
find a way of removing all the apples except 
one by jumping over one at a time to the next 
vacant square, as in draughts; or, better, as 
in solitaire, for you are not allowed to make 



" to remove eight of the almonds and leave th 
ninth in the central square. You make the re 
movals by jumping one almond over anotbe 
to the vacant square beyond and taking off th 
one jumped over — just as in draughts, only her 
you can jump in any direction, and not diagc 
nally only. The point is to do the thing in tb 
fewest possible moves." 

The following specimen attempt will mai 
everything clear. Jump 4 over i, 5 over 9 
over 6, 5 over 3, 7 over 5 and 2, 4 over 7, 8 ove 
4. But 8 is not left in the central square, as 



MOVING COUNTER PROBLEMS. 



65 




tion round the table, take up one penny, pass it 
over two other pennies, and place it in the next 
plate. Go on again ; take up another penny and, 
having passed it over two pennies, place it in a 
plate; and so continue your journey. Six coins 
only are to be removed, and when these have 
been placed there should be two coins in each 
of six plates and six plates empty. An impor- 
tant point of the puzzle is to go round the table 



should be. Remember to remove those 
you jump over. Any number of jmnps 
in succession with the same almond count 
as one move. 

230.— THE TWELVE PENNIES, 

Here is a pretty little puzzle that only 
requires twelve pennies or counters. Ar- 
range them in a circle, as shown in the 
illustration. Now take up one penny at 
a time and, passing it over two pennies, 
place it on the third penny. Then take 
up another single penny and do the same thing, 
and so on, until, in six such moves, you have the 





coins in six pairs in the positions 1, 2, 3, 4, 5, 6. 
You can move in either direction round the 
circle at every play, and it does not matter 
whether the two jumped over are separate or a 
pair. This is quite easy if you use just a little 
thought. 

231.— PLATES AND COINS. 

; Place twelve plates, as shown, on a round table, 
: with a penny or orange in every plate. Start from 
any plate you like and, always going in one direc- 
(1,926) 



as few times as possible. It does not matter 
whether the two coins passed over are in one 
or two plates, nor how many empty plates you 
pass a coin over. But you must always go in 
one direction round the table and end at the 
point from which you set out. Your hand, 
that is to say, goes steadily forward in one 
direction, without ever moving backwards. 

232.— CATCHING THE MICE. 




" Play fair ! " said the mice. " You know the 
rules of the game." 

" Yes, I know the rules," said the cat. " I've 
got to go round and round the circle, in the 



66 



AMUSEMENTS IN MATHEMATICS. 



direction that you are looking, and eat every 
thirteenth mouse, but I must keep the white 
mouse for a tit-bit at the finish. Thirteen is an 
unlucky number, but I will do my best to oblige 
you." 

" Hurry up, then ! ** shouted the mice. 

" Give a fellow time to think," said the cat. 
" I don't know which of you to start at. I 
must figure it out." 

While the cat was working out the puzzle he 
feU asleep, and, the spell being thus broken, the 
mice returned home in safety. At which mouse 
should the cat have started the count in order 
that the white mouse should be the last eaten ? 

When the reader has solved that little puzzle, 
here is a second one for him. What is the 
smallest number that the cat can count round 
and round the circle, if he must start at the 
white mouse (calling that " one " in the count) 
and stiU eat the white mouse last of all ? 

And as a third puzzle try to discover what is 
the smallest number that the cat can count 
round and round if she must start at the white 
mouse (calling that " one ") and make the white 
mouse the third eaten. 



He places sixteen cheeses on the floor in a 
straight row and then makes them into four 
piles, with four cheeses in every pile, by always 
passing a cheese over four others. If you use 
sixteen counters and number them in order 
from I to i6, then you may place i on 6, ii on i, 
7 on 4, and so on, imtil there are four in every 
pile. It will be seen that it does not matter 
whether the four passed over are standing alone ■ 
or piled ; they count just the same, and you can . 
always carry a cheese in either direction. There 
are a great many different ways of doing it in . 
twelve moves, so it makes a good game of t 
" patience " to try to solve it so that the four i 
piles shall be left in different stipulated places. 
For example, try to leave the pil^ at the ex- 
treme ends of the row, on Nos. i, 2, 15 and 16 ; 
this is quite easy. Then try to leave three piles ♦ 
together, on Nos. 13, 14, and 15. Then again 
play so that they shall be left on Nos. 3, 5, 12, 
and 14. 

234.— THE EXCHANGE PUZZLE. 
Here is a rather entertaining little puzzle withii 




233 —THE ECCENTRIC CHEESEMONGER. 

The cheesemonger depicted in the illustration 
is an inveterate puzzle lover. One of his fa- 
vourite puzzles is the piling of cheeses in his 
warehouse, an amusement that he finds good 
exercise for the body as well as for the mind. 



moving counters. You only need twelve couii 
ters — six of one colour, marked A, C, E, G, 
and K, and the other six marked B, D, F, H, 
and L. You first place them on the diagranj 
as shown in the illustration, and the puzzle 
to get them into regular alphabetical order, i 
follows : — 



MOVING COUNTER PROBLEMS. 



67 



A 


B 


C 


D 


E 


F 


G 


H 


I 


J 


K 


L 



The moves are made by exchanges of opposite 
colours standing on the same line. Thus, G and 
J may exchange places, or F and A, but you 
cannot exchange G and C, or F and D, because 




in one case they are both white and in the other 
case both black. Can you bring about the re- 
quired arrangement in seventeen exchanges ? 



vessels and sank the fourth ? In the diagram 
we have arranged the fleet in square formation, 
where it will be seen that as many as seven ships 
may be sunk (those in the top row and first 
column) by firing the torpedoes indicated by 
arrows. Anchoring the fleet as we like, to 
what extent can we increase this number ? 
Remember that each successive ship is sunk 
before another torpedo is launched, and that 
every torpedo proceeds in a different direction ; 
otherwise, by placing the ships in a straight 
line, we might sink as many as thirteen ! It 
is an interesting little study in naval warfare, 
and eminently practical — provided the enemy 
will allow you to arrange his fleet for your 
convenience and promise to lie still and do 
nothing ! 

236.— THE HAT PUZZLE. 

Ten hats were hung on pegs as shown in the 
illustration^ — five silk hats and five felt " bowl- 
ers," alternately silk and felt. The two pegs 
at the end of the row were empty. 

The puzzle is to remove two contiguous hats 
to the vacant pegs, then two other adjoining 
hats to the pegs now unoccupied, and so on 
until five pairs have been moved and the hats 
again hang in an imbrokeh row, but with all 




It cannot be done in fewer moves. The puzzle 
is reaUy much easier than it looks, if properly 
attacked. 



235.— TORPEDO PRACTICE. 



IL IL H 



M, 



-^ 1 



•«-« z 



a 



it 



M. 



M <^3 



X 

7 



t 

6 



% 
5 



♦« 4 



If a fleet of sixteen men-of-war were lying at 
anchor and surrounded by the enemy, how 
many ships might be sunk if every torpedo, 
projected in a straight line, passed under three 



the silk ones together and all the felt hats 
together. 

Remember, the two hats removed must alwajrs 
be contiguous ones, and you must take one in 
each hand and place them on their new pegs 
without reversing their relative position. You 
are not allowed to cross your hands, nor to hang 
up one at a time. 

Can you solve this old puzzle, which I give as 
introductory to the next ? Try it with counters 
of two colours or with coins, and remember that 
the two empty pegs must be left at one end of 
the row. 

237.— BOYS AND GIRLS. 

If you mark oS ten divisions on a sheet of paper 
to represent the chairs, and use eight numbered 
counters for the children, you will have a fas- 
cinating pastime. Let the odd numbers repre- 
sent boys and even numbers girls, or you can 
use counters of two colours, or coins. 

The puzzle is to remove two children who are 
occupjdng adjoining chairs and place them in 
two empty chairs, making them first change sides ; 
then remove a second pair of children from 
adjoining chairs and place them in the two now 
vacant, making them change sides ; and so on, 
imtil aU the boys are together and aU the girls 
together, with the two vacant chairs at one 
end as at present. To solve the puzzle you 
must do this in five moves. The two children 
must always be taken from chairs that are next 
to one another ; and remember the important 
point of making the two children change sides, 



68 



AMUSEMENTS IN MATHEMATICS. 




as this latter is the distinctive feature of the 
puzzle. By " change sides " I simply mean that 
if, for example, you first move i and 2 to the 
vacant chairs, then the first (the outside) chair 
will be occupied by 2 and the second one by i. 

238.— ARRANGING THE JAMPOTS. 

I HAPPENED to see a little girl sorting out some 
jam in a cupboard for her mother. She was 
putting each different kind of preserve apart on 
the shelves. I noticed that she took a pot of 
damson in one hand and a pot of gooseberry in 
the other and made them change places ; then 
she changed a strawberry with a raspberry, 
and so on. It was interesting to observe what 
a lot of unnecessary trouble she gave herself by 
making more interchanges than there was any 
need for, and I thought it would work into a 
good puzzle. 

It wiU be seen in the illustration that little 
Dorothy has to manipulate twenty-four large 
jampots in as many pigeon-holes. She wants 
to get them in correct numerical order — that is, 
I. 2> 3, 4> 5i 6 on the top shelf, 7, 8, 9, 10, 11, 12 
on the next shelf, and so on. Now, if she always 
takes one pot in the right hand and another in 
the left and makes them change places, how 
many of these interchanges wiU be necessary 



to get all the jampots in proper order ? She 
would naturally first change the i and the 3, 
then the 2 and the 3, when she would have the 
first three pots in their places. How would you 
advise her to go on then ? Place some num- 







bered counters on a sheet of paper divided into 
squares for the pigeon-holes, and you will find 
it an amusing puzzle. 



UNICURSAL AND ROUTE PROBLEMS. 



" I see them on their winding way." 

Reginald Heber. 

It is reasonable to suppose that from the earliest 
ages one man has asked another such questions 
as these : " Which is the nearest way home ? " 
" Which is the easiest or pleasantest way ? " 
" How can we find a way that will enable us to 
dodge the mastodon and the plesiosaurus ? " 
" How can we get there without ever crossing 
the track of the enemy ? " All these are ele- 
mentary route problems, and they can be turned 
into good puzzles by the introduction of some 
conditions that complicate matters. A variety 
of such complications will be found in the fol- 
lowing examples. I have also included some 
enumerations of more or less difficulty. These 
afford excellent practice for the reasoning facul- 
ties, and enable one to generalize in the case of 
symmetrical forms in a manner that is most 
instructive. 



239._A JUVENILE PUZZLE. 

For years I have been perpetually consulted 
by my juvenUe friends about this little puzzle. 
Most children seem to know it, and yet, curi- 
ously enough, they are invariably unacquainted 
with the answer. The question they always 
ask is, " Do, please, tell me whether it is really 
possible." I believe Houdin the conjurer used 
to be very fond of giving it to his child friends, 
but I cannot say whether he invented the little 
puzzle or not. No doubt a large number of my 
readers wiU be glad to have the mystery of the 
solution cleared up, so I make no apology for 
introducing this old " teaser." 

The puzzle is to draw with three strokes of 
the pencil the diagram that the little girl is 
exhibiting in the illustration. Of course, you 
must not remove your pencil from the paper 
during a stroke or go over the same line a 
second time. You will find that you can get 



UNICURSAL AND ROUTE PROBLEMS. 



69 



in a good deal of the figure with one continu- 
ous stroke, but it will always appear as if four 
strokes are necessary. 




Another form of the puzzle is to draw the 
diagram on a slate and then rub it out in three 
rubs. 

240.— THE UNION JACK. 




The illustration is a rough sketch somewhat 
resembling the British flag, the Union Jack. It 
is not possible to draw the whole of it without 
lifting the pencil from the paper or going over 
the same line twice. The puzzle is to find out 
just how much of the drawing it is possible to 
make without lifting your pencil or going twice 
over the same line. Take your pencil and see 
what is the best you can do. 

^ 241.— THE DISSECTED CIRCLE. 

How many continuous strokes, without lifting 
your pencil from the paper, do you require to 
draw the design shown in our illustration ? 
Directly you change the direction of your pencil 
it begins a new stroke. You may go over the 
same line more than onqe if you like. It re- 




quires just a little care, or you may find yourselt 
beaten by one stroke. 

242.— THE TUBE INSPECTOR'S PUZZLE. 

The man in our illustration is in a little dilemma. 
He has just been appointed inspector of a certain 
system of tube railways, and it is his duty to 
inspect regularly, within a stated period, all the 
company's seventeen lines connecting twelve 
stations, as shown on the big poster plan that 
he is contemplating. Now he wants to arrange 
his route so that it shall take him over aU the 
lines with as little travelling as possible. He 
may begin where he likes and end where he 
likes. What is his shortest route ? 

Could anything be simpler ? But the reader 
will soon find that, however he decides to pro 
ceed, the inspector must go over some of the 
lines more than once. In other words, if we 
say that the stations are a mile apart, he will 




have to travel more than seventeen miles to 
inspect every line. There is the little difficulty. 
How far is he compelled to travel, and which 
route do you recommend ? 



70 



AMUSEMENTS IN MATHEMATICS. 



243.— VISITING THE TOWNS. 




A TRAVETLLER, Starting from town No. i, wishes 
to visit every one of the towns once, and once 
only, going only by roads indicated by straight 
lines. How many different routes are there 
from which he can select ? Of course, he must 
end his journey at No. i, from which he started, 
and must take no notice of cross roads, but go 
straight from town to town. This is an ab- 
surdly easy puzzle, if you go the right way to 
work. 

244.— THE FIFTEEN TURNINGS. 

Here is another queer travelling puzzle, the 
solution of which calls for ingenuity. In this 
case the traveller starts from the black town 
and wishes to go as far as possible while making 
only fifteen turnings and never going along the 
same road twice. The towns are supposed to 
be a mile apart. Supposing, for example, that 
he went straight to A, then straight to B, then 



c9--<p---9--o-H>-o 



9-<>--<>--o~<>--<> 



6~cW> 





Y^'Y^V'Y^VK 



<>-<)—(>-<> 



<V-<>-<>-<> 



6— 6-<>-<!> 




to C, D, E, and F, you will then find that he 
has travelled thirty-seven miles in five turnings. 
Now, how far can he go in fifteen turnings ? 

245— THE FLY ON THE OCTAHEDRON. 

" Look here," said the professor to his colleague, 
" I have been watching that fly on the octa- 



hedron, and it confines its walks entirely to the 
edges. What can be its reason for avoiding the 
sides ? " 

" Perhaps it is tr3ring to solve some route 
problem," suggested the other. " Supposing 
it to start from the top point, how many differ- 
ent routes are there by which it may walk over 
all the edges, without ever going twice along 
the same edge in any route ? " 

The problem was a harder one than they ex- 
pected, £md after working at it during leisure 
moments for several days their results did not 
agree — in fact, they were both wrong. If the 




reader is svtrprised at their failure, let him at- 
tempt the little puzzle himself. I will just ex- 
plain that the octahedron is one of the five 
regular, or Platonic, bodies, and is contained 
under eight equal and equilateral triangles. 
If you cut out the two pieces of cardboard of : 
the shape shown in the margin of the illustra- 
tion, cut half through along the dotted lines 
and then bend them and put them together, you 
will have a perfect octahedron. In any route 
over aU the edges it will be found that the fly 
must end at the point of departure at the top. 

246.— THE ICOSAHEDRON PUZZLE. 

The icosahedron is another of the five regular, 
or Platonic, bodies having all their sides, angles, 
and planes similar and equal. It is bounded 
by twenty similar equilateral triangles. If you 
cut out a piece of cardboard of the form shown 
in the smaller diagram, and cut half through i 
along the dotted lines, it will fold up and form 
a perfect icosahedron. 

Now, a Platonic body does not mean a 



UNICURSAL AND ROUTE PROBLEMS. 



71 



heavenly body; but it will suit the purpose of 
our puzzle if we suppose there to be a habitable 
planet of this shape. We will also suppose that, 
owing to a superfluity of water, the only dry 
land is along the edges, and that the inhabitants 
have no knowledge of navigation. If every one 




of those edges is 10,000 miles long and a solitary 
traveller is placed at the North Pole (the highest 
point shown), how far will he have to travel 
before he will have visited every habitable part 
of the planet — that is, have traversed every one 
of the edges ? 

247.— INSPECTING A MINE. 

The diagram is supposed to represent the pas- 
jsages or galleries in a mine. We will assume 
jthat every passage, A to B, B to C, C to H, H to 
I, and so on, is one furlong in length. It will be 
seen that there are thirty-one of these passages. 
Now, an official has to inspect all of them, and 
he descends by the shaft to the point A. How far 
must he travel, and what route do you recom- 
mend ? The reader may at first say, " As there 
are thirty-one passages, each a furlong in length, 
he will have to travel just thirty-one furlongs." 
But this is assuming that he need never go along 
a passage more than once, which is not the case. 
Take your pencil and try to find the shortest 
route. You will soon discover that there is 



A 




B 




c 


^ 


X> 




e 


F 




G 




H 




1 




J 




























K 




L 




M 




N 















P 




9 




R 




S 




T 



room for considerable judgment, 
is a perplexing puzzle. 



In fact, it 



248.— THE CYCLISTS' TOUR. 

Two cyclists were consulting a road map in 
preparation for a little tour together. The 
circles represent towns, and all the good roads 
are represented by lines. They are starting 
from the town with a star, and must complete 
their torn: at E. But before arriving there they 
want to visit every other town once, and only 
once. That is the difficulty. Mr. Spicer said, 
" I am certain we can find a way of doing it ; " 
but Mr. Maggs replied, "No way, I'm sure." 
Now, which of them was correct ? Take your 















C,^^- 


5^V~ 


—> 






f/5 




p 






^**^ ^h- 















pencil and see if you can find any way of doing 
it. Of course you must keep to the roads indi- 
cated. 

249.— THE SAILOR'S PUZZLE. 

The sailor depicted in the illustraticai stated 
that he had since his boyhood been engaged in 
trading with a small vessel among some twenty 
little islands in the Pacific. He supplied the 
rough chart of which I have given a copy, and 
explained that the lines from island to island 
represented the only routes that he ever adopted. 
He always started from island A at the begin- 
ning of the season, and then visited every island 



7a 



AMUSEMENTS IN MATHEMATICS. 




once, and once only, finishing up his tour at the 
starting-point A. But he always put ofiE his 
visit to C as long as possible, for trade reasons 
that I need not enter into. The puzzle is to 
discover his exact route, and this can be done 
with certainty. Take your pencil and, starting 
at A, try to trace it out. If you write down 
the islands in the order in which you visit them 
— thus, for example, A, I, O, L, G, etc. — you 
can at once see if you have visited an island 
twice or omitted any. Of course, the crossings 
of the lines must be ignored — that is, you must 
continue your route direct, and you are not 
allowed to switch off at a crossing and proceed 
in another direction. There is no trick of this 
kind in the puzzle. The sailor knew the best 
route. Can you find it ? 

250.— THE GRAND TOUR. 

One of the everyday puzzles of life is the work- 
ing out of routes. If you are taking a holiday 
on your bicycle, or a motor tour, there always 
arises the question of how you are to make the 
best of your time and other resources. You 



have determined to get as far as some particular 
place, to include visits to such-and-such a town, 
to try to see something of special interest else- 
where, and perhaps to try to look up an old 
friend at a spot that wiU not take you much 
out of your way. Then you have to plan your 
route so as to avoid bad roads, uninteresting 
country, and, if possible, the necessity of a 1 
return by the same way that you went. With 
a map before you, the interesting puzzle is 
attacked and solved. I will present a little 
poser based on these lines. 

I give a rough map of a country — it is not 
necessary to say what particular country — the 
circles representing towns and the dotted lines 
the railways connecting them. Now there lived i 
in the town marked A a man who was born 
there, and during the whole of his life had never 
once left his native place. From his youth up- 
wards he had been very industrious, sticking! 
incessantly to his trade, and had no desire what- 
ever to roam abroad. However, on attaining,] 
his fiftieth birthday he decided to see something j 
of his country, and especially to pay a visit tot 
a very old friend living at the town marked Z. 



UNICURSAL AND ROUTE PROBLEMS. 



7S 




What he proposed was this : that he would 
start from his home, enter every town once and 
only once, and finish his journey at Z. As he 
made up his mind to perform this grand tour 
by rail only, he found it rather a puzzle to work 



of the three houses, A, B, and C, without any 
pipe crossing another. Take your pencil and 



xL 





& 





draw lines showing how this should be done. 
You will soon find yourself landed in difficulties. 

252.— A PUZZLE FOR MOTORISTS. 

Eight motorists drove to church one morning. 
Their respective houses and churches, together 
with the only roads available (the dotted lines), 
are shown. One went from his house A to his 






•-• ! — r-Hll/^-r— 






•V 



^-■f 



.-.w. '. 



m-'" 






J...LJ 



out his route, but he at length succeeded in 
doing so. How did he manage it ? Do not 
forget that every town has to be visited once, 
and not more than once. 

251.— WATER, GAS, AND ELECTRICITY. 

There are some half-dozen puzzles, as old as 
the hills, that are perpetually cropping up, and 
there is hardly a month in the year that does 
not bring inquiries as to their solution. Occa- 
sionally one of these, that one had thought was 
an extinct volcano, bursts into eruption in a 
surprising manner. I have received an extraor- 
dinary number of letters respecting the ancient 
puzzle that I have called " Water, Gas, and 
Electricity." It is much older than electric 
lighting, or even gas, but the new dress brings 
it up to date. The puzzle is to lay on water, 
gas, and electricity, from W, G, and E, to each 



f-T 



church A, another from his house B to his church 
B, another from C to C, and so on, but it was 
afterwards found that no driver ever crossed 
the track of another car. Take your pencil and 
try to trace out their various routes. 

253.— A BANK HOLIDAY PUZZLE. 

Two friends were spending their bank holiday 
on a cycling trip. Stopping for a rest at a vil- 
lage inn, they consulted a route map, which is 
represented in our illustration in an exceedingly 
simplified form, for the puzzle is interesting 
enough without all the original complexities. 
They started from the town in the top left- 
hand corner marked A. It will be seen that 
there are one hundred and twenty such towns, 
aU connected by straight roads. Now they dis- 
covered that there are exactly 1,365 different 
routes by which they may reach their destina- 



74 



AMUSEMENTS IN MATHEMATICS. 



tion, always travelling either due south or due 
east. The puzzle is to discover which town is 
their destination. 

N 




Of course, if you find that there are more 
than 1,365 different routes to a town it cannot 
be the right one. 

254.— THE MOTOR-CAR TOUR. 




In the above diagram the circles represent 
towns and the lines good roads. In just how 
many different ways can a motorist, starting 
from London (marked with an L), make a tour 
of all these towns, visiting every town once, and 
only once, on a tour, and always coming back 
to London on the last ride ? The exact reverse 
of any route is not counted as different. 

255 —THE LEVEL PUZZLE. 

This is a simple counting puzzle. In how many 
different ways can you spell out the word 
LEVEL by placing the point of your pencil 
on an L and then passing along the lines from 




letter to letter. You may go in any direction, 
backwards or forwards. Of comrse you are not 
allowed to miss letters — that is to say, if you 
come to a letter you must use it. 

256.— THE DIAMOND PUZZLE. 

In how many different ways may the word 
DIAMOND be read in the arrangement shown ? 
You may start wherever you like at a D and 
go up or down, backwards or forwards, in and 
out, in any direction you like, so long as you 




always pass from one letter to another that 
adjoins it. How many ways are there ? 

257 —THE DEIFIED PUZZLE. 

In how many different ways may the word 
DEIFIED be read in this arrangement under 



UNIGURSAL AND ROUTE PROBLEMS. 



75 




the same conditions as in the last puzzle, with 
the addition that you can use any letters twice 
in the same reading ? 

258.— THE VOTERS' PUZZLE. 




Here we have, perhaps, the most interesting 
form of the puzzle. In how many different ways 
can you read the political injunction, " RISE 
TO VOTE, SIR," under the same conditions 
as before ? In this case every reading of the 
palindrome requires the use of the central V as 
the middle letter. 

259.— HANNAH'S PUZZLE. 

A MAN was in love with a yoimg lady whose 
Christian name was Hannah. When he asked 
her to be his wife she wrote down the letters 
of her name in this manner : — 



H H H H H H 
H A A A A H 
HANNAH 
HANNAH 
H A A A A H 
H H H H H H 

and promised that she would be his if he could 
tell her correctly in how many different ways 
it was possible to spell out her name, always 
passing from one letter to another that was 
adjacent. Diagonal steps are here allowed. 
Whether she did this merely to tease him or to 
test his cleverness is not recorded, but it is 
satisfactory to know that he succeeded. Would 
you have been equally successful ? Take your 
pencil and try. You may start from any of the 
H's and go backwards or forwards and in any 
direction, so long as aU the letters in a spelling 
are adjoining one another. How many ways 
are there, no two exactly alike ? 

260. —THE HONEYCOMB PUZZLE. 




Here is a little puzzle with the simplest possible 
conditions. Place the point of your pencil on 
a letter in one of the cells of the honeycomb, 
and trace out a very familiar proverb by passing 
always from a cell to one that is contiguous to 
it. If you take the right route you will have 
visited every cell once, and only once. The 
puzzle is much easier than it looks. 

261.— THE MONK AND THE BRIDGES. 

In this case I give a rough plan of a river with 
an island and five bridges. On one side of the 
river is a monastery, and on the other side is 
seen a monk in the foreground. Now, the monk 
has decided that he wiU cross every bridge once, 
and only once, on his return to the monastery. 
This is, of course, quite easy to do, but on the 
way he thought to himself, " I wonder how 
many different routes there are from which I 
might have selected." Could you have told 
him ? That is the puzzle. Take your pencil 
and trace out a route that will take you once 



76 



AMUSEMENTS IN MATHEMATICS. 





over all the five bridges. Then trace out a 
second route, then a third, and see if you can 
count aU the variations. You will find that 




the difficulty is twofold : you have to avoid 
dropping routes on the one hand and counting 
the same routes more than once on the other. 



COMBINATION AND GROUP PROBLEMS. 



" A combination and a form indeed." 

Hamlet, iii. 4. 

Various puzzles in this class might be termed 
problems in the " geometry of situation," but 
their solution really depends on the theory 
of combinations which, in its turn, is derived 
directly from the theory of permutations. It 
has seemed convenient to include here certain 
group puzzles and enumerations that might, 
perhaps, with equal reason have been placed 
elsewhere ; but readers are again asked not to 
be too critical about the classification, which 
is very difficult and arbitrary. As I have in- 
cluded my problem of " The Round Table " 
(No. 273), perhaps a few remarks on another 
well-known problem of the same class, known 
by the French as La Probleme des Menages, 
may be interesting. If n married ladies are 
seated at a round table in any determined order, 
in how many different ways may their n hus- 
bands be placed so that every man is between 
two ladies but never next to his own wife ? 

This difficult problem was first solved by 
Laisant, and the method shown in the following 
table is due to Moreau : — 



4 





2 


5 
6 


3 
13 


13 

80 


7 
8 

9 


83 

592 
4821 


579 

4738 

43387 


10 


43979 


439792 



The first column shows the number of married 
couples. The numbers in the second coluimi 
are obtained in this way : 5X3-Fo — 2 = 13; 
6 X 13 -H 3 + 2 = 83 ; 7 X 83 -h 13 - 2 = 592 ; 8 X 592 
-f 83 -f2 = 4821 ; and so on. Find all the num- 
bers, except 2, in the table, and the method will 
be evident. It will be noted that the 2 is sub- 
tracted when the first number (the number of 
couples) is odd, and added when that num- 
ber is even. The numbers in the third column 
are obtained thus : 13-0 = 13; 83 - 3 = 80 ; 
592-13=579 ; 4821-83 = 4738 ; and so on. 
The numbers in this last column give the re- 
quired solutions. Thus, four husbands may be 
seated in two ways, five husbands may be placed 
in thirteen ways, and six husbands in eighty 
ways. 

The following method, by Lucas, will show 
the remarkable way in which chessboard ar r>l- 
ysis may be applied to the solution of a cir- 
cular problem of this kind. Divide a square 
into thirty-six cells, six by six, and strike out 
all the cells in the long diagonal from the 
bottom left-hand corner to the top right-hand 
corner, also the five cells in the diagonal next 
above it and the cell in the bottom right-hand 
corner. The answer for six couples will be the 
same as the number of ways in which you can 
place six rooks (not using the cancelled cells) 
so that no rook shall ever attack another rook. 
It will be found that the six rooks may be 
placed in eighty different ways, which agrees 
with the above table. 



COMBINATION AND GROUP PROBLEMS. 



77 



262.— THOSE FIFTEEN SHEEP. 

A CERTAIN cyclopaedia has the following curious 
problem, I am told : " Place fifteen sheep in 
four pens so that there shall be the same number 
of sheep in each pen." No answer whatever 
is vouchsafed, so I thought I would investigate 
the matter. I saw that in dealing with apples 
or bricks the thing would appear to be quite 
impossible, since four times any number must 
be an even number, while fifteen is an odd num- 
ber. I thought, therefore, that there must be 
some quality peculiar to the sheep that was not 
generally known. So I decided to interview 
some farmers on the subject. The first one 
pointed out that if we put one pen inside an- 



263.— KING ARTHUR'S KNIGHTS. 

King Arthur sat at the Round Table on three 
successive evenings with his knights — Beleobus, 
Caradoc, Driam, Eric, FloU, and Galahad — but 
on no occasion did any person have as his 
neighbour one who had before sat next to him. 
On the first evening they sat in alphabetical 
order round the table. But afterwards King 
Arthur arranged the two next sittings so that 
he might have Beleobus as near to him as 
possible and Galahad as far away from him as 
could be managed. How did he seat the knights 
to the best advantage, remembering that rule 
that no knight may have the same neighbour 
twice ? 







other, like the rings of a target, and placed all 
sheep in the smallest pen, it would be all right. 
But I objected to this, because you admittedly 
place all the sheep in one pen, not in four pens. 
The second man said that if I placed four sheep 
in each of three pens and three sheep in the last 
pen (that is fifteen sheep in all), and one of the 
ewes in the last pen had a lamb during the 
night, there would be the same number in each 
pen in the morning. This also failed to satisfy 
me. 

The third farmer said, "I've got four hurdle 
pens down in one of my fields, and a small flock 
of wethers, so if you will just step down with 
me I will show you how it is done." The 
illustration depicts my friend as he is about to 
demonstrate the matter to me. His lucid ex- 
planation was evidently that which was in the 
mind of the writer of the article in the cyclo- 
paedia. What was it ? Can you place those 
fifteen sheep ? 



264.— THE CITY LUNCHEONS. 

Twelve men connected with a large firm in 
the City of London sit down to luncheon to- 
gether every day in the same room. The tables 
are small ones that only accommodate two 
persons at the same time. Can you show how 
these twelve men may lunch together on eleven 
days in pairs, so that no two of them shall ever 
sit twice together ? We will represent the men 
by the first twelve letters of the alphabet, and 
suppose the first day's pairing to be as follows — 

(A B) (C D) (E F) (G H) (I J) (K L). 

Then give any pairing you like for the next 
day, say — ■ 

(A C) (B D) (E G) (F H) (I K) (J L), 

and so on, until you have completed your 
eleven lines, with no pair ever occurring twice. 
There are a good many different arrakngeinents 
possible. Try to find one of them. 



78 



AMUSEMENTS IN MATHEMATICS. 



265.— A PUZZLE FOR CARD-PLAYERS. 

Twelve members of a club arranged to play 
bridge together on eleven evenings, but no 
player was ever to have the same partner more 
than once, or the same opponent more than 
twice. Can you draw up a scheme showing how 
they may aU sit down at three tables every 
evening ? Call the twelve players by the first 
twelve letters of the alphabet and try. to group 
them. 

266.— A TENNIS TOURNAMENT. 

Four married couples played a "mixed double " 
tennis toiunament, a man and a lady always 
playing against a man and a lady. But no 
person ever played with or agaiast any other 
person more than once. Can you show how 
they aU could have played together ia the two 
courts on three successive days ? This is a 
little puzzle of a quite practical kind, and it is 
just perplexing enough to be interesting. 

267.—THE WRONG HATS. 

" One of the most perplexing things I have come 
across lately," said Mr. Wilson, " is this. Eight 
men had been dining not wisely but too well at 
a certain London restaurant. They were the 
last to leave, but not one man was in a condition 
to identify his own hat. Now, considering that 
they took their hats at random, what axe the 
chances that every man took a hat that did not 
belong to him ? " 

" The first thing," said Mr. Waterson, " is to 
see in how many different ways the eight hats 
could be taken." 

" That is quite easy," Mr. Stubbs explained. 
" Multiply together the numbers, i, 2, 3, 4, 5, 6, 
7, and 8. Let me see — ^half a minute — ^yes ; 
there are 40,320 different ways." 

" Now aU you've got to do is to see in how 
many of these cases no man has his own hat," 
said Mr. Waterson. 

" Thank you, I'm not taking any," said Mr. 
Packhurst. " I don't envy the man who at- 
tempts the task of writing out aU those forty- 
thousand-odd cases and then picking out the 
ones he wants." 

They all agreed that life is not long enough for 
that sort of amusement ; and as nobody saw 
any other way of getting at the answer, the 
matter was postponed indefinitely. Can you 
solve the puzzle ? 

268.--THE PEAL OF BELLS. 

A CORRESPONDENT, who is apparently much 
interested in campanology, asks me how he is 
to construct what he calls a " true and correct " 
peal for four beUs. He says that every possible 
permTitation of the four beUs must be rung 
once, and once only. He adds that no beU 
must move more than one place at a time, that 
no beU must make more than two successive 
strokes in either the first or the last place, and 
that the last change must be able to pass into 
the first. These fantastic conditions will be 



found to be observed in the little peal for three 
bells, as follows : — 

123 

213 

231 

321 

312 

132 

How are we to give him a correct solution for 
his four bells ? 

269.— THREE MEN IN A BOAT. 

A CERTAIN generous London manufacturer gives 
his workmen every year a week's holiday at the 
seaside at his own expense. One year fifteen 
of his men paid a visit to Heme Bay. On the 
morning of their departure from London they 
were addressed by their employer, who ex- 
pressed the hope that they would have a very 
pleasant time. 

" I have been given to understand," he added, 
" that some of you fellows are very fond of row- 
ing, so I propose on this occasion to provide you 
with this recreation, and at the same time give 
you an amusing little puzzle to solve. During 
the seven days that you are at Heme Bay every 
one of you will go out every day at the same 
time for a row, but there must always be three 
men in a boat and no more. No two men may 
ever go out in a boat together more than once, 
and no man is allowed to go out twice in the 
same boat. If you can manage to do this, and 
use as few different boats as possible, you may 
charge the firm with the expense." 

One of the men teUs me that the exp^ience 
he has gained in such matters soon enabled 
him to work out the answer to the entire satis- 
faction of themselves and their employer. But 
the amusing part of the thing is that they never 
really solved the little mystery. I find their 
method to have been quite incorrect, and I 
think it will amuse my readers to discover how 
the men should have been placed in the boats. 
As their names happen to have been Andrews, 
Baker, Carter, Danby, Edwards, Frith, Gay, 
Hart, Isaacs, Jackson, Kent, Lang, Mason, 
Napper, and Onslow, we can Ccdl them by their 
initials and write out the five groups for each of 
the seven days in the following simple way : — 

12345 
First Day: (ABC) (DEF) (GHI) (JKL) (MNO). 

The men within each pair of brackets are here 
seen to be in the same boat, and therefore A 
can never go out with B or with C again, and C 
can never go out again with B. The same 
applies to the other four boats. The figures 
show the number on the boat, so that A, B, or 
C, for example, can never go out in boat No. i 
again. 

270.— THE GLASS BALLS. 

A NUMBER of clever marksmen were staying at 
a country house, and the host, to provide a 
little amusement, suspended strings of glass 
balls, as shown in the illustration, to be fired 



COMBINATION AND GROUP PROBLEMS. 



79 



at. After they had all put their skill to a 
sufficient test, somebody asked the following 
question : " What is the total number of differ- 
ent ways in which these sixteen balls may be 
broken, if we must always break the lowest 
ball that remains on any string ? " Thus, one 
way would be to break all the four balls on 
each string in succession, taking the strings 
from left to right. Another would be to break 
all the fourth balls on the four strings first, then 



K, L, M, N, and O, and with them form thirty- 
five groups of three letters so that the com- 
binations should include the greatest number 
possible of common English words. No two 
letters may appear together in a group more 
than once. Thus, A and L having been to- 
gether in ALE, must never be found together 
again ; nor may A appear again in a group with 
E, nor L with E. These conditions will be 
found complied with in the above solution, and 




break the three remaining on the first string, 
then take the balls on the three other strings 
alternately from right to left, and so on. There 
is such a vast number of different ways (since 
every little variation of order makes a different 
way) that one is apt to be at first impressed by 
the great difficulty of the problem. Yet it is 
really quite simple when once you have hit on 
the proper method of attacking it. How many 
different ways are there ? 

271.— FIFTEEN LETTER PUZZLE. 



ALE 


FOE 


HOD 


BGN 


CAB 


HEN 


JOG 


KFM 


HAG 


GEM 


MOB 


BFH 


FAN 


KIN 


JEK 


DFL 


JAM 


HIM 


GCL 


LJH 


AID 


JIB 


FCJ 


NJD 


OAK 


FIG 


HCK 


MLN 


BED 


OIL 


MCD 


BLK 


ICE 


CON 


DGK 





The above is the solution of a puzzle I gave in 
Tit-bits in the summer of 1896. It was required 
to take the letters, A, B, C, D, E, F, G, H, I, J, 



the number of words formed is twenty-one. 
Many persons have since tried hard to beat 
this number, but so far have not succeeded. 

More than thirty-five combinations of the 
fifteen letters cannot be formed within the con- 
ditions. Theoretically, there cannot possibly 
be more than twenty-three words formed, be- 
cause only this number of combinations is pos- 
sible with a vowel or vowels in each. And as 
no English word can be formed from three of 
the given vowels (A, E, I, and O), we must 
reduce the number of possible words to twenty- 
two. This is correct theoretically, but practi- 
cally that twenty-second word cannot be got 
in. If JEK, shown above, were a word it 
would be aU right ; but it is not, and no amount 
of juggling with the other letters has resulted 
in a better answer than the one shown. ^ I 
should say that proper nouns and abbrevia- 
tions, such as Joe, Jim, Alf, Hal, Flo, Ike, etc., 
are disallowed. 

Now, the present puzzle is a variation of the 
above. It is simply this : Instead of using the 
fifteen letters given, the reader is allowed to 
select any fifteen different letters of the alphabet 
that he may prefer. Then construct thirty-five 



8o 



AMUSEMENTS IN MATHEMATICS. 



groups in accordance with the conditions, and 
show as many good English words as possible. 

272.— THE NINE SCHOOLBOYS. 

This is a new and interesting companion puzzle 
to the " Fifteen Schoolgirls " (see solution of 
No. 269), and even in the simplest possible 
form in which I present it there are unques- 
tionable difficulties. Nine schoolboys walk out 
in triplets on the six week days so that no boy 
ever walks side by side with any other boy 
more than once. How would you arrange 
them ? 

If we represent them by the first nine letters 
of the alphabet, they might be grouped on the 
first day as follows : — 



A 


B 


C 


D 


E 


F 


G 


H 


I 



Then A can never walk again side by side with 
B, or B with C, or D with E, and so on. But A 
can, of course, walk side by side with C. It is 
here not a question of being together in the 
same triplet, but of walking side by side in a 
triplet. Under these conditions they can walk 
out on six days ; vmder the " Schoolgirls " 
conditions they can only walk on four days. 

273.— THE ROUND TABLE. 

Seat the same n persons at a round table on 

(n — i)(n — 2) . ,, . , „ 

^ occasions so that no person shall 

ever have the same two neighbours twice. 
This is, of course, equivalent to saying that 
every person must sit once, and once only, 
between every possible pair. 

274-— THE MOUSE-TRAP PUZZLE. 




This is a modern version, with a difference, of 
an old puzzle of the same name. Number 
twenty-one cards, i, 2, 3, etc., up to 21, and 
place them in a circle in the particular order 
shown in the illustration. These cards repre- 
sent mice. You start from any card, calling 
that card " one," and count, " one, two, three," 



etc., in a clockwise direction, and when your 
count agrees with the number on the card, you 
have made a " catch," and you remove the 
card. Then start at the next card, calling that 
" one," and try again to make another " catch." 
And so on. Supposing you start at 18, calling 
that card " one," your first " catch " will be 
19. Remove 19 and your next " catch " is 10. 
Remove 10 and your next "catch " is i. Re- 
move the I, and if you count up to 21 (you 
must never go beyond), you cannot make an- 
other " catch." Now, the ideal is to " catch " 
all the twenty-one mice, but this is not here 
possible, and if it were it would merely require 
twenty-one different trials, at the most, to suc- 
ceed. But the reader may make any two cards 
change places before he begins. Thus, you 
can change the 6 with the 2, or the 7 with the 
II, or any other pair. This can be done in 
several ways so as to enable you to " catch " 
all the twenty-one mice, if you then start at 
the right place. You may never pass over a 
" catch " ; you must always remove the card 
and start afresh. 

275.— THE SIXTEEN SHEEP. 




OOIOKDJ 



Here is a new puzzle with matches and counters 
or coins. In the illustration the matches rep- 
resent hurdles and the counters sheep. The 
sixteen hurdles on the outside, and the sheep, 
must be regarded as immovable ; the puzzle 
has to do entirely with the nine hurdles on the 
inside. It will be seen that at present these 
nine hurdles enclose four groups of 8, 3, 3, and 
2 sheep. The farmer requires to readjust some 
of the hurdles so as to enclose 6, 6, and 4 sheep. 
Can you do it by only replacing two hurdles ? 
When you have succeeded, then try to do it by 
replacing three hurdles ; then four, five, six, 
and seven in succession. Of course, the hurdles 
must be legitimately laid on the dotted lines, 
and no such tricks are allowed as leaving un- 
connected ends of hurdles, or two hurdles placed 
side by side, or merely making hurdles change 
places. In fact, the conditions are so simple 
that any farm labourer will vmderstand it 
directly. 

276.— THE EIGHT VILLAS. 

In one of the outlying suburbs of London a 
man had a square plot of ground on which hs 



COMBINATION AND GROUP PROBLEMS. 



8i 




1 


6 


2 
6 


6 


1 


£ 


6 


1 



2 


6 


1 


6 


P. 


6 


1 


6 


2 



1 


6 


2 


4 


y/// 


4 


4 


2 


5 



B C 

decided to build eight villas, as shown in the 
illustration, with a common recreation ground 
in the middle. After the houses were com- 
pleted, and all or some of them let, he discovered 
that the number of occupants in the three houses 
forming a side of the square was in every case 
nine. He did not state how the occupants were 
distributed, but I have shown by the numbers 
on the sides of the houses one way in which it 
might have happened. The puzzle is to dis- 



turned round in front of a mirror, four other 
arrangements. All eight must be counted. 

277.— COUNTER CROSSES. 

All that we need for this puzzle is nine counters, 
numbered i, 2, 3, 4, 5, 6, 7, 8, and 9. It will be 
seen that in the illustration A these are arranged 
so as to form a Greek cross, while in the case of 
B they form a Latin cross. In both cases the 
reader will find that the sum of the numbers in 
the upright of the cross is the same as the sum 
of the numbers in the horizontal arm. It is 
quite easy to hit on such an arrangement by 
trial, but the problem is to discover in exactly 
how many different ways it may be done in 
each case. Remember that reversals and re- 
flections do not count as different. That is to 
say, if you turn this page round you get four 
arrangements of the Greek cross, and if you 
turn it round again in front of a mirror you 
will get four more. But these eight are all 
regarded as one and the same. Now, how 
many different ways are there in each case ? 

278.— A DORMITORY PUZZLE. 

In a certain convent there were eight large dor- 
mitories on one floor, approached by a spiral 
staircase in the centre, as shown in our plan. 
On an inspection one Monday by the abbess it 
was found that the south aspect was so much 
preferred that six times as many nuns slept on 
the south side as on each of the other three 
sides. She objected to this overcrowding, and 
ordered that it should be reduced. On Tues- 




cover the total number of ways in which all or 
any of the houses might be occupied, so that 
there should be nine persons on each side. In 
order that there may be no misunderstanding, 
I will explain that although B is what we call 
a reflection of A, these would count as two 
different arrangements, while C, if it is turned 
roimd, will give four arrangements ; and if 
(1,926) I 



day she found that five times as many slept on 
the south side as on each of the other sides. 
Again she complained. On Wednesday she 
found four times as many on the south side, on 
Thursday three times as many, and on Friday 
twice as many. Urging the nuns to further 
efforts, she was pleased to find on Saturday 
that an equal number slept on each of the four 



82 



AMUSEMENTS IN MATHEMATICS. 




sides of the house. What is the smallest num- 
ber of nims there could have been, and how 
might they have arranged themselves on each 
of the six nights? No room may ever be 
unoccupied. 



value. So that the best quality was numbered 
" I " and the worst numbered " lo," and all 
the other numbers of graduating values. Now, 
the rule of Ahmed Assan, the merchant, was 
that he never put a barrel either beneath or 
to the right of one of less value. The ar- 
rangement shown is, of course, the simplest 
way of complying with this condition. But 
there are many other ways — such, for example, 
as this : — 

12578 

3 4 6 9 10 

Here, again, no barrel has a smaller number 
than itself on its right or beneath it. The 
puzzle is to discover in how many different 
ways the merchant of Bagdad might have 
arranged his barrels in the two rows without 
breaking his rule. Can you count the number 
of ways ? 

280.— BUILDING THE TETRAHEDRON. 

I POSSESS a tetrahedron, or triangular pyramid, 
formed of six sticks glued together, as shown in 
the illustration. Can you count correctly the 
number of different ways in which these six 
sticks might have been stuck together so as to 
form the pyramid ? 




279.— THE BARRELS OF BALSAM. 

A MERCHANT of Bagdad had ten barrels of 
precious balsam for sale. They were numbered, 
and were arranged in two rows, one on top of 
the other, as shown in the picture. The smaller 
the number on the barrel, the greater was its 



Some friends worked at it together one even- 
ing, each person providing himself with six 
lucifer matches to aid his thoughts ; but it was 
found that no two results were the same. You 
see, if we remove one of the sticks and turn it 
round the other way, that will be a different 
pyramid. If we make two of the sticks change 



COMBINATION AND GROUP PROBLEMS. 



83 



places the result will again be dififerent. But 
remember that every pyramid may be made to 




stand on either of its four sides without being a 
different one. How many ways are there alto- 
gether ? 

281.— PAINTING A PYRAMID. 

This puzzle concerns the painting of the four 
sides of a tetrahedron, or triangular pyramid. 
If you cut out a piece of cardboard of the tri- 
angular shape shown in Fig. i, and then cut half 
through along the dotted lines, it will fold up 
and form a perfect triangular pyramid. And 
I would first remind my readers that the pri- 
mary colours of the solar spectrum are seven — 
violet, indigo, blue, green, yellow, orange, and 
red. When I was a child I was taught to re- 
member these by the ungainly word formed by 
the initials of the colours, " Vibgyor." 



other side that is out of view is yellow), and 
then paint another in the order shown in Fig. 3, 
these are really both the same and coimt as one 
way. For if you tilt over No. 2 to the right it 
wiU so fall as to represent No. 3. The avoid- 
ance of repetitions of this kind is the real puzzle 
of the thing. If a coloured pyramid cannot be 
placed so that it exactly resembles in its colours 
and their relative order another pyramid, then 
they are different. Remember that one way 
would be to colour all the four sides red, another 
to colour two sides green, and the remaining 
sides yellow and blue ; and so on. 

282.— THE ANTIQUARY'S CHAIN. 

An antiquary possessed a number of curious 
old links, which he took to a blacksmith, and 
told him to join together to form one straight 
piece of chain, with the sole condition that the 
two circular links were not to be together. 
The following illustration shows the appearance 
of the chain and the form of each link. Now, 
supposing the owner should separate the links 




again, and then take them to another smith 
and repeat his former instructions exactly, what 
are the chances against the links being put to- 
gether exactly as they were by the first man ? 




In how many different ways may the tri- 
angular pyramid be coloured, using in every 
case one, two, three, or four colours of the solar 
spectrum ? Of course a side can only receive 
a single colour, and no side can be left uncol- 
oured. But there is one point that I must 
make quite clear. The four sides are not to be 
regarded as individually distinct. That is to 
say, if you paint your pyramid as shown in 
Fig. 2 (where the bottom side is green and the 



Remember that every successive link can be 
joined on to another in one of two ways, just 
as you can put a ring on your finger in two ways, 
or link your forefingers and thmnbs in two ways. 

283.— THE FIFTEEN DOMINOES. 

In this case we do not use the complete set of 
twenty-eight dominoes to be found in the ordi- 
nary box. We dispense with all those dominoes 



84 



AMUSEMENTS IN MATHEMATICS. 



that have a five or a six on them and limit 
ourselves to the fifteen that remain, where the 
double-four is the highest. 

In how many different ways may the fifteen 
dominoes be arranged in a straight line in ac- 
cordance with the simple rule of the game that 
a number must always be placed against a 
similar number — that is, a four against a four, a 
blank against a blank, and so on ? Left to 
right and right to left of the same arrangement 
are to be counted as two difierent ways. 

284.— THE CROSS TARGET. 




In the illustration we have a somewhat curious 
target designed by an eccentric sharpshooter. 
His idea was that in order to score you must 
hit four circles in as many shots so that those 
four shots shall form a square. It will be seen 
by the results recorded on the target that two 
attempts have been successful. The first man 
hit the four circles at the top of the cross, and 
thus formed his square. The second man in- 
tended to hit the four in the bottom arm, but 
his second shot, on the left, went too high. 
This compelled him to complete his four in a 
difierent way than he intended. It will thus 
be seen that though it is immaterial which 
circle you hit at the first shot, the second shot 
may commit you to a definite procedure if you 
are to get your square. Now, the puzzle is to 
say in just how many different ways it is pos- 
sible to form a square on the target with four 
shots. 



285.— THE FOUR POSTAGE STAMPS. 

" It is as easy as counting," is an expression one 
sometimes hears. But mere counting may be 



■ 


2 


3 


4 


6 


6 


7 


8 


9 


10 


II 


12 



puzzling at times. Take the following simple 
example. Suppose you have just bought twelve 
postage stamps, in this form — three by four — 
and a friend asks you to oblige him with four 
stamps, all joined together — no stamp hanging 
on by a mere corner. In how many difierent 
ways is it possible for you to tear ofi those four 
stamps ? You see, you can give him i, 2, 3, 4, 
or 2, 3, 6, 7, or i, 2, 3, 6, or i, 2, 3, 7, or 2, 3, 4, 8, 
and so on. Can you count the number of differ- 
ent ways in which those four stamps might be 
delivered ? There are not many more than fifty 
ways, so it is not a big coimt. Can you get the 
exact number ? 

286.--PAINTING THE DIE. 

In how many different ways may the numbers 
on a single die be marked, with the only condi- 
tion that the i and 6, the 2 and 5, and the 3 and 
4 must be on opposite sides ? It is a simple 
enough question, and yet it will puzzle a good 
many people. 

287.— AN ACROSTIC PUZZLE. 

In the making or solving of double acrostics, 
has it ever occurred to you to consider the 
variety and limitation of the pair of initial 
and final letters available for cross words ? 
You may have to find a word beginning with 
A and ending with B, or A and C, or A and D, 
and so on. Some combinations are obviously 
impossible — such, for example, as those with 
Q at the end. But let us assume that a good 
English word can be found for every case. 
Then how many possible pairs of letters are 
available ? 



CHESSBOARD PROBLEMS. 



" You and I wiU goe to the chesse." 

Greene's Groatsworth of Wit. 

During a heavy gale a chimney-pot was hurled 
through the air, and crashed upon the pavement 
just in front of a pedestrian. He quite calmly 
said, " I have no use for it : I do not smoke." 
Some readers, when they happen to see a puzzle 
represented on a chessboard with chess pieces, 
are apt to make the equally inconsequent re- 
mark, " I have no use for it : I do not play 



chess." This is largely a result of the common, 
but erroneous, notion that the ordinary chess 
puzzle with which we are familiar in the press 
(dignified, for some reason, with the name 
" problem ") has a vital connection with the 
game of chess itself. But there is no condition 
in the game that you shall checkmate your 
opponent in two moves, in three moves, or in 
four moves, while the majority of the positions 
given in these puzzles are such that one player 
would have so great a superiority in pieces that 



CHESSBOARD PROBLEMS. 



85 



the other would have resigned before the situa- 
tions were reached. And the solving of them 
helps you but little, and that quite indirectly, 
in playing the game, it being well known that, 
as a rule, the best " chess problemists " are 
indifferent players, and vice versa. Occasion- 
ally a man will be found strong on both subjects, 
but he is the exception to the rule. 

Yet the simple chequered board and the 
characteristic moves of the pieces lend them- 
selves in a very remarkable manner to the de- 
vising of the most entertaining puzzles. There 
is room for such infinite variety that the true 
puzzle lover cannot afford to neglect them. It 
was with a view to securing the interest of 
readers who are frightened off by the mere 
presentation of a chessboard that so many 
puzzles of this class were originally published 
by me in various fanciful dresses. Some of 
these posers I still retain in their disguised 
form ; others I have translated into terms of 
the chessboard. In the majority of cases the 
reader will not need any knowledge whatever 
of chess, but I have thought it best to assume 
throughout that he is acquainted with the 
terminology, the moves, and the notation of 
the game. 

I first deal with a few questions affecting the 
chessboard itself; then with certain statical 
puzzles relating to the Rook, the Bishop, the 
Queen, and the Knight in turn ; then dynamical 
puzzles with the pieces in the same order ; and, 
finally, with some miscellaneous puzzles on the 
chessboard. It is hoped that the formulae and 
tables given at the end of the statical puzzles 
will be of interest, as they are, for the most 
part, published for the first time. 



THE CHESSBOARD. 

" Good company's a chessboard." 

Byron's Don Juan, xiii. 89. 

A CHESSBOARD is essentially a square plane 
divided into sixty -four smaller squares by 
straight lines at right angles. Originally it 
was not chequered (that is, made with its rows 
and columns alternately black and white, or of 
any other two colours), and this improvement 
was introduced merely to help the eye in actual 
play. The utility of the chequers is unques- 
tionable. For example, it facilitates the opera- 
tion of the bishops, enabling us to see at the 
merest glance that our king or pawns on black 
squares are not open to attack from an oppo- 
nent's bishop running on the white diagonals. 
Yet the chequering of the board is not essential 
to the game of chess. Also, when we are pro- 
pounding puzzles on the chessboard, it is often 
well to remember that additional interest may 
result from " generalizing " for boards contain- 
ing any number of squares, or from limiting 
ourselves to some particular chequered arrange- 
ment, not necessarily a square. We will give 
a few puzzles dealing with chequered boards in 
this general way. 



288.— CHEQUERED BOARD DIVISIONS. 

I RECENTLY asked myself the question : In how 
many different ways may a chessboard be di- 
vided into two parts of the same size and shape 
by cuts along the lines dividing the squares ? 
The problem soon proved to be both fascinating 
and bristling with difficulties. I present it in 
a simplified form, taking a board of smaller 
dimensions. 

It is obvious that a board of four squares 
can only be so divided in one way — by a straight 
cut down the centre — because we shall not count 
reversals and reflections as different. In the 
case of a board of sixteen squares — four by 
four — there are just six different ways. I have 
given all these in the diagram, and the reader 




will not find any others. Now, take the larger 
board of thirty-six squares, and try to discover 
in how many ways it may be cut into two parts 
of the same size and shape. 

289.— LIONS AND CROWNS. 

The young lady in the illustration is confronted 
with a little cutting-out difficulty in which the 
reader may be glad to assist her. She wishes, 
for some reason that she has not communi- 



86 



AMUSEMENTS IN MATHEMATICS. 



cated to me, to cut that square piece of valuable 
material into four parts, all of exactly the same 
size and shape, but it is important that every 
piece shall contain a lion and a crown. As she 




insists that the cuts can only be made along the 
lines dividing the squares, she is considerably 
perplexed to find out how it is to be done. Can 
you show her the way ? There is only one pos- 
sible method of cutting the stuff. 

290.— BOARDS WITH AN ODD NUMBER 
OF SQUARES. 

We will here consider the question of those 
boards that contain an odd number of squares. 
We wiU suppose that the central square is first 
cut out, so as to leave an even number of squares 
for division. Now, it is obvious that a square 
three by three can only be divided in one way, as 
shown in Fig. i. It will be seen that the pieces 
A and B are of the same size and shape, and that 
any other way of cutting would only produce the 
same shaped pieces, so remember that these vari- 
ations are not counted as different ways. The 
puzzle I propose is to cut the board five by five 
(Fig. 2) into two pieces of the same size a^d 




Fig. I. 



Fig. 2. 



shape in as many different ways as possible. I 
have shown in the illustration one way of doing 
it. How many different ways are there alto- 
gether ? A piece which when tmned over 
resembles another piece is not considered to be 
of a different shape. 

291.— THE GRAND LAMA'S PROBLEM. 

Once upon a time there was a Grand Lama who 
had a chessboard made of pure gold, magnifi- 
cently engraved, and, of course, of great value. 
Every year a tournament was held at Lhassa 
among the priests, and whenever any one beat 
the Grand Lama it was considered a great 
honour, and his name was inscribed on the back 
of the board, and a costly jewel set in the par- 
ticular square on which the checkmate had been 
given. After this sovereign pontiff had been 
defeated on four occasions he died — ^possibly 
of chagrin. , 

Now the new Grand Lama was an inferior 
chess-player, and preferred other forms of inno- 
cent amusement, such as cutting off people's 
heads. So he discouraged chess as a degrading 
game, that did not improve either the mind or 
the morsds, and abolished the tournament sum- 
marily. Then he sent for the four priests who 




had had the effrontery to play better than a 
Grand Lama, and addressed them as follows : 



CHESSBOARD PROBLEMS. 



87 



" Miserable and heathenish men, calling your- 
selves priests ! Know ye not that to lay claim 
to a capacity to do anything better than my 
predecessor is a capital offence ? Take that 
chessboard and, before day dawns upon the 
torture chamber, cut it into four equal parts of 
the same shape, each containing sixteen perfect 
squares, with one of the gems in each part ! 
If in this you fail, then shall other sports be 
devised for your special delectation. Go ! " 
The four priests succeeded in their apparently 
hopeless task. Can you show how the board 
may be divided into four equal parts, each 
of exactly the same shape, by cuts along the 
lines dividing the squares, each part to contain 
one of the gems ? 

292.— THE ABBOT'S WINDOW. 



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Once upon a time the Lord Abbot of St. Ed- 
mondsbury, in consequence of " devotions too 
strong for his head," fell sick and was imable 
to leave his bed. As he lay awake, tossing his 
head restlessly from side to side, the attentive 
monks noticed that something was disturbing 
his mind ; but nobody dared ask what it might 
be, for the abbot was of a stem disposition, 
and never would brook inquisitiveness. Sud- 
denly he called for Father John, and that 
venerable monk was soon at the bedside. 



no love for the things that are odd ? Look 
there ! " 

The Abbot pointed to the large dormitory 
window, of which I give a sketch. The monk 
looked, and was perplexed. 

" Dost thou not see that the sixty-four lights 
add up an even number vertically and hori- 
zontally, but that all the diagonal lines, except 
fourteen are of a number that is odd ? Why 
is this ? " 

" Of a truth, my Lord Abbot, it is of the very 
nature of things, and cannot be changed." 

" Nay, but it shall be changed. I command 
thee that certain of the lights be closed this day, 
so that every line shall have an even number 
of lights. See thou that this be done without 
delay, lest the cellars be locked up for a month 
and other grievous troubles befall thee." 

Father John was at his wits' end, but after 
consultation with one who was learned in strange 
mysteries, a way was found to satisfy the whim 
of the Lord Abbot. Which lights were blocked 
up, so that those which remained added up an 
even number in every line horizontally, verti- 
cally, and diagonally, while the least possible 
obstruction of light was caused ? 

293.— THE CHINESE CHESSBOARD. 

Into how large a number of different pieces may 
the chessboard be cut (by cuts along the lines 
only), no two pieces being exactly alike ? Re- 
member that the arrangement of black and 
white constitutes a difference. Thus, a single 
black square will be different from a single 
white square, a row of three containing two 
white squares will differ from a row of three 
containing two black, and so on. If two pieces 
cannot be placed on the table so as to be ex- 
actly alike, they count as different. And as 
the back of the board is plain, the pieces cannot 
be turned over. 

294._THE CHESSBOARD SENTENCE. 

I ONCE set myself the amusing task of so dis- 
secting an ordinary chessboard into letters of 
the alphabet that they would form a complete 
sentence. It will be seen from the illustration 
that the pieces assembled give the sentence, 
" CUT THY LIFE," with the stops between." 
The ideal sentence would, of course, have only 




■DD 



" Father John," said the Abbot, " dost thou 
know that I came into this wicked world on a 
Christmas Even ? " 

The monk nodded assent. 

" And have I not often told thee that, 
having been bom on Christmas Even, I have 



one fuU stop, but that I did not succeed in 
obtaining. 

The sentence is an appeal to the transgressor 
to cut himself adrift from the evil life he is 
living. Can you fit these pieces together to 
form a perfect chessboard ? 



88 



AMUSEMENTS IN MATHEMATICS. 



STATICAL CHESS PUZZLES. 

They also serve who only stand and wait." 

Milton. 

295.— THE EIGHT ROOKS. 




Fig. I. 




Fig. 2. 

It will be seen in the first diagram that every 
square on the board is either occupied or at- 
tacked by a rook, and that every rook is 
" guarded " (if they were alternately black and 
white rooks we should say " attacked ") by 
another rook. Placing the eight rooks on any 
row or file obviously will have the same effect. 
In diagram 2 every square is again either occu- 
pied or attacked, but in this case every rook is 
imguarded. Now, in how many different ways 
can you so place the eight rooks on the board 
that every square shall be occupied or attacked 
and no rook ever guarded by another ? I do 
not wish to go into the question of reversals 
and reflections on this occasion, so that placing 



the rooks on the other diagonal will count as 
different, and similarly with other repetitions 
obtained by turning the board round. 

296.— THE FOUR LIONS. 

The puzzle is to find in how many different ways 
the four lions may be placed so that there shall 
never be more than one lion in any row or 
column. Mere reversals and reflections wiU not 
cotmt as different. Thus, regarding the ex- 
ample given, if we place the lions in the other 




diagonal, it wiU be considered the same arrange- 
ment. For if you hold the second arrangement 
in front of a mirror or give it a quarter turn, 
you merely get the first arrangement. It is a 
simple little puzzle, but requires a certain 
amount of careful consideration. 

297.— BISHOPS— UNGUARDED. 

Place as few bishops as possible on an ordinary 
chessboard so that every square of the board 
shaU be either occupied or attacked. It will be 
seen that the rook has more scope than the 
bishop : for wherever you place the former, it 
will always attack fourteen other squares ; where- 
as the latter will attack seven, nine, eleven, or 
thirteen squares, according to the position of 
the diagonal on which it is placed. And it is weU 
here to state that when we speak of "diagonals" 
i in connection with the chessboard, we do not 
1 limit ourselves to the two long diagonals from 
I corner to corner, but include all the shorter lines 
that are parallel to these. To prevent misunder- 
standing on future occasions, it will be well for 
the reader to note carefully this fact. 

298.— BISHOPS— GUARDED. 

Now, how many bishops are necessary in order 
that every square shaU be either occupied or 
attacked, and every bishop guarded by another 
bishop ? And how may they be placed ? 



CHESSBOARD PROBLEMS. 



89 



299.— BISHOPS IN CONVOCATION. 




The greatest number of bishops that can be 
placed at the same time on the chessboard, 
without any bishop attacking another, is four- 
teen. I show, in diagram, the simplest way of 
doing this. In fact, on a square chequered 
board of any number of squares the greatest 
number of bishops that can be placed without 
attack is always two less than twice the number 
of squares on the side. It is an interesting 
puzzle to discover in just how many different 
ways the fourteen bishops may be so placed 
without mutual attack. I shall give an ex- 
ceedingly simple rule for determining the num- 
ber of ways for a square chequered board of any 
number of squares. 

300.— THE EIGHT QUEENS. 




The queen is by far the strongest piece on the 
chessboard. If you place her on one of the 
four squares in the centre of the board, she 
attacks no fewer than twenty-seven other 
squares ; and if you try to hide her in a corner, 



she still attacks twenty-one squares. Eight 
queens may be placed on the board so that no 
queen attacks another, and it is an old puzzle 
(first proposed by Nauck in 1850, and it has 
quite a little literature of its own) to discover 
in just how many different ways this may be 
done. I show one way in the diagram, and 
there are in all twelve of these fundamentally 
different ways. These twelve produce ninety- 
two ways if we regard reversals and reflections 
as different. The diagram is in a way a sym- 
metrical arrangement. If you turn the page 
upside down, it will reproduce itself exactly; 
but if you look at it with one of the other sides 
at the bottom, you get another way that is not 
identical. Then if you reflect these two ways 
in a mirror you get two more ways. Now, all 
the other eleven solutions are non-symmetrical, 
and therefore each of them may be presented 
in eight ways by these reversals and reflections. 
It wiU thus be seen why the twelve funda- 
mentally different solutions produce only 
ninety-two arrangements, as I have said, and 
not ninety-six, as would happen if all twelve 
were non-symmetrical. It is well to have a 
clear understanding on the matter of reversals 
and reflections when dealing with puzzles on 
the chessboard. 

Can the reader place the eight queens on the 
board so that no queen shall attack another 
and so that no three queens shall be in a straight 
line in any oblique direction ? Another glance 
at the diagram will show that this arrangement 
will not answer the conditions, for in the two 
directions indicated by the dotted lines there 
are three queens in a straight line. There is 
only one of the twelve fundamental ways that 
will solve the puzzle. Can you find it ? 

301.— THE EIGHT STARS. 

The puzzle in this case is to place eight stars in 
the diagram so that no star shall be in line with 




another star horizontally, vertically, or diago- 
nally. One star is already placed, and that 
must not be moved, so there are only seven 
for the reader now to place. But you must 
not place a star on any one of the shaded 
squares. There is only one way of solving this 
little puzzle. 



90 



AMUSEMENTS IN MATHEMATICS. 



302.— A PROBLEM IN MOSAICS. 

The art of producing pictures or designs by 
meaas of joining together pieces of hard sub- 
stances, either naturally or artificiaUy coloured, 
is of very great antiquity. It was certadnly 
known in the time of the Pharaohs, and we find 
a reference in the Book of Esther to " a pave- 
ment of red, and blue, and white, and black 
marble." Some of this ancient work that has 
come down to us, especially some of the Roman 
mosaics, would seem to show clearly, even where 
design is not at first evident, that much thought 
was bestowed upon apparently disorderly ar- 
rangements. Where, for example, the work has 
been produced with a very limited number of 
colours, there are evidences of great ingenuity 
in preventing the same tints coming in close 
proximity. Lady readers who are familiar with 
the construction of patchwork quilts will know 
how desirable it is sometimes, when they are 
limited in the choice of material, to prevent 
pieces of the same stuff coming too near to- 
gether. Now, this puzzle will apply equally to 
patchwork quilts or tesselated pavements. 

It wiU be seen from the diagram how a 
square piece of flooring may be paved with 
sixty-two square tiles of the eight colours 
violet, red, yellow, green, orange, purple, white, 
and blue (indicated by the initial letters), so 
that no tile is in line with a similarly coloured 
tile, vertically, horizontally, or diagonally. Sixty- 
four such tiles could not possibly be placed under 
these conditions, but the two shaded squares 
happen to be occupied by iron ventilators. 

The puzzle is this. These two ventilators 



no letter is in line with a similar one horizontally, 
vertically, or diagonally. Thus, no V is in line 
with another V, no E with another E, and so 



V 


R 


y 


^ 





P 


w 


B 


w 


B 





p 


Y 


G 


V 


R 


a 


pRv 


BlBlY 





R 


Y 


6 





Gc 


V 


P 


W 


B 


G 


R 


Y 


P 


w 





V 





V 


P 


W 


R 


Y 


B 


G 


p. 


w 


a 


B 


V 





R 


Y 


W//, 





V 


R 


W 


B 


Q 


Wa 



have to be removed to the positions indicated 
by the darkly bordered tiles, and two tiles placed 
in those bottom comer squares. Can you re- 
adjust the thirty-two tiles so that no two of the 
same colour shall still be in line ? 

303.— UNDER THE VEIL. 

If the reader will examine the above diagram, 
he will see that I have so placed eight V's, eight 
E*s, eight I's, and eight L's in the diagram that 







V 


E 


1 


L 










1 


L 


V 


E 






1 


V 










L 


E 


L 


E 










1 


V 


V 


! 










ROWMB 


L 


E 


L 










V 


1 






E 


V 


L 


1 










L 


1 


E 


V 







on. There are a great many different ways of 
arranging the letters imder this condition. The 
puzzle is to find an arrangement that produces 
the greatest possible number of four-letter 
words, reading upwards and downwards, back- 
wards and forwards, or diagonally. All repeti- 
tions count as different words, and the five 
variations that may be used are : VEIL, VILE, 
LEVI, LIVE, and EVIL. 

This win be made perfectly clear when I say 
that the above arrangement scores eight, because 
the top and bottom row both give VEIL; the 
second and seventh coliunns both give VEIL; 
and the two diagonals, starting from the L in 
the 5 th row and E in the 8 th row, both give 
LIVE and EVIL. There are therefore eight 
different readings of the words in all. 

This difi&cult word puzzle is given as an 
example of the use of chessboard analysis in 
solving such things. Only a person who is 
familiar with the " Eight Queens " problem 
could hope to solve it. 

304.— BACHET'S SQUARE. 

One of the oldest card puzzles is by Claude 
Caspar Bachet de Meziriac, first published, I 
believe, in the 1624 edition of his work. Re- 
arrange the sixteen court cards (including the 
aces) in a square so that in no row of four cards, 
horizontal, vertical, or diagonal, shall be found 
two cards of the same suit or the same value. 
This in itself is easy enough, but a point of the 
puzzle is to find in how many different ways 
this may be done. The eminent French mathe- 
matician A. Labosne, in his modem edition of 
Bachet, gives the answer incorrectly. And yet 
the puzzle is really quite easy. Any arrange- 
ment produces seven more by turning the square 
round and reflecting it in a mirror. These are 
counted as different by Bachet. 



CHESSBOARD PROBLEMS. 



91 



Note " row of four cards," so that the only 
diagonals we have here to consider are the two 
long ones. 

305.— THE THIRTY-SIX LETTER- 
BLOCKS. 



A 


B 


C 


D 


E 


F 


A 


B 


C 


D 


E 


F 


A 


B 


c 


J> 


E 


F 


A 


B 


c 


D 


£ 


F 


A 


B 


c 


D 


E 


F 


A 


B 


c 


D 


E 


F 



The illustration represents a box containing 
thirty-six letter-blocks. The puzzle is to re- 
arrange these blocks so that no A shall be in a 
line vertically, horizontally, or diagonally with 
another A, no B with another B, no C with 
another C, and so on. You will find it impos- 
sible to get all the letters into the box under 
these conditions, but the point is to place as 
many as possible. Of course no letters other 
than those shown may be used. 

306.— THE CROWDED CHESSBOARD. 




The puzzle is to rearrange the fifty-one pieces 
on the chessboard so that no queen shall attack 



another queen, no rook attack another rook, no 
bishop attack another bishop, and no knight 
attack another knight. No notice is to be 
taken of the intervention of pieces of another 
type from that under consideration — that is, 
two queens will be considered to attack one 
another although there may be, say, a rook, a 
bishop, and a knight between them. And so 
with the rooks and bishops. It is not difficult 
to dispose of each type of piece separately ; 
the difficulty comes in when you have to find 
room for aU the arrangements on the board 
simultaneously. 

307.— THE COLOURED COUNTERS. 




The diagram represents twenty-five coloured 
counters. Red, Blue, Yellow, Orange, and Green 
(indicated by their initials), and there are five 
of each colour, numbered i, 2, 3, 4, and 5. The 
problem is so to place them in a square that 
neither colour nor number shall be found re- 
peated in any one of the five rows, five columns, 
and two diagonals. Can you so rearrange them? 

308.— THE GENTLE ART OF STAMP- 
LICKING. 

The Insurance Act is a most prolific source 
of entertaining puzzles, particularly entertain- 
ing if you happen to be among the exempt. 
One's initiation into the gentle art of stamp- 
licking suggests the following little poser : If 
you have a card divided into sixteen spaces 
(4x4), and are provided with plenty of stamps 
of the values id., 2d., sd., ^d., and 5d., what is 
the greatest value that you can stick on the 
card if the Chancellor of the Exchequer forbids 
you to place any stamp in a straight line (that is, 
horizontally, vertically, or diagonally) with an- 
other stamp of similar value ? Of coiurse, only 
one stamp can be affixed in a space. The reader 
will probably find, when he sees the solution, 
that, like the stamps themselves, he is licked. 



92 



AMUSEMENTS IN MATHEMATICS. 



He will most likely be twopence short of the 
maximum. A friend asked the Post Of&ce how 
it was to be done ; but they sent him to the 
Customs and Excise officer, who sent him to 
the Insurance Commissioners, who sent him to 
an approved society, who profanely sent him 
— ^but no matter. 

309.— THE FORTY-NINE COUNTERS. 

@ (g)@(g)@ ©) 
m (D3) 64^ (gs) ($6) ^7) 




Can you rearrange the above forty-nine counters 
in a square so that no letter, and also no number, 
shall be in line with a similar one, vertically, 
horizontally, or diagonally ? Here I, of course, 
mean in the lines parallel with the diagonals, in 
the chessboard sense. 

310.— THE THREE SHEEP. 









1 



























A FARMER had three sheep and an arrangement 
of sixteen pens, divided off by hurdles in the 
manner indicated in the illustration. In how 
many diSerent ways could he place those sheep, 
each in a separate pen, so that every pen should 
be either occupied or in line (horizontally, verti- 
cally, or diagonally) with at least one sheep ? 
I have given one arrangement that fulfils the 
conditions. How many others can you find ? 



Mere reversals and reflections must not be 
counted as different. The reader may regard 
the sheep as queens. The problem is then to 
place the three queens so that every square 
shall be either occupied or attacked by at least 
one queen — in the maximum number of differ- 
ent ways. 

311.— THE FIVE DOGS PUZZLE. 

In 1863, C. F. de Jaenisch first discussed the 
" Five Queens Puzzle " — to place five queens 
on the chessboard so that every square shall be 
attacked or occupied — which was propounded 
by his friend, a " Mr. de R." Jaenisch showed 
that if no queen may attack another there are 
ninety-one different ways of placing the five 
queens, reversals and reflections not counting as 
different. If the queens may attack one an- 
other, I have recorded hundreds of ways, but 
it is not practicable to enumerate them exactly. 
The illustration is supposed to represent an 

















I 






















%. 
















h\ 




h. 












hi 


















h. 











































arrangement of sixty-four kennels. It will be 
seen that five kennels each contain a dog, and 
on further examination it wiU be seen that every 
one of the sixty-four kennels is in a straight line 
with at least one dog — either horizontally, verti- 
cally, or diagonally. Take any kennel you like, 
and you will find that you can draw a straight 
line to a dog in one or other of the three ways 
mentioned. The puzzle is to replace the five 
dogs and discover in just how many different 
ways they may be placed in five kennels in a 
straight row, so that every kennel shall always 
be in line with at least one dog. Reversals and 
reflections are here counted as different. 



312. 



-THE FIVE CRESCENTS OF 
BYZANTIUM. 



When Philip of Macedon, the father of Alex- 
ander the Great, found himself confronted with 
great difficulties in the siege of Byzantium, he 



CHESSBOARD PROBLEMS. 



93 



set his men to undermine the walls. His desires, 
however, miscarried, for no sooner had the 
operations been begun than a crescent moon 
suddenly appeared in the heavens and dis- 
covered his plans to his adversaries. The By- 
zantines were naturally elated, and in order to 
show their gratitude they erected a statue to 
Diana, and the crescent became thenceforward 
a symbol of the state. In the temple that con- 
tained the statue was a square pavement com- 
posed of sixty-four large and costly tiles. These 
were all plain, with the exception of five, which 
bore the symbol of the crescent. These five 
were for occult reasons so placed that every tile 
should be watched over by (that is, in a straight 
line, vertically, horizontally, or diagonally with) 
at least one of the crescents. The arrangement 
adopted by the Byzantine architect was as 
follows : — 




Now, to cover up one of these five crescents 
was a capital offence, the death being something 
very painful and lingering. But on a certain 
occasion of festivity it was necessary to lay 
down on this pavement a square carpet of the 
largest dimensions possible, and I have shown 
in the illustration by dark shading the largest 
dimensions that would be available. 

The puzzle is to show how the architect, if he 
had foreseen this question of the carpet, might 
have so arranged his five crescent tiles in accord- 
ance with the required conditions, and yet have 
allowed for the largest possible square carpet 
to be laid down without any one of the five 
crescent tiles being covered, or any portion of 
them. 

313.— QUEENS AND BISHOP PUZZLE. 

It will be seen that every square of the board is 
either occupied or attacked. The puzzle is to 
substitute a bishop for the rook on the same 
square, and then place the four queens on other 
squares so that every square shall again be 
either occupied or attacked. 




314.— THE SOUTHERN CROSS. 



-i^ ^ i^ ^ i^ ti t!^ 

A- -iV ^ T^ ^^i!k ^ 

^ tx :Ck i^^i^--^ ^ 

1^ ^ ^^-^^-^ 

^ ^ ^ ^^-^ ^ 

t:i A A A-.t^t:^ ^ 

^ ^ 1^ ^ t^ ^ 1:^ 

^ <:i ^ i^ ^ t^ ^ 

<^ ^ ^ 1^ ^ -C^ ^ 






In the above illustration we have five Planets 
and eighty-one Fixed Stars, five of the latter 
being hidden by the Planets. It will be found 
that every Star, with the exception of the ten 
that have a black spot in their centres, is in a 
straight line, vertically, horizontally, or diago- 
nally, with at least one of the Planets. The 
puzzle is so to rearrange the Planets that all the 
Stars shall be in line with one or more of them. 
In rearranging the Planets, each of the five 
may be moved once in a straight line, in either 
of the three directions mentioned. They will, 
of course, obscure five other Stars in place of 
those at present covered. 

315.— THE HAT-PEG PUZZLE. 

Here is a five-queen puzzle that I gave in a 
fanciful dress in 1897. As the queens were 



94 



AMUSEMENTS IN MATHEMATICS. 



there represented as hats on sixty-four pegs, 
I will keep to the title, " The Hat-Peg Puzzle." 
It will be seen that every square is occupied or 
attacked. The puzzle is to remove one queen 





board that are not attacked. The removal of 
the three queens need not be by " queen moves." 
You may take them up and place them any- 
where. There is only one solution. 

317.— A PUZZLE WITH PAWNS. 

Place two pawns in the middle of the chess- 
board, one at Q 4 and the other at K 5. Now, 
place the remaining fourteen pawns (sixteen 
in all) so that no three shall be in a straight line 
in any possible direction. 

Note that I purposely do not say queens, 
because by the words " any possible direction " 
I go beyond attacks on diagonals. The pawns 
must be regarded as mere points in space — at 
the centres of the squares. See dotted lines 
in the case of No. 300, "The Eight Queens." 

318.— LION-HUNTING. 



to a different square so that still every square 
is occupied or attacked, then move a second 
queen under a similar condition, then a third 
queen, and finally a fourth queen. After the 
fourth move every square must be attacked j 
or occupied, but no queen must then attack 
another. Of course, the moves need not be 
" queen moves ; " you can move a queen to 
any part of the board. 

316.— THE AMAZONS. 



This puzzle is based on one by Captain Turton. 
Remove three of the queens to other squares 
so that there shall be eleven squares on the 




My friend Captain Potham Hall, the renowned 
hunter of big game, says there is nothing more 
exhilarating than a brush with a herd — ^a 
pack — a team — a flock — a swarm (it has taken 
mc a full quarter of an hour to recall the right 
word, but I have it at last) — a pride of lions. 
Why a number of lions are called a " pride," a 
number of whales a " school," and a number of 
foxes a " skulk " are mysteries of philology into 
which I will not enter. 

Well, the captain says that if a spirited lion 
crosses your path in the desert it becomes 
lively, for the lion has generally been looking 
for the man just as much as the man has sought 
the king of the forest. And yet when they 
meet they always quarrel and fight it out. A 
little contemplation of this unfortunate and 
long-standing feud between two estimable fam- 
ilies has led me to figure out a few calculations 
as to the probability of the man and the lion 
crossing one another's path in the jungle. In 
aU these cases one has to start on certain more 



CHESSBOARD PROBLEMS. 



95 



or less arbitrary assumptions. That is why in 
the above illustration I have thought it neces- 
sary to represent the paths in the desert with 
such rigid regularity. Though the captain 
assures me that the tracks of the lions usually 
run much in this way, I have doubts. 

The puzzle is simply to find out in how many 
different ways the man and the lion may be 
placed on two different spots that are not on 
the same path. By " paths " it must be under- 
stood that I only refer to the ruled lines. Thus, 
with the exception of the four comer spots, 
each combatant is always on two paths and no 
more. It will be seen that there is a lot of 
scope for evading one another in the desert, 
which is just what one has always imderstood. 

319.— THE KNIGHT-GUARDS. 




The knight is the irresponsible low comedian 
of the chessboard. " He is a very uncertain, 
sneaking, and demoralizing rascal," says an 
American writer. " He can only move two 
squares, but makes up in the quality of his 
locomotion for its quantity, for he can spring 
one square sideways and one forward simul- 
taneously, like a cat ; can stand on one leg in 
the middle of the board and jump to any one 
of eight squares he chooses ; can get on one 
side of a fence and blackguard three or four 
men on the other ; has an objectionable way 
of inserting himself in safe places where he can 
scare the king and compel him to move, and 
then gobble a queen. For pure cussedness the 
knight has no equal, and when you chase him 
out of one hole he skips into another. ' ' Attempts 
have been made over and over again to obtain 
a short, simple, and exact definition of the move 
of the knight — ^without success. It really con- 
sists in moving one square hke a rook, and then 
another square like a bishop — the two opera- 
tions being done in one leap, so that it does not 
matter whether the first square passed over is 
occupied by another piece or not. It is, in 



fact, the only leaping move in chess. But 
difficult as it is to define, a child can learn it by 
inspection in a few minutes. 

I have shown in the diagram how twelve 
knights (the fewest possible that will perform the 
feat) may be placed on the chessboard so that 
every square is either occupied or attacked by 
a knight. Examine every square in turn, and 
you will find that this is so. Now, the puzzle 
in this case is to discover what is the srnallQpt 
possible number of knights that is required in 
order that every square shall be either occupied 
or attacked, and every knight protected by 
another knight. And how would you arrange 
them ? It wiU be found that of the twelve 
shown in the diagram only four are thus pro- 
tected by being a knight's move from another 
knight. 



THE GUARDED CHESSBOARD. 

On an ordinary chessboard, 8 by 8, every square 
can be guarded — that is, either occupied or 
attacked — by 5 queens, the fewest possible. 
There are exactly 91 fundamentally different 
arrangements in which no queen attacks an- 
other queen. If every queen must attack (or 
be protected by) another queen, there are at 
fewest 41 arrangements, and I have recorded 
some 150 ways in which some of the queens 
are attacked and some not, but this last case 
is very difficult to enumerate exactly. 

On an ordinary chessboard every square can 
be guarded by 8 rooks (the fewest possible) in 
40,320 ways, if no rook may attack another 
rook, but it is not known how many of these 
are fundamentally different. (See solution to 
No. 295, "The Eight Rooks.*') I have not 
enumerated the ways in which every rook shall 
be protected by another rook. 

On an ordinary chessboard every square can 
be guarded by 8 bishops (the fewest possible), 
if no bishop may attack another bishop. Ten 
bishops are necessary if every bishop is to be 
protected. (See Nos. 297 and 298, " Bishops 
unguarded " and " Bishops guarded.") 

On an ordinary chessboard every square can 
be guarded by 12 knights if all but 4 are im- 
protected. But if every knight must be pro- 
tected, 14 are necessary. (See No. 319, "The 
Knight-Guards.") 

Dealing with the queen on n^ boards gener- 
ally, where n is less than 8, the following results 
wiU be of interest : — 



22 board in i fundamental 

32 board in i fundamental 

42 board in 3 fundamental 

3 queens guard 42 board in 2 fundamental 

fundamental 
fundamental 



queen guards 

way. 
queen guards 

way. 
queens guard 

ways (protected). 



ways (not protected), 
queens guard 52 board in 37 

ways (protected), 
queens guard 52 board in 2 

ways (not protected). 



96 



AMUSEMENTS IN MATHEMATICS. 



queens guard 6^ board in i fundamental n^ — n^ 

way (protected). j ^ 

queens guard 6^ board in 17 fundamental 

ways (not protected), 
queens guard 72 board in 5 fundamental 

ways (protected), 
queens guard 7^ board in i fundamental 

way (not protected). 



NON-ATTACKING CHESSBOARD 
ARRANGEMENTS. 

We know that n queens may always be placed 
on a square board of n^ squares (if n be greater 
than 3) without any queen attacking another 
queen. But no general formula for enumerat- 
ing the number of different ways in which it 
may be done has yet been discovered ; probably 
it is undiscoverable. The known results are as 
follows : — 

Where n = 4 there is i fundamental solution 
and 2 in all. 

Where n = 5 there are 2 fundamental solu- 
tions and 10 in aU. 

Where n = 6 there is i fundamental solution 
and 4 in all. 

Where n = 7 there are 6 fundamental solu- 
tions and 40 in all. 

Where n = 8 there are 12 fundamental solu- 
tions and 92 in all. 

Where n = g there are 46 fundamental solu- 
tions. 

Where w = 10 there are 92 fundamental solu- 
tions. 

Where m = 11 there are 341 fundamental solu- 
tions. 

Obviously n rooks may be placed without at- 
tack on an n^ board in \n ways, but how many 
of these are fundamentally different I have only 
worked out in the four cases where n equals 2, 
3, 4, and 5. The answers here are respectively 
1, 2, 7, and 23. (See No. 296, " The Four 
Lions.") 

We can place 2« — 2 bishops on an n'^ board 
in 2" ways. (See No. 299, " Bishops in Con- 
vocation.") For boards containing 2, 3, 4, 5, 
6, 7, 8 squares on a side there are respectively 
I, 2, 3, 6, 10, 20, 36 fundamentally different 
arrangements. Where n is odd there are 
2i(«-i) such arrangements, each giving 4 by re- 
versals and reflections, and 2" -3 - 24(«-3) giving 
8. Where « is even there are 2^^**-^^ each 
giving 4 by reversals and reflections, and 
2M-3 — 2i(«-'«), each giving 8. 

We can place ^(n'^ + i) knights on an n^ board 
without attack, when n is odd, in i fundamental 
way ; and ^n^ knights on an n^ board, when n 
is even, in i fundamental way. In the first 
case we place all the knights on the same colour 
as the central square ; in the second case we 
place them all on black, or all on white, squares. 

THE TWO PIECES PROBLEM. 

On a board of n^ squares, two queens, two 
rooks, two bishops, or two knights can always 
be placed, irrespective of attack or not, in 



ways. The following formulae will show 

in how many of these ways the two pieces may 
be placed with attack and without : — 



With Attack. Without Attack. 

3 6 
2 Kooks w — n^ ! — . 



2 Queens 



2 Ri<;h <; 4^3 — Gw^ + 2W 3^* — 4»3 -)- 3n2 — 2n 
6 6 



2 Knights 4m2_i2m-|-8 
(See No. 318, " Lion Hunting.") 



n'^ — gn^+24n 



DYNAMICAL CHESS PUZZLES. 

" Push on — ^keep moving." 
Thos. Morton : Cure for the Heartache 

320.— THE ROOK'S TOUR. 




The puzzle is to move the single rook over the 
whole board, so that it shall visit every square 
of the board once, and only once, and end its 
tour on the square from which it starts. You 
have to do this in as few moves as possible, and 
unless you are very careful you wiU take just 
one move too many. Of course, a square is 
regarded equally as " visited " whether you 
merely pass over it or make it a stopping-place, 
and we wiU not quibble over the point whether 
the original square is actually visited twice. 
We will assume that it is not. 

321.— THE ROOK'S JOURNEY. 

This puzzle I call " The Rook's Journey," be- 
cause the word " tour " (derived from a turner's 
wheel) implies that we return to the point from 
which we set out, and we do not do this in the 
present case. We should not be satisfied with 



CHESSBOARD PROBLEMS. 



97 



a personally conducted holiday tour that ended 
by leaving us, say, in the middle of the Sahara. 
The rook here makes twenty-one moves, in the 




course of which journey it visits every square 
of the board once and only once, stopping at 
the square marked lo at the end of its tenth 
move, and ending at the square marked 21. 
Two consecutive moves cannot be made in the 
same direction — that is to say, you must make 
a turn after every move. 

322.— THE LANGUISHING MAIDEN. 



XI — I — I — r 

-^ « I » ■ j » * I » l i t 

i*^^ ^■■••^ ■^•♦"^W ^*^'^^'% ^■•^■% 

"*+ i" "i" *i" 

> I ' ■ > • [ » « I * ■ I < 

■"+ + + + 



"T~T — r 
'i" "hi" 

• I * t I ■ t j ■< 

^^■■^ ^"^■^* ^"^T^*^ 



t r 1 L-J I L 



A WICKED baron in the good old days impris- 
oned an innocent maiden in one of the deepest 
dungeons beneath the castle moat. It wiU be 
seen from our illustration that there were sixty- 
three cells in the dungeon, all connected by 
open doors, and the maiden was chained in 
the cell in which she is shown. Now, a valiant 
knight, who loved the damsel, succeeded in 
(1,926) 



rescuing her from the enemy. Having gained 
an entrance to the dungeon at the point where 
he is seen, he succeeded in reaching the maiden 
after entering every cell once and only once. 
Take your pencil and try to trace out such a 
route. When you have succeeded, then try to 
discover a route in twenty-two straight paths 
through the cells. It can be done in this num- 
ber without entering any cell a second time. 

323 —A DUNGEON PUZZLE. 



—I 1 1 1 1 1 1 1 

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A French prisoner, for his sins (or other 
people's), was confined in an underground dun- 
geon containing sixty-four cells, aU communi- 
cating with open doorways, as shown in our 
illustration. In order to reduce the tedium of 
his restricted life, he set himself various puzzles, 
and this is one of them. Starting from the 
cell in which he is shown, how could he visit 
every cell once, and only once, and make as 
many turnings as possible ? His first attempt 
is shown by the dotted track. It will be found 
that there are as many as fifty-five straight lines 
in his path, but after many attempts he im- 
proved upon this. Can you get more than fifty- 
five ? You may end your path in any cell you 
like. Try the puzzle with a pencil on chess- 
board diagrams, or you may regard them as 
rooks' moves on a board. 

324.— THE LION AND THE MAN. 

In a public place in Rome there once stood a 
prison divided into sixty-four cells, all open to 
the sky and all communicating with one an- 
other, as shown in the illustration. The sports 
that here took place were watched from a high 
tower. The favourite game was to place a 
Christian in one corner ceU and a lion in the 
diagonally opposite comer and then leave them 
with all the inner doors open. The consequent 
effect was sometimes most laughable. On one 
occasion the man was given a sword. He was 



98 



AMUSEMENTS IN MATHEMATICS. 



no coward, and was as anxious to find the lion 
as the lion undoubtedly was to find him. 

The man visited every cell once and only 
once in the fewest possible straight lines until 
he reached the lion's cell. The lion, curiously 
enough, also visited every cell once and only 
once in the fewest possible straight lines until 
he finally reached the man's cell. They started 
together and went at the same speed ; yet, 



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although they occasionally got glimpses of one 
another, they never once met. The puzzle is 
to show the route that each happened to take. 

325.— AN EPISCOPAL VISITATION. 

The white squares on the chessboard repre- 
sent the parishes of a diocese. Place the bishop 
on any square you like, and so contrive that 
(using the ordinary bishop's move of chess) he 
shall visit every one of his parishes in the 
fewest possible moves. Of course, aU the 
parishes passed through on any move are re- 
garded as " visited." You can visit any 
squares more than once, but you are not al- 
lowed to move twice between the same two 
adjoining squares. What are the fewest pos- 
sible moves ? The bishop need not end his 
visitation at the parish from which he first set 
out. 

326.— A NEW COUNTER PUZZLE. 

Here is a new puzzle with moving counters, or 
coins, that at first glance looks as if it must be 
absurdly simple. But it will be found quite a 
little perplexity. I give it in this place for a 
reason that I will explain when we come to the 
next puzzle. Copy the simple diagram, en- 
larged, on a sheet of paper ; then place two 
white counters on the points i and 2, and two 
red coimters on 9 and 10. The puzzle is to 
make the red and white change places. You 
may move the counters one at a time in 




any order you like, along the lines from point 
to point, with the only restriction that a red 
and a white counter may never stand at once 
on the same straight line. Thus the first move 
can only be from i or 2 to 3, or from 9 or 10 to 7. 

327.— A NEW BISHOP'S PUZZLE. 




This is quite a fascinating little puzzle. Place 
eight bishops (four black and four white) on the 
reduced chessboard, as shown in the illustration. 
The problem is to make the black bishops 
change places with the white ones, no bishop 
ever attacking another of the opposite colour. 
They must move alternately — first a white, 
then a black, then a white, and so on. When 
you have succeeded in doing it at aU, try to 
find the fewest possible moves. 

If you leave out the bishops standing on 
black squares, and only play on the white 
squares, you will discover my last puzzle turned 1 
on its side. 

328.— THE QUEEN'S TOUR. 

The puzzle of making a complete tour of the i 
chessboard with the queen in the fewest possible 
moves (in which squares may be visited more ' 
than once) was first given by the late Sam Loyd 



CHESSBOARD PROBLEMS. 



99 



in his Chess Strategy. But the solution shown 
below is the one he gave in American Chess- 
Nuts in 1868. I have recorded at least six 



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different solutions in the minimum number of 
moves — fourteen — ^but this one is the best of 
all, for reasons I will explain. 

If you will look at the lettered square you 
will understand that there are only ten really 
differently placed squares on a chessboard — 
those enclosed by a dark line — all the others 
are mere reversals or reflections. For example, 
every A is a corner square, and every J a central 
square. Consequently, as the solution shown 
has a turning-point at the enclosed D square, 
we can obtain a solution starting from and 



A 


B 


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ending at any square marked D — by just 
turning the board about. Now, this scheme 
will give you a tour starting from any A, B, C, 
D, E, F, or H, while no other route that I 
know can be adapted to more than five differ- 
ent starting-points. There is no Queen's Tour 
in fourteen moves (remember a tour must be 



re-entrant) that may start from a G, I, or J. But 
we can have a non-re-entrant path over the 
whole board in fourteen moves, starting from 
any given square. Hence the following puzzle : — 
Start from the J in the enclosed part of the 
lettered diagram and visit every square of the 
board in fourteen moves, ending wherever you 
like. 

329.— THE STAR PUZZLE. 






Put the point of your pencil on one of the white 
stars and (without ever lifting your pencil 
from the paper) strike out all the stars in four- 
teen continuous straight strokes, ending at the 
second white star. Your straight strokes may 
be in any direction you like, only every turn- 
ing must be made on a star. There is no 
objection to striking out any star more than 
once. 

In this case, where both your starting and 
ending squares are fixed inconveniently, you 
cannot obtain a solution by breaking a Queen's 
Tour, or in any other way by queen moves 
alone. But you are allowed to use oblique 
straight lines — such as from the upper white 
star direct to a corner star. 

330.— THE YACHT RACE. 

Now then, ye land-lubbers, hoist your baby- 
jib-topsails, break out your spinnakers, ease 
off your balloon sheets, and get your head-sails 
set! 

Our race consists in starting from the point 
at which the yacht is lying in the illustration 
and touching every one of the sixty-four buoys 
in fourteen straight courses, returning in the 
final tack to the buoy from which we start. 
The seventh course must finish at the buoy 
from which a flag is flying. 

This puzzle will caU for a lot of skilful sea- 
manship on account of the sharp angles at 
which it will occasionally be necessary to tack. 



lOO 



AMUSEMENTS IN MATHEMATICS. 



The point of a lead pencil and a good nautical 
eye are all the outfit that we require. 




^ ^ 4 



d 
d 



5 



d ^ 



This is difficult, because of the condition as 
to the flag-buoy, and because it is a re-entrant 
tour. But again we are allowed those oblique 
lines. 

331— THE SCIENTIFIC SKATER. 







It will be seen that this skater has marked on 
the ice sixty-four points or stars, and he pro- 
poses to start from his present position near the 
corner and enter every one of the points in 
fourteen straight lines. How will he do it ? 
Of course there is no objection to his passing 
over any point more than once, but his last 
straight stroke must bring him back to the 
position from which he started. 

It is merely a matter of taking your pencil 
and starting from the spot on which the skater's 
foot is at present resting, and striking out all 



the stars in fourteen continuous straight lines, 
returning to the point from which you set out. 

332.— THE FORTY-NINE STARS. 




The puzzle in this case is simply to take your 
pencil and, starting from one black star, strike 
out all the stars in twelve straight strokes, 
ending at the other black star. It wiU be seen 
that the attempt shown in the illustration re- 
quires fifteen strokes. Can you do it in twelve ? 
Every turning must be made on a star, and the 
lines must be parallel to the sides and diagonals 
of the square, as shown. In this case we are 
dealing with a chessboard of reduced dimensions, 
but only queen moves (without going outside 
the boundary as in the last case) are required. 

333.— THE QUEEN'S JOURNEY. 




Place the queen on her own square, as shown 
in the illustration, and then try to discover the 



CHESSBOARD PROBLEMS. 



lOI 



greatest distance that she can travel over the 
board in five queen's moves without passing 
over any square a second time. Mark the 
queen's path on the board, and note carefully 
also that she must never cross her own track. 
It seems simple enough, but the reader may 
find that he has tripped. 

334.— ST. GEORGE AND THE DRAGON. 




Here is a little puzzle on a reduced chessboard 
of forty-nine squares. St. George wishes to 
kill the dragon. Killing dragons was a well- 
known pastime of his, and, being a knight, it 
was only natural that he should desire to per- 
form the feat in a series of knight's moves. 
Can you show how, starting from that central 
square, he may visit once, and only once, every 
square of the board in a chain of chess knight's 
moves, and end by capturing the dragon on his 
last move ? Of course a variety of different 
ways are open to him, so try to discover a route 
that forms some pretty design when you have 
marked each successive leap by a straight line 
from square to square. 

335.— FARMER LAWRENCE'S CORN- 
FIELDS. 

One of the most beautiful districts within easy 
distance of London for a sunmaer ramble is 
that part of Buckinghamshire known as the 
Valley of the Chess — at least, it was a few years 
ago, before it was discovered by the specula- 
tive builder. At the beginning of the present 
century there lived, not far from Latimers, a 
worthy but eccentric farmer named Lawrence. 
One of his queer notions was that every person 
who lived near the banks of the river Chess 
ought to be in some way acquainted with the 
noble game of the same name, and in order to 
impress this fact on his men and his neighbours 
he adopted at times strange terminology. For 
example, when one of his ewes presented him 



with a lamb, he would say that it had " queened 
a pawn " ; when he put up a new barn against 
the highway, he called it " castling on the king's 
side " ; and when he sent a man with a gun to 
keep his neighbour's birds off his fields, he spoke 
of it as " attacking his opponent's rooks." 
Everybody in the neighbourhood used to be 
amused at Farmer Lawrence's little jokes, and 
one boy (the wag of the village) who got his ears 
pulled by the old gentleman for stealing his 
" chestnuts " went so far as to call him " a 
silly old chess-protector ! " 

One year he had a large square field divided 
into forty-nine square plots, as shown in the 
illustration. The white squares were sown with 
wheat and the black squares with barley. 
When the harvest time came round he gave 
orders that his men were first to cut the corn 
in the patch marked i, and that each successive 
cutting should be exactly a knight's move from 
the last one, the thirteenth cutting being in 




the patch marked 13, the twenty-fifth in the 
patch marked 25, the thirty-seventh in the one 
marked 37, and the last, or forty-ninth cutting, 
in the patch marked 49. This was too much 
for poor Hodge, and each day Farmer Lawrence 
had to go down to the field and show which 
piece had to be operated upon. But the prob- 
lem will perhaps present no difficulty to my 
readers. 

336.— THE GREYHOUND PUZZLE. 

In this puzzle the twenty kennels do not com- 
municate with one another by doors, but arc 
divided off by a low wall. The solitary occu- 
pant is the greyhound which lives in the kennel 
in the top left-hand corner. When he is allowed 
his liberty he has to obtain it by visiting every 
kennel once and only once in a series of knight's 



I02 



AMUSEMENTS IN MATHEMATICS. 



moves, aiding at the bottom right-hand corner, 
which is open to the world. The lines in the 




above diagram show one solution. The puzzle 
is to discover in how many different ways the 
greyhound may thus make his exit from his 
comer kennel. 

337.--THE FOUR KANGAROOS. 



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In introducing a little Commonwealth problem, 
I must first explain that the diagram represents 
the sixty-four fields, all properly fenced off 
from one another, of an Australian settlement, 
though I need hardly say that our kith and kin 
" down imder " always cU> set out their land in 
this methodical and exact manner. It will be 
seen that in every one of the four comers is a 
kangaroo. Why kangaroos have a marked pref- 
erence for comer plots has never been satis- 
factorily explained, and it would be out of place 
to discuss the point here. I should also add 
that kangaroos, as is well known, always leap 
in what we caU " knight's moves." In fact, 



chess players would probably have adopted the 
better term " kangaroo's move " had not chess 
been invented before kangaroos. 

The puzzle is simply this. One morning each 
kangaroo went for his morning hop, and in 
sixteen consecutive knight's leaps visited just 
fifteen different fields and jumped back to his 
comer. No field was visited by more than one 
of the kangaroos. The diagram shows how 
they arranged matters. What you are asked 
to do is to show how they might have per- 
formed the feat without any kangaroo ever 
crossing the horizontal line in the middle of 
the square that divides the board into two 
equal parts. 

338.— THE BOARD |N COMPARTMENTS. 

We cannot divide the ordinary chessboard into 
four equal square compartments, and describe a 
complete tour, or even path, in each compart- 
ment. But we may divide it into four com- 
partments, as in the illustration, two containing 



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each twenty squares, and the other two each 
twelve squares, and so obtain an interesting 
puzzle. You are asked to describe a complete 
re-entrant tour on this board, starting where 
you like, but visiting every square in each 
successive compartment before passing into 
emother one, and making the final leap back 
to the square from which the knight set out. 
It is not difficult, but wiU be found very enter- 
taining and not uninstructive. 

Whether a re-entrant " tour " or a complete 
knight's " path " is possible or not on a rectan- 
gular board of given dimensions depends not 
only on its dimensions, but also on its shape. 
A tour is obviously not possible on a board 
containing an odd number of cells, such as 5 by 
5 or 7 by 7, for this reason : Every successive 
leap of the knight must be from a white square 
to a black and a black to a white alternately. 
But if there be an odd number of cells or squares 
there must be one more square of one colour 
than of the other, therefore the path must begin 
from a square of the colour that is in excess, 
and end on a similar colour, and as a knight's 



CHESSBOARD PROBLEMS. 



103 



move from one colour to a similar colour is 
impossible the path cannot be re-entrant. But 
a perfect tour may be made on a rectangular 
board of any dimensions provided the number 
of squares be even, and that the number of 
squares on one side be not less than 6 and on 
the other not less than 5. In other words, the 
smallest rectangular board on which a re- 
entrant tour is possible is one that is 6 by 5. 

A complete knight's path (not re-entrant) 
over all the squares of a board is never possible 
if there be only two squares on one side ; nor 
is it possible on a square board of smaller di- 
mensions than 5 by 5. So that on a board 
4 by 4 we can neither describe a knight's tour 
nor a complete knight's path; we must leave 
one square unvisited. Yet on a board 4 by 3 
(containing four squares fewer) a complete path 
may be described in sixteen different ways. It 
may interest the reader to discover all these. 
Every path that starts from and ends at differ- 
ent squares is here counted as a difierent solu- 
tion, and even reverse routes are called difierent. 

339.— THE FOUR KNIGHTS' TOURS. 




I WILL repeat that if a chessboard be cut into 
four equal parts, as indicated by the dark lines 
in the illustration, it is not possible to perform 
a knight's tour, either re-entrant or not, on one 
of the parts. The best re-entrant attempt is 
shown, in which each knight has to trespass 
twice on other parts. The puzzle is to cut the 
board difierently into four parts, each of the 
same size and shape, so that a re-entrant 
knight's tour may be made on each part. Cuts 
along the dotted lines will not do, as the four 
central squares of the board would be either 
detached or hanging on by a mere thread. 

340.— THE CUBIC KNIGHT'S TOUR. 

Some few years ago I happened to read some- 
where that Abnit Vandermonde, a clever mathe- 



matician, who was bom in 1736 and died in 1793, 
had devoted a good deal of study to the ques- 
tion of knight's tours. Beyond what may be 
gathered from a few fragmentary references, I 
am not aware of the exact nature or results of 
his investigations, but one thing attracted my 
attention, and that was the statement that he 
had proposed the question of a tour of the 
knight over the six surfaces of a cube, each 
surface being a chessboard. Whether he ob- 
tained a solution or not I do not know, but I 
have never seen one published. So I at once 
set to work to master this interesting problem. 
Perhaps the reader may like to attempt it. 

341.— THE FOUR FROGS. 




In the illustration we have eight toadstools, 
with white frogs on i and 3 and black frogs on 
6 and 8. The puzzle is to move one frog at a 
time, in any order, along one of the straight 
lines from toadstool to toadstool, until they 
have exchanged places, the white frogs being 
left on 6 and 8 and the black ones on i and 3. 
If you use four counters on a simple diagram, 
you will find this quite easy, but it is a little 
more puzzling to do it in only seven plays, any 
number of successive moves by one frog count- 
ing as one play. Of course, more than one frog 
cannot be on a toadstool at the same time. 

342.— -THE MANDARIN'S PUZZLE. 

The following puzzle has an added interest 
from the circumstance that a correct solution:of 
it secured for a certain young Chinaman the 
hand of his charming bride. The wealthiest 
mandarin within a radius of a hundred miles 
of Peking was Hi-Chum-Chop, and his beauti- 
ful daughter, Peeky-Bo, had innumerable ad- 
mirers. One of her most ardent lovers was 
Winky-Hi, and when he asked the old mandarin 
for his consent to their marriage, Hi-Chum-Chop 
presented him with the following puzzle and 
promised his consent if the youth brought him 
the correct answer within a week. Winky-Hi, 
following a habit which obtains among certain 



I04 



AMUSEMENTS IN MATHEMATICS. 



solvers to this day, gave it to all his friends, 
and when he had compared their solutions he 
handed in the best one as his own. Luckily it 
was quite right. The mandarin thereupon ful- 
filled his promise. The fatted pup was killed 
for the wedding feast, and when Hi-Chum-Chop 
passed Winky-Hi the liver wing all present 
knew that it was a token of eternal goodwill, in 
accordance with Chinese custom from time 
immemorial. 

The mandarin had a table divided into 
twenty-five squares, as shown in the diagram. 
On each of twenty-four of these squares was 
placed a numbered counter, just as I have in- 
dicated. The puzzle is to get the counters in 
numerical order by moving them one at a time 
in what we call " knight's moves." Counter i 
should be where i6 is, 2 where 11 is, 4 where 13 
now is, and so on. It wiU be seen that all the 
coimters on shaded squares are in their proper 
positions. Of course, two counters may never 
be on a square at the same time. Can you 
perform the feat in the fewest possible moves ? 

In order to make the manner of moving per- 
fectly clear I will point out that the first knight's 
move can only be made by i or by 2 or by 10. 
Supposing I moves, then the next move must 




be by 23, 4, 8, or 21. As there is never more 
than one square vacant, the order in which the 
counters move may be written out as follows : 
I — 21 — 14 — 18 — 22, etc. A rough diagram 
should be made on a larger scale for practice, 
and numbered counters or pieces of cardboard 
used. 

343.— EXERCISE FOR PRISONERS. 

The following is the plan of the north wing 
of a certain gaol, showing the sixteen ceUs all 
communicating by open doorways. Fifteen pris- 
oners were numbered and arranged in the cells 
as shown. They were allowed to change their 
c«^,lls as much as they liked, b.ut if two prisoners 



were ever in the same cell together there was ft 
severe pimishment promised them. 

Now, in order to reduce their growing obesity, 
and to combine physical exercise with mental 




recreation, the prisoners decided, on the sug- 
gestion of one of their number who was inter- 
ested in knight's tours, to try to form them- 
selves into a perfect knight's path without 
breaking the prison regulations, and leaving 
the bottom right-hand corner cell vacant, as 
originally. The joke of the matter is that the 
arrangement at which they arrived was as 
follows : — 



8 


3 


12 


I 


II 


14 


9 


6 


4 


7 


2 


13 


15 


10 


5 





The warders failed to detect the important 
fact that the men could not possibly get into 
this position without two of them having been 
at some time in the same cell together. Make 
the attempt with counters on a ruled diagram, 
and you will find that this is so. Otherwise the 
solution is correct enough, each member being, 
as required, a knight's move from the pre- 
ceding number, and the original comer cell 
vacant. 

The puzzle is to start with the men placed 
as in the illustration and show how it might 
have been done in the fewest moves, while 
giving a complete rest to as many prisoners as 
possible. 

As there is never more than one vacant cell 
for a man to enter, it is only necessary to write 
down the numbers of the men in the order in 
which they move. It is clear that very few 
men can be left throughout in their cells un- 
disturbed, but I will leave the solver to discover ( 
just how many, as this is a very essential parf 
of the puzzle. 



CHESSBOARD PROBLEMS. 



105 



344.— THE KENNEL PUZZLE. 




2r 22^ 23 

T T I 



A MAN has twenty-five dog kennels all com- 
municating with each other by doorways, as 
shown in the illustration. He wishes to arrange 
his twenty dogs so that they shall form a knight's 
string from dog No. i to dog No. 20, the bottom 
row of five kennels to be left empty, as at 
present. This is to be done by moving one dog 
at a time into a vacant kennel. The dogs are 
well trained to obedience, and may be trusted 
to remain in the kennels in which they are 
placed, except that if two are placed in the 
same kennel together they will fight it out to 
the death. How is the puzzle to be solved in 
the fewest possible moves without two dogs 
ever being together ? 

345.— THE TWO PAWNS. 

BLACK 



Here is a neat little puzzle in counting. In 
how many different ways may the two pawns 
advance to the eighth square ? You may move 
them in any order you like to form a different 
sequence. For example, you may move the 
Q R P (one or two squares) first, or the K R P 
first, or one pawn as far as you like before 
touching the other. Any sequence is permis- 
sible, only in this puzzle as soon as a pawn 
reaches the eighth square it is dead, and re- 
mains there unconverted. Can you count the 
number of different sequences ? At first it will 
strike you as being very dif&cult, but I will 
show that it is really quite simple when properly 
attacked. 



VARIOUS CHESS PUZZLES. 

" Chesse-play is a good and wittie exercise of 
the minde for some kinde of men." 

Burton's Anatomy of Melancholy. 

346.— SETTING THE BOARD. 

I HAVE a single chessboard and a single set of 
chessmen. In how many different ways may 
the men be correctly set up for the beginning 
of a game ? I fimd that most people slip at a 
particular point in making the calculation. 

347.— COUNTING THE RECTANGLES. 

Can you say correctly just how many squares 
and other rectangles the chessboard contains ? 
In other words, in how great a number of 
different ways is it possible to indicate a square 
or other rectangle enclosed by lines that sepa- 
rate the squares of the board ? 



THE ROOKERY. 





WHITE 



The White rooks cannot move outside the little 
square in which they are enclosed except on the 
final move, in giving checkmate. The puzzle 



io6 



AMUSEMENTS IN MATHEMATICS. 



is how to checkmate Black in the fewest pos- 
sible moves with No. 8 rook, the other rooks 
being left in numerical order round the sides of 
their square with the break between i and 7. 

349.— STALEMATE. 

Some years ago the puzzle was proposed to 
construct an imaginary game of chess, in which 
White shall be stalemated in the fewest possible 
moves with all the thirty-two pieces on the 
board. Can you build up such a position in 
fewer than twenty moves ? 

350.— THE FORSAKEN KING. 

Set up the position shown in the diagram. 
Then the condition of the puzzle is — ^White to 
play and checkmate in six moves. Notwith- 
standing the complexities, I wiU show how the 




manner of play may be condensed into quite a 
few lines, merely stating here that the first 
two moves of White cannot be varied. 

351.— THE CRUSADER. 

The following is a prize puzzle propoxmded by 
me some years ago. Produce a game of chess 
which, after sixteen moves, shall leave White 
with all his sixteen men on their original squares 
and Black in possession of his king alone (not 
necessarily on his own square). White is then 
to ioru mate in three moves. 

352.— IMMOVABLE PAWNS. 

Starting from the ordinary arrangement of 
the pieces as for a game, what is the smallest 
possible number of moves necessary in order 
to arrive at the following position ? The moves 
for both sides must, of course, be played strictly 
in accordance with the rules of the game, 
though the result wiU necessarily be a very 
weird kind of chess. 



BLACK 




WHITE. 



353.— THIRTY-SIX MATES. 

Place the remaining eight White pieces in 
such a position that White shall have the choice 
of thirty-six different mates on the move. 




i 



Every move that checkmates and leaves a dif- 
ferent position is a different mate. The pieces 
already placed must not be moved. 

354.— AN AMAZING DILEMMA. 

In a game of chess between Mr. Black and Mr. 
White, Black was in difficulties, and as usual 
was obliged to catch a train. So he proposed 
that White should complete the game in his 
absence on condition that no moves whatever 



CHESSBOARD PROBLEMS. 



107 



should be made for Black, but only with the 
White pieces. Mr. White accepted, but to his 
dismay found it utterly impossible to win the 
game under such conditions. Try as he would, 
he could not checkmate his opponent. On 

BLACK 




VrfHITE 



which square did Mr. Black leave his king ? 
The other pieces are in their proper positions 
in the diagram. White may leave Black in 
check as often as he likes, for it makes no 
difference, as he can never arrive at a check- 
mate position. 



355.— CHECKMATE ! 
BLACK 




who had gone. This position is shown in the 
diagram. It is evident that White has check- 
mated Black. But how did he do it ? That 
is the puzzle. 

356.— QUEER CHESS. 

Can you place two White rooks and a White 
knight on the board so that the Black king (who 
must be on one of the four squares in the 
middle of the board) shall be in check with no 
possible move open to him ? "In other 
words," the reader wiU say, " the king is to be 
shown checkmated." Well, you can use the 
term if you wish, though I intentionally do 
not employ it myself. The mere fact that 
there is no White king on the board would 
be a sufficient reason for my not doing so. 

357.— ANCIENT CHINESE PUZZLE. 
BLACK 




WHITE. 



Strolling into one of the rooms of a London 
club, I noticed a position left by two players 



WHITE. 



My next puzzle is supposed to be Chinese, many 
hundreds of years old, and never fails to interest. 
White to play and mate, moving each of the 
three pieces once, and once only. 

358.— THE SIX PAWNS. 

In how many different ways may I place six 
pawns on the chessboard so that there shall be 
an even number of unoccupied squares in every 
row and every column ? We are not here con- 
sidering the diagonals at all, and every different 
six squares occupied makes a different solution, 
so we have not to exclude reversals or reflections. 

359.— COUNTER SOLITAIRE. 

Here is a little game of solitaire that is quite 
easy, but not so easy as to be uninteresting. 
You can either rule out the squares on a sheet 
of cardboard or paper, or you can use a portion 



io8 



AMUSEMENTS IN MATHEMATICS. 



of your chessboard. I have shown numbered 
counters in the illustration so as to make the 



















(9)@@@(g@(g@ 


© 


© 


© 


© 


© 


© 


© 


© 



solution easy and intelligible to all, but chess 
pawns or draughts will serve just as well in 
practice. 

The puzzle is to remove all the coimters 
except one, and this one that is left must be 
No. I. You remove a cotmter by jumping 
over another counter to the next space beyond, 
if that square is vacant, but you cannot make 
a leap in a diagonal direction. The following 
moves will make the play quite clear : i — 9, 
2 — 10, I — 2, and so on. Here i jumps over g, 
and you remove 9 from the board ; then 2 jumps 
over 10, and you remove 10 ; then i jumps over 
2, and you remove 2. Every move is thus a 
captiure, until the last capture of aU is made 
by No. 1. 

360.— CHESSBOARD SOLITAIRE. 




Here is an extension of the last game of soli- 
taire. All you need is a chessboard and the 
thirty-two pieces, or the same number of 
draughts or counters. In the illustration num- 
bered counters are used. The puzzle is to re- 
move all the counters except two, and these 
two must have originally been on the same 
side of the board ; that is, the two left must 
either belong to the group i to 16 or to the 
other group, 17 to 32. You remove a counter 
by jumping over it with another counter to the 



next square beyond, if tuat square is vacant, 
but you cannot make a leap in a diagonal 
direction. The following moves wiU make the 
play quite clear : 3 — 11, 4 — 12, 3 — 4, 13 — 3- 
Here 3 jumps over 11, and you remove 11 ; 
4 jumps over 12, and you remove 12 ; and so 
on. It will be found a fascinating little game 
of patience, and the solution requires the exer- 
cise of some ingenuity. 

361.— THE MONSTROSITY. 

One Christmas Eve I was travelling by rsiil to 
a little place in one of the southern counties. 
The compartment was very full, and the pas- 
sengers were wedged in very tightly. My 
neighbour in one of the corner seats was closely 
studying a position set up on one of those little 
folding chessboards that can be carried con- 
veniently in the pocket, and I could scarcely 
avoid looking at it myself. Here is the posi- 
tion : — 

BLACK 




WHITE. 

My fellow-passenger suddenly turned his 
head and caught the look of bewilderment on 
my face. 

" Do you play chess ? " he asked. 

" Yes, a little. What is that ? A prob- 
lem ? " 

" Problem ? No ; a game." 

" Impossible ! " I exclaimed rather rudely. 
" The position is a perfect monstrosity ! " 

He took from his pocket a postcard and 
handed it to me. It bore an address at one 
side and on the other the words " 43. K to Kt 8." 

" It is a correspondence game," he exclaimed. 
" That is my friend's last move, and I am con- 
sidering my reply." 

" But you really must excuse me ; the posi- 
tion seems utterly impossible. How on earth, 
for example " 

'* Ah ! " he broke in smilingly. " I see ; you 
are a beginner ; you play to win." 



MEASURING, WEIGHING, AND PACKING PUZZLES. 



109 



" Of course you wouldn't play to lose or 
draw ! " 

He laughed aloud. 

" You have much to learn. My friend and 
myself do not play for results of that anti- 
quated kind. We seek in chess the wonderful, 
the whimsical, the weird. Did you ever see a 
position like that ? " 

I inwardly congratulated myself that I never 
had. 

" That position, sir, materializes the sinuous 
evolvements and syncretic, synthetic, and syn- 
chronous concatenations of two cerebral indi- 
vidualities. It is the product of an amphoteric 
and intercalatory interchange of " 

" Have you seen the evening paper, sir ? " 
interrupted the man opposite, holding out a 



newspaper. I noticed on the margin beside 
his thumb some pencilled writing. Thanking 
him, I took the paper and read — " Insane, but 
quite harmless. He is in my charge." 

After that I let the poor fellow run on in 
his wild way until both got out at the next 
station. 

But that queer position became fixed in- 
delibly in my mind, with Black's last move 43. 
K to Kt 8 ; and a short time afterwards I 
found it actually possible to arrive at such a 
position in forty-three moves. Can the reader 
construct such a sequence ? How did White 
get his rooks and king's bishop into their present 
positions, considering Black can never have 
moved his king's bishop ? No odds were given, 
and every move was perfectly legitimate. 



MEASURING, WEIGHING, AND PACKING PUZZLES. 



" Measure still for measure." 

Measure for Measure, v. i. 

Apparently the first printed puzzle involving 
the measuring of a given quantity of liquid by 
pouring from one vessel to others of known 
capacity was that propounded by Niccola Fon- 
tana, better known as " Tartaglia " (the 
stammerer), 1500-1559. It consists in dividing 
24 oz. of valuable balsam into three equal parts, 
the only measures available being vessels hold- 
ing 5, II, and 13 ounces respectively. There 
are many difierent solutions to this puzzle in 
six manipulations, or pourings from one vessel 
to another. Bachet de Meziriac reprinted this 
and other of Tartaglia's puzzles in his Prohlemes 
plaisans et dilectahles (161 2). It is the general 
opinion that puzzles of this class can only be 
solved by trial, but I think formula9 can be 
constructed for the solution generally of certain 
related cases. It is a practically unexplored 
field for investigation. 

The classic weighing problem is, of course, 
that proposed by Bachet. It entails the de- 
termination of the least number of weights that 
would serve to weigh any integral number of 
pounds from i lb. to 40 lbs. inclusive, when 
we are allowed to put a weight in either of the 
two pans. The answer is i, 3, 9, and 27 lbs. 
Tartaglia had previously propounded the same 
puzzle with the condition that the weights may 
only be placed in one pan. The answer in that 
case is i, 2, 4, 8, 16, 32 lbs. Major Mac- 
Mahon has solved the problem quite generally. 
A full account will be found in Ball's Mathe- 
matical Recreations (5th edition). 

Packing puzzles, in which we are required to 
pack a maximum number of articles of given 
dimensions into a box of known dimensions, 
are, I believe, of quite recent introduction. 
At least I cannot recall any example in the 
books of the old writers. One would rather 
expect to find in the toy shops the idea pre- 
sented as a mechanical puzzle, but I do not 
think I have ever seen such a thing. The 
nearest approach to it would appear to be the 



puzzles of the jig-saw character, where there is 
only one depth of the pieces to be adjusted. 

362.— THE WASSAIL BOWL. 

One Christmas Eve three Weary Willies came 
into possession of what was to them a veritable 
wassail bowl, in the form of a small barrel, con- 
taining exactly six quarts of fine ale. One of the 
men possessed a five-pint jug and another a three- 
pint jug, and the problem for them was to 
divide the liquor equally amongst them with- 
out waste. Of course, they are not to use 
any other vessels or measures. If you can 
show how it was to be done at aU, then try to 
find the way that requires the fewest possible 
manipulations, every separate pouring from one 
vessel to another, or down a man's throat, 
counting as a manipulation. 

363.— THE DOCTOR'S QUERY. 

" A CURIOUS little point occurred to me in my 
dispensary this morning," said a doctor. " I 
had a bottle containing ten ounces of spirits 
of wine, and another bottle containing ten 
ounces of water. I poured a quarter of an 
ounce of spirits into the water and shook them 
up together. The mixture was then clearly 
forty to one. Then I poured back a quarter- 
ounce of the mixture, so that the two bottles 
should again each contain the same quantity 
of fluid. What proportion of spirits to water 
did the spirits of wine bottle then contain ? " 

364.— THE BARREL PUZZLE. 

The mfen in the illustration are disputing over 
the liquid contents of a barrel. What the parti- 
cular liquid is it is impossible to say, for we are 
unable to look into the barrel ; so we will call it 
water. One man says that the barrel is more than 
half full, while the other insists that it is not half 
full. What is their easiest way of settling the 
point ? It is not necessary to use stick, string, 
or implement of any kind for measuring. I 



no 



AMUSEMENTS IN MATHEMATICS. 




give this merely as one of the simplest possible 
examples of the value of ordinary sagacity in 
the solving of puzzles. What are apparently 
very dif&cult problems may frequently be 
solved in a similarly easy manner if «re only 
use a little common sense. 

365.— NEW MEASURING PUZZLE. 

Here is a new poser in measuring liquids that 
will be found interesting. A man has two ten- 
quart vessels full of wine, and a five-quart and 
a four-quart measure. He wants to put ex- 
actly three quarts into each of the two mea- 
sures. How is he to do it ? And how many 
manipulations (pourings from one vessel to 
another) do you require ? Of course, waste of 
wine, tilting, and other tricks are not allowed. 

366.--THE HONEST DAIRYMAN. 

An honest dairyman in preparing his milk for 
public consumption employed a can marked B, 
containing milk, and a can marked A, containing 
water. From can A he poured enough to 
double the contents of can B, Then he poured 
from can B into can A enough to double its 
contents. Then he finally poured from can 
A into can B until their contents were exactly 
equal. After these operations he would send 
the can A to London, and the puzzle is to dis- 
cover what are the relative proportions of milk 
and water that he provides for the Londoners* 
breakfast-tables. Do they get equal propor- 
tions of milk and water — or two parts of milk 



and one of water — or what ? It is an interest- 
ing question, though, curiously enough, we are 
not told how much milk or water he puts into 
the cans at the start of his operations. 

367.— WINE AND WATER. 

Mr. Goodfellow has adopted a capital idea of 
late. When he gives a little dinner party and 
the time arrives to smoke, after the departure 
of the ladies, he sometimes finds that the con- 
versation is apt to become too political, too 
personal, too slow, or too scandalous. Then 
he always manages to introduce to the company 
some new poser that he has secreted up his 
sleeve for the occasion. This invariably results 
in no end of interesting discussion and debate, 
and puts everybody in a good humour. 

Here is a little puzzle that he propounded the 
other night, and it is extraordinary how the 
company differed in their answers. He fiUed 
a wine-glass half fuU of wine, and another glass 
twice the size one-third full of wine. Then he 
fiUed up each glass with water and emptied the 
contents of both into a tumbler. " Now," he 
said, " what part of the mixture is wine and 
what part water ? " Can you give the correct 
answer ? 

368.— THE KEG OF WINE. 

Here is a curious little problem. A man had a 
ten-gaUon keg fuU of wine and a jug. One day 
he drew off a jugful of wine and fiUed up the 
keg with water. Later on, when the wine and 
water had got thoroughly mixed, he drew off 



MEASURING, WEIGHING, AND PACKING PUZZLES. 



Ill 



another jugful and again filled up the keg with 
water. It was then found that the keg con- 
tained equal proportions of wine and water. 
Can you find from these facts the capacity of 
the jug ? 

369.— MIXING THE TEA. 

*' Mrs. Spooner called this morning," said the 
honest grocer to his assistant. " She wants 
twenty pounds of tea at 2s. 4jd. per lb. Of 
course we have a good 2s. 6d. tea, a slightly 
inferior at 2s. 3d., and a cheap Indian at is. gd., 
but she is very particular always about her 
prices." 

" What do you propose to do ? " asked the 
innocent assistant. 

" Do ? " exclaimed the grocer. " Why, just 
mix up the three teas in different proportions 
so that the twenty pounds will work out fairly 
at the lady's price. Only don't put in more of 
the best tea than you can help, as we make less 
profit on that, and of course you wiU use only 
our complete pound packets. Don't do any 
weighing." 

How was the poor feUow to mix the three 
teas ? Could you have shown him how to 
doit? 

370.— A PACKING PUZZLE. 

As we all know by experience, considerable in- 
genuity is often required in packing articles 
into a box if space is not to be unduly wasted. 
A man once told me that he had a large nmnber 
of iron balls, aU exactly two inches in diameter, 
and he wished to pack as many of these as pos- 
sible into a rectangular box 24^ inches long, 
22^ inches wide, and 14 inches deep. Now, 
what is the greatest number of_the balls that he 
could pack into that box ? 

371.— GOLD PACKING IN RUSSIA. 

The editor of the Times newspaper was invited 
by a high Russian of&cial to inspect the gold 
stored in reserve at St. Petersburg, in order that 
he might satisfy himself that it was not another 
" Humbert safe." He replied that it would be 
of no use whatever, for although the gold might 
appear to be there, he would be quite unable 
from a mere inspection to declare that what he 
saw was really gold. A correspondent of the 
Daily Mail thereupon took up the challenge, 
but, although he was greatly impressed by what 
he saw, he was compelled to confess his in- 
competence (without empt)dng and counting 
the contents of every box and sack, and assaying 
every piece of gold) to give any assurance on 
the subject. In presenting the following little 
puzzle, I wish it to be also imderstood that I do 
not guarantee the real existence of the gold, 
and the point is not at aU material to our 



piurpose. Moreover, if the reader szys that 
gold is not usually " put up " in slabs of the 
dimensions that I give, I can only claim proble- 
matic licence. 

Russian ofi&cials were engaged in packing 
800 gold slabs, each measuring 12^ inches long, 
II inches wide, and i inch deep. What are 
the interior dimensions of a box of equal 
length and width, and necessary depth, that 
wiU exactly contain them without any space 
being left over ? Not more than twelve slabs 
may be laid on edge, according to the rules of 
the government. It is an interesting little 
problem in packing, and not at all difficult. 

372.— THE BARRELS OF HONEY. 




Once upon a time there was an aged merchant 
of Bagdad who was much respected by all who 
knew him. He had three sons, and it was a 
rule of his life to treat them all exactly alike. 
Whenever one received a present, the other two 
were each given one of equal value. One day 
this worthy man fell sick and died, bequeathing 
aU his possessions to his three sons in equal 
shares. 

The only difficulty that arose was over the 
stock of honey. There were exactly twenty- 
one barrels. The old man had left instructions 
that not only should every son receive an equal 
quantity of honey, but should receive exactly 
the same number of barrels, and that no honey 
should be transferred from barrel to barrel on 
account of the waste involved. Now, as seven 
of these barrels were fuU of honey, seven were 
half-fuU, and seven were empty, this was found 
to be quite a puzzle, especially as each brother 
objected to taking more than four barrels of 
the same description — fuU, half-full, or empty. 
Can you show how they succeeded in making 
a correct division of the property ? 



112 



AMUSEMENTS IN MATHEMATICS. 



CROSSING RIVER PROBLEMS. 



" My boat is on the shore." 

Byron. 

This is another mediaeval class of puzzles. 
Probably the earliest example was by Abbot 
Alcuin, who was bom in Yorkshire in 735 and 
died at Tours in 804. And everybody knows 
the story of the man with the wolf, goat, and 
basket of cabbages whose boat would only 
take one of the three at a time with the man 
himself. His difficulties arose from his being 
unable to leave the wolf alone with the goat, 
or the goat alone with the cabbages. These 
puzzles were considered by Tartaglia and Bachet, 
and have been later investigated by Lucas, De 
Fonteney, Delannoy, Tarry, and others. In 



sight, because she had to come back again with 
the boat, so nothing was gained by that opera- 
tion. How did they all succeed in getting 
across ? The reader will find it much easier 
than the Softleigh family did, for their greatest 
enemy could not have truthfully called them a 
brilliant quartette — ^while the dog was a perfect 
fool. 

374.— CROSSING THE RIVER AXE. 

Many years ago, in the days of the smuggler 
known as " Rob Roy of the West," a piratical 
band buried on the coast of South Devon a 
quantity of treasure which was, of course, 
abandoned by them in the usual inexplicable 
way. Some time afterwards its whereabouts 




the puzzles I give there will be found one or 
two new conditions which add to the complexity 
somewhat. I also include a pulley problem that 
practically involves the same principles. 

373.— CROSSING THE STREAM. 

During a country ramble Mr. and Mrs. Soft- 
leigh found themselves in a pretty little dilemma. 
They had to cross a stream in a small boat which 
was capable of carrying only 150 lbs. weight. 
But Mr. Softleigh and his wife each weighed 
exactly 150 lbs., and each of their sons weighed 
75 lbs. And then there was the dog, who could 
not be induced on any terms to swim. On the 
principle of " ladies first," they at once sent 
Mrs. Softleigh over ; but this was a stupid over- 



was discovered by three countrymen, who 
visited the spot one night and divided the spoil 
between them, Giles taking treasure to the 
value of £800, Jasper £500 worth, and Timothy 
£300 worth. In returning they had to cross 
the river Axe at a point where they had left a 
small boat in readiness. Here, however, was 
a difficulty they had not anticipated. The boat 
would only carry two men, or one man and a 
sack, and they had so little confidence in one 
another that no person could be left alone on 
the land or in the boat with more than his 
share of the spoil, though two persons (being 
a check on each other) might be left with more 
than their shares. The puzzle is to show how 
they got over the river in the fewest possible 
crossings, taking their treasure with them. No 



CROSSING RIVER PROBLEMS. 



"3 



tricks, such as ropes, " flying bridges," currents, 
swimming, or similar dodges, may be employed. 

375 .—FIVE JEALOUS HUSBANDS. 

During certain local floods five married couples 
found themselves surrounded by water, and had 
to escape from their unpleasant position in a 
boat that would only hold three persons at a 
time. Every husband was so jealous that he 
would not allow his wife to be in the boat or 
on either bank with another man (or with other 
men) unless he was himself present. Show the 
quickest way of getting these five men and their 
wives across into safety. 

Call the men A, B, C, D, E, and their respec- 
tive wives a, b, c, d, e. To go over and return 
counts as two crossings. No tricks such as 
ropes, swimming, currents, etc., are permitted. 



THE FOUR ELOPEMENTS. 



376 

Colonel B was a widower of a very taci- 
turn disposition. His treatment of his four 
daughters was unusually severe, almost cruel, 
and they not unnaturally felt disposed to resent 
it. Being charming girls with every virtue and 
many accomplishments, it is not surprising 
that each had a fond admirer. But the father 
forbade the young men to call at his house, 
intercepted all letters, and placed his daughters 
under stricter supervision than ever. But love, 
which scorns locks and keys and garden walls, 
was equal to the occasion, and the fom: youths 
conspired together and planned a general elope- 
ment. 

At the foot of the tennis lawn at the bottom 
of the garden ran the silver Thames, and one 
night, after the four girls had been safely con- 
ducted from a dormitory window to Una firma, 
they all crept softly down to the bank of the 
river, where a small boat belonging to the 
Colonel was moored. With this they proposed 
to cross to the opposite side and make their way 
to a lane where conveyances were waiting to 
carry them in their flight. Alas ! here at the 
water's brink their difficulties already began. 

The yoTing men were so extremely jealous 
that not one of them would allow his prospec- 
tive bride to remain at any time in the com- 
pany of another man, or men, unless he himself 
were present also. Now, the boat would only 
hold two persons, though it could, of course, be 
rowed by one, and it seemed impossible that 
the four couples would ever get across. But 
midway in the stream was a small island, and 
this seemed to present a way out of the difficulty, 
because a person or persons could be left there 
while the boat was rowed back or to the oppo- 
site shore. If they had been prepared for their 
difficulty they could have easily worked out a 
solution to the little poser at any other time. 
But they were now so hurried and excited in 
their flight that the confusion they soon got into 
was exceedingly amusing — or would have be^n 
to any one except themselves. 

As a consequence they took twice as long and 
crossed the river twice as often as was really 
(1,926) 8 



necessary. Meanwhile, the Colonel, who was 
a very light sleeper, thought he heard a splash 
of oars. He quickly raised the alarm among his 
household, and the young ladies were found to 
be missing. Somebody was sent to the police- 
station, and a number of officers soon aided in 
the pursuit of the fugitives, who, in consequence 
of that delay in crossing the river, were quickly 
overtaken. The four girls returned sadly to 
their homes, and afterwards broke oS their 
engagements in disgust. 

For a considerable time it was a mystery 
how the party of eight managed to cross the 
river in that little boat without any girl being 
ever left with a man, unless her betrothed was 
also present. The favourite method is to take 
eight counters or pieces of cardboard and mark 
them A, B, C, D, a, b, c, d, to represent the 
four men and their prospective brides, and 
carry them from one side of a table to the 
other in a matchbox (to represent the boat), 
a penny being placed in the middle of the table 
as the island. 

Readers are now asked to find the quickest 
method of getting the party across the river. 
How many passages are necessary from land 
to land ? By " land " is imderstood either 
shore or island. Though the boat woifld not 
necessarily call at the island every time of 
crossing, the possibility of its doing so must 
be provided for. For example, it would not 
do for a man to be alone in the boat (though it 
were understood that he intended merely to 
cross from one bank to the opposite one) if 
there happened to be a girl alone on the island 
other than the one to whom he was engaged. 

377.__STEALING THE CASTLE 
TREASURE. 

The ingenious manner in which a box of trea- 
sure, consisting principally of jewels and precious 
stones, was stolen from Gloomhurst Castle has 
been handed down as a tradition in the De 
Gourney family. The thieves consisted of a 
man, a youth, and a small boy, whose only 
mode of escape with the box of treasure was 
by means of a high window. Outside the win- 
dow was fixed a pulley, over which ran a rope 
with a basket at each end. When one basket 
was on the ground the other was at the window. 
The rope was so disposed that the persons in 
the basket could neither help themselves by 
means of it nor receive help from others. In 
short, the only way the baskets could be used 
was by placing a heavier weight in one than 
in the other. 

Now, the man weighed 195 lbs., the youth 
105 lbs., the boy 90 lbs., and the box of treasure 
75 lbs. The weight in the descending basket 
could not exceed that in the other by more than 
15 lbs. without causing a descent so rapid as to 
be most dangerous to a human being, though it 
would not injure the stolen property. Only two 
persons, or one person and the treasure, could 
be placed in the same basket at one time. How 
did they all manage to escape and take the box 
of treasure with them ? 



114 



AMUSEMENTS IN MATHEMATICS. 



The puzzle is to find the shortest way of 
performing the feat, which in itself is not dif&- 
cult. Remember, a person cannot help himself 



by hanging on to the rope, the only way being 
to go down " with a bump," with the weight 
in the other basket as a counterpoise. 



PROBLEMS CONCERNING GAMES. 



** The little pleasure of the game." 

Matthew Prior. 

Every game lends itself to the propounding of 
a variety of puzzles. They can be made, as we 
have seen, out of the chessboard and the peculiar 
moves of the chess pieces. I will now give just 
a few examples of puzzles with playing cards 
and dominoes, and also go out of doors and 
consider one or two little posers in the cricket 
field, at the football match, and the horse race 
and motor-car race. 

378.— DOMINOES IN PROGRESSION. 

4 ff 6 



— — — —■ ———————— ' 

• • • • • • 

I • • • • ' • • 



9 

It will be seen that I have played six dominoes, 
in the illustration, in accordance with the ordin- 
ary rules of the game, 4 against 4, i against i, 
and so on, and yet the sum of the spots on the 
successive dominoes, 4, 5, 6, 7, 8, 9, are in 
arithmetical progression ; that is, the numbers 
taken in order have a common difference of i. 
In how many different ways may we play six 
dominoes, from an ordinary box of twenty- 
eight, so that the numbers on them may lie in 
arithmetical progression ? We must always 
play from left to right, and numbers in decreas- 
ing arithmetical progression (such as 9, 8, 7, 6, 
5, 4) are not admissible. 

379.— THE FIVE DOMINOES. 



• 
• 


• 


• 

1 • 


• 

• < 


1 


•• 

• 















(that is, with i against i, 2 against 2, and so 
on), that the total number of pips on the two 
end dominoes is five, and the sum of the pips 
on the three dominoes in the middle is also five. 
There are just three other arrangements giving 
five for the additions. They are : — 

(I— o) (0—0) (0—2) (2—1) (1—3) 
(4—0) (0—0) (0—2) (2—1) (I— o) 
(2—0) (0—0) (o— I) (1—3) (3—0) 

Now, how many similar arrangements are there 
of five dominoes that shall give six instead of 
five in the two additions ? 

380.— THE DOMINO FRAME PUZZLE. 



• 
• 

• 








• 




V 


• *• 


THE 


• 
• 


• • 

• • 

« • 

• • 

• • 


DOMINO FRAME 


m 

:< 
••• 


• 
• 

•• 


PUZZLE. 


- 


••. 




■■** 


• 

••• 




• • 

• m 

• • 

• • 

• • 

• • 

• • 


\::::::::X/:vv.\ %..!..- 



Here is a new little puzzle that is not difficult, 
but will probably be found entertaining by 
my readers. It will be seen that the five 
dominoes are so arranged in proper sequence 



It will be seen in the illustration that the full 
set of twenty-eight dominoes is arranged in 
the form of a square frame, with 6 against 6, 
2 against 2, blank against blank, and so on, as 
in the game. It will be found that the pips in 
the top row and left-hand column both add up 
44. The pips in the other two sides sum to 
59 and 32 respectively. The puzzle is to re- 
arrange the dominoes in the same form so that 
aU of the four sides shall sum to 44. Re- 
member that the dominoes must be correctly 
placed one against another as in the game. 

381.— THE CARD FRAME PUZZLE. 

In the illustration we have a frame constructed 
from the ten playing cards, ace to ten of dia- 
monds. The children who made it wanted the 
pips on all four sides to add up alike, but they 
failed in their attempt and gave it up as im- 
possible. It wiU be seen that the pips in the 



PROBLEMS CONCERNING GAMES. 



"5 



top row, the bottom row, and the left-hand side 
all add up 14, but the right-hand side sums to 
23. Now, what they were trying to do is quite 




possible. Can you rearrange the ten cards in 
the same formation so that all four sides shall 
add up alike ? Of course they need not add up 
14, but any number you choose to select. 

382.— THE CROSS OF CARDS. 



♦ ♦ ♦ ♦ ♦ 

V \\ — I ) 



4 ♦ 

♦ ^^ 

♦ 

■ 

♦ ♦ 

♦ ♦ ♦ ♦ ♦ 



In this case we use only nine cards — the ace 
to nine of diamonds. The puzzle is to arrange 
them in the form of a cross, exactly in the way 



shown in the illustration, so that the pips in 
the vertical bar and in the horizontal bar add 
up alike. In the example given it will be found 
that both directions add up 23. What I want 
to know is, how many different ways are there 
of rearranging the cards in order to bring about 
this result ? It will be seen that, without 
affecting the solution, we may exchange the 5 
with the 6, the 5 with the 7, the 8 with the 3, 
and so on. Also we may make the horizontal 
and the vertical bars change places. But such 
obvious manipulations as these are not to be 
regarded as different solutions. They are all 
mere variations of one fundamental solution. 
Now, how many of these fundamentally differ- 
ent solutions are there ? The pips need not, 
of course, always add up 23. 

383.— THE "T" CARD PUZZLE. 



♦ ♦](♦ ^^ 1 
« ♦ ♦ ♦ 

, 

♦ 

♦ 

♦ ♦ 

♦ ♦ 

♦ ♦ 

♦ ♦ 



♦ ♦ ♦ 
♦ ♦ 



♦ ♦♦J 



An entertaining little puzzle with cards is to 
take the nine cards of a suit, from ace to nine 
inclusive, and arrange them in the form of the 
letter " T," as shown in the illustration, so that 
the pips in the horizontal line shall <Jount the 
same as those in the column. In the example 
given they add up twenty-three both ways. 
Now, it is quite easy to get a single correct 
arrangement. The puzzle is to discover in just 
how many different ways it may be done. 
Though the number is high, the solution is not 
really difiScult if we attack the puzzle in the 
right manner. The reverse way obtained by 
reflecting the illustration in a mirror we wiU 
not count as different, but all other changes 
in the relative positions of the cards will here 
count. How many different ways are there ? 

384.— CARD TRIANGLES. 

Here you pick out the nine cards, ace to nine of 
diamonds, and arrange them in the form of a 
triangle, exactly as shown in the illustration, so 
that the pips add up' the same on the three 
sides. In the example given it wiU be seen 
that they sum to 20 on each side, but the par- 
ticular number is of no importsmce so long as 
it is the same on all thrQ.e sides. The puzzle 



ii6 



AMUSEMENTS IN MATHEMATICS. 



is to find out in just how many different ways 
this can be done. 

If you simply turn the cards round so that 
one of the other two sides is nearest to you this 
will not count as different, for the order will be 
the same. Also, if you make the 4, 9, 5 change 
places with the 7, 3, 8, and at the same time 
exchange the i and the 6, it will not be differ- 
ent. But if you only change the i and the 6 it 




will be different, because the order round the 
triangle is not the same. This explanation will 
prevent any doubt arising as to the conditions. 

385.—" STRAND " PATIENCE. 

The idea for this came to me when considering 
the game of Patience that I gave in the Strand 
Magazine for December, 1910, which has been 
reprinted in Ernest Bergholt's Second Book of 
Patience Games, under the new name of " King 
Albert." 

Make two piles of cards as follows : 9 D, 8 S, 
7 D, 6 S, 5 D, 4 S, 3 D, 2 S, I D, and 9 H, 8 C, 
7 H, 6 C, 5 H, 4 C, 3 H, 2 C, I H, with the 9 of 
diamonds at the bottom of one pile and the 9 of 
hearts at the bottom of the other. The point is 
to exchange the spades with the clubs, so that 
the diamonds and clubs are still in numerical 
order in one pile and the hearts and spades in 
the other. There are four vacant spaces in 
addition to the two spaces occupied by the 
piles, and any card may be laid on a space, but 
a card can only be laid on another of the next 
higher value — an ace on a two, a two on a three, 
and so on. Patience is required to discover the 
shortest way of doing this. When there are four 
vacant spaces you can pile fom: cards in seven 
moves, with only three spaces you can pile them 
in nine moves, and with two spaces you cannot 
pile more than two cards. When you have a 
grasp of these and similar facts you wiU be able 
to remove a number of cards bodily and write 
down 7, 9, or whatever the number of moves may 
be. The gradual shortening of play is fascinat- 
ing, and first attempts ^re surprisingly lengthy. 



386.— A TRICK WITH DICE. 




Here is a neat little trick with three dice. I 
ask you to throw the dice without my seeing 
them. Then I tell you to multiply the points 
of the first die by 2 and add 5 ; then multiply 
the result by 5 and add the points of the second 
die ; then multiply the result by 10 and add 
the points of the third die. You then give me 
the total, and I can at once tell you the points 
thrown with the three dice. How do I do it ? 
As an example, if you threw i, 3, and 6, as in 
the illustration, the result you would give me 
would be 386, from which I could at once say 
what you had thrown. 

387.— THE VILLAGE CRICKET MATCH. 

In a cricket match, Dingley Dell v. All Muggle- 
ton, the latter had the first innings. Mr. 
Dumkins and Mr. Podder were at the wickets, 
when the wary Dumkins made a splendid late 
cut, and Mr. Podder called on him to run. Four 
runs were apparently completed, but the vigi- 
lant umpires at each end caUed, " three short," 
making six short runs in all. What number did 
Mr. Dumkins score ? When Dingley Dell took 
their turn at the wickets their champions were 
Mr. Luffey and Mr. Struggles. The latter made 
a magnificent off-drive, and invited his colleague 
to " come along," with the result that the obser- 
vant spectators applauded them for what was 
supposed to have been three sharp runs. But 
the umpires declared that there had been two 
short runs at each end — four in all. To what 
extent, if any, did this manoeuvre increase Mr. 
Struggles's total ? 

388.— SLOW CRICKET. 

In the recent county match between Wessex 
and Nincomshire the former team were at the 
wickets all day, the last man being put out a 
few minutes before the time for drawing stumps. 
The play was so slow that most of the spec- 
tators were fast asleep, and, on being awakened 
by one of the officials clearing the ground, we 
learnt that two men had been put out leg- 
before- wicket for a combined score of 19 runs ; 
four men were caught for a combined score of 
17 runs ; one man was run out for a duck's 
egg ; and the others were all bowled for 3 runs 
each. There were no extras. We were not 
told which of the men was the captain, but he 
made exactly 15 more than the average of his 
team. What was the captain's score ? 

389.— THE FOOTBALL PLAYERS. 

"It is a glorious game ! " an enthusiast was 
heard to exclaim. " At the close of last season, 



PUZZLE GAMES. 



117 



of the footballers of my acquaintance four had 
broken their left arm, five had broken their 
right arm, two had the right arm sound, and 
three had sound left arms." Can you discover 
from that statement what is the smallest number 
of players that the speaker could be acquainted 
with ? 

It does not at all follow that there were as 
many as fourteen men, because, for example, 
two of the men who had broken the left arm 
might also be the two who had sound right arms. 

390.— THE HORSE-RACE PUZZLE. 

There are no morals in puzzles. When we are 
solving the old puzzle of the captain who, 
having to throw half his crew overboard in a 
storm, arranged to draw lots, but so placed the 
men that only the Turks were sacrificed, and 
aU the Christians left on board, we do not stop 
to discuss the questionable morality of the pro- 
ceeding. And when we are dealing with a 
measuring problem, in which certain thirsty 
pilgrims are to make an equitable division of a 
barrel of beer, we do not object that, as total 
abstainers, it is against our conscience to have 
anything to do with intoxicating liquor. There- 
fore I make no apology for introducing a puzzle 
that deals with betting. 

Three horses — Acorn, Bluebottle, and Cap- 
sule — start in a race. The odds are 4 to i, 
Acorn; 3 to i. Bluebottle; 2 to i. Capsule. 
Now, how much must I invest on each horse in 
order to win £13, no matter which horse comes 



in first ? Supposing, as an example, that I 
betted £5 on each horse. Then, if Acorn won, 
I should receive £20 (four times £5), and have 
to pay £5 each for the other two horses ; 
thereby winning £10. But it will be found 
that if Bluebottle was first I should only win 
£5, and if Capsule won I should gain nothing 
and lose nothing. This will make the question 
perfectly clear to the novice, who, like myself, 
is not interested in the calling of the fraternity 
who profess to be engaged in the noble task of 
" improving the breed of horses." 

391.— THE MOTOR-CAR RACE. 

Sometimes a quite simple statement of fact, 
if worded in an unfamiliar manner, will cause 
considerable perplexity. Here is an example, 
and it will doubtless puzzle some of my more 
youthful readers just a little. I happened to be 
at a motor-car race at Brooklands, when one 
spectator said to another, while a number of 
cars were whirling round and round the circular 
track : — 

"There's Gogglesmith — that man in the 
white car ! " 

" Yes, I see," was the reply ; *' but how many 
cars are running in this race ? " 

Then came this curious rejoinder : — 

" One-third of the cars in front of Goggle- 
smith added to three-quarters of those behind 
him will give you the answer." 

Now, can you tell how many cars were running 
in the race ? 



PUZZLE GAMES. 



" He that is beaten may be said 
To lie in honour's truckle bed." 

HUDIBRAS. 

It may be said generally that a game is a con- 
test of skill for two or more persons, into which 
we enter either for amusement or to win a prize. 
A puzzle is something to be done or solved 
by the individual. For example, if it were pos- 
sible for us so to master the complexities of the 
game of chess that we coiild be assured of always 
winning with the first or second move, as the 
case might be, or of always drawing, then it 
would cease to be a game and would become a 
puzzle. Of course among the young and un- 
informed, when the correct winning play is not 
understood, a puzzle may well make a very 
good game. Thus there is no doubt children 
wiU continue to play " Noughts and Crosses," 
though I have shown (No. 109, " Canterbury 
Puzzles ") that between two players who both 
thoroughly understand the play, every game 
should be drawn. Neither player could ever 
win except through the blundering of his oppon- 
ent. But I am writing from the point of view 
of the student of these things. 

The examples that I give in this class are 
apparently games, but, since I show in every 
case how one player may win if he only play 



correctly, they are in reality puzzles. Their in- 
terest, therefore, lies in attempting to discover 
the leading method of play. 

392.— THE PEBBLE GAME. 

Here is an interesting little puzzle game that I 
used to play with an acquaintance on the beach 
at Slocomb-on-Sea. Two players place an odd 
number of pebbles, we will say fifteen, between 
them. Then each takes in turn one, two, or 
three pebbles (as he chooses), and the winner is 
the one who gets the odd number. Thus, if you 
get seven and your opponent eight, you win. 
If you get six and he gets nine, he wins. Ought 
the first or second player to win, and how ? 
When you have settled the question with fifteen 
pebbles try again with, say, thirteen. 

393.— THE TWO ROOKS. 

This is a puzzle game for two players. Each 
player has a single rook. The first player places 
his rook on any square of the board that he may 
choose to select, and then the second player does 
the same. They now play in turn, the point of 
each play being to capture the opponent's rook. 
But in this garne you cannot play through a 
line of attack without being captured. That is 
to say, if in the diagram it is Black's turn to 



Ii8 



AMUSEMENTS IN MATHEMATICS. 



play, he cannot move his rook to his king's 

knight's square, or to his long's rook's square, 

because he would enter the " line of fire " when 

. passing his king's bishop's square. For the 



BLACK. 




WHITE. 

same reason he cannot move to his queen's 
rook's seventh or eighth squares. Now, the 
game can never end in a draw. Sooner or later 
one of the rooks must fall, unless, of course, 
both players commit the absurdity of not try- 
ing to win. The trick of winning is ridiculously 
simple when you know it. Can you solve the 
puzzle ? 

394.— PUSS IN THE CORNER. 




and the other puts one on No. 55, and they play 
alternately by removing the counter to any 
other number in a line. If your opponent 
moves at any time on to one of the lines you 
occupy, or even crosses one of your lines, you 
immediately capture him and win. We will 
take an illustrative game. 

A moves from 55 to 52 ; B moves from 6 to 
13 ; A advances to 23 ; B goes to 15 ; A re- 
treats to 26 ; B retreats to 13 ; A advances to 
21 ; B retreats to 2 ; A advances to 7 ; B goes 
to 3 ; A moves to 6 ; B must now go to 4 ; A 
establishes himself at 11, and B must be cap- 
tured next move because he is compelled to 
cross a line on which A stands. Play this over 
and you will understand the game directly. 
Now, the puzzle part of the game is this : Which 
player should win, and how many moves are 
necessary ? 

395.--A WAR PUZZLE GAME. 




This variation of the last puzzle is also played 
by two persons. One puts a coimter on No. 6, 



Here is another puzzle game. One player, 
representing the British general, places a coun- 
ter at B, and the other player, representing the 
enemy, places his coimter at E. The Britisher 
makes the first advance along one of the roads 
to the next town, then the enemy moves to one 
of his nearest towns, and so on in turns, imtil 
the British general gets into the same town as 
the enemy and captures him. Although each 
must always move along a road to the next 
town only, and the second player may do his 
utmost to avoid capture, the British general 
(as we should suppose, from the analogy of real 
life) must infallibly win. But how ? That is 
the question. 

396.— A MATCH MYSTERY. 

Here is a little game that is childishly simple 
in its conditions. But it is well worth investi- 
gation. 

Mr. Stubbs pulled a small table between him- 
self and his friend, Mr. Wilson, and took a box 
of matches, from which he counted out thirty. 

" Here are thirty matches," he said. " I 



MAGIC SQUARE PROBLEMS, 



119 



divide them into three unequal heaps. Let me 
see. We have 14, 11, and 5, as it happens. 
Now, the two players draw alternately any 
number from any one heap, and he who draws 
the last match loses the game. That's all ! I 
will play with you, Wilson. I have formed 
the heaps, so you have the first draw." 

" As I can draw any number," Mr. Wilson 
said, " suppose I exhibit my usual moderation 
and take all the 14 heap." 

" That is the worst you could do, for it loses 
right away. I take 6 from the 11, leaving two 
equal heaps of 5, and to leave two equal heaps 
is a certain win (with the single exception of 
I, i), because whatever you do in one heap I 
can repeat in the other. If you leave 4 in one 
heap, I leave 4 in the other. If you then leave 
2 in one heap, I leave 2 in the other. If you 
leave only i in one heap, then I take aU the 
other heap. If you take all one heap, I take 
all but one in the other. No, you must never 
leave two heaps, tmless they are equal heaps 
and more than i, i. Let's begin again." 

"Very well, then," said Mr. Wilson. " I will 
take 6 from the 14, and leave you 8, 11, 5." 

Mr. Stubbs then left 8, 11, 3 ; Mr. Wilson, 
8, 5, 3 ; Mr. Stubbs, 6, 5, 3 ; Mr. Wilson, 4, 5,3 ; 
Mr. Stubbs, 4, 5, 1 ; Mr. Wilson, 4, 3, i ; Mr. 
Stubbs, 2, 3, I ; Mr. Wilson, 2, i, i ; which Mr. 
Stubbs reduced to i, i, i. 

" It is now quite clear that I must win," said 
Mr. Stubbs, because you must take i, and then 
I take I, leaving you the last match. You never 
had a chance. There are just thirteen difEerent 
ways in which the matches may be grouped at 
the start for a certain win. In fact, the groups 
selected, 14, 11, 5, are a certain win, because 
for whatever your opponent may play there is 
another winning group you can secure, and so 
on and on down to the last match." 

397.— THE MONTENEGRIN DICE GAME. 
It is said that the inhabitants of Montenegro 



have a little dice game that is both ingenious 
and well worth investigation. The two players 
first select two different pairs of odd numbers 
(always higher than 3) and then alternately 
toss three dice. Whichever first throws the 
dice so that they add up to one of his selected 
numbers wins. If they are both successful in 
two successive throws it is a draw and they try 
again. For example, one player may select 
7 and 15 and the other 5 and 13. Then if the 
first player throws so that the three dice add 
up 7 or 15 he wins, unless the second man gets 
either 5 or 13 on his throw. 

The puzzle is to discover which two pairs 
of numbers should be selected in order to give 
both players an exactly even chance. 

398.— THE CIGAR PUZZLE. 

I ONCE propounded the following puzzle in a 
London club, and for a considerable period it 
absorbed the attention of the members. They 
could make nothing of it, and considered it 
quite impossible of solution. And yet, as I 
shall show, the answer is remarkably simple. 

Two men are seated at a square-topped table. 
One places an ordinary cigar (flat at one end, 
pointed at the other) on the table, then the 
other does the same, and so on alternately, a 
condition being that no cigar shall touch an- 
other. Which player should succeed in placing 
the last cigar, assuming that they each will 
play in the best possible manner ? The size of 
the table top and the size of the cigar are not 
given, but in order to exclude the ridiculous 
answer that the table might be so diminutive 
as only to take one cigar, we wiU say that the 
table must not be less than 2 feet square and 
the cigar not more than 4I inches long. With 
those restrictions you may take any dimen- 
sions you like. Of course we assume that all 
the cigars are exactly alike in every respect. 
Should the first player, or the second player, 
win ? 



MAGIC SQUARE PROBLEMS. 



" By magic numbers." 
CoNGREVE, The Mourning Bride. 

This is a very ancient branch of mathematical 
puzzledom, and it has an immense, though 
scattered, literature of its own. In their simple 
form of consecutive whole numbers arranged in 
a square so that every column, every row, and 
each of the two long diagonals shall add up 
alike, these magic squares offer three main 
lines of investigation : Construction, Enumera- 
tion, and Classification. Of recent years many 
ingenious methods have been devised for the 
construction of magics, and the law of their for- 
mation is so well understood that all the ancient 
mystery has evaporated and there is no longer 
any difficulty in making squares of any dimen- 
sions. Almost the last word has been said on 
this subject. 
The question of the enumeration of all the 



possible squares of a given order stands just 
where it did over two hundred years ago. 
Everybody knows that there is only one solution 
for the third order, three cells by three; and 
Fr6nicle published in 1693 diagrams of all the 
arrangements of the fourth order — 880 in 
number — and his results have been verified 
over and over again. I may here refer to the 
general solution for this order, for numbers not 
necessarily consecutive, by E. Bergholt in 
Nature, May 26, 1910, as it is of the greatest 
importance to students of this subject. The 
enumeration of the examples of any higher 
order is a completely unsolved problem. 

As to classification, it is largely a matter of 
individual taste — perhaps an aesthetic question, 
for there is beauty in the law and order of 
numbers. A man once said that he divided the 
human race into two great classes : those who 
take snufi and those who do not. I am not 



I20 



AMUSEMENTS IN MATHEMATICS. 



sure that some of our classifications of magic 
squares are not almost as valueless. However, 
lovers of these things seem somewhat agreed 
that Nasik magic squares (so named by Mr. 



SIMPLE. 



SEMI-NASlk 



1 


12 


14 


7 


4 


15 


9 


6 


13 


2 


8 


II 


16 


5 


5 


lo 



1 


14 


\2. 


7 


4 


15 


9 


6 


13 


Z 


8 


'1 


16 


3 


S 


(O 



Frost, a student of them, after the town in 
India where he lived, and also called Dia- 
bolique and Pandiagonal) and Associated magic 
squares are of special interest, so I will just 



TYP 


1 1- 




> 


< 


p 




\^ 


^ 


Y 




^ 


^ 


V 




^ 

•^ 


< 


* 



TYPE JL 


«'V .^V 


^^A^ 


•VV^ 



19 1 o, an article that would enable the reader to 
write out, if he so desired, all the 880 magics of 
the fourth order, and the following is the com- 
plete classification that I gave. The first ex- 



ASSOCIATED 



NASIK 



1 


14 


12 


7 


8 


II 


13 


Z 


IS 


4 


6 


9 


10 


5 


3 


t6 



1 


14 


7 


IZ 


IS 


4 


9 


6 


10 


S 


16 


3 


S 


M 


2 


13 



ample is that of a Simple square that fulfils the 
simple conditions and no more. The second 
example is a Semi-Nasik, which has the addi- 
tional property that the opposite short diag- 



TYPE. HL 




Tvpi: w 



f 



i 



f 



TYP 


E. V 


r 


/• 


f 


# 


h 


K 


K 


h 


H 


M 


H 


1 


J 


•/ 


•/ 


J 



TYPE "Srr TYP6. "VIL TYPE, "^on 




1 



i 



TYPE. 




TYPE X 



explain what these are for the benefit of the 
novice. 

I published in The Queen for January 15, 



% 


PYP 


& 30: 


# 


«k 


^1- 


^ 







^ 


^ 


9 


«s 


^ 


^ 


& 


«^ 


^ 


^ 





rypi xn 




/* 


\ 


/» 




K 


V 


^ 




I 


•^ 


r 


J 


y 


N. 



onals of two cells each together sum to 34. 
Thus, 14 + 4 4- II 4- 5 = 34 and 12 + 6 4- 13 -|-3=34- 
The third example is not only Semi-Nasik but 



MAGIC SQUARE PROBLEMS. 



121 



also Associated, because in it every number, if 
added to the number that is equidistant, in a 
straight line, from the centre gives 17. Thus, 
1+ 16, 2 +15, 3 +14, etc. The fourth example, 
considered the most " perfect " of all, is a 
Nasik. Here aU the broken diagonals sum to 
34. Thus, for example, 15 +14 + 2 + 3, and 
10 + 4 + 7 + 13, and 15 + 5 + 2 + 12. As a con- 
sequence, its properties are such that if you 
repeat the square in all directions you may 
mark off a square, 4X4, wherever you please, 
and it will be magic. 

The following table not only gives a complete 
enumeration under the four forms described, 
but also a classification under the twelve graphic 
types indicated in the diagrams. The dots at 
the end of each line represent the relative posi- 
tions of those complementary pairs, i + 16, 
2 + 15, etc., which sum to 17. For example, it 
will be seen that the first and second magic 
squares given are of Type VI., that the third 
square is of Type HI., and that the fourth is of 
Type I. fidouard Lucas indicated these t5rpes, 
but he dropped exactly half of them and did not 
attempt the classification. 

Nasik (Type I.) 48 

Semi-Nasik (T3rpe II., Transposi- 
tions of Nasik) . 48 
„ (Type III., Associated) 48 

(Type IV.) ... 96 
„ (Type v.). ... 96 192 



Simple. - 



(Type VI.) 

(Type VI.) 
(Type VII.) 
(Type VIII.) 
(Type IX.) 
(Type X.) 

(Type XI.) 
(Type XII.) 



96 384 
208 



56 

56 
56 
56 224 



16 448 
880 



It is hardly necessary to say that every one 
of these squares will produce seven others by 
mere reversals and reflections, which we do not 
count as different. So that there are 7,040 
squares of this order, 880 of which are funda- 
mentally different. 

An infinite variety of puzzles may be made 
introducing new conditions into the magic 
square. In The Canterbury Puzzles I have 
given examples of such squares with coins, 
with postage stamps, with cutting-out condi- 
tions, and other tricks. I wiU now give a few 
variants involving further novel conditions. 

399.— THE TROUBLESOME EIGHT. 

Nearly everybody knows that a " magic 
square " is an arrangement of numbers in the 
form of a square so that every row, every 
column, and each of the two long diagonals 
adds up alike. For example, you would find 
little difficulty in merely placing a different 



number in each of the nine cells in the illustra- 
tion so that the rows, columns, and diagonals 
shall all add up 15. And at your first attempt 
you will probably find that you have an 8 in 





8 

















one of the corners. The puzzle is to construct 
the magic square, under the same conditions, 
with the 8 in the position shown. 



400.— THE 


MAGIC SI 


-RlPi 


3. 




2 


3 


4 


5 


6 


7 






2 


3 


4 


5 


6 


/ 






2 


3 


4 


5 


6 


7 


. 




2 


3 


4 


5 


6 


7 






2 


3 


4 


5 


6 


r 






2 


3 


4 


5 


6 


7 


' 




2 


3 


4 


5 


6 


r 



I HAPPENED to have lying on my table a number 
of strips of cardboard, with numbers printed on 
them from i upwards in numerical order. The 
idea suddenly came to me, as ideas have a way 
of unexpectedly coming, to make a little puzzle 
of this. I wonder whether many readers will 
arrive at the same solution that I did. 

Take seven strips of cardboard and lay them 
together as above. Then write on each of them 
the numbers i, 2, 3, 4, 5, 6, 7, as shown, so that 
the numbers shaU form seven rows and seven 
columns. 

Now, the puzzle is to cut these strips into 
the fewest possible pieces so that they may be 
placed together and form a magic square, the 
seven rows, seven columns, and two diagonals 
adding up the same number. No figures may 



122 



AMUSEMENTS IN MATHEMATICS. 



be turned upside down or placed on their sides — 
that is, all the strips must lie in their original 
direction. 

Of course you could cut each strip into seven 
separate pieces, each piece containing a number, 
and the puzzle would then be very easy, but I 
need hardly say that forty-nine pieces is a long 
way from being the fewest possible. 

401.— EIGHT JOLLY GAOL BIRDS. 




The illustration shows the plan of a prison of 
nine cells all communicating with one another 
by doorways. The eight prisoners have their 
numbers on their backs, and any one of them 
is allowed to exercise himself in whichever cell 
may happen to be vacant, subject to the rule 
that at no time shall two prisoners be in the 
same cell. The merry monarch in whose do- 
minions the prison was situated offered them 
special comforts one Christmas Eve if, without 
breaking that rule, they could so place them- 
selves that their numbers should form a magic 
square. 

Now, prisoner No. 7 happened to know a good 
deal about magic squares, so he worked out a 
scheme and naturally selected the method that 
was most expeditious — that is, one involving 
the fewest possible moves from cell to cell. 
But one man was a surly, obstinate fellow (quite 
tmfit for the society of his jovial companions), 
and he refused to move out of his cell or take 
any part in the proceedings. But No. 7 was 
quite equal to the emergency, and found that 
he could still do what was required in the 
fewest possible moves without troubling the 
brute to leave his cell. The puzzle is to show 
how he did it and, incidentally, to discover 
which prisoner was so stupidly obstinate. Can 
you find the fellow ? 

402.— NINE JOLLY GAOL BIRDS. 

Shortly after the episode recorded in the last 
puzzle occurred, a ninth prisoner was placed in 




the vacant cell, and the merry monarch then 
offered them aU complete liberty on the follow- 
ing strange conditions. They were required so 
to rearrange themselves in the ceUs that their 
nmnbers formed a magic square without their 
movements causing any two of them ever to 
be in the same ceU together, except that at the 
start one man was allowed to be placed on the 
shoulders of another man, and thus add their 
numbers together, and move as one man. For 
example, No. 8 might be placed on the shoul- 
ders of No. 2, and then they would move about 
together as 10. The reader should seek first to 
solve the puzzle in the fewest possible moves, 
and then see that the man who is burdened has 
the least possible amoimt of work to do. 

403.— THE SPANISH DUNGEON. 

Not fifty miles from Cadiz stood in the middle 
ages a castle, all traces of which have for cen- 
turies disappeared. Among other interesting 
features, this castle contained a particularly 
unpleasant dimgeon divided into sixteen cells, 
aU communicating with one another, as shown 
in the illustration. 

Now, the governor was a merry wight, and 
very fond of puzzles withal. One day he went 
to the dungeon and said to the prisoners, " By 
my halidame ! " (or its equivalent in Spanish) 
" you shall aU be set free if you can solve this 
puzzle. You must so arrange yourselves in the 
sixteen cells that the numbers on your backs 
shall form a magic square in which every 
column, every row, and each of the two dia- 
gonals shall add up the same. Only remember 
this : that in no case may two of you ever be 
together in the same cell." 

One of the prisoners, after working at the 
problem for two or three days, with a piece of 
chalk, imdertook to obtain the liberty of him- 
self and his fellow-prisoners if they would follow 
his directions and move through the doorways 



MAGIC SQUARE PROBLEMS. 



123 



from cell to cell in the order in which he should 
call out their numbers. 

He succeeded in his attempt, and, what is 



mv^p ^^^^^^^^^V ^^^^^^^^^^ ^^^^^^^^W ^F*^ 



more remarkable, it would seem from the 
account of his method recorded in the ancient 
manuscript lying before me, that he did so in 
the fewest possible moves. The reader is asked 
to show what these moves were. 

404.— THE SIBERIAN DUNGEONS. 

L _J I 

21 22 23 ' 2^ 



The above is a trustworthy plan of a certain 
Russian prison in Siberia. All the cells are 
numbered, and the prisoners are numbered the 
same as the ceUs they occupy. The prison diet 
is so fattening that these political prisoners are 
in perpetual fear lest, should their pardon arrive, 
they might not be able to squeeze themselves 
through the narrow doorways and get out. And 
of course it would be an unreasonable thing to 
ask any government to pull down the walls of 
a prison just to liberate the prisoners, however 
innocent they might be. Therefore these men 
take all the healthy exercise they can in order 
to retard their increasing obesity, and one of 
their recreations will serve to furnish us with 
the following puzzle. 

Show, in the fewest possible moves, how the 
sixteen men may form themselves into a magic 
square, so that the numbers on their backs shall 



add up the same in each of the four columns, 
four rows, and two diagonals without two pris- 
oners having been at any time in the same cell 
together. I had better say, for the information 
of those who have not yet been made acquainted 
with these places, that it is a peculiarity of 
prisons that you are not allowed to go outside 
their walls. Any prisoner may go any distance 
that is possible in a single move. 

405---CARD MAGIC SQUARES. 



0^0 




■' ■ ' " --N 




<► ♦ 




4k 




« ♦ 


<> 








4r « 


^ ^ 








-^ 





r^ ^ 




\ 


r ■-> 


C? <5? 





0^0 


Q? 


a 




00 






Take an ordinary pack of cards and throw out 
the twelve court cards. Now, with nine of the 
remainder (different suits are of no consequence) 
form the above magic square. It will be seen 
that the pips add up fifteen in every row in every 
column, and in each of the two long diagonals. 
The puzzle is with the remaining cards (without 
disturbing this arrangement) to form three more 
such magic squares, so that each of the four 
shall add up to a different sum. There will, of 
course, be four cards in the reduced pack that 
wiU not be used. These four may be any that 
you choose. It is not a difficult puzzle, but 
requires just a little thought. 

406.— THE EIGHTEEN DOMINOES. 

The illustration shows eighteen dominoes ar- 
ranged in the form of a square so that the pips 
in every one of the six columns, six ro\\^, and 
two long diagonals add up 13. This is the 
smallest summation possible with any selection 
of dominoes from an ordinary box of twenty- 
eight. The greatest possible summation is 23, 
and a solution for this number may be easily 
obtained by substituting for every number its 
complement to 6. Thus for every blank sub- 
stitute a 6, for every i a 5, for every 2 a 4, for 



124 



AMUSEMENTS IN MATHEMATICS. 



3 a 3, for 4 a 2, for 5 a i, and for 6 a blank. 
But the puzzle is to make a selection of eighteen 
dominoes and arrange them (in exactly the form 






9 9 

• 



9 



d e 



m 



9 • 



=1r 



• • 



« • 






shown) so that the summations shall be i8 in 
all the fourteen directions mentioned. 



SUBTRACTING, MULTIPLYING, 
AND DIVIDING MAGICS. 

Although the adding magic square is of such 
great antiquity, curiously enough the multiply- 
ing magic does not appear to have been men- 
tioned until the end of the eighteenth century, 
when it was referred to slightly by one writer 
and then forgotten until I revived it in Tit-Bits 
in 1897. The dividing magic was apparently 
first discussed by me in The Weekly Dispatch in 
June 1898. The subtracting magic is here in- 
troduced for the first time. It will now be 
convenient to deal with all four kinds of magic 
squares together. 



AODIN Gl 



svBTR/Kcrma 



second, and the result from the third. You 
can, of course, perform the operation in either 
direction ; but, in order to avoid negative num- 
bers, it is more convenient simply to deduct 
the middle number from the sum of the two 
extreme numbers. This is, in effect, the same 
thing. It will be seen that the constant of the 
adding square is n times that of the subtracting 
square derived from it, where n is the number 
of cells in the side of square. And the manner 
of derivation here is simply to reverse the two 
diagonals. Both squares are " associated " — 
a term I have explained in the introductory 
article to this department. 

The third square is a multiplying magic. 
The constant, 216, is obtained by miiltiplying 
together the three numbers in any line. It is 
" associated " by multiplication, instead of by 
addition. It is here necessary to remark that 
in an adding square it is not essential that the 
nine numbers should be consecutive. Write 
down any nine numbers in this way — 



I 3 
4 6 
7 9 



5 

8 

II 



so that the horizontal differences are all alike 
and the vertical differences also alike (here 
2 and 3), and these numbers will form an 
adding magic square. By making the differ- 
ences I and 3 we, of course, get consecutive 
numbers — a particular case, and nothing more. 
Now, in the case of the multiplying square we 
must take these numbers in geometrical instead 
of arithmetical progression, thus — 



I 


3 


9 


2 


6 


18 


4 


12 


36 



Here each successive number in the rows is 
multiplied by 3, and in the columns by 2. 
Had we multiplied by 2 and 8 we should 
get the regular geometrical progression, i, 2, 4, 
8, 16, 32, 64, 128, and 256, but I wish to avoid 
high numbers. The numbers are arranged in 
the square in the same order as in the adding 
square. 



MULTIPLY IN(5: 



DIVIDING. 



8 


1 


6 




2 


1 


4 




la 


1 


is 




3 


1 


2, 


3 


5 


7 


3 


5 


7 


9 


6 


4 


9 


6 


,4 


^ 


9 


2 


6 


9 


8 


2 


36 


3' 


IS 


36 


12 



In these four diagrams we have examples 
in the third order of adding, subtracting, multi- 
plying, and dividing squares. In the first the 
constant, 15, is obtained by the addition of the 
rows, columns, and two diagonals. In the 
second case you get the constant, 5, by sub- 
tracting the first number in a line from the 



The fourth diagram is a dividing magic 
square. The constant 6 is here obtained by 
dividing the second number in a line by the 
first (in either direction) and the third number 
by the quotient. But, again, the process is 
simplified by dividing the product of the two 
extreme numbers by the middle number. This 



MAGIC SQUARE PROBLEMS. 



125 



square is also " associated " by multiplication. 
It is derived from the multiplying square by 
merely reversing the diagonals, and the constant 
of the multiplying square is the cube of that of 
the dividing square derived from it. 

The next set of diagrams shows the solutions 
for the fifth order of square. They are all " asso- 
ciated " in the same way as before. The sub- 
tracting square is derived from the adding 
square by reversing the diagonals and exchang- 
ing opposite numbers in the centres of the 
borders, and the constant of one is again n 
times that of the other. The dividing square 
is derived from the multiplying square in the 
same way, and the constant of the latter is the 
5th power (that is the nth) of that of the former. 





ApOlNCw 






SOSTI^ACTlNa 


17 


24 


1 


8 


IS 




9 


24 


ZS 


S 


n 


23> 


-5 


7 


1 + 


;6 


2i 


Zi 


1 


12 


}6 


4 


6 


J3 


20 


22 


22 


6 


)3 


20 


^ 


10 


12 


19 


Zi 


3 


10 


14- 


/9 


5 


3 


/i 


IS 


ZS 


Z 


9 


15 


18 


1 


Z 


'7 



NIOLTIPLVlNt^ 



CXVtOINCc 



54 


648 


1 


»Z 


144 




24 


C^S 


ngd 


/2 


9 


524 


t6 


6 


7i 


ZJ 


3^4 


81 


6 


IS 


27 


S 


3 


36 


^32 


t6l 


ihZ 


3 


36 


4^2 


5 


4-S 


IS 


2/6 


Si 


4- 


4-S 


7^ 


2/6 


IG 


4- 


9 


I OS 


1296 


Z 


24 


144 


108 


1 


2 


5-4 



These squares are thus quite easy for odd 
orders. But the reader will probably find some 
difficulty over the even orders, concerning which 
I will leave him to make his own researches, 
merely propounding two little problems. 

407.— TWO NEW MAGIC SQUARES. 

Construct a subtracting magic square with 
the first sixteen whole numbers that shall be 
" associated " by subtraction. The constant 
is, of course, obtained by subtracting the first 
number from the second in line, the result from 
the third, and the result again from the fourth. 
Also construct a dividing magic square of the 
same order that shall be " associated" by divi- 
sion. The constant is obtained by dividing the 
second number in a line by the first, the third 
by the quotient, and the fourth by the next 
quotient. 

408.— MAGIC SQUARES OF TWO 
DEGREES. 

While reading a French mathematical work I 
happened to come across the following state- 



ment : "A very remarkable magic square of 8, 
in two degrees, has been constructed by M. 
PfeSermann. In other words, he has managed 
to dispose the sixty-four fijst numbers on the 
squares of a chessboard in such a way that the 
sum of the numbers in every line, every column, 
and in each of the two diagonals, shall be the 
same ; and more, that if one substitutes for aU 
the numbers their squares, the square still re- 
mains magic." I at once set to work to solve 
this problem, and, although it proved a very 
hard nut, one was rewarded by the discovery 
of some curious and beautiful laws that govern 
it. The reader may like to try his hand at the 
puzzle. 

MAGIC SQUARES OF PRIMES. 

The problem of constructing magic squares 
with prime numbers only was first discussed by 
myself in The Weekly Dispatch for 22nd July and 
5 th August 1900 ; but during the last three or 
four years it has received great attention from 
American mathematicians. First, they have 
sought to form these squares with the lowest 
possible constants. Thus, the first nine prime 
numbers, i to 23 inclusive, sum to 99, which 
(being divisible by 3) is theoretically a suitable 
series ; yet it has been demonstrated that the 
lowest possible constant is in, and the re- 
quired series as follows : i, 7, 13, 31, 37, 43, 61, 
67, and 73. Similarly, in the case of the fourth 
order, the lowest series of primes that are 
" theoretically suitable " will not serve. But 
ia every other order, up to the 12th inclusive, 
magic squares have been constructed with the 
lowest series of primes theoretically possible. 
And the 12 th is the lowest order in which a 
straight series of prime numbers, unbroken, 
from I upwards has been made to work. In 
other words, the first 144 odd prime numbers 
have actually been arranged ia magic form. 
The following summary is taken from The 
Monist (Chicago) for October 191 3 : — 



Order of 


Totals of 


Lowest 


Squares 


Square. 


Series. 


Constants. 


made by — 
fHenry E. 


3rd 


333 


in 


Dudeney 

(1900). 
/- Ernest Berg- 
J holt and C 


4th 


408 


102 


" D. Shuld- 
ham. 


5th 


1065 


213 


H. A. Sayles 
(C. D. Shuld- 


6th 


2448 


408 


ham and J 
N. Muncey 


7th 


4893 


699 


do. 


8th 


8912 


III4 


do. 


9th 


15129 


1681 


do. 


loth 


24160 


2416 


J. N. Muncey 


nth 


36095 


3355 


do. 


i2th 


54168 


4514 


do. 



For further details the reader should consult 
the article itself, by W. S. Andrews and H. A. 
Sayles. 



126 



AMUSEMENTS IN MATHEMATICS. 




These same investigators have also performed 
notable feats in constructing associated and 
bordered prime magics, and Mr. Shuldham has 
sent me a remarkable paper in which he gives 
examples of Nasik squares constructed with 
primes for all orders from the 4th to the loth, 
with the exception of the 3rd (which is clearly 
impossible) and the 9th, which, up to the time 
of writing, has baffled all attempts. 

409.— THE BASKETS OF PLUMS. 




This is the form in which I first introduced the 



question of magic squares with prime numbers. 
I will here warn the reader that there is a little 
trap. 

A fruit merchant had nine baskets. Every 
basket contained plums (all sound and ripe), 
and the number in every basket was different. 
When placed as shown in the illustration they 
formed a magic square, so that if he took any 
three baskets in a line in the eight possible 
directions there would always be the same 
number of plums. This part of the puzzle is 
easy enough to understand. But what follows 
seems at first sight a little queer. 

The merchant told one of his men to distribute 
the contents of any basket he chose among some 
children, giving plums to every child so that 
each should receive an equal number. But the 
man found it quite impossible, no matter which 
basket he selected and no matter how many 
children he included in the treat. Show, by 
giving contents of the nine baskets, how this 
could come about. 

410.— THE MANDARIN'S "T" PUZZLE. 

Before Mr. Beauchamp Cholmondely Marjori- 
banks set out on his tour in the Far East, he 
prided himself on his knowledge of magic 
squares, a subject that he had made his special 
hobby; but he soon discovered that he had 
never really touched more than the fringe of 
the subject, and that the wily Chinee could 



MAZES AND HOW TO THREAD THEM. 



127 



beat him easily. I present a little problem 
that one learned mandarin propounded to our 
traveller, as depicted on the last page. 

The Chinaman, after remarking that the con- 
struction of the ordinary magic square of 
twenty-five cells is " too velly muchee easy," 
asked our countryman so to place the numbers 
I to 25 in the square that every column, every 
row, and each of the two diagonals should add 
up 65, with only prime numbers on the shaded 
" T." Of course the prime numbers available 
are i, 2, 3, 5, 7, n, 13, i7, 19, and 23, so you 
are at liberty to select any nine of these that 
will serve your purpose. Can you construct 
this curious little magic square ? 

411.— A MAGIC SQUARE OF 
COMPOSITES. 

As we have just discussed the construction of 



magic squares with prime numbers, the follow- 
ing forms an interesting companion problem. 
Make a magic square with nine consecutive 
composite numbers — the smallest possible. 

412.— THE MAGIC KNIGHT'S TOUR. 

Here is a problem that has never yet been 
solved, nor has its impossibility been demon- 
strated. Play the knight once to every square 
of the chessboard in a complete tour, numbering 
the squares in the order visited, so that when 
completed the square shall be " magic," adding 
up to 260 in every column, every row, and each 
of the two long diagonals. I shall give the 
best answer that I have been able to obtain, in 
which there is a slight error in the diagonals 
alone. Can a perfect solution be foimd ? I 
am convinced that it cannot, but it is ordy a 
" pious opinion." 



MAZES AND HOW TO THREAD THEM. 



*• In wandering mazes lost." 

Paradise Lost. 

The Old English word " maze," signifjdng a 
labyrinth, probably comes from the Scandi- 
navian, but its origin is somewhat uncertain. 
The late Professor Skeat thought that the sub- 
stantive was derived from the verb, and as in 
old times to be mazed or amazed was to be 
" lost in thought," the transition to a maze in 
whose tortuous windings we are lost is natural 
and easy. 

The word " labyrinth " is derived from a 
Greek word signifying the passages of a mine. 
The ancient mines of Greece and elsewhere 
inspired fear and awe on account of their dark- 
ness and the danger of getting lost in their in- 
tricate passages. Legend was afterwards built 
round these mazes. The most familiar instance 
is the labyrinth made by Daedalus in Crete for 
King Minos. In the centre was placed the 
Minotaur, and no one who entered could find 
his way out again, but became the prey of the 
monster. Seven youths and seven maidens 
were sent regularly by the Athenians, and were 
duly devoured, until Theseus slew the monster 
and escaped from the maze by aid of the clue 
of thread provided by Ariadne ; which accounts 
for our using to-day the expression " threading 
a maze." 

The various forms of construction of mazes 
include complicated ranges of caverns, archi- 
tectural labyrinths, or sepulchral buildings, tor- 
tuous devices indicated by coloured marbles and 
tiled pavements, winding paths cut in the turf, 
and topiary mazes formed by clipped hedges. 
As a matter of fact, they may be said to have 
descended to us in precisely this order of variety. 

Mazes were used as ornaments on the state 
robes of Christian emperors before the ninth 
century, and were soon adopted in the decora- 
tion of cathedrals and other chiurches. The 
original idea was doubtless to employ them as 



symbols of the complicated folds of sin by which 
man is surrounded. They began to abound in 
the early part of the twelfth century, and I give 
an illustration of one of this period in the parish 
church at St. Quentin (Fig. i). It formed a pave- 




FiG. I. — Maze at St. Quentin. 

ment of the nave, and its diameter is 34^ feet. 
The path here is the line itself. If you place 
your pencil at the point A and ignore the en- 
closing hne, the line leads you to the centre by 
a long route over the entire area ; but you 
never have any option as to direction during 
your course. As we shall find in similar cases, 
these early ecclesiastical mazes were generally 
not of a puzzle nature, but simply long, winding 
paths that took you over practically all the 
greund enclosed. 



128 



AMUSEMENTS IN MATHEMATICS. 



In the abbey church of St. Bertin, at St. 
Omer, is another of these curious floors, repre- 
senting the Temple of Jerusalem, with stations 
for pilgrims. These mazes were actually visited 




Fig. 2. — Maze in Chartres Cathedral. 

and traversed by them as a compromise 
for not going to the Holy Land in fulfilment 
of a vow. They were also used as a means of 
penance, the penitent frequently being directed 
to go the whole course of the maze on hands 
and knees. 

The maze in Chartres Cathedral, of which I 
give an illustration (Fig. 2), is 40 feet across, 
and was used by penitents following the pro- 




FiG. 3. — Maze in Lucca Cathedral. 

cession of Calvary. A labyrinth in Amiens 
Cathedral was octagonal, similar to that at St. 



Quentin, measuring 42 feet across. It bore the 
date 1288, but was destroyed in 1708. In the 
chapter-house at Bayeux is a laibyrmth formed 
of tiles, red, black, and encaustic, with a pattern 
of brown and yellow. Dr. Ducarel, in his. Tour 
through Part of Normandy " (printed in 1767), 
mentions the floor of the great guard-chamber 
in the abbey of St. Stephen, at Caen, '* the 
middle whereof represents a maze or labyrinth 
about 10 feet diameter, and so artfully contrived 
that, were we to suppose a man following all 
the intricate meanders of its volutes, he could 
not travel less than a mile before he got from 
one end to the other." 

Then these mazes were sometimes reduced 
in size and represented on a single tile (Fig. 3). 
I give an example from Lucca Cathedral. It 
is on one of the porch piers, and is 19 J inches in 
diameter. A writer in 1858 says that, " from 
the continual attrition it has received from 
thousands of tracing fingers, a central group of 
Theseus and the Minotaur has now been very 




Fig. 4. — Maze at Safiron Walden, Essex. 

nearly effaced." Other examples were, and 
perhaps still are, to be found in the Abbey of 
Toussarts, at Chalons-siur-Mame, in the very 
ancient church of St. Michele at Pavia, at Aix 
in Provence, in the cathedrals of Poitiers, 
Rheims, and Arras, in the church of Santa 
Maria in Aquiro in Rome, in San Vitale at 
Ravenna, in the Roman mosaic pavement foimd 
at Salzburg, and elsewhere. These mazes were 
sometimes called " Chemins de Jerusalem," as 
being emblematical of the dififtculties attending 
a journey to the earthly Jerusalem and of those 
encountered by the Christian before he can 
reach the heavenly Jerusalem — where the centre 
was frequently called " Ciel." 

Common as these mazes were upon the Con- 
tinent, it is probable that no example is to be 
found in any English chiurch ; at least I am not 
aware of the existence of any. But almost 
every county has, or has had, its specimens of 



MAZES AND HOW TO THREAD THEM. 



129 



mazes cut in the turf. Though these are fre- 
quently known as " miz-mazes " or " mize- 
mazes," it is not uncommon to find them locally 
called " Troy-towns," " shepherds' races," or 
" Julian's Bowers " — ^names that are misleading, 
as suggesting a false origin. From the facts 
alone that many of these English turf mazes are 
clearly copied from those in the Continental 
churches, and practically all are found close to 
some ecclesiastical building or near the site of 
an ancient one, we may regard it as certain that 
they were of church origin and not invented by 
the shepherds or other rustics. And curiously 
enough, these turf mazes are apparently un- 




to give illustrations. I shall therefore write of 
them all in the past tense, retaining the hope 
that some are still preserved. 



Fig. 5. — Maze at Sneinton, Nottinghamshire. 

known on the Continent. They are distinctly 
mentioned by Shakespeare : — 

" The nine men's morris is filled up with mud. 
And the quaint mazes in the wanton green 
For lack of tread are undistinguishable." 

A Midsummer Night's Dream, ii. i. 

" My old bones ache : here's a maze trod indeed, 
Through forth-rights and meanders ! " 

The Tempest, iii. 3. 

There was such a maze at Comberton, in 
Cambridgeshire, and another, locally called the 
" miz-maze," at Leigh, in Dorset. The latter 
was on the highest part of a field on the top of 
a hill, a quarter of a mile from the village, and 
was slightly hollow in the middle and enclosed by 
a bank about 3 feet high. It was circular, and 
was thirty paces in diameter. In 1868 the turf 
had grown over the little trenches, and it was 
then impossible to trace the paths of the maze. 
The Comberton one was at the same date be- 
lieved to be perfect, but whether either or both 
have now disappeared I cannot say. Nor have 
I been able to verify the existence or non-exist- 
ence of the other examples of which I am able 
(1,926) 




Fig. 6. — Maze at Alkborough, Lincolnshire. 

In the next two mazes given — that at Saffron 

Walden, Essex (no feet in diameter, Fig. 4), 

and the one near St. Anne's Well, at Sneinton, 

Nottinghamshire (Fig. 5), which was ploughed 

up on February 27th, 1797 (51 feet in diameter, 

. with a path 535 yards long) — the paths must in 

j each case be understood to he on the lines, black 

or white, as the case may be. 

I give in Fig. 6 a maze that was at Alk- 




FiG. 7. — Maze at Boughton Green, 
Nottinghamshire. 

borough, Lincolnshire, overlooking the Humber. 
This was 44 feet in diameter, and the resem- 



I30 



AMUSEMENTS IN MATHEMATICS. 



blance between it and the mazes at Chartre-s and 
Lucca (Figs. 2 and 3) will be at once perceived. 




line itself from end to end. This maze 
was 86 feet square, cut in the turf, and was 
locally known as the " Mize-maze." It be- 
came very indistinct about 1858, and was then 
recut by the Warden of Winchester, with the 
aid of a plan possessed by a lady living in the 
neighbourhood. 



Fig. 8. — Maze at Wing, Rutlandshire. 

A maze at Boughton Green, in Nottingham- 
shire, a place celebrated at one time for its fair 
(Fig. 7), was 37 feet in diameter. I also include 
the plan (Fig. 8) of one that used to be on the 
outskirts of the village of Wing, near Upping- 
ham, Rutlandshire. This maze was 40 feet in 
diameter. 

The maze that was on St. Catherine's Hill, 
Winchester, in the parish of Chilcombe, was a 
poor specimen (Fig. 9), since, as will be seen, 





Fig. 10. — Maze on Ripon Common. 

A maze formerly existed on Ripon Common, 
in Yorkshire (Fig. 10). It was ploughed up in 
1827, but its plan was fortimately preserved. 
This example was 20 yards in diameter, and its 
path is said to have been 407 yards long. 



Fig. 9. — Maze on St, Catherine's Hill, 
Winchester. 

there was one short direct route to the centre, 
unless, as in Fig. 10 again, the path is the 



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n 



Fig. II. — Maze at Theobalds, Hertfordshire. 

In the case of the maze at Theobalds, Hert- 
fordshire, after you have found the entrance 
within the four enclosing hedges, the path is 



MAZES AND HOW TO THREAD THEM. 



131 





1 
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Fig. 12. — Italian Maze of Sixteenth Century. 

forced (Fig. 11). As further illustrations of 
this class of maze, I give one taken from an 
Italian work on architecture by Serlio, pub- 




FiG. 13. — By the Designers of Hampton 
Court Maze. 

lished in 1537 (Fig. 12), and one by London and 
Wise, the designers of the Hampton Court maze, 
from their book, The Retired Gard'ner, published 



in 1706 (Fig. 13). Also, I add a Dutch maze 
(Fig. 14). 

So far our mazes have been of historical 
interest, but they have presented no difficulty 
in threading. After the Reformation period 
we find mazes converted into mediums for re- 
creation, and they generally consisted of laby- 
rinthine paths enclosed by thick and carefully 
trimmed hedges. These topiary hedges were 
known to the Romans, with whom the topiarius 
was the ornamental gardener. This type of 
maze has of late years degenerated into the 
seaside " Puzzle Gardens. Teas, sixpence, in- 
cluding admission to the Maze." The Hampton 




Fig. 14. — A Dutch Maze. 

Court Maze, sometimes called the " Wilderness," 
at the royal palace, was designed, as I have said, 
by London and Wise for William III., who had 
a liking for such things (Fig. 15). I have before 
me some three or four versions of it, all slightly 
different from one another ; but the plan I select 
is taken from an old guide-book to the palace, 
and therefore ought to be trustworthy. The 
meaning of the dotted lines, etc., will be ex- 
plained later on. 

The maze at Hatfield House (Fig. 16), the 
seat of the Marquis of Salisbury, like so many 
labyrinths, is not difficult on paper ; but both 




Fig. 15. — Maze at Hampton Court Palace. 



132 



AMUSEMENTS IN MATHEMATICS. 









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Fig. i6. — Maze at Hatfield House, Herts. 



this and the Hampton Court Maze may prove 
very puzzling to actually thread without know- 




FiG. 17. — Maze formerly at South Kensington. 



ing the plan. One reason is that one is so apt 
to go down the same blind alleys over and over 
again, if one proceeds without method. The 
maze planned by the desire of the Prince Con- 
sort for the Royal Horticultural Society's Gar- 
dens at South Kensington was allowed to go to 
ruin, and was then destroyed — ^no great loss, 
for it was a feeble thing. It will be seen that 
there were three entrances from the outside 
(Fig. 17), but the way to the centre is very 
easy to discover. I include a German maze 
that is curious, but not difficult to thread on 
paper (Fig. 18). The example of a labjnrinth 




Fig. 18. — A German Maze. 

formerly existing at Pimpeme, in Dorset, is in a 
class by itself (Fig. 19). It was formed of small 
ridges about a foot high, and covered nearly an 



MAZES AND HOW TO THREAD THEM. 



133 



acre of ground; but it was, unfortunately, 
ploughed up in 1730. 




Fig. 19. — Maze at Pirapeme, Dorset. 

We will now pass to the interesting subject 
of how to thread any maze. While being neces- 
sarily brief, I will try to make the matter clear 
to readers who have no knowledge of mathe- 
matics. And first of all we will assume that we 
are trying to enter a maze (that is, get to the 
" centre ") of which we have no plan and about 
which we know nothing. The first rule is this : 
If a maze has no parts of its hedges detached 



from the rest, then if we always keep in touch 
with the hedge with the right hand (or always 
touch it with the left), going down to the stop 
in every blind alley and coming back on the 
other side, we shall pass through every part of 




Fig. 20. — M. Tremaux's Method of Solution. 

the maze and make our exit where we went in. 
Therefore we must at one time or another enter 
the centre, and every alley will be traversed 
twice. 

Now look at the Hampton Court plan. Fol- 
low, say to the right, the path indicated by the 




Fig. 21. — How to thread the Hatfield Maze. 



134 



AMUSEMENTS IN MATHEMATICS. 



dotted line, and what I have said is clearly 
correct if we obliterate the two detached parts, 
or " islands," situated on each side of the star. 
But as these islands are there, you cannot by 
this method traverse every part of the maze; 
and if it had been so planned that the " centre " 
was, like the star, between the two islands, you 
«rould never pass through the " centre " at all. 
A glance at the Hatfield maze will show that 
there are three of these detached hedges or 
islands at the centre, so this method will never 



This blunder happened to me a few years ago 
in a little maze on the isle of Caldy, South 
Wales. I knew the maze was a small one, but 
after a very long walk I was amazed to find that 
I did not either reach the " centre " or get out 
again. So I threw a piece of paper on the 
ground, and soon came round to it ; from which 
I knew that I had blundered over a supposed 
blind aUey and was going round and round an 
island. Crossing to the opposite hedge and 
using more care, I was quicMy at the centre and 




Fig. 22. — The Philadelphia Maze, and its Solution. 



take you to the " centre " of that one. But 
the rule will at least always bring you safelj' 
out again unless you blunder in the following 
way. Suppose, when you were going in the 
direction of the arrow in the Hampton Court 
Maze, that you could not distinctly see the turn- 
ing at the bottom, that you imagined you were 
in a blind alley and, to save time, crossed at 
once to the opposite hedge, then you would go 
roimd and round that U-shaped island with 
your right hand stiU always on the hedge — for 
ever after 



out again. Now, if I had made a similar mis- 
take at Hampton Court, and discovered the 
error when at the star, I should merely have 
passed from one island to another ! And if I 
had again discovered that I was on a detached 
part, I might with ill luck have recrossed to the 
first island again ! We thus see that this 
" touching the hedge " method should always 
bring us safely out of a maze that we have 
entered ; it may happen to take us through the 
" centre," and if we miss the centre we shall 
know there must be islands. But it has to be 



MAZES AND HOW TO THREAD THEM. 



135 



done with a little care, and in no case can we be 
sure that we have traversed every alley or that 
there are no detached parts. 




Fig. 23. — Simplified Diagram of Fig. 22. 

If the maze has many islands, the traversing 
of the whole of it may be a matter of consider- 
able difficulty. Here is a method for solving 



character that will serve our purpose just as 
well as something more complex (Fig. 20). The 
circles at the regions where we have a choice 
of turnings we may call nodes. A " new " path 
or node is one that has not been entered before 
on the route ; an " old " path or node is one that 
has already been entered, i. No path may be 
traversed more than twice. 2. When you come 
to a new node, take any path you like. 3. When 
by a new path you come to an old node or to the 
stop of a blind alley, return by the path you 
came. 4. When by an old path you come to 
an old node, take a new path if there is one ; if 
not, an old path. The route indicated by the 




Fig. 24. — Can you find the Shortest Way to Centre ? 



any maze, due to M. Tr^maux, but it necessitates 
carefully marking in some way your entrances 
and exits where the galleries fork. I give a dia- 
gi'am of an imaginary maze of a very simple 



dotted line in the diagram is taken in accord- 
ance with these simple rules, and it will be seen 
that it leads us to the centre, although the maze 
consists of four islands. 



136 



AMUSEMENTS IN MATHEMATICS. 




Fig. 25. — Rosamund's Bower. 



Neither of the methods I have given will dis- 
close to us the shortest way to the centre, nor 
the number of the different routes. But we 
can easily settle these points with a plan. Let 
us take the Hatfield maze (Fig. 21). It will be 
seen that I have suppressed all the blind alleys 
by the shading. I begin at the stop and work 
backwards until the path forks. These shaded 
parts, therefore, can never be entered without 
oiur having to retrace our steps. Then it is very 
clearly seen that if we enter at A we must come 



out at B ; if we enter at C we must come out at 
D. Then we have merely to determine whether 
A, B, E, or C, D, E, is the shorter route. As a 
matter of fact, it will be foimd by rough mea- 
surement or calculation that the shortest route 
to the centre is by way of C, D, E, F. 

I wiU now give three mazes that are simply 
puzzles on paper, for, so far as I know, they 
have never been constructed in any other way. 
The first I will caU the Philadelphia maze (Fig. 
22). Fourteen years ago a travelling salesman. 



THE PARADOX PARTY. 



137 



living in Philadelphia, U.S.A., developed a curi- 
ously unrestrained passion for puzzles. He 
neglected his business, and soon his position 
was taken from him. His days and nights 
were now passed with the subject that fasci- 
nated him, and this little maze seems to have 
driven him into insanity. He had been puz- 
zling over it for some time, and finally it sent 
him mad and caused him to fire a bullet through 
his brain. Goodness knows what his difficulties 
could have been ! But there can be little doubt 
that he had a disordered mind, and that if this 
little puzzle had not c^sed him to lose his 
mental balance some other more or less trivial 
thing would in time have done so. There is no 
moral in the story, imless it be that of the Irish 
maxim, which applies to every occupation of 
life as much as to the solving of puzzles : " Take 
things aisy ; if you can't take them aisy, take 
them as aisy as you can." And it is a bad and 
empirical way of solving any puzzle — by blowing 
your brains out. 

Now, how many different routes are there 
from A to B in this maze if we must never in 
any route go along the same passage twice ? 
The four open spaces where four passages end 
are not reckoned as " passages." In the 
diagram (Fig. 22) it will be seen that I have 
again suppressed the blind alleys. It will be 
found that, in any case, we must go from 
A to C, and also from F to B. But when we 



have arrived at C there are three ways, marked 
I, 2, 3, of getting to D. Similarly, when we get 
to E there are three ways, marked 4, 5, 6, of 
getting to F. We have also the dotted route 
from C to E, the other dotted route from D to 
F, and the passage from D to E, indicated by 
stars. We can, therefore, express the position 
of afiairs by the little diagram annexed (Fig. 23). 
Here every condition of route exactly corre- 
sponds to that in the circular maze, only it is 
much less confusing to the eye. Now, the num- 
ber of routes, under the conditions, from A to B 
on this simplified diagram is 640, and that is the 
required answer to the maze puzzle. 

Finally, I will leave two easy maze puzzles 
(Figs. 24, 25) for my readers to solve for them- 
selves. The puzzle in each case is to find the 
shortest possible route to the centre. Every- 
body knows the story of Fair Rosamund and the 
Woodstock maze. What the maze was like or 
whether it ever existed except in imagination is 
not known, many writers believing that it was 
simply a badly-constructed house with a large 
number of confusing rooms and passages. At 
any rate, my sketch lacks the authority of the 
other mazes in this article. My " Rosamund's 
Bower " is simply designed to show that where 
you have the plan before you it often happens 
that the easiest way to find a route into a maze 
is by working backwards and first finding a way 
out. 



THE PARADOX PARTY. 



** Is not life itself a paradox ? " 

C. L. DoDGSON, Pillow Problems. 

" It is a wonderful age ! " said Mr. Allgood, and 
everybody at the table turned towards him and 
assumed an attitude of expectancy. 

This was an ordinary Christmas dinner of the 
Allgood family, with a sprinkling of local friends. 
Nobody would have supposed that the above 
remark would lead, as it did, to a succession of 
curious puzzles and paradoxes, to which every 
member of the party contributed something of 
interest. The little symposium was quite un- 
premeditated, so we must not be too critical 
respecting a few of the posers that were forth- 
coming. The varied character of the contribu- 
tions is just what we would expect on such an 
occasion, for it was a gathering not of expert 
mathematicians and logicians, but of quite 
ordinary folk. 

" It is a wonderful age ! " repeated Mr. All- 
good. " A man has just designed a square 
house in such a cunning manner that all the 
windows on the four sides have a south aspect." 

" That would appeal to me," said Mrs. All- 
good, " for I cannot endure a room with a north 
aspect." 

" I cannot conceive how it is done," Uncle 
John confessed. " I suppose he puts bay win- 
dows on the east and west sides ; but how on 
earth can he contrive to look south from the 



north side ? Does he use mirrors, or something 
of that kind ? " 

" No," replied Mr. Allgood, " nothing of the 
sort. AU the windows are flush with the walls, 
and yet you get a southerly prospect from 
every one of them. You see, there is no real 
difficulty in designing the house if you select 
the proper spot for its erection. Now, this 
house is designed for a gentleman who proposes 
to build it exactly at the North Pole. If you 
think a moment you will realize that when you 
stand at the North Pole it is impossible, no 
matter which way you may turn, to look else- 
where than due south ! There are no such 
directions as north, east, or west when you are 
exactly at the North Pole. Everything is due 
south ! " 

" I am afraid, mother," said her son George, 
after the laughter had subsided, " that, how- 
ever much you might like the aspect, the situa- 
tion would be a little too bracing for you." 

" Ah, well ! " she replied. " Your Uncle 
John fell also into the trap. I am no good at 
catches and puzzles. I suppose I haven't the 
right sort of brain. Perhaps some one will ex- 
plain this to me. Only last week I remarked 
to my hairdresser that it had been said that 
there are more persons in the world than any 
one of them has hairs on his head. He replied, 
' Then it follows, madam, that two persons, at 
least, must have exactly the same number of 



138 



AMUSEMENTS IN MATHEMATICS. 



hairs on their heads.' If this is a fact, I confess 
I cannot see it." 

" How do the bald-headed affect the ques- 
tion ? " asked Uncle John. 

"If there are such persons in existence," 
replied Mrs. AUgood, " who haven't a solitary 
hair on their heads discoverable under a magni- 
fying-glass, we wiU leave them out of the ques- 
tion. Still, I don't see how you are to prove 
that at least two persons have exactly the same 
number to a hair." 

" I think I can make it clear," said Mr. 
Filkins, who had dropped in for the evening. 
" Assume the population of the world to be 
only one million. Any number wiU do as well 
as another. Then your statement was to the 
effect that no person has more than nine 
hundred and ninety-nine thousand nine hundred 
and ninety-nine hairs on his head. Is that so ? " 

" Let me think," said Mrs. Allgood. " Yes 
— yes — that is correct." 

" Very well, then. As there are only nine 
hundred and ninety-nine thousand nine hundred 
and ninety-nine different ways of bearing hair, 
it is clear that the millionth person must repeat 
one of those ways. Do you see ? " 

" Yes ; I see that — at least I think I see it." 

" Therefore two persons at least must have 
the same number of hairs on their heads ; and 
as the number of people on the earth so greatly 
exceeds the number of hairs on any one person's 
head, there must, of course, be an immense 
number of these repetitions." 

" But, Mr. Filkms," said little Willie Allgood, 
" why could not the millionth man have, say, 
ten thousand hairs and a half ? " 

" That is mere hair-splitting, Willie, and does 
not come into the question." 

" Here is a curious paradox," said George. 
"If a thousand soldiers are drawn up in battle 
array on a plane " — they imderstood him to 
mean " plain " — " only one man will stand up- 
right." 

Nobody could see why. But George ex- 
plained that, according to Euclid, a plane can 
touch a sphere only at one point, and that per- 
son only who stands at that point, with respect 
to the centre of the earth, will stand upright. 

" In the same way," he remarked, " if a 
billiard-table were qviite level — that is, a perfect 
plane — the balls ought to roll to the centre." 

Though he tried to explain this by placing a 
visiting-card on an orange and expounding the 
law of gravitation, Mrs. Allgood declined to 
accept the statement. She could not see that 
the top of a true billiard-table must, theoreti- 
cally, be spherical, just like a portion of the 
orange-peel that George cut out. Of course, 
the table is so small in proportion to the surface 
of the earth that the curvature is not appreci- 
able, but it is nevertheless true in theory. A 
surface that we call level is not the same as our 
idea of a true geometrical plane. 

" Uncle John," broke in Willie Allgood, 
" there is a certain island situated between 
England and France, and yet that island is 
farther from France than England is. What 
is the island ? " 



"That seems absurd, my boy ; because if I 
place this tumbler, to represent the island, be- 
tween these two plates, it seems impossible that 
the txmibler can be farther from either of the 
plates than they are from each other." 

" But isn't Guernsey between England and 
France ? " asked Willie. 

" Yes, certainly." 

" Well, then, I think you will find, uncle, 
that Guernsey is about twenty-six miles from 
France, and England is only twenty-one miles 
from France, between Calais and Dover." 

" My mathematical master," said George, 
" has been trying to induce me to accept the 
axiom that ' if equals be multiplied by equals 
the products are equal.* " 

" It is self-evident," pointed out Mr. Filkins. 
"For example, if 3 feet equal i yard, then 
twice 3 feet will equal 2 yards. Do you 
see ? " 

" But, Mr. Filkins," asked George, " is this 
tumbler half full of water equal to a similar 
glass half empty ? " 

" Certainly, George." 

" Then it follows from the axiom that a glass 
full must equal a glass empty. Is that 
correct ? " 

" No, clearly not. I never thought of it 
in that light." 

" Perhaps," suggested Mr. Allgood, " the rule 
does not apply to liquids." 

" Just what I was thinking, Allgood. It 
would seem that we must make an exception 
in the case of liquids." 

" But it would be awkward," said George, with 
a smile, " if we also had to except the case of 
solids. For instance, let us take the solid earth. 
One mile square equals one square mile. There- 
fore two miles square must equal two square 
miles. Is this so ? " 

" Well, let me see ! No, of course not," Mr. 
Filkins replied, "because two miles square is 
four square miles." 

" Then," said George, " if the axiom is not 
true in these cases, when is it true ? " 

Mr. Filkins promised to look into the matter, 
and perhaps the reader wiU also like to give it 
consideration at leisure. 

" Look here, George," said his cousin Regi- 
nald WooUey : " by what fractional part does 
four- fourths exceed three-fourths ? " 

" By one-fourth ! " shouted everybody at 
once. 

" Try another one," George suggested. 

" With pleasure, when you have answered 
that one correctly," was Reginald's reply. 

" Do you mean to say that it isn't one- 
fourth ? " 

" Certainly I do." 

Several members of the company failed to see 
that the correct answer is " one-third," although 
Reginald tried to explain that three of anything, 
if increased by one-third, becomes four. 

" Uncle John, how do you pronoimce 
' t-o-o ' ? " asked WiUie. 

" • Too,' my boy." 

" And how do you pronounce * t-w-o ' ? " 

" That is also ' too.^ " 



THE PARADOX PARTY. 



139 



" Then how do you pronounce the second day 
of the week ? " 

" Well, that I should pronounce ' Tuesday,' 
not ' Toosday.' " 

" Would you really ? I should pronounce it 
' Monday.* " 

" If you go on like this, Willie," said Uncle 
John, with mock severity, " you wiU soon be 
without a friend in the world." 

" Can any of you write down quickly in 
figures ' twelve thousand twelve hundred and 
twelve pounds ' ? " asked Mr. AUgood. 

His eldest daughter, Miss Mildred, was the 
only person who happened to have a pencil at 
hand. 

" It can't be done," she declared, after mak- 
ing an attempt on the white table-cloth; but 
Mr. AUgood showed her that it should be 
written, "£13,212." 

" Now it is my turn," said Mildred. " I 
have been waiting to ask you all a question. 
In the Massacre of the Innocents under Herod, 
a nimiber of poor little children were buried in 
the sand with only their feet sticking out. How 
might you distinguish the boys from the girls ? " 

" I suppose," said Mrs. AUgood, " it is a 
conundrum — something to do with their poor 
little • souls.' " 

But after everybody had given it up, Mildred 
reminded the company that only boys were put 
to death. 

" Once upon a time," began George, " AchiUes 
had a race with a tortoise " 

"Stop, George ! " interposed Mr. AUgood. 
" We won't have that one. I knew two men in 
my youth who were once the best of friends, 
but they quarreUed over that infernal thing of 
Zeno's, and they never spoke to one another 
again for the rest of their lives. I draw the 
line at that, and the other stupid thing by Zeno 
about the flying arrow. I don't believe any- 
body understands them, because I could never 
do so myself." 

" Oh, very weU, then, father. Here is an- 
other. The Post-Office people were about to 
erect a line of telegraph-posts over a high hiU 
from TurmitviUe to Wurzleton; but as it was 
found that a railway company was making a 
deep level cutting in the same direction, they 
arranged to put up the posts beside the line. 
Now, the posts were to be a hundred yards apart, 
the length of the road over the hill being five 
mUes, and the length of the level cutting only 
four and a half mUes. How many posts did 
they save by erecting them on the level ? " 

" That is a very simple matter of calculation," 
said Mr. Filkins. " Find how many times one 
himdred yards wiU go in five miles, and how 
many times in four and a half miles. Then 
deduct one from the other, and you have the 
number of posts saved by the shorter route." 

" Quite right," confirmed Mr. AUgood. 
" Nothing could be easier." 

" That is just what the Post-Office people 
said," replied George, " but it is quite wrong. 
If you look at this sketch that I have just made, 
you wiU see that there is no difference whatever. 
If the posts are a hundred yards apart, just the 



same number wiU be required on the level as 
over the surface of the hiU." 




" Surely you must be wrong, George," said 
Mrs. AUgood, " for if the posts are a hundred 
yards apart and it is half a mile farther over 
the hill, you have to put up posts on that extra 
half-mile." 

" Look at the diagram, mother. You wiU 
see that the distance from post to post is not 
the distance from base to base measured along 
the ground. I am just the same distance from 
you if I stand on this spot on the carpet or stand 
immediately above it on the chair." 

But Mrs. AUgood was not convinced. 

Mr. Smoothly, the curate, at the end of the 
table, said at this point that he had a little 
question to ask. 

" Suppose the earth were a perfect sphere 
with a smooth surface, and a girdle of steel were 
placed roimd the Equator so that it touched at 
every point." 

" ' I'U put a girdle round about the earth in 
forty minutes,' " muttered George, quoting the 
words of Puck in A Midsummer Night's Dream. 

" Now, if six yards were added to the length 
of the girdle, what would then be the distance 
between the girdle and the earth, supposing 
that distance to be equal aU round ? " 

" In such a great length," said Mr. AUgood, 
" I do not suppose the distance would be worth 
mentioning." 

" What do you say, George ? " asked Mr. 
Smoothly. 

" WeU, without calculating I should imagine 
it would be a very minute fraction of an 
inch." 

Reginald and Mr. FUkins were of the same 
opinion. 

" I think it will surprise you aU," said the 
curate, " to learn that those extra six yards 
would make the distance from the earth aU 
round the girdle very nearly a yard ! " 

" Very nearly a yard ! " everybody ex- 
claimed, with astonishment ; but Mr. Smoothly 
was quite correct. The increase is independ- 
ent of the original length of the girdle, which 
may be roimd the earth or round an orange ; 
in any case the additional six yards wiU give a 
distance of nearly a yard aU round. This is apt 
to surprise the non-mathematical mind. 

" Did you hear the story of the extraordinary 
precocity of Mrs. Perkins's baby that died last 
week ? " asked Mrs. AUgood. " It was only 
three months old, and lying at the point of 
death, when the grief-stricken mother asked the 
doctor if nothing could save it. ' Absolutely 
nothing ! ' said the doctor. Then the infant 
looked up pitif uUy into its mother's face and said 
— absolutely nothing ! " 

" Impossible ! " insisted Mildred. " And only 
three months old ! " 



140 



AMUSEMENTS IN MATHEMATICS. 



" There have been extraordinary cases of in- 
fantile precocity," said Mr. Filkins, " the truth 
of which has often been carefully attested. But 
are you sure this reaUy happened, Mrs. All- 
good ? " 

" Positive," replied the lady. " But do you 
reaUy think it astonishing that a child of three 
months should say absolutely nothing ? What 
would you expect it to say ? " 

" Speaking of death," said Mr. Smoothly, 
solemnly, " I knew two men, father and son, 
who died in the same battle during the South 
African War. They were both named Andrew 
Johnson and buried side by side, but there was 
some difficulty in distinguishing them on the 
headstones. What would you have done ? " 

" Quite simple," said Mr, AUgood. " They 
should have described one as ' Andrew Johnson, 
Senior,' and the other as * Andrew Johnson, 
Junior.' " 

" But I forgot to tell you that the father died 
first." 

" What difierence can that make ? " 

" Well, you see, they wanted to be absolutely 
exact, and that was the difficulty." 

" But I don't see any difficulty," said Mr. 
Allgood, nor could anybody else. 

" Well," explained Mr. Smoothly, " it is like 
this. If the father died first, the son was then 
no longer ' Junior.' Is that so ? " 

"To be strictly exact, yes." 

" That is just what they wanted — to be 
strictly exact. Now, if he was no longer 
' Jimior,* then he did not die ' Jvmior,' Con- 
sequently it must be incorrect so to describe 
him on the headstone. Do you see the 
point ? " 

" Here is a rather curious thing," said Mr. 
Filkins, " that I have just remembered. A man 
wrote to me the other day that he had recently 
discovered two old coins while digging in his 
garden. One was dated ' 51 b.c.,' and the other 
one marked ' George I.' How do I know that 
he was not writing the truth ? " 

" Perhaps you know the man to be addicted 
to lying," said Reginald. 

" But that wovild be no proof that he was 
not telling the truth in this instance." 

" Perhaps," suggested Mildred, " you know 
that there were no coins made at those dates." 

" On the contrary, they were made at both 
periods." 

" Were they silver or copper coins ? " asked 
WiUie. 

" My friend did not state, and I really cannot 
see, Willie, that it makes any difference." 

" I see it ! " shouted Reginald. *' The letters 
' B.C.' would never be used on a coin made 
before the birth of Christ. They never antici- 
pated the event in that way. The letters were 
only adopted later to denote dates previous to 
those which we call ' a.d.' That is very good ; 
but I cannot see why the other statement could 
not be correct." 

" Reginald is quite right," said Mr. Filkins, 
* about the first coin. The second one could 
not exist, because the first George would never 
be described in his lifetime as ' George I.' " 



" Why not ? " asked Mrs. Allgood. " He was 
George I." 

" Yes ; but they would not know it until 
there was a George II." 

" Then there was no George II. until George 
III. came to the throne ? " 

" That does not follow. The second George 
becomes * George II.' on accoimt of there having 
been a ' George I.'" 

" Then the first George was ' George I.' on 
accovmt of there having been no king of that 
name before him." 

" Don't you see, mother," said George All- 
good, " we did not caU Queen Victoria * Vic- 
toria I. ; ' but if there is ever a ' Victoria II.,' 
then she will be known that way." 

" But there have been several Georges, and 
therefore he was ' George I.' There haven't 
been several Victorias, so the two cases are not 
similar." 

They gave up the attempt to convince Mrs. 
Allgood, but the reader will, of course, see the 
point clearly. 

" Here is a question," said Mildred AUgood, 
" that I should like some of you to settle for 
me. I am accustomed to buy from our green- 
grocer bundles of asparagus, each 12 inches in 
circumference. I always put a tape measure 
round them to make sure I am getting the full 
quantity. The other day the man had no 
large bundles in stock, but handed me instead 
two small ones, each 6 inches in circumference. 
' That is the same thing,' I said, ' and, of course, 
the price wiU be the same ; ' but he insisted that 
the two bundles together contained more than 
the large one, and charged me a few pence extra. 
Now, what I want to know is, which of us was 
correct ? Would the two small bundles contain 
the same quantity as tLe large one ? Or would 
they contain more ? " 

" That is the ancient puzzle," said Reginald, 
laughing, " of the sack of com that Sempronius 
borrowed from Caius, which your greengrocer, 
perhaps, had been reading about somewhere. 
He caught you beautifully." 

" Then they were equal ? " 

" On the contrary, you were both wrong, and 
you were badly cheated. You only got half the 
quantity that would have been contained in a 
large bundle, and therefore ought to have been 
charged half the original price, instead of more." 

Yes, it was a bad swindle, undoubtedly. A 
circle with a circumference half that of another 
must have its area a quarter that of the other. 
Therefore the two small bundles contained to- 
gether only half as much asparagus as a large 
one. 

" Mr. Filkins, can you answer this ? " asked 
WiUie. "There is a man in the next village 
who eats two eggs for breakfast every morning." 

" Nothing very extraordinary in that," George 
broke in. "If you told us that the two eggs 
ate the man it would be interesting." 

" Don't interrupt the boy, George," said his 
mother. 

" Well," Willie continued, " this man neither 
buys, borrows, barters, begs, steals, nor finds 
the eggs. He doesn't keep hens, and the eggs 



THE PARADOX PARTY. 



141 



are not given to him. How does he get the 
eggs ? " 

" Does he take them in exchange for some- 
thing else ? " asked Mildred. 

" That would be bartering them," WiUie 
replied. 

Perhaps some friend sends them to him," 
suggested Mrs. Allgood. 

" I said that they were not given to him." 

" I know," said George, with confidence. 
" A strange hen comes into his place and lays 
them." 

" But that would be finding them, wouldn't 
it?" 

" Does he hire them ? " asked Reginald. 

" If so, he could not return them after they 
were eaten, so that would be stealing them." 



All agreed that Willie's answer was quite 
satisfactory. Then Uncle John produced a 
little fallacy that " brought the proceedings 
to a close," as the newspapers say. 

413.— A CHESSBOARD FALLACY. 

" Herb is a diagram of a chessboard," he said. 
" You see there are sixty-four squares — eight by 
eight. Now I draw a straight line from the top 
left-hand corner, where the first and second 
squares meet, to the bottom right-hand comer. 
I cut along this line with the scissors, slide up 
the piece that I have marked B, and then clip 
ofE the little corner C by a cut along the first 
upright line. This little piece will exactly fit 
into its place at the top, and we now have 





\ 
















^ 


\ 
















\ 


\ 




B 












\ 


V 












A 




\ 


\ 
















\ 


















\ 


















K 






" Perhaps it is a pun on the word ' lay,' " 
Mr. Filkins said. " Does he lay them on the 
table ? " 

" He would have to get them first, wouldn't 
he ? The question was, How does he get 
them ? " 

" Give it up ! " said everybody. Then little 
WiUie crept round to the protection of his 
mother, for George was apt to be rough on such 
occasions. 

" The man keeps ducks ! " he cried, " and 
his servant collects the eggs every morning." 

" But you said he doesn't keep birds ! " 
George protested. 

"I didn't, did I, Mr. Filkins? I said he 
doesn't keep hens." 

" But he finds them," said Reginald. 

" No ; I said his servant finds them." 

" Well, then," Mildred interposed, "his serv- 
ant gives them to him." 

" You cannot give a man his own property, 
can you ? " 



an oblong with seven squares on one side and 
nine squares on the other. There are, therefore, 
now only sixty-three squares, because seven 
multiplied by nine makes sixty-three. Where 
on earth does that lost square go to ? I have 
tried over and over again to catch the little 
beggar, but he always eludes me. For the life 
of me I cannot discover where he hides himself." 

" It seems to be like the other old chess- 
board fallacy, and perhaps the explanation is 
the same," said Reginald — " that the pieces do 
not exactly fit." 

" But they do fit," said Uncle John. " Try 
it, and you will see." 

Later in the evening Reginald and George 
were seen in a comer with their heads together, 
trying to catch that elusive little square, and 
it is only fair to record that before they retired 
for the night they succeeded in securing their 
prey, though some others of the company failed 
to see it when captured. Can the reader solve 
the little mystery ? 



142 



AMUSEMENTS IN MATHEMATICS. 



UNCLASSIFIED PROBLEMS. 



" A snapper up of unconsidered trifles." 
Winter's Tale, iv. 2. 

414.— WHO WAS FIRST? 

Anderson, Biggs, and Carpenter were staying 
together at a place by the seaside. One day 
they went out in a boat and were a mile at sea 
when a rifle was fired on shore in their direction. 
Why or by whom the shot was fired fortunately 
does not concern us, as no information on these 
points is obtainable, but from the facts I picked 
up we can get material for a curious little puzzle 
for the novice. 

It seems that Anderson only heard the report 
of the gun, Biggs only saw the smoke, and Car- 
penter merely saw the bullet strike the water 
near them. Now, the question arises : Which 
of them first knew of the discharge of the rifle ? 

415.— A WONDERFUL VILLAGE. 

There is a certain village in Japan, situated in 
a very low valley, and yet the sim is nearer to 
the inhabitants every noon, by 3,000 miles and 
upwards, than when he either rises or sets to 
these people. In what part of the country is 
the village situated ? 

416.— A CALENDAR PUZZLE. 

If the end of the world should come on the first 
day of a new century, can you say what are 
the chances that it will happen on a Sunday ? 

417.— THE TIRING IRONS. 



The puzzle wiU be seen to consist of a simple 
loop of wire fixed in a handle to be held in the 
left hand, and a certain number of rings secured 
by wires which pass through holes in the bar and 
are kept there by their blimted ends. The 
wires work freely in the bar, but cannot come 
apart from it, nor can the wires be removed 
from the rings. The general puzzle is to de- 
tach the loop completely from all the rings, and 
then to put them all on again. 

Now, it wiU be seen at a glance that the first 
ring (to the right) can be taken off at any time 
by sliding it over the end and dropping it 
through the loop ; or it may be put on by re- 
versing the operation. With this exception, 
the only ring that can ever be removed is the 
one that happens to be a contiguous second on 
the loop at the right-hand end. Thus, with all 
the rings on, the second can be dropped at once ; 
with the first ring down, you cannot drop the 
second, but may remove the third ; with the 
first three rings down, you cannot drop the 
fourth, but may remove the fifth ; and so on. 
It wiU be foimd that the first and second rings 
can be dropped together or put on together ; 
but to prevent confusion we will throughout 
disallow this exceptional double move, and say 
that only one ring may be put on or removed 
at a time. 

We can thus take off one ring in i move ; 
two rings in 2 moves ; three rings in 5 moves ; 
four rings in 10 moves ; five rings in 21 moves ; 
and if we keep on doubling (and adding one 
where the number of rings is odd) we may easily 
ascertain the number of moves for completely 
removing any number of rings. To get off all 




The illustration represents one of the most 
ancient of aU mechanical puzzles. Its origin is 
unknown. Cardan, the mathematician, wrote 
about it in 1550, and Wallis in 1693 ; while it is 
said still to be found in obscure English villages 
(sometimes deposited in strange places, such as 
a church belfry), made of iron, and appropriately 
called " tiring-irons," and to be used by the 
Norwegians to-day as a lock for boxes and bags. 
In the toyshops it is sometimes called the 
" Chinese rings," though there seems to be no 
authority for the description, and it more fre- 
quently goes by the unsatisfactory name of 
the puzzling rings." The French call it 
" Baguenaudier." 



the seven rings requires 85 moves. Let us look 
at the five moves made in removing the first 
three rings, the circles above the line standing 
for rings on the loop and those under for rings 
off the loop. 

Drop the first ring ; drop the third ; put up 
the first ; drop the second ; and drop the first — 
5 moves, as shown clearly in the diagrams. The 
dark circles show at each stage, from the start- 
ing position to the finish, which rings it is possible 
to drop. After move 2 it will be noticed that 
no ring can be dropped until one has been put 
on, because the first and second rings from the 
right now on the loop are not together. After 
the fifth move, if we wish to remove all seven 



UNCLASSIFIED PROBLEMS. 



143 



rings we must now drop the fifth. But before 
we can then remove the fourth it is necessary 
to put on the first three and remove the first 



00000 



000000 



0000 O 



O 



O O 



0000 



o 



4^ 



0000 



o o 



o o # o 



000 



two. We shall then have 7, 6, 4, 3 on the loop, 
£uid may therefore drop the fourth. When we 
have put on 2 and i and removed 3, 2, i, we 
may <kop the seventh ring. The next operation 
then will be to get 6, 5, 4, 3, 2, i on the loop and 
remove 4, 3, 2, i, when 6 will come off ; then 
get 5, 4, 3, 2, I on the loop, and remove 3> 2, i, 
when 5 will come off ; then get 4, 3, 2, i on the 
loop and remove 2, i, when 4 will come off ; 
then get 3, 2, I on the loop and remove i, when 
3 will come off ; then get 2, i on the loop, when 
2 will come off ; and i will fall through on the 
85 th move, leaving the loop quite free. The 
reader should now be able to understand the 
puzzle, whether or not he has it in his hand in 
a practical form. 

The particular problem I propose is simply 
this. Suppose there are altogether fourteen 
rings on the tiring-irons, and we proceed to take 
them all off in the correct way so as not to waste 
any moves. What wiU be the position of the 
rings after the 9,999th move has been made ? 



41B.— SUCH A GETTING UPSTAIRS. 

In a suburban villa there is a small staircase 
with eight risers, not counting the landing. 
The little puzzle with which Tommy Smart 
perplexed his family is this. You are required 
to start from the bottom and land twice on the 
floor above (stopping there at the finish), having 
returned once to the ground floor. But you 
must be careful to use every riser the same 
number of times. In how few steps can you 
make the ascent ? It seems a very simple 
matter, but it is more than likely that at your 
first attempt you will make a great many more 
steps than are necessary. Of course you must 
not go more than one riser at a time. 

Tommy knows the trick, and has shown it to 
his father, who professes to have a contempt 
for such things ; but when the children are in 
bed the pater will often take friends out into 
the hall and enjoy a good laugh at their be- 
wilderment. And yet it is all so very simple 
when you know how it is done. 

419.— THE FIVE PENNIES. 

Here is a reaUy hard puzzle, and yet its con- 
ditions are so absurdly simple. Every reader 
knows how to place four pennies so that they 
are equidistant from each other. All you have 
to do is to arrange three of them flat on the 
table so that they touch one another in the 
form of a triangle, and lay the fourth penny on 
top in the centre. Then, as every penny touches 
every other penny, they are all at equal dis- 
tances from one another. Now try to do the 
same thing with five pennies- — ^place them so that 
every penny shall touch every other penny — and 
you will find it a different matter altogether. 

420.— THE INDUSTRIOUS BOOKWORM. 




Our friend Professor Rackbrane is seen in the 
illustration to be propounding another of his 



144 



AMUSEMENTS IN MATHEMATICS. 



little posers. He is explaining that since he 
last had occasion to take down those three 
volumes of a learned book from their place on 
his shelves a bookworm has actually bored a 
hole straight through from the first page to the 
last. He says that the leaves are together three 
inches thick in each volume, and that every cover 
is exactly one-eighth of an inch thick, and he 
asks how long a tunnel had the industrious 
worm to bore in preparing his new tube railway. 
Can you teU him ? 

421.— A CHAIN PUZZLE. 



wood's rubies. There have, of course, been 
many greater robberies in point of value, but 
few so artfully conceived. Lady Littlewood, 
of Romley Manor, had a beautiful but rather 
eccentric heirloom in the form of a ruby brooch. 
While staying at her town house early in the 
eighties she took the jewel to a shop in Bromp- 
ton for some slight repairs. 

" A fine collection of rubies, madam," said 
the shopkeeper, to whom her ladyship was a 
stranger. 

" Yes," she replied ; " but curiously enough 
I have never actually counted them. My 




This is a puzzle based on a pretty little idea 
first dealt with by the late Mr. Sam Loyd. A 
man had nine pieces of chain, as shown in the 
illustration. He wanted to join these fifty links 
into one endless chain. It will cost a penny to 
open any link and twopence to weld a link to- 
gether again, but he could buy a new endless 
chain of the same character and quality for 
2S. 2d. What was the cheapest course for him 
to adopt ? Unless the reader is cunning he 
may find himself a good way out in his answer. 

422.— THE SABBATH PUZZLE. 

I HAVE come across the following little poser 
in an old book. I wonder how many readers 
will see the author's intended solution to the 
riddle. 

Christians the week's first day for Sabbath hold ; 
The Jews the seventh, as they did of old ; 
The Turks the sixth, as we have oft been told. 
How can these three, in the same place and 

day, 
Have each his own true Sabbath ? teU, I pray. 

423.— THE RUBY BROOCH. 

The annals of Scotland Yard contain some re- 
markable cases of jewel robberies, but one of 
the most perplexing was the theft of Lady Little- 



mother once pointed out to me that if you start 
from the centre and count up one line, along 
the outside and down the next line, there are 




always eight rubies. So I should always know 
if a stone were missing." 

Six months later a brother of Lady Little- 



UNCLASSIFIED PROBLEMS. 



145 



wood's, who had returned from his regiment 
in India, noticed that his sister was wearing the 
ruby brooch one night at a county ball, and on 
their return home asked to look at it more 
closely. He immediately detected the fact that 
four of the stones were gone. 

" How can that possibly be ? " said Lady 
Littlewood. " If you count up one line from 
the centre, along the edge, and down the next 
line, in any direction, there are always eight 
stones. This was always so and is so now. 
How, therefore, would it be possible to remove 
a stone without my detecting it ? " 

" Nothing could be simpler," replied the 
brother. " I know the brooch well. It origi- 
nally contained forty-five stones, and there are 
now only forty-one. Somebody has stolen four 
rubies, and then reset as small a number of the 
others as possible in such a way that there shall 
always be eight in any of the directions you 
have mentioned." 

There was not the slightest doubt that the 
Brompton jeweUer was the thief, and the matter 
was placed in the hands of the police. But the 
man was wanted for other robberies, and had 
left the neighbourhood some time before. To 
this day he has never been found. 

The interesting little point that at first baffled 
the police, and which forms the subject of our 
puzzle, is this : How were the forty-five rubies 
originally arranged on the brooch ? The illus- 
tration shows exactly how the forty-one were 
arranged after it came back from the jeweller ; 
but although they count eight correctly in any 
of the directions mentioned, there are four 
stones missing. 

424.— THE DOVETAILED BLOCK. 




Here is a curious mechanical puzzle that was 
given to me some years ago, but I cannot 
say who first invented it. It consists of two 
solid blocks of wood securely dovetailed to- 
gether. On the other two vertical sides that 
are not visible the appearance is precisely the 
same as on those shown. How were the pieces 
put together ? When I published this little 
puzzle in a London newspaper I received (though 
they were unsolicited) quite a staek of models, 
(1.926) I 



in oak, in teak, in mahogany, rosewood, satin- 
wood, elm, and deal ; some half a foot in length, 
and others varying in size right down to a deli- 
cate little model about half an inch square. It 
seemed to create considerable interest. 

425.— JACK AND THE BEANSTALK. 



' — Wl/7* 




The illustration, by a British artist, is a sketch 
of Jack climbing the beanstalk. Now, the 
artist has made a serious blunder in this draw- 
ing. Can you find out what it is ? 

426.— THE HYMN-BOARD POSER. 

The worthy vicar of Chumpley St. Winifred is 
in great distress. A little church difficulty has 
arisen that all the combined intelligence of the 
parish seems unable to surmoimt. What this 
difficulty is I will state hereafter, but it may add 
to the interest of the problem if I first give a 
short account of the curious position that has 
been brought about. It all has to do with the 
church hymn-boards, the plates of which have 
become so damaged that they have ceased to 
fulfil the purpose for which they were devised. 
A generous parishioner has promised to pay for 
a new set of plates at a certain rate of cost ; but 
strange as it may seem, no agreement can be 
come to as to what that cost should be. 

The proposed maker of the plates has named 



146 



AMUSEMENTS IN MATHEMATICS. 



a price which the donor declares to be absurd. 
The good vicar thinks they are both wrong, so 
he asks the schoolmaster to work out the little 
sum. But this individual declares that he can 
find no rule bearing on the subject in any of his 




arithmetic books. An application having been 
made to the local medical practitioner, as a man 
of more than average intellect at Chumpley, he 
has assured the vicar that his practice is so 
heavy that he has not had time even to look at 
it, though his assistant whispers that the doctor 
has been sitting up unusually late for several 
nights past. Widow Wilson has a smart son, 
who is reputed to have once won a prize for 
puzzle-solving. He asserts that as he cannot 
find any solution to the problem it must have 
something to do with the squaring of the circle, 
the duplication of the cube, or the trisection of 
an angle ; at any rate, he has never before seen 
a puzzle on the principle, and he gives it up. 

This was the state of afiairs when the assist- 
ant curate (who, I should say, had frankly con- 
fessed from the first that a profoxmd study of 
theology had knocked out of his head all the 
knowledge of mathematics he ever possessed) 
kindly sent me the puzzle. 

A church has three hymn-boards, each to in- 
dicate the numbers of five different hyinns to 
be sung at a service. All the boards are in use 
at the same service. The hymn-book contains 
700 hymns. A new set of numbers is required, 
and a kind parishioner oflers to present a set 



painted on metal plates, but stipulates that only 
the smallest number of plates necessary shaU 
be purchased. The cost of each plate is to be 
6d., and for the painting of each plate the 
charges are to be : For one plate, is. ; for two 
plates alike, ii|d. each ; for three plates alike, 
iijd. each, and so on, the charge being one 
farthing less per plate for each similarly painted 
plate. Now, what should be the lowest cost ? 
Readers wiU note that they are required to 
use every legitimate and practical method of 
economy. The illustration will make clear the 
nature of the three hymn-boards and plates. 
The five hjnnns are here indicated by means of 
twelve plates. These plates slide in separately 
at the back, and in the illustration there is room, 
of course, for three more plates. 

427.— PHEASANT-SHOOTING. 

A Cockney friend, who is very apt to draw the 
long bow, and is evidently less of a sportsman 
than he pretends to be, relates to me the follow- 
ing not very credible yam : — 

" I've just been pheasant -shooting with my 
friend the duke. We had splendid sport, and- 
I made some wonderful shots. What do you 
think of this, for instance ? Perhaps you can 
twist it into a puzzle. The duke and I were 
crossing a field when suddenly twenty-four 
pheasants rose on the wing right in front of us. 
I fired, and two-thirds of them dropped dead 1 
at my feet. Then the duke had a shot at what 
were left, and brought down three-twenty- 
fomrths of them, wounded in the wing. Now, 
out of those twenty-four birds, how many still 
remained ? " 

It seems a simple enough question, but can 
the reader give a correct answer ? 

428.— THE GARDENER AND THE COOK. 

A CORRESPONDENT, Signing himself " Simple 
Simon," suggested that I should give a special 
catch puzzle in the issue of The Weekly Dispatch 
for All Fools' Day, 1900. So I gave the follow- 
ing, and it caused considerable amusement ; for 
out of a very large body of competitors, many 
quite expert, not a single person solved it, 
though it ran for nearly a month. 




" The illustration is a fancy sketch of my 
correspondent, ' Simple Simon,' in the act of 
trying to solve the following innocent little 



UNCLASSIFIED PROBLEMS. 



147 



arithmetical puzzle. A race between a man and 
a woman that I happened to witness one All 
Fools' Day has fixed itself indelibly on my 
memory. It happened at a country-house, 
where the gardener and the cook decided to 
run a race to a point 100 feet straight away and 
return. I foimd that the gardener ran 3 feet 
at every bound and the cook only 2 feet, but 
then she made three bounds to his two. Now, 
what was the result of the race ? " 

A fortnight after publication I added the 



your second coin at exactly the distance of an 
inch from the first, the third an inch distance 
from the second, and so on. No halfpenny may 
touch another halfpenny or cross the boundary. 
Our illustration will make the matter perfectly 
clear. No. 2 coin is an inch from No. i ; No. 3 
an inch from No. 2 ; No. 4 an inch from No. 3 ; 
but after No. 10 is placed we can go no further 
in this attempt. Yet several more halfpennies 
might have been got in. How many can the 
reader place ? 




following note : " It has been suggested that 
perhaps there is a catch in the ' return,' but 
there is not. The race is to a point 100 feet 
away and home again — that is, a distance of 
200 feet. One correspondent asks whether they 
take exactly the same time in turning, to which 
I reply that they do. Another seems to sus- 
pect that it is really a conundrum, and that 
the answer is that ' the result of the race was 
a (matrimonial) tie.' But I had no such inten- 
tion. The puzzle is an arithmetical one, as it 
purports to be." 

429.— PLACING HALFPENNIES. 




Here is an interesting little puzzle suggested 
to me by Mr. W. T. Whyte. Mark off on a 
sheet of paper a rectangular space 5 inches by 
3 inches, and then find the greatest number of 
halfpennies that can be placed within the en- 
closure under the following conditions. A half- 
penny is exactly an inch in diameter. Place 
your first halfpenny where you like, then place 



430.— FIND THE MAN'S WIFE. 

One summer day in 1903 I was loitering on the 
Brighton front, watching the people strolling 
about on the beach, when the friend who was 
with me suddenly drew my attention to an indi- 
vidual who was standing alone, and said, " Can 
you point out that man's wife ? They are stop- 
ping at the same hotel as I am, and the lady is 
one of those in view." After a few minutes' 
observation, I was successful in indicating the 
lady correctly. My friend was curious to know 
by what method of reasoning I had arrived at 
the result. This was my answer : — 

" We may at once exclude that Sister of 
Mercy and the girl in the short frock ; also the 
woman selling oranges. It cannot be the lady 
in widows' weeds. It is not the lady in the 
bath chair, because she is not staying at your 
hotel, for I happened to see her come out of 
a private house this morning assisted by her 
maid. The two ladies in red breakfasted at my 
hotel this morning, and as they were not wear- 
ing outdoor dress I conclude they are staying 
there. It therefore rests between the lady in 
blue and the one with the green parasol. But 
the left hand that holds the parasol is, you see, 
ungloved and bears no wedding-ring. Conse- 
quently I am driven to the conclusion that the 
lady in blue is the man's wife — and you say 
this is correct." 

Now, as my friend was an artist, and as I 
thought an amusing puzzle might b';. devised on 
the lines of his question, I asked him to make 
me a drawing according to some directions that 
I gave him, and I have pleasure in presenting 
his production to my readers. It wiU be seen 
that the picture shows six men and six ladies : 
Nos. I, 3, 5, 7, 9, and 11 are ladies, and Nos. 2, 



148 



AMUSEMENTS IN MATHEMATICS. 



4, 6, 8, 10, and 12 are men. These twelve in- 
dividuals represent six married couples, all 
strangers to one another, who, in walking aim- 
lessly about, have got mixed up. But we are 
only concerned with the man that is wearing a 
straw hat — Number 10. The puzzle is to find 
this man's wife. Examine the six ladies care- 
fully, and see if you can determine which one of 
them it is. 

I showed the picture at the time to a few 
friends, and they expressed very different opin- 
ions on the matter. One said, " I don't believe 
he would marry a girl like Number 7." An- 
other said, " I am sure a nice girl like Number 3 
would not marry such a fellow ! " Another 
said, " It must be Number i, because she has 



got as far away as possible from the brute ! " 
It was suggested, again, that it must be Number 
II, because "he seems to be looking towards 
her;" but a cynic retorted, "For that very 
reason, if he is really looking at her, I should 
say that she is not his wife ! " 

I now leave the question in the hands of my 
readers. Which is really Number lo's wife ? 

The illustration is of necessity considerably 
reduced from the large scale on which it origi- 
nally appeared in The Weekly Dispatch (24th 
May 1903), but it is hoped that the details will 
be sufficiently clear to allow the reader to de- 
rive entertainment from its examination. In 
any case the solution given will enable him to 
follow the points with interest. 



SOLUTIONS. 



1.— A POST-OFFICE PERPLEXITY. 

The yotmg lady supplied 5 twopenny stamps, 
30 penny stamps, and 8 twopence-halfpenny 
stamps, which delivery exactly fulfils the condi- 
tions and represents a cost of five shillings. 

2.— YOUTHFUL PRECOCITY. 

The price of the banana must have been one 
penny farthing. Thus, 960 bananas would cost 
£5, and 480 sixpences would buy 2,304 bananas. 

3.— AT A CATTLE MARKET. 

Jakes must have taken 7 animals to market, 
Hodge must have taken 11, and Durrant must 
have taken 21. There were thus 39 animals 
altogether. 

4.— THE BEANFEAST PUZZLE. 

The cobblers spent 35s., the tailors spent also 
35s., the hatters spent 42s., and the glovers 
spent 21S. Thus, they spent altogether £6, 13s., 
while it will be foimd that the five cobblers spent 
as much as four tailors, twelve tailors as much 
as nine hatters, and six hatters as much as eight 
glovers. 

5.— A QUEER COINCIDENCE. 

Puzzles of this class are generally solved in 
the old books by the tedious process of " work- 
ing backwards." But a simple general solution 
is as follows : If there are n players, the amount 
held by every player at the end wiU be w(2"), 
the last winner must have held w(n + 1) at the 
start, the next w(2«-f-i), the next w(4n + i), 
the next w(8n + 1), and so on to the first player, 
who must have held fn(2'»-^M4-i). 

Thus, in this case, n=7, and the amount held 
by every player at the end was 2' farthings. 
Therefore m=x, and G started with 8 farthings, 
F with 15, £ with 29, D with 57, C with 113, 
B with 225, and A with 449 farthings. 

6.— A CHARITABLE BEQUEST. 

There are seven different ways in which the 
money may be distributed : 5 women and 19 



men, 10 women and 16 men, 15 women and 13 
men, 20 women and 10 men, 25 women and 7 
men, 30 women and 4 men, and 35 women and 
I man. But the last case must not be counted, 
because the condition was that there should be 
" men," and a single man is not men. There- 
fore the answer is six years. 

7.— THE WIDOW'S LEGACY. 

The widow's share of the legacy must be 
£205, 2S. 6d. and ^ of a penny. 

8.— INDISCRIMINATE CHARITY. 

The gentleman must have had 3s. 6d. in his 
pocket when he set out for home. 

9.— THE TWO AEROPLANES. 

The man must have paid £500 and £750 for the 
two machines, making together £1,250; but as 
he sold them for only £1,200, he lost £50 by the 
transaction. 

10.— BUYING PRESENTS. 

JoRKiNS had originally £19, i8s. in his pocket, 
and spent £9, 19s. 

II.— THE CYCLISTS' FEAST. 

There were ten cyclists at the feast. They 
should have paid 8s. each ; but, owing to the 
departure of two persons, the remaining eight 
would pay los. each. 

12.— A QUEER THING IN MONEY. 

The answer is as foUows : £44,444, 4s. 4d.= 
28, and, reduced to pence, 10,666,612=28. 

It is a curious little coincidence that in the 
answer 10,666,612 the four central figures indi- 
cate the only other answer, £66, 6s. 6d. 

i3._A NEW MONEY PUZZLE. 

The smallest sum of money, in pounds, shillings, 
pence, and farthings, containing aU the nine 
digits once, and once only, is £2,567, i8s. 9|d. 



SOLUTIONS. 



149 



I4._SQUARE MONEY. 

The answer is i|d. and 3d. Added together 
they make 4id., and i|d. multiplied by 3 is 
also 4|d. 

1 5. —POCKET MONEY. 

The largest possible sum is 15s. gd., composed 
of a crown and a half-crown (or three half- 
crowns), four florins, and a threepenny piece. 

16.— THE MILLIONAIRE'S PERPLEXITY. 

The answer to this quite easy puzzle may, of 
course, be readily obtained by trial, deducting 
the largest power of 7 that is contained in one 
million doUars, then the next largest power 
from the remainder, and so on. But the little 
problem is intended to illustrate a simple direct 
method. The answer is given at once by con- 
verting 1,000,000 to the septenary scale, and it 
is on this subject of scales of notation that I 
propose to write a few words for the benefit of 
those who have never sufficiently considered 
the matter. 

Our manner of figuring is a sort of perfected 
arithmetical shorthand, a system devised to 
enable us to manipulate numbers as rapidly 
and correctly as possible by means of symbols. 
If we write the number 2,341 to represent two 
thousand three hundred and forty-one dollars, 
we wish to imply i doUar, added to four times 
10 dollars, added to three times 100 dollars, 
added to two times 1,000 dollars. From the 
number in the units place on the right, every 
figure to the left is understood to represent a 
multiple of the particular power of 10 that its 
position indicates, while a cipher (o) must be 
inserted where necessary in order to prevent 
confusion, for if instead of 207 we wrote 27 it 
would be obviously misleading. We thus only 
require ten figures, because directly a number 
exceeds 9 we put a second figure to the left, 
directly it exceeds 99 we put a third figure to 
the left, and so on. It will be seen that this is 
a purely arbitrary method. It is working in 
the denary (or ten) scale of notation, a system 
undoubtedly derived from the fact that our 
forefathers who devised it had ten fingers upon 
which they were accustomed to count, like our 
children of to-day. It is unnecessary for us 
ordinarily to state that we are using the denary 
scale, because this is always understood in the 
common affairs of life. 

But if a man said that he had 6,553 doUars in 
the septenary (or seven) scale of notation, you 
wiU find that this is precisely the same amount 
as 2,341 in our ordinary denary scale. Instead 
of using powers of ten, he uses powers of 7, so 
that he never needs any figure higher than 6, 
and 6,553 really stands for 3, added to five 
times 7, added to five times 49, added to six 
times 343 (in the ordinary notation), or 2,341. 
To reverse the operation, and convert 2,341 
from the denary to the septenary scale, we 
divide it by 7, and get 334 and remainder 3 ; 
divide 334 by 7, and get 47 and remainder 5 ; 
and so keep on dividing by 7 as long as there 
is anything to divide. The remainders, read 



backwards, 6, 5, 5, 3, give us the answer, 

6,553. 

Now, as I have said, our puzzle may be solved 
at once by merely converting 1,000,000 doUars 
to the septenary scale. Keep on dividing this 
number by 7 until there is nothing more left 
to divide, and the remainders will be found to be 
i^3333ii» which is 1,000,000 expressed in the 
septenary scale. Therefore, i gift of i dollar, 
I gift of 7 dollars, 3 gifts of 49 dollars, 3 gifts 
of 343 doUars, 3 gifts of 2,401 dollars, 3 gifts 
of 16,807 dollars, i gift of 117,649 dollars, 
and one substantial gift of 823,543 dollars, 
satisfactorily solves our problem. And it is 
the only possible solution. It is thus seen that 
no " trials " are necessary ; by converting to 
the septenary scale of notation we go direct to 
the answer. 

17.— THE PUZZLING MONEY BOXES. 

The correct answer to this puzzle is as foUows : 
John put into his money-box two double florins 
fSs.), WiUiam a half-sovereign and a florin 
(i2s.), Charles a crown (5s,), and Thomas a 
sovereign (20s.). There are six coins in aU, 
of a total value of 45s. If John had 2s. more, 
William 2S. less, Charles twice as much, and 
Thomas half as much as they reaUy possessed, 
they would each have had exactly los. 

18.— THE MARKET WOMEN. 

The price received was in every case 105 far- 
things. Therefore the greatest number of 
women is eight, as the goods could only be sold 
at the foUowing rates : 105 lbs. at i farthing, 
35 at 3, 21 at 5, 15 at 7, 7 at 15, 5 at 21, 3 at 35, 
and I lb. at 105 farthings. 

19.— THE NEW YEAR'S EVE SUPPERS. 

The company present on the occasion must 
have consisted of seven pairs, ten single men, 
and one single lady. Thus, there were twenty- 
five persons in aU, and at the prices stated they 
would pay exactly £5 together. 

20.— BEEF AND SAUSAGES. 

The lady bought 48 lbs. of beef at 2S., and the 
same quantity of sausages at is. 6d., thus spend- 
ing £8, 8s. Had she bought 42 lbs. of beef and 
56 lbs. of sausages she would have spent £4, 4s. 
on each, and have obtained 98 lbs. instead of 
96 lbs. — a gain in weight of 2 lbs. 

21.— A DEAL IN APPLES. 

f 

I WAS first offered sixteen apples for my shiUing, 
which would be at the rate of ninepence a dozen. 
The two extra apples gave me eighteen for a 
shilling, which is at the rate of eightpence a 
dozen, or one penny a dozen less than the first 
price asked. 

22.— A DEAL IN EGGS. 

The man must have bought ten eggs at five- 
pence, ten eggs at one penny, and eighty eggs 



I50 



AMUSEMENTS IN MATHEMATICS. 



at a halfpenny. He woiild then have one hun- 
dred eggs at a cost of eight shillings and four- 
pence, and the same number of eggs of two of 
the qualities. 

23.— THE CHRISTMAS-BOXES. 

The distribution took place " some years ago," 
when the fourpenny-piece was in circulation. 
Nineteen persons must each have received nine- 
teen pence. There are five different ways in 
which this sum may have been paid in silver 
coins. We need only use two of these ways. 
Thus if fourteen men each received four four- 
penny-pieces and one threepenny-piece, and five 
men each received five threepenny-pieces and 
one fourpenny-piece, each man would receive 
nineteen pence, and there would be exactly one 
hundred coins of a total value of £1, los. id. 

24.— A SHOPPING PERPLEXITY. 

The first purchase amounted to is. 5|d., the 
second to is. iijd., and together they make 
3s. 5jd. Not one of these three amounts can 
be paid in fewer than six current coins of the 
realm. 

25.— CHINESE MONEY. 

As a ching-chang is worth twopence and foUr- 
fifteenths of a ching-chang, the remaining 
eleven-fifteenths of a ching-chang must be worth 
twopence. Therefore eleven ching-changs are 
worth exactly thirty pence, or half a crown. 
Now, the exchange must be made with seven 
roimd-holed coins and one square-holed coin. 
Thus it will be seen that 7 round-holed coins 
are worth seven-elevenths of 15 ching-changs, 
and I square-holed coin is worth one-eleventh 
of 16 ching-changs — that is, 'j'j rounds equal 
105 ching-changs and 11 squares equal 16 ching- 
changs. Therefore "jy rounds added to 11 
squares equal 121 ching-changs ; or 7 rounds 
and I square equal 11 ching-changs, or its 
equivalent, half a crown. This is more simple 
in practice than it looks here. 

26.— THE JUNIOR CLERKS' PUZZLE. 

Although Snoggs's reason for wishing to take 
his rise at £2, los. half-yearly did not concern 
our puzzle, the jad that he was duping his 
employer into paying him more than was in- 
tended did concern it. Many readers will be 
surprised to find that, although Moggs only 
received ^(^350 in five years, the artful Snoggs 
actually obtained £362, los. in the same time. 
The rest is simplicity itself. It is evident that 
if Moggs saved £87, los. and Snoggs £181, 5s., 
the latter would be saving twice as great a 
proportion of his salary as the former (namely, 
one-half as against one-quarter), and the two 
simis added together make £268, 15s. 

27.— GIVING CHANGE. 

The way to help the American tradesman out 
of his dilemma is this. Describing the coins by 
the number of cents that they represent, the 



tradesman puts on the counter 50 and 25 ; the 
buyer puts down 100, 3, and 2 ; the stranger 
adds his 10, 10, 5, 2, and i. Now, considering 
that the cost of the purchase amounted to 
34 cents, it is clear that out of this pooled 
money the tradesman has to receive 109, the 
buyer 71, and the stranger his 28 cents. There- 
fore it is obvious at a glance that the loo-piece 
must go to the tradesman, and it then follows 
that the 50-piece must go to the buyer, and 
then the 25 -piece can only go to the stranger. 
Another glance will now make it clear that the 
two lo-cent pieces must go to the buyer, be- 
cause the tradesman now only wants 9 and the 
stranger 3. Then it becomes obvious that the 
buyer must take the i cent, that the stranger 
must take the 3 cents, and the tradesman the 
5, 2, and 2. To sum up, the tradesman takes 
100, 5, 2, and 2 ; the buyer, 50, 10, 10, and i ; 
the stranger, 25 and 3. It wiU be seen that not 
one of the three persons retains any one of his 
own coins. 

28.— DEFECTIVE OBSERVATION. 

Of course the date on a penny is on the same 
side as Britannia — the " tail " side. Six pen- 
nies may be laid around another penny, all flat 
on the table, so that every one of them touches 
the central one. The number of threepenny- 
pieces that may be laid on the surface of a half- 
crown, so that no piece lies on another or over- 
laps the edge of the half-crown, is one. A 
second threepenny-piece will overlap the edge 
of the larger coin. Few people guess fewer 
than three, and many persons give an absurdly 
high number. 

29.— THE BROKEN COINS. 

If the three broken coins when perfect were 
worth 253 pence, and are now in their broken 
condition worth 240 pence, it should be obvi- 
ous that ^ of the original value has been lost. 
And as the same fraction of each coin has been 
broken away, each coin has lost ^^ of its 
original bulk. 

30._TW0 QUESTIONS IN PROBA- 
BILITIES. 

In tossing with the five pennies aU at the same 
time, it is obvious that there are 32 different 
ways in which the coins may fall, because the 
first coin may fall in either of two ways, then 
the second coin may also fall in either of two 
ways, and so on. Therefore five 2's multiplied 
together make 32. Now, how are these 32 
ways made up ? Here they are : — 

(a) 5 heads i way 

(6) 5 tails I way 

(c) 4 heads and i tail ... 5 ways 

\A) 4 tails and i head ... 5 ways 

\e) 3 heads and 2 tails ... 10 ways 

(/) 3 tails and 2 heads ... 10 ways 

Now, it will be seen that the only favourable 
cases are a, b, c, and d — 12 cases. The remain- 
ing 20 cases are unfavourable, because they do 



SOLUTIONS. 



151 



not give at least four heads or four tails. 
ITierefore the chances are only 12 to 20 in 
your favour, or (which is the same thing) 3 to 
5. Put another way, you have only 3 chances 
out of 8. 

The amount that should be paid for a draw 
from the bag that contains three sovereigns 
and one shilling is 153. 3d. Many persons will 
say that, as one's chances of drawing a sovereign 
were 3 out of 4, one should pay three-fourths of 
a pound, or 15s., overlooking the fact that one 
must draw at least a shilling — there being no 
blanks. 

31.— DOMESTIC ECONOMY. 

Without the hint that I gave, my readers would 
probably have been unEmimous in deciding that 
Mr. Perkins's income must have been £1,710. 
But this is quite wrong. Mrs. Perkins says, 
" We have spent a third of his yearly income in 
rent," etc., etc. — that is, in two years they have 
spent an amount in rent, etc., equal to one- 
third of his yearly income. Note that she does 
not say that they have spent each year this sum, 
whatever it is, but that during the two years that 
amount has been spent. The only possible 
answer, according to the exact reading of her 
words, is, therefore, that his income was £180 
per annum. Thus the amount ^ spent in two 
years, during which his income has amounted 
to £360, will be £60 in rent, etc., £90 in domestic 
expenses, £20 in other ways, leaving the balance 
of £190 in the bank as stated. 

32.— THE EXCURSION TICKET PUZZLE. 

Nineteen shillings and ninepence may be paid 
in 458,908,622 difierent ways. 

I do not propose to give my method of solu- 
tion. Any such explanation would occupy an 
amount of space out of proportion to its interest 
or value. If I could give within reasonable 
limits a general solution for all money payments, 
I would strain a point to find room ; but such a 
solution would be extremely complex and cum- 
bersome, and I do not consider it worth the 
labour of working out. 

Just to give an idea of what such a solution 

would involve, I will merely say that I find 

that, dealing only with those sums of money 

that are multiples of threepence, if we only use 

bronze coins any sum can be paid in (» -f i)2 ways 

where n always represents the number of pence. 

If threepenny-pieces are admitted, there are 

2n3+i5n2-f-33» , , ^, . 

^g — hi ways. If sixpences are also 

used there are ^^+22n3+i59n=^-f4i4n-f-2i6 

216 
ways, when the sum is a multiple of sixpence, 
and the constant, 216, changes to 324 when the 
money is not such a multiple. And so the 
formulas increase in complexity in an acceler- 
ating ratio as we go on to the other coins. 

I will, however, add an interesting little table 
of the possible ways of changing our current 
coins which I believe has never been given in a 
book before. Change may be given for a 



Farthing in 
Halfpenny in . 
Penny in . . . 
Threepenny-piece in 
Sixpence in 
Shilling in . 
Florin in . . . 
Half-crown in . 
Double florin in . 
Crown in . . . 
Half-sovereign in . 
Sovereign in . 



way. 

1 way. 
3 ways. 
16 ways. 
66 ways. 
402 ways. 
3,818 ways. 
8,709 ways, 
60,239 ways. 
166,651 ways. 
6,261,622 ways. 
500,291,833 ways. 



It is a little surprising to find that a sovereign 
may be changed in over five hundred million 
different ways. But I have no doubt as to 
the correctness of my figures. 

33.— A PUZZLE IN REVERSALS. 

(i) £i3' (2) £23, 19s. iid. The words " the 
number of pounds exceeds that of the pence " 
exclude such sums of money as £2, i6s. 2d. and 
all sums under £1. 

34.— THE GROCER AND DRAPER. 

The grocer was delayed half a minute and the 
draper eight minutes and a half (seventeen 
times as long as the grocer), making together 
nine minutes. Now, the grocer took twenty- 
four minutes to weigh out the sugar, and, with 
the half-minute delay, spent 24 min. 30 sec. 
over the task ; but the draper had only to make 
forty-seven cuts to divide the roll of cloth, 
containing forty-eight yards, into yard pieces ! 
This took him 15 min. 40 sec, and when we add 
the eight minutes and a half delay we get 
24 min. 10 sec, from which it is clear that the 
draper won the race by. twenty seconds. The 
majority of solvers make forty-eight cuts to 
divide the roU into forty-eight pieces ! 

35.— JUDKINS'S CATTLE. 

As there were five droves with an equal number 
of animals in each, drove, the number must be 
divisible by 5 ; and as every one of the eight 
dealers bought the same number of animals, 
the number must be divisible by 8. Therefore 
the number must be a multiple of 40. The 
highest possible multiple of 40 that will work 
will be foimd to be 120, and this number could 
be made up in one of two ways — i ox, 23 pigs, 
and 96 sheep, or 3 oxen, 8 pigs, and 109 sheep. 
But the first is excluded by the statement that 
the animals consisted of " oxen, pigs, and 
sheep," because a single ox is not oxen. There- 
fore the second grouping is the correct answer. 

36.— BUYING APPLES. 

As there were the same number of boys as girls, 
it is clear that the number of children must be 
even, and, apart from a careful and exact read- 
ing of the question, there would be three differ- 
ent answers. There might be two, six, or four- 
[ teen children. In the first of these cases there 



152 



AMUSEMENTS IN MATHEMATICS. 



are ten different ways in which the apples could 
be bought. But we were told there was an 
equal number of " boys and girls," and one boy 
and one girl are not boys and girls, so this case 
has to be excluded. In the case of fourteen 
children, the only possible distribution is that 
each child receives one halfpenny apple. But 
we were told that each child was to receive an 
equal distribution of " apples," and one apple 
is not apples, so this case has also to be excluded. 
We are therefore driven back on our third case, 
which exactly fits in with all the conditions. 
Three boys and three girls each receive i half- 
penny apple and 2 third-penny apples. The 
value of these 3 apples is one penny and one- 
sixth, which multiplied by six makes seven- 
pence. Consequently, the correct answer is 
that there were six children — three girls and 
three boys. 

37._BUYING CHESTNUTS. 

In solving this little puzzle we are concerned 
with the exact interpretation of the words used 
by the buyer and seller. I wiU give the question 
again, this time adding a few words to make 
the matter more clear. The added words are 
printed in italics. 

" A man went into a shop to buy chestnuts. 
He said he wanted a pennyworth, and was given 
five chestnuts. * It is not enough ; I ought to 
have a sixth of a chestnut more,^ he remarked. 
' But if I give you one chestnut more,' the 
shopman replied, " you will have f\.vQ-sixths too 
many.' Now, strange to say, they were both 
right. How many chestnuts should the buyer 
receive for half a crown ? " 

The answer is that the price was 155 chest- 
nuts for half a crown. Divide this number by 
30, and we find that the buyer was entitled to 
5^ chestnuts in exchange for his penny. He 
was, therefore, right when he said, after receiv- 
ing five only, that he stiU wanted a sixth. And 
the salesman was also correct in saying that if 
he gave one chestnut more (that is, six chestnuts 
in all) he would be giving five-sixths of a chest- 
nut in excess. 

38.— THE BICYCLE THIEF. 

People give all sorts of absurd answers to this 
question, and yet it is perfectly simple if one 
just considers that the salesman cannot possibly 
have lost more than the cyclist actually stole. 
The latter rode away with a bicycle which cost 
the salesman eleven pounds, and the ten poimds 
" change ; " he thus made off with twenty-one 
pounds, in exchange for a worthless bit of paper. 
This is the exact amount of the salesman's loss, 
and the other operations of changing the cheque 
and borrowing from a friend do not affect the 
question in the slightest. The loss of prospec- 
tive profit on the sale of the bicycle is, of course, 
not direct loss of money out of pocket. 

39.— THE COSTERMONGER'S PUZZLE. 

Bill must have paid 8s. per hundred for his 
oranges — that is, 125 for los. At 8s. 4d. per 



hundred, he would only have received 120 
oranges for los. This exactly agrees with Bill's 
statement. 

40.— MAMMA'S AGE. 

The age of Manmaa must have been 29 years 
2 months ; that of Papa, 35 years ; and that 
of the child, Tommy, 5 years 10 months. 
Added together, these make seventy years. 
The father is six times the age of the son, and, 
after 23 years 4 months have elapsed, their united 
ages will amount to 140 years, and Tommy will 
be just half the age of his father. 

41.— THEIR AGES. 

The gentleman's age must have been 54 years 
and that of his wife 45 years. 

42.— THE FAMILY AGES. 

The ages were as follows : Billie, 3^ years ; 
Gertrude, I J year ; Henrietta, 5 J years ; Charlie, 
10^ years ; and Janet, 21 years. 

43.— MRS. TIMPKINS'S AGE. 

The age of the younger at marriage is always 
the same as the num^ber of years that expire 
before the elder becomes twice her age, if he 
was three times as old at marriage. In our 
case it was eighteen years afterwards ; there- 
fore Mrs. Timpkins was eighteen years of age 
on the wedding-day, and her husband fifty-four. 

44.— A CENSUS PUZZLE. 

Miss Ada Jorkins must have been twenty- 
four and her little brother Johnnie three years 
of age, with thirteen brothers and sisters be- 
tween. There was a trap for the solver in the 
words " seven times older than little Johnnie." 
Of course, " seven times older " is equal to 
eight times as old. It is surprising how many 
people hastily assume that it is the same as 
" seven times as old." Some of the best writers 
have committed this blunder. Probably many 
of my readers thought that the ages 24 i and 3 J 
were correct. 

45.— MOTHER AND DAUGHTER. 

In four and a half years, when the daughter will 
be sixteen years and a half and the mother forty- 
nine and a half years of age. 

46.— MARY AND MARMADUKE. 

Marmaduke's age must have been twenty-nine 
years and two-fifths, and Mary's nineteen years 
and three-fifths. When Marmaduke was aged 
nineteen and three-fifths, Mary was only nine 
and four-fifths ; so Marmaduke was at that 
time twice her age. 

47.— ROVER'S AGE. 

Rover's present age is ten years and Mildred's 
thirty years. Five years ago their respective 



SOLUTIONS. 



153 



ages were five and twenty-five. Remember 
that we said " four times older than the dog," 
which is the same as " five times as old." (See 
answer to No. 44.) 

48.— CONCERNING TOMMY'S AGE. 

Tommy Smart's age must have been nine years 
and three-fifths. Ann's age was sixteen and 
four-fifths, the mother's thirty-eight and two- 
fifths, and the father's fifty and two-fifths. 

49.— NEXT-DOOR NEIGHBOURS. 

Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 
13 ; Mr. Simkin 42 ; Mrs. Simkin 40 ; Sophy 
10 ; and Sammy 8. 

50.— THE BAG OF NUTS. 

It wiU be found that when Herbert takes twelve, 
Robert and Christopher will take nine and four- 
teen respectively, and that they will have to- 
gether taken thirty-five nuts. As 35 is con- 
tained in 770 twenty-two times, we have merely 
to multiply 12, 9,'and 14 by 22 to discover that 
Herbert's share was 264, Robert's 198, and 
Christopher's 308. Then, as the total of their 
ages is 17J years or half the sum of 12, 9, and 14, 
their respective ages must be 6, 4J, and 7 years. 

51.— HOW OLD WAS MARY ? 

The age of Mary to that of Ann must be as 
5 to 3. And as the sum of their ages was 44, 
Mary was 27^ and Ann i6|-. One is exactly 11 
years older than the other. I will now insert in 
brackets in the original statement the various 
ages specified : " Mary is (27!) twice as old as 
Ann was (13I) when Mary was half as old (24I) as 
Ann will be (49 1) when Ann is three times as old 
(49 J) as Mary was (16 J) when Mary was (i6|) 
three times as old as Ann (5i)." Now, check 
this backwards. When Mary was three times 
as old as Ann, Mary was i6| and Ann 5^ (11 
years younger). Then we get 49 1 for the age 
Ann will be when she is three times as old as 
Mary was then. When Mary was half this she 
was 24|. And at that time Ann must have 
been 13! (11 years younger). Therefore Mary 
is now twice as old — 27 J, and Ann 11 years 
younger — 16 J. 

52.— QUEER RELATIONSHIPS. 

If a man marries a woman, who dies, and he 
then marries his deceased wife's sister and him- 
self dies, it may be correctly said that he had 
(previously) married the sister of his widow. 

The youth was not the nephew of Jane Brown, 
because he happened to be her son. Her sur- 
name was the same as that of her brother, 
because she had married a man of the same 
name as herself. 

53.— HEARD ON THE TUBE RAILWAY. 
The gentleman was the second lady's uncle. 



54.— A FAMILY PARTY. 

The party consisted of two little girls and a boy, 
their father and mother, and their father's 
father and mother. 

55.— A MIXED PEDIGREE. 

Thos. Bloggs m 



W. Snoggs m Kate Bloggs. 



in Henry Bloggs. 



Joseph Bloggs m 



Jane John 

Bloggs m Snoggs 



Alf. Mary 

Snoggs fn Bloggs 



The letter m stands for " married." It wiU 
be seen that John Snoggs can say to Joseph 
Bloggs, " You are my father's brother-in-law, 
because my father married your sister Kate ; 
you are my brother's father-in-law, because my 
brother Alfred married your daughter Mary ; 
and you are my father-in-law's brother, because 
my wife Jane was your brother Henry's 
daughter." 

56.— WILSON'S POSER. 

If there are two men, each of whom marries 
the mother of the other, and there is a son of 
each marriage, then each of such sons will be 
at the same time uncle and nephew of the other. 
There are other ways in which the relationship 
may be brought about, but this is the simplest. 

57.— WHAT WAS THE TIME ? 

The time must have been 9.36 p.m. A quarter 
of the time since noon is 2 hr. 24 min., and a half 
of the time till noon next day is 7 hr. 12 min. 
These added together make 9 hr. 36 min. 

58.— A TIME PUZZLE. 
Twenty-six minutes. 

59.— A PUZZLING WATCH. 

If the 65 minutes be counted on the face of the 
same watch, then the problem would be im- 
possible : for the hands must coincide every 
65^^ minutes as shown by its face, and it 
matters not whether it runs fast or slow ; but 
if it is measured by true time, it gains ^ of a 
minute in 65 minutes, or -^^ of a minute per 
hour. 

60.— THE WAPSHAW'S WHARF MYSTERY. 

There are eleven different times in twelve hours 
when the hour and minute hands of a clock are 
exactly one above the other. If we divide 12 



154 



AMUSEMENTS IN MATHEMATICS. 



hours by II we get i hr. 5 min. 2y^j sec, and 
this is the time after twelve o'clock when they 
are first together, and also the time that elapses 
between one occasion of the hands being to- 
gether and the next. They are together for 
the second time at 2 hr. 10 min. 54^ sec. (twice 
the above time) ; next at 3 hr. 16 min. 21^^ sec. ; 
next at 4 hr. 21 min. 49^ sec. This last is the 
only occasion on which the two hands are to- 
gether with the second hand " just past the 
forty-ninth second." This, then, is the time 
at which the watch must have stopped. Guy 
Boothby, in the opening sentence of his Across 
the World for a Wife, says, "It was a cold, dreary 
winter's afternoon, and by the time the hands 
of the clock on my mantelpiece joined forces and 
stood at twenty minutes past four, my cham- 
bers were well-nigh as dark as midnight." It is 
evident that the author here made a slip, for, 
as we have seen above, he is i min. 49^ sec. 
out in his reckoning. 

61.— CHANGING PLACES. 

There are thirty-six pairs of times when the 
hands exactly change places between three p.m. 
and midnight. The number of pairs of times 
from any hour (m) to midnight is the sum of 12 
— («-l-i) natural nimibers. In the case of the 
puzzle «=3 ; therefore 12 — (3-fi) = 8 and 1 + 
2 -f3 -1-4-1-5+6-1-7-1-8 = 36, the required answer. 
The first pair of times is 3 hr. 21^^^ min, and 
4 hr. 16^ min., and the last pair is 10 hr. 593^ 
min. and 11 hr. 54^-|| min. I wiU not give all 
the remainder of the thirty-six pairs of times, 
but supply a formula by which any of the sixty- 
six pairs that occur from midday to midnight 
may be at once found : — 

a hr. ■ mm. and hr. — ! mm. 

143 143 

For the letter a may be substituted any hour 
from o, I, 2, 3 up to 10 (where nought stands for 
12 o'clock midday) ; and h may represent any 
hour, later than a, up to 11. 

By the aid of this formiila there is no difficulty 
in discovering the answer to the second ques- 
tion : a=8 and &=ii will give the pair 8 hr. 
58^1 min, and 11 hr. 44^1 min,, the latter being 
the time when the minute hand is nearest of 
all to the point IX — in fact, it is only ^^ of a 
minute distant. 

Readers may find it instructive to make a 
table of aU the sixty-six pairs of times when the 
hands of a clock change places. An easy way 
is as follows : Make a column for the first times 
and a second column for the second times of the 
pairs. By making a=o and 6=1 in the above 
expressions we find the first case, and enter o hr. 
SyI^ min. at the head of the first column, and 
I hr. o^^ min. at the head of the second column. 
Now, by successively adding Sy^ min. in the 
first, and x hr. ofi^ min. in the second column, 
we get all the eleven pairs in which the first time 
is a certain number of minutes after nought, or 
mid-day. Then there is a "jump" in the times, 
but you can find the next pair by making a=i 
and 6=2, and then by successively adding these 
two times as before you will get all the ter^ pairs 



after i o'clock. Then there is another " jump," 
and you wiU be able to get by addition all the 
nine pairs after 2 o'clock. And so on to the end. 

I wiU leave readers to investigate for them- 
selves the nature and cause of the " jumps." 
In this way we get xmder the successive hours, 

II + 10 + 9 + 8 -t- 7-1- 6 + 5 -I- 4 -i- 3 4- 2 -f I = 66 
pairs of times, which result agrees with the 
formula in the first paragraph of this article. 

Some time ago the principal of a Civil Service 
Training CoUege, who conducts a " Civil Service 
Column " in one of the periodicals, had the 
query addressed to him, " How soon after XII 
o'clock wiU a clock with both hands of the same 
length be ambiguous ? " His first answer was, 
" Some time past one o'clock," but he varied the 
answer from issue to issue. At length some of 
his readers convinced him that the answer is, 
"At 5t|^ min. past XII;" and this he finally 
g^ve as correct, together with the reason for it 
that at that time the time indicated is the same 
whichever hand you may assume as hour hand ! 

62.— THE CLUB CLOCK. 

The positions of the hands shown in the illustra- 
tion could only indicate that the clock stopped 
at 44 min. 5iH|| sec. after eleven o'clock. 
The second hand would next be " exactly 
midway between the other two hands " at 45 
min. 52^fj sec. after eleven o'clock. If we had 
been dealmg with the points on the circle to 
which the three hands are directed, the answer 
would be 45 min. 22^'^^ sec. after eleven ; but 
the question applied to the hands, and the 
second hand would not be between the others 
at that time, but outside them. 

63.— THE STOP-WATCH. 

The time indicated on the watch was 5^ min. 
past 9, when the second hand would be at 27^ 
sec. The next time the hands would be similar 
distances apart would be 54tx min, past 2, when 
the second hand would be at 32^^ sec. But 
you need only hold the watch (or our previous 
illustration of it) in front of a mirror, when you 
will see the second time reflected in it ! Of 
course, when reflected, you wiU read XI as I, 
X as II, and so on. 

64.— THE THREE CLOCKS. 

As a mere arithmetical problem this question 
presents no difficulty. In order that the hands 
shaU all point to twelve o'clock at the same 
time, it is necessary that B shall gain at least 
twelve hours and that C shall lose twelve homrs. 
As B gains a minute in a day of twenty-four 
hours, and C loses a minute in precisely the same 
time, it is evident that one will have gained 
720 minutes (just twelve hours) in 720 days, 
and the other will have lost 720 minutes in 720 
days. Clock A keeping perfect time, all three 
clocks must indicate twelve o'clock simultane- 
ously at noon on the 720th day from April i, 
1898. What day of the month will that be ? 
I published this little puzzle in 1898 to see 



SOLUTIONS. 



^55 



how many people were aware of the fact that 
1900 would not be a leap year. It was surpris- 
ing how many were then ignorant on the point. 
Every year that can be divided by four without 
a remainder is bissextile or leap year, with the 
exception that one leap year is cut off in the 
century. 1800 was not a leap year, nor was 
1900. On the other hand, however, to make 
the calendar more nearly agree with the sun's 
course, every fourth hundred year is still con- 
sidered bissextile. Consequently, 2000, 2400, 
2800, 3200, etc., wiU all be leap years. May 
my readers live to see them. We therefore find 
that 720 days from noon of April i, 1898, brings 
us to noon of March 22, 1900. 

65.— THE RAILWAY STATION CLOCK. 

The time must have been 43^ min. past two 
o'clock. 

66.— THE VILLAGE SIMPLETON. 

The day of the week on which the conversation 
took place was Sunday. For when the day 
after to-morrow (Tuesday) is " yesterday," 
" to-day " will be Wednesday ; and when the 
day before yesterday (Friday) was " to- 
morrow," " to-day " was Thursday. There are 
two days between Thursday and Sunday, and 
between Sunday and Wednesday. 

67.— AVERAGE SPEED. 

The average speed is twelve miles an hour, not 
twelve and a half, as most people will hastily 
declare. Take any distance you like, say sixty 
miles. This would have taken six hours going 
and four hours returning. The double Journey 
of 120 miles would thus take ten hours, and the 
average speed is clearly twelve miles an hour. 

68.— THE TWO TRAINS. 

One train was running just twice as fast as the 
other. 

69.— THE THREE VILLAGES. 

Calling the three villages by their initial letters, 
it is clear that the three roads form a triangle, 
A, B, C, with a perpendicular, measuring twelve 
miles, dropped from C to the base A, B. This 
divides our triangle into two right-angled tri- 
angles with a twelve-mile side in common. It 
is then found that the distance from A to C is 
15 miles, from C to B 20 miles, and from A to B 
25 (that is 9 and 16) miles. These figures are 
easily proved, for the square of 12 added to the 
square of 9 equals the square of 15, and the 
square of 12 added to the square of 16 equals 
the square of 20. 

70.— DRAWING HER PENSION. 
The distance must be 6| miles. 

71.— SIR EDWYN DE TUDOR. 

The distance must have been sixty miles. If 
Sir Edwyn left at noon and rode 15 miles an 



hour, he would arrive at four o'clock — an hour 
too soon. If he rode 10 miles an hour, he would 
arrive at six o'clock — an hour too late. But if 
he went at 12 miles an hour, he would reach the 
castle of the wicked baron exactly at five o'clock 
— ^the time appointed. 

72.— THE HYDROPLANE QUESTION. 

The machine must have gone at the rate of 
seven-twenty-fourths of a mile per minute and 
the wind travelled five-twenty-fourths of a mile 
per minute. Thus, going, the wind would help, 
and the machine would do twelve-twenty- 
fourths, or half a mile a minute, and returning 
only two-twenty-fourths, or one-twelfth of a 
mile per minute, the wind being against it. 
The machine without any wind could therefore 
do the ten miles in thirty- four and two-sevenths 
minutes, since it could do seven miles in twenty- 
four minutes. 

73.— DONKEY RIDING. 

The complete mile was run in nine minutes. 
From the facts stated we cannot determine the 
time taken over the first and second quarter- 
miles separately, but together they, of course, 
took four and a half minutes. The last two 
quarters were run in two and a quarter minutes 
each. 

74.— THE BASKET OF POTATOES. 

Multiply together the number of potatoes, the 
number less one, and twice the number less one, 
then divide by 3. Thus 50, 49, and 99 multi- 
plied together make 242,550, which, divided by 
3, gives us 80,850 yards as the correct answer. 
The boy would thus have to travel 45 miles 
and fifteen-sixteenths — a nice little recreation 
after a day's work. 

75.— THE PASSENGER'S FARE. 

Mr. Tompkins should have paid fifteen shillings 
as his correct share of the motor-car fare. He 
only shared half the distance travelled for £3, 
and therefore should pay half of thirty shillings, 
or fifteen shillings. 

76.— THE BARREL OF BEER. 

Here the digital roots of the six numbers are 
6, 4, 1, 2, 7, 9, which together sum to 29, whose 
digital root is 2. As the contents of the barrels 
sold must be a number divisible by 3, if one 
buyer purchased twice as much as the other, 
we must find a barrel with root 2, 5, or 8 to set 
on one side. There is only one barrel, that 
containing 20 gallons, that fulfils these condi- 
tions. So the man must have kept these 20 
gallons of beer for his own use and sold one 
man 33 gallons (the 18 -gallon and 15 -gallon 
barrels) and sold the other man 66 gallons (the 
16, 19, and 31 gallon barrels). 

77.— DIGITS AND SQUARES. 

The top row must be one of the four following 
numbers : 192, 219, 273, 327. The first was 
the example given. 



156 



AMUSEMENTS IN MATHEMATICS. 



78.— ODD AND EVEN DIGITS. 

As we have to exclude complex and improper 
fractions and recurring decimals, the simplest 
solution is this : 79 + 5 J and 84 + f, both equal 
84J. Without any use of fractions it is obvi- 
ously impossible. 

79.— THE LOCKERS PUZZLE. 

The smallest possible total is 356=107 + 249, 
and the largest sum possible is 981 = 235+746, 
or 657 + 324. The middle sum may be either 720 
= 134 + 586, or7o2 = 134 + 568, or 407= 138 + 269. 
The total in this case must be made up of three 
of the figures o, 2, 4, 7, but no sum other than 
the three givai can possibly be obtained. We 
have therefore no choice in the case of the first 
locker, an alternative in the case of the third, 
and any one of three arrangements in the case 
of the middle locker. Here is one solution : — 



107 
249 

356 



134 
586 

720 



235 



981 



Of course, in each case figures in the first two 
lines may be exchanged vertically without alter- 
ing the total, and as a result there are just 3,072 
different ways in which the figures might be 
actually placed on the locker doors. I must 
content myself with showing one little prin- 
ciple involved in this puzzle. The sum of the 
digits in the total is always governed by the 
digit omitted. f-^-J^-^^-^-^-^y-,^ 
— ^ — Ts- Whichever digit shown here in the 
upper line we omit, the sum of the digits in the 
total will be fovmd beneath it. Thus in the 
case of locker A we omitted 8, and the figures 
in the total sum up to 14, If, therefore, we 
wanted to get 356, we may know at once to a 
certainty that it can only be obtained (if at all) 
by dropping the 8. 

80.-— THE THREE GROUPS. 

There are nine solutions to this puzzle, as 
follows, and no more : — 

12x483 = 5,796 27x198 = 5,346 

42x138 = 5,796 39x186=7,254 

18x297=5,346 48x159=7,632 

28 X 157=4,396 

4x1,738 = 6,952 

4x1,963 = 7,852 

The seventh answer is the one that is most 
likely to be overlooked by solvers of the puzzle. 

81.— THE NINE COUNTERS. 

In this case a certain amount of mere " trial " 
is unavoidable. But there are two kinds of 
" trials " — those that are purely haphazard, 
and those that are methodical. The true 
puzzle lover is never satisfied with mere hap- 
hazard trials. The reader will find that by just 
reversing the figures in 23 and 46 (making the 
mviltipliers 32 and 64) both products will be 



5,056. This is an improvement, but it is not 
the correct answer. We can get as large a 
product as 5,568 if we multiply 174 by 32 and 
96 by 58, but this solution is not to be found 
without the exercise of some judgment and 
patience. 

82.— THE TEN COUNTERS. 

As I pointed out, it is quite easy so to arrange 
the counters that they shall form a pair of 
simple multiplication sums, each of which wiU 
give the same product — in fact, this can be done 
by anybody in five minutes with a little patience. 
But it is quite another matter to find that pair 
which gives the largest product and that which 
gives the smallest product. 

Now, in order to get the smallest product, it 
is necessary to select as multipliers the two 
smallest possible numbers. If, therefore, we 
place I and 2 as multipliers, all we have to do 
is to arrange the remaining eight counters in 
such a way that they shall form two numbers, 
one of which is just double the other ; and in 
doing this we must, of course, try to make the 
smaller number as low as possible. Of course 
the lowest number we could get would be 
3,045 ; but this will not work, neither will 
3,405, 3,450, etc., and it may be ascertained 
that 3,485 is the lowest possible. One of the 
required answers is 3,485 x 2 = 6,970, and 
6,970 X 1 = 6,970. 

The other part of the puzzle (finding the pair 
with the highest product) is, however, the real 
knotty point, for it is not at all easy to discover 
whether we should let the multiplier consist 
of one or of two figures, though it is clear that 
we must keep, so far as we can, the largest 
figures to the left in both multiplier and multi- 
plicand. It will be seen that by the follow- 
ing arrangement so high a number as 58,560 
may be obtained. Thus, 915 x 64=58,560, and 
732X80=58,560. 

83.— DIGITAL MULTIPLICATION. 

The solution that gives the smallest possible 
sum of digits in the common product is 23 
X 174=58 X 69=4,002, and the solution that 
gives the largest possible sum of digits, 9 X 654 
= 18x327=5,886. In the first case the digits 
sum to 6 and in the second case to 27. There is 
no way of obtaining the solution but by actual 
trial. 

84.— THE PIERROT'S PUZZLE. 

There are just six different solutions to this 
puzzle, as follows : — 

8 multiplied by 473 equals 3784 

9 
15 
21 

27 
35 

It will be seen that in every case the two multi- 
pliers contain exactly the same figures as the 
product. 



93 




1395 


87 




1827 


81 




2187 


41 




1435 



SOLUTIONS. 



157 



85.— THE CAB NUMBERS. 

The highest product is, I think, obtained 
by multiplying 8,745,231 by 96 — namely, 
839.542,176. 

Dealing here with the problem generally, 
I have shown in the last puzzle that with three 
digits there are only two possible solutions, 
and with four digits only six different solu- 
tions. 

These cases have all been given. With five 
digits there are just twenty-two solutions, as 
follows : — 

3 X 4128 = 12384 

3 X 4281 = 12843 

3 X 7125 = 21375 

3 X 7251 = 21753 

2541 X 6 = 15246 

651 X 24 = 15624 

678 X 42 = 28476 

246 X 51 = 12546 

57 X 834 = 47538 

75 X 231 = 17325 

624 X 78 = 48672 

435 X 87 = 37845 



9 
72 


X 
X 


7461 
936 


= 


67149 
67392 


2 

2 

65 
65 


X 
X 
X 
X 


8714 
8741 
281 

983 


= 


17428 

17482 
18265 
63895 


4973 
6521 

14 
86 


X 
X 
X 
X 


8 
8 

926 
251 


= 


39784 
52168 
12964 
21586 



Now, if we took every possible combination 
and tested it by multiplication, we should need 
to make no fewer than 30,240 trials, or, if we at 
once rejected the number i as a multiplier, 
28,560 trials — a task that I think most people 
would be inclined to shirk. But let us consider 
whether there be no shorter way of getting at the 
results required, I have already explained that 
if you add together the digits of any number 
and then, as often as necessary, add the digits of 
the result, you must ultimately get a number 
composed of one figure. This last number I 
call the " digital root." It is necessary in every 
solution of our problem that the root of the sum 
of the digital roots of our multipliers shall be 
the same as the root of their product. There 
are only four ways in which this can happen : 
when the digital roots of the multipliers are 3 
and 6, or 9 and 9, or 2 and 2, or 5 and 8. I have 
divided the twenty-two answers above into these 
four classes. It is thus evident that the digital 
root of any product in the first two classes must 
be 9, and in the second two classes 4. 

Owing to the fact that no number of five 
figures can have a digital sum less than 15 or 
more than 35, we find that the figures of our 
product must sum to either 18 or 27 to produce 
the root 9, and to either 22 or 31 to produce 
the root 4. There are 3 ways of selecting five 



different figures that add up to 18, there are 11 
ways of selecting five figures that add up to 27, 
there are 9 ways of selecting five figures that 
add up to 22, and 5 ways of selecting five figures 
that add up to 31. There are, therefore, 28 
different groups, and no more, from any one of 
which a product may be formed. 

We next write out in a column these 28 sets 
of five figures, and proceed to tabulate the pos- 
sible factors, or multipliers, into which they 
may be split. Roughly speaking, there would 
now appear to be about 2,000 possible cases to 
be tried, instead of the 30,240 mentioned above ; 
but the process of elimination now begins, and 
if the reader has a quick eye and a clear head 
he can rapidly dispose of the large bulk of these 
cases, and there will be comparatively few test 
multiplications necessary. It would take far 
too much space to explain my own method in 
detail, but I will take the first set of figures in 
my table and show how easily it is done by 
the aid of little tricks and dodges that should 
occur to everybody as he goes along. 

My first product group of five figures is 
84,321. Here, as we have seen, the root of 
each factor must be 3 or a multiple of 3. As 
there is no 6 or 9, the only single multiplier is 3. 
Now, the remaining four figures can be arranged 
in 24 different ways, but there is no need to 
make 24 multiplications. We see at a glance 
that, in order to get a five-figure product, either 
the 8 or the 4 must be the first figure to the left. 
But unless the 2 is preceded on the right by 
the 8, it will produce when multiplied either a 6 
or a 7, which must not occur. We are, there- 
fore, reduced at once to the two cases, 3x4,128 
and 3x4,281, both of which give correct solu- 
tions. Suppose next that we are trying the 
two-figure factor, 21. Here we see that if the 
number to be multiplied is under 500 the pro- 
duct will either have only four figures or begin 
with 10. Therefore we have only to examine 
the cases 21 X 843 and 21 x 834. But we know 
that the first figure will be repeated, and that 
the second figure wiU be twice the first figure 
added to the second. Consequently, as twice 3 
added to 4 produces a nought in our product, the 
first case is at once rejected. It only remains to 
try the remaining case by multiplication, when 
we find it does not give a correct answer. If we 
are next trying the factor 12, we see at the start 
that neither the 8 nor the 3 can be in the units 
place, because they would produce a 6, and so 
on. A sharp eye and an alert judgment will 
enable us thus to run through our table in a 
much shorter time than would be expected. 
The process took me a little more than three 
hours. 

I have not attempted to enumerate the solu- 
tions in the cases of six, seven, eight, and nine 
digits, but I have recorded nearly fifty examples 
with nine digits alone. 

86.— QUEER MULTIPLICATION. 

If we multiply 32547891 by 6, we get the pro- 
duct, 195287346. In both cases all the nine 
digits are used once and once only. 



158 



AMUSEMENTS IN MATHEMATICS. 



87.— THE NUMBER CHECKS PUZZLE. 

Divide the ten checks into the following three 
groups: 7 I 5 — 4 6 — 3 2890, and the first 
multiplied by the second produces the third. 

88.— DIGITAL DIVISION. 

It is convenient to consider the digits as ar- 
ranged to form fractions of the respective values, 
one-half, one-third, one-fourth, one-fifth, one- 
sixth, one-seventh, one-eighth, and one-ninth. 
I will first give the eight answers, as follows : — 

^ms=h Tm,=h ^W^=i» 



2 5 40 8 ■ 



U t 



■AV» 



The sum of the nmnerator digits and the 
denominator digits will, of course, always be 
45, and the " digital root " is 9. Now, if we 
separate the nine digits into any two groups, 
the sum of the two digital roots will always be 
9. In fact, the two digital roots must be either 
9 — 9, 8 — I, 7 — 2, 6 — 3, or 5 — 4. In the first 
case the actual sum is 18, but then the digital 
root of this number is itself 9. The solutions 
in the cases of one-third, one-fourth, one-sixth, 
one-seventh, and one-ninth must be of the form 
9 — 9 ; that is to say, the digital roots of both 
numerator and denominator wiU be 9. In the 
cases of one-half and one-fifth, however, the 
digital roots are 6 — 3, but of course the higher 
root may occur either in the numerator or in 
the denominator; thus ^Wt, ^^> ^Z^, 
■ri^^^t where, in the first two arrangements, 
the roots of the numerator and denominator 
are respectively 6 — 3, and in the last two 
3 — 6. The most curious case of all is, perhaps, 
one-eighth, for here the digital roots may be of 
any one of the five forms given above. 

The denominators of the fractions being re- 
garded as the numerators multiplied by 2, 3, 4, 
5, 6, 7, 8, and 9 respectively, we must pay 
attention to the " carryings over." In order to 
get five figures in the product there will, of 
course, always be a carry-over after multiplying 
the last figmre to the left, and in every case higher 
than 4 we must carry over at least three times. 
Consequently in cases from one-fifth to one- 
ninth we cannot produce different solutions by 
a mere change of position of pairs of figures, 
as, for example, we may with iW^^ ^^'^ 
tWA> where the f and f change places. It is 
true that the same figures may often be dif- 
ferently arranged, as shown in the two pairs of 
values for one-fifth that I have given in the 
last paragraph, but here it will be found there 
is a general readjustment of figures and not a 
simple changing of the positions of pairs. There 
are other little points that would occur to every 
solver — such as that the figvire 5 cannot ever 
appear to the extreme right of the numerator, 
as this would result in our getting either a nought 
or a second 5 in the denominator. Similarly i 
cannot ever appear in the same position, nor 
6 in the fraction one-sixth, nor an even figure 
in the fraction one-fifth, and so on. The pre- 



liminary consideration of such points as I have 
touched upon wiU not only prevent our wasting 
a lot of time in trying to produce impossible 
forms, but wiU lead us more or less directly to 
the desired solutions. 

89.— ADDING THE DIGITS. 

The smallest possible sum of money is £1, 
8s. 9|d., the digits of which add to 25. 

90.— THE CENTURY PUZZLE. 

The problem of expressing the number 100 as 
a mixed number or fraction, usmg all the nine 
digits once, and once only, has, like all these 
digital puzzles, a fascinating side to it. The 
merest t5T:o can by patient trial obtain correct 
results, and there is a singular pleasure in dis- 
covering and recording each new arrangement 
akin to the delight of the botanist in finding 
some long-sought plant. It is simply a matter 
of arranging those nine figures correctly, and 
yet with the thousands of possible combinations 
that confront us the task is not so easy as might 
at first appear, if we are to get a considerable 
number of results. Here are eleven answers, 
including the one I gave as a specimen : — 

96V^«, 96¥5fe», 96S^, 94¥/^, 9iW^, 

91 W^, 91 5^, 82'WV, SiWirS 

81 W/, 3*MF. 

Now, as all the fractions necessarily repre- 
sent whole numbers, it wiU be convenient to 
deal with them in the following form : 96 -|- 4, 
94 + 6, 914-9, 82-1-18, 81 + 19, and 3 + 97. 

With any whole number the digital roots of 
the fraction that brings it up to 100 will always 
be of one particular form. Thus, in the case of 
96 + 4, one can say at once that if any answers 
are obtainable, then the roots of both the 
numerator and the denominator of the fraction 
wiU be 6. Examine the first three arrangements 
given above, and you will find that this is so. 
In the case of 94 + 6 the roots of the numerator 
and denominator wiU be respectively 3 — 2, in 
the case of 91+9 and of 82-i-i8 they wiU be 
9 — 8, in the case of 8i + ig they will be 9—9, 
and in the case of 3+97 they will be 3-— 3. 
Every fraction that can be employed has, there- 
fore, its particular digital root form, and you are 
only wasting your time in unconsciously at- 
tempting to break through this law. 

Every reader will have perceived that certain 
whole numbers are evidently impossible. Thus, 
if there is a 5 in the whole number, there 
will also be a nought or a second 5 in the 
fraction, which are barred by the conditions. 
Then multiples of 10, such as 90 and 80, 
cannot of course occur, nor can the whole 
number conclude with a 9, like 89 and 79, 
because the fraction, equal to 11 or 21, will 
have I in the last place, and will therefore 
repeat a figure. Whole numbers that repeat 
a figure, such as 88 and 77, are also clearly 
useless. These cases, as I have said, are all 
obvious to every reader. But when I declare 



SOLUTIONS. 



159 



that such combinations as 98+2, 92 + 8, 86 + 14, 
83 + 17, 74 + 26, etc., etc., are to be at once dis- 
missed as impossible, the reason is not so evi- 
dent, and I mifortunately cannot spare space 
to explain it. 

But when aU those combinations have been 
struck out that are known to be impossible, it 
does not follow that aU the remaining " possible 
forms " wiU actually work. The elemental 
form may be right enough, but there are other 
and deeper considerations that creep in to 
defeat our attempts. For example, 98 + 2 is 
an impossible combination, because we are able 
to say at once that there is no possible form for 
the digital roots of the fraction equal to 2. 
But in the case of 97+3 there is a possible 
form for the digital roots of the fraction, namely, 
6—5, and it is only on further investigation 
that we are able to determine that this form 
cannot in practice be obtained, owing to curious 
considerations. The working is greatly simpli- 
fied by a process of elimination, based on such 
considerations as that certain multiplications 
produce a repetition of figures, and that the 
whole number cannot be from 12 to 23 inclusive, 
since in every such case su£&ciently smaU de- 
nominators are not available for forming the 
fractional part. 

91.— MORE MIXED FRACTIONS. 

The point of the present puzzle lies in the fact 
that the numbers 15 and 18 are not capable of 
solution. There is no way of determining this 
without trial. Here are answers for the ten 
possible numbers : — 

9mf=i3; 9HU=u; 

I2\«^ = i6; 6ifffs=2o; 
i5Y^=27; 24Vi¥=36; 

27%%* = 40; 65^3'^ = 69; 

59^^W=72; 75Wi^ = 94. 

I have only found the one arrangement for 
each of the numbers 16, 20, and 27; but the 
other numbers are all capable of being solved 
in more than one way. As for 15 and 18, 
though these may be easily solved as a simple 
fraction, yet a " mixed fraction " assumes the 
presence of a whole number ; and though my 
own idea for dodging the conditions is the fol- 
lowing, where the fraction is both complex and 
mixed, it wiU be fairer to keep exactly to the 
form indicated : — 

3^=15; 9^#= 18. 

I have proved the possibility of solution for 
all numbers up to 100, except i, 2, 3, 4, 15, and 
18. The first three are easily shown to be im- 
possible. I have also noticed that numbers 
whose digital root is 8 — such as 26, 35, 44, 53, 
etc. — seem to lend themselves to the greatest 
number of answers. For the number 26 alone 
I have recorded no fewer than twenty-five 
different arrangements, and I have no doubt 
that there are many more. 



92.— DIGITAL SQUARE NUMBERS. 

So far as I know, there are no published tables 
of square numbers that go sufficiently high to 
be available for the purposes of this puzzle. 
The lowest square number containing aU the 
nine digits once, and once only, is 139,854,276, 
the square of 11,826. The highest square num- 
ber under the same conditions is, 923,187,456, 
the square of 30,384. 

93.— THE MYSTIC ELEVEN. 

Most people know that if the sum of the digits 
in the odd places of any number is the same as 
the siHn of the digits in the even places, then 
the number is divisible by 1 1 without remainder. 
Thus in 896743012 the odd digits, 20468, add up 
20, and the even digits, 1379, also add up 20. 
Therefore the number may be divided by 11. 
But few seem to know that if the difference 
between the sum of the odd and the even digits 
is II, or a multiple of 11, the rule equally applies. 
This law enables us to find, with a very little 
trial, that the smallest number containing nine 
of the ten digits (calling nought a digit) that is 
divisible by 11 is 102,347,586, and the highest 
number possible, 987,652,413. 

94.— THE DIGITAL CENTURY. 

There is a very large number of different ways 
in which arithmetical signs may be placed be- 
tween the nine digits, arranged in numerical 
order, so as to give an expression equal to 100. 
In fact, imless the reader investigated the 
matter very closely, he might not suspect that 
so many ways are possible. It was for this 
reason that I added the condition that not only 
must the fewest possible signs be used, but also 
the_ fewest possible strokes. In this way we 
limit the problem to a single solution, and arrive 
at the simplest and therefore (in this case) the 
best result. 

Just as in the case of magic squares there are 
methods by which we may write down with 
the neatest ease a large number of solutions, 
but not all the solutions, so there are several 
ways in which we may quickly arrive at 
dozens of arrangements of the " Digital Cen- 
tury," without finding aU the possible arrange- 
ments. There is, in fact, very little principle 
in the thing, and there is no certain way of 
demonstrating that we have got the best pos- 
sible solution. All I can say is that the ar- 
rangement I shall give as the best is the best 
I have up to the present succeeded in discover- 
ing. _ I wiU give the reader a few interesting 
specimens, the first being the solution usually 
published, and the last the best solution that I 
Icnow. 

Signs. Strokes. 

i+2+3+4+5 + 6 + 7 + (8X9)=ioo ( 9 .. 18) 

-(iX2)-3-4-5 + (6X7) + (8X9) 
= 100 (12 . . 20) 

i + (2X3) + (4X5)-6 + 7 + (8X9) 

= 100 (11 . . 21) 

(1+3-3-4) (5-6-7-8-9)=ioo ( 9 •• 12) 



i6o 



AlVIUSEMENTS IN MATHEMATICS. 



Signs. 


Strokes. 


COO . (8 


.. i6) 


^ . ( 7 


.. 13) 


. . ( 6 


.. II) 


. . ( 6 


.• 7) 


. . ( 4 


.. 6) 


. . ( 4 


.. 6) 


. . { 3 


.. 4) 



i + (2X 3) + 4 + 5 + 67 + 8 + 9 = 100 
(IX 2) + 34+56+7- 8+9=100 

12 + 3-4 + 5 + 67 + 8 + 9 = 100 
■123-4-5-6-7 + 8-9 = 100 
123 + 4-5 + 67-8-9 = 100 
123 + 45-67 + 8-9 = 100 . 
123-45-67 + 89 = 100 . . 

It will be noticed that in the above I have 
counted the bracket as one sign and two strokes. 
The last solution is singvilarly simple, and I do 
not think it will ever be beaten. 

95.— THE FOUR SEVENS. 

The way to write four sevens with simple arith- 
metical signs so that they represent 100 is as 
follows : — 

7 7 

-^ X - = 100. 

•7 7 

Of course the fraction, 7 over decimal 7, 
equals 7 divided by ^, which is the same 
as 70 divided by 7, or 10. Then 10 multiplied 
by 10 is 100, and there you are ! It will be seen 
that this solution applies equally to any aumber 
whatever that you may substitute for 7. 

96.— THE DICE NUMBERS. 

The sum of all the numbers that can be formed 
with any given set of four different figures is 
always 6,666 multiplied by the sum of the four 
figures. Thus, i, 2, 3, 4 add up 10, and ten 
times 6,666 is 66,660. Now, there are thirty- 
five different ways of selecting four figures from 
the seven on the dice — remembering the 6 and 
9 trick. The figures of all these thirty-five 
groups add up to 600. Therefore 6,666 multi- 
plied by 600 gives us 3,999,600 as the correct 
answer. 

Let us discard the dice and deal with the 
problem generally, using the nine digits, but ex- 
cluding nought. Now, if you were given simply 
the sum of the digits — that is, if the condition 
were that you could use any four figures so long as 
they summed to a given amount — then we have 
to remember that several combinations of four 
digits will, in many cases, make the same sum. 

10 II 12 13 14 15 16 17 18 19 20 
I I 2 3 5 6 8 9 II II 12 

21 22 23 24 25 26 27 28 29 30 
111198653211 

Here the top row of numbers gives all the 
possible sums of four different figures, and the 
bottom row the number of different ways in 
which each sum may be made. For example, 
13 may be made in three ways : 1237, 1246, and 
1345. It will be found that the numbers in the 
bottom row add up to 126, which is the number 
of combinations of nine figures taken four at 
a time. From this table we may at once cal- 
culate the answer to such a question as this : 



What is the sum of all the numbers composed of 
four different digits (nought excluded) that add 
up to 14 ? Multiply 14 by the number beneath 
it in the table, 5, and multiply the result by 
6,666, and you will have the answer. It follows 
that, to know the sum of all the numbers com- 
posed of four different digits, if you multiply 
all the pairs in the two rows and then add the 
results together, you will get 2,520, which, 
multiplied by 6,666, gives the answer 16,798,320. 
The following general solution for any number 
of digits will doubtless interest readers. Let n 
represent number of digits, then 5 (10" — i) |8 
divided by [ 9 — n equals the required sum. 
Note that |o^ equals i. This may be reduced 
to the following practical rule : Multiply to- 
gether 4x7x6x5... to (w— i) factors; now 
add (m + i) ciphers to the right, and from this 
result subtract the same set of figures with a 
single cipher to the right. Thus for m=4 (as 
in the case last mentioned), 4 x 7 X 6=168. 
Therefore 16,800,000 less 1,680 gives us 
16,798,320 in another way. 

97.— THE SPOT ON THE TABLE. 

The ordinary schoolboy would correctly treat 
this as a quadratic equation. Here is the 
actual arithmetic. Double the product of the 
two distances from the walls. This gives us 
144, which is the square of 12. The sum of the 
two distances is 17. If we add these two 
numbers, 12 and 17, together, and also subtract 
one from the other, we get the two answers 
that 29 or 5 was the radius, or half-diameter, of 
the table. Consequently, the full diameter was 
58 in. or 10 in. But a table of the latter dimen- 
sions would be absurd, and not at all in accord- 
ance with the illustration. Therefore the table 
must have been 58 in. in diameter. In this 
case the spot was on the edge nearest to the 
comer of the room — to which the boy was 
pointing. If the other answer were admissible, 
the spot would be on the edge farthest from the 
comer of the room. 

98.— ACADEMIC COURTESIES. 

There must have been ten boys and twenty 
girls. The number of bows girl to girl was 
therefore 380, of boy to boy 90, of girl with boy 
400, and of boys and girls to teacher 30, making 
together 900, as stated. It will be remembered 
that it was not said that the teacher himself 
returned the bows of any child, 

99.— THE THIRTY-THREE PEARLS. 

The value of the large central pearl must have 
been £3,000. The pearl at one end (from which 
they increased in value by £100) was £1,400 ; 
the pearl at the other end, £600. 

100.— THE LABOURER'S PUZZLE. 

The man said, " I am going twice as deep," not 
" as deep again." That is to say, he was still 
going twice as deep as he had gone already, sc 



SOLUTIONS. 



i6i 



that when finished the hole would be three 
times its present depth. Then the answer is 
that at present the hole is 3 ft. 6 in. deep and 
the man 2 ft. 4 in. above ground. When com- 
pleted the hole will be 10 ft. 6 in. deep, and 
therefore the man will then be 4 ft. 8 in. below 
the surface, or twice the distance that he is 
now above ground. 

loi.— THE TRUSSES OF HAY. 

Add together the ten weights and divide by 4, 
and we get 289 lbs. as the weight of the five 
trusses together. If we call the five trusses in 
the order of weight A, B, C, D, and E, the 
lightest being A and the heaviest E, then the 
lightest, no lbs., must be the weight of A and 
B ; and the next lightest, 112 lbs., must be the 
weight of A and C. Then the two heaviest, 
D and E, must weigh 121 lbs., and C and E 
must weigh 120 lbs. We thus know that A, B, 
D, and E weigh together 231 lbs., which, de- 
ducted from 289 lbs. (the weight of the five 
trusses), gives us the weight of C as 58 lbs. 
Now, by mere subtraction, we find the weight 
of each of the five trusses — 54 lbs., 56 lbs., 58 
lbs., 59 lbs., and 62 lbs. respectively. 

102.— MR. GUBBINS IN A FOG. 

The candles must have burnt for three hours 
and three-quarters. One candle had one- 
sixteenth of its total length left and the other 
four-sixteenths. 

103.— PAINTING THE LAMP-POSTS. 

Pat must have painted six more posts than 
Tim, no matter how many lamp-posts there 
were. For example, suppose twelve on each 
side ; then Pat painted fifteen and Tim nine. 
If a hundred on each side, Pat painted one 
himdred and three, and Tim only ninety-seven 

104.— CATCHING THE THIEF. 

The constable took thirty steps. In the same 
time the thief would take forty-eight, which, 
added to his start of twenty-seven, carried him 
seventy-five steps. This distance would be ex- 
actly equal to thirty steps of the constable. 

105.— THE PARISH COUNCIL ELECTION. 

The voter can vote for one candidate in 23 wa3rs, 
for two in 253 ways, for three in 1,771, for four 
in 8,855, for five in 33,649, for six in 100,947, for 
seven in 245,157, for eight in 490,314, and for 
nine candidates in 817,190 different ways. Add 
these together, and we get the total of 1,698,159 
ways of voting. 

106.— THE MUDDLETOWN ELECTION. 

The numbers of votes polled respectively by the 
Liberal, the Conservative, the Independent, and 
the Socialist were i,553, i,535, i,4o7, and 978. 
All that was necessary was to add the sum of 
the three majorities (739) to the total poll of 



(1.928) 



5,473 (making 6,212) and divide by 4, which 
gives us 1,553 as the poll of the Liberal. Then 
the polls of the other three candidates can, of 
course, be found by deducting the successive 
majorities from the last-mentioned number. 

107.— THE SUFFRAGISTS' MEETING. 

Eighteen were present at the meeting and 
eleven left. If twelve had gone, two-thirds 
would have retired. If only nine had gone, the 
meeting would have lost half its members. 

108.— THE LEAP-YEAR LADIES. 

Tnfe correct and only answer is that 11,616 
ladies made proposals of marriage. Here are 
aU the details, which the reader can check for 
himself with the original statements. Of 
10,164 spinsters, 8,085 married bachelors, 627 
married widowers, 1,221 were declined by 
bachelors, and 231 declined by widowers. Of 
the 1,452 widows, 1,155 married bachelors, 
and 297 married widowers. No widows were 
declined. The problem is not difficult, by 
algebra, when once we have succeeded in 
correctly stating it. 

109.— THE GREAT SCRAMBLE. 

The smallest number of sugar plums that will 
fulfil the conditions is 26,880. The five boys 
obtained respectively : Andrew, 2,863 ; Bob, 
6,335 ; Charlie, 2,438 ; David, 10,294 ; Edgar, 
4,950. There is a little trap concealed in the 
words near the end, " one-fifth of the same," 
that seems at first sight to upset the whole 
account of the affair. But a little thought 
will show that the words could only mean " one- 
fifth of five-eighths, the fraction last men- 
tioned " — that is, one-eighth of the three- 
quarters that Bob and Andrew had last ac- 
quired. 

no.— THE ABBOT'S PUZZLE. 

The only answer is that there were 5 men, 25 
women, and 70 children. There were thus 100 
persons in all, 5 times as many women as men, 
and as the men would together receive 15 
bushels, the women 50 bushels, and the children 
35 bushels, exactly 100 bushels would be dis- 
tributed. 

III.— REAPING THE CORN. 

The whole field must have contained 46.626 
square rods. The side of the central square, 
left by the farmer, is 4.8 284 rods, so it contains 
23.313 square rods. The area of the field was 
thus something more than a quarter of an acre 
and less than one-third ; to be more precise, 
.2 914 of an acre. 

112.— A PUZZLING LEGACY. 

As the share of Charles falls in through his 
death, we have merely to divide the whole 
hundred acres between Alfred and Benjamin in 
the proportion of one-third to one-fourth — that 
is in the proportion of four-twelfths to three- 



II 



1 62 



AMUSEMENTS IN MATHEMATICS. 



twelfths, which is the same as four to three. 
Therefore Alfred takes four-sevenths of the 
hundred acres and Benjamin three-sevenths. 

113.— THE TORN NUMBER. 

The other number that answers all the require- 
ments of the puzzle is 9,801. If we divide this 
in the middle into two numbers and add them 
together we get 99, which, multiplied by itself, 
produces 9,801. It is true that 2,025 may be 
treated in the same way, only this number is 
excluded by the condition which requires that 
no two figures should be alike. 

The general solution is curious. Call the 
number of figures in each half of the torn label 
«. Then, if we add i to each of the exponents 
of the prime factors (other than 3) of 10" — i 
(i being regarded as a factor with the constant 
exponent, i), their product will be the number 
of solutions. Thus, for a label of six figmres, 
n=3. The factors of lo^-i are iiX37^ (not 
considering the 3^), and the product of 2 X 2=4, 
the number of solutions. This always includes 
the special cases 98 — 01, 00 — 01, 998 — 01, 
000—001, etc. The solutions are obtained as 
follows : — Factorize 10' — i in all possible 
ways, always keeping the powers of 3 together, 
thus, 37x27, 999x1. Then solve the equa- 
tion s7x=27y + i. Here ;i;=i9 and y=26. 
Therefore, 19x37=703, the square of which 
gives one label, 494,209. A complementary 
solution (through 27«=37y+i) can at once be 
foxmd by 10" — 703=297, the square of which 
gives 088,209 for second label. (These non- 
significant noughts to the left must be included, 
though they lead to peculiar cases like 00238 — 
04641 = 48792, where 0238 — 4641 would not 
work.) The special case 999 X i we can write 
at once 998,001, according to the law shown 
above, by adding nines on one half and noughts 
on the other, and its complementary will be i 
preceded by five noughts, or 00000 1. Thus we 
get the squares of 999 and i. These are the four 
solutions. 

114.— CURIOUS NUMBERS. 

The three smallest numbers, in addition to 48, 
are 1,680, 57,120, and 1,940,448. It wiU be 
found that 1,681 and 841, 57,121 and 28,561, 
1,940,449 and 970,225, are respectively the 
squares of 41 and 29, 239 and 169, 1,393 and 
985. 

115.— A PRINTER'S ERROR. 

The answer is that 2^ .92 is the same as 2592, 
and this is the only possible solution to the 
puzzle. 

116.— THE CONVERTED MISER. 

As we are not told in what year Mr. Jasper 
Bullyon made the generous distribution of his 
accumulated wealth, but are required to find 
the lowest possible amount of money, it is clear 
that we must look for a year of the most favour- 
able form. 

There are four cases to be considered — an 
ordinary year with fifty-two Sundays and with 
fifty-three Svmdays, and a leap-year with fifty- 



two and fifty-three Sundays respectively. Here 
are the lowest possible amounts in each case : — 



313 weekdays, 52 Sundays 

312 weekdays, 53 Sundays 

314 weekdays, 52 Sundays 

313 weekdays, 53 Sundays 



. £112,055 
19,345 
( No solution 
( possible. 

£69,174 



The lowest possible amount, and therefore 1 
the correct answer, is £19,345, distributed in an \ 
ordinary year that began on a Sunday. The 
last year of this kind was 191 1. He would 
have paid £53 on every day of the year, or £62 
on every weekday, with £1 left over, as required, 
in the latter event. 

117.— A FENCE PROBLEM. 

Though this puzzle presents no great difficulty 
to any one possessing a knowledge of algebra, 
it has perhaps rather interesting features. 

Seeing, as one does in the illustration, just 
one corner of the proposed square, one is 
scarcely prepared for the fact that the field, in 
order to comply with the conditions, must 
contain exactly 501,760 acres, the fence re- 
quiring the same number of rails. Yet this is 
the correct answer, and the only answer, euid 
if that gentleman in Iowa carries out his in- 
tention, his field wiU be twenty-eight miles 
long on each side, and a little larger than the 
county of Westmorland. I am not aware that 
any limit has ever been fixed to the size of a 
" field," though they do not run so large as 
this in Great Britain. Still, out in Iowa, where 
my correspondent resides, they do these things 
on a very big scale. I have, however, reason 
to believe that when he finds the sort of task 
he has set himself, he -wiU decide to abandon it; 
for if that cow decides to roam to fresh woods 
and pastures new, the milkmaid may have to 
start out a week in advance in order to obtain 
the morning's milk. 

Here is a little rule that will always apply 
where the length of the rail is half a pole. 
Multiply the number of rails in a hmrdle by 
four, and the result is the exact number of 
miles in the side of a square field containing 
the same number of acres as there are rails in 
the complete fence. Thus, with a one-rail 
fence the field is four miles square ; a two-rail 
fence gives eight miles square; a three-rail 
fence, twelve miles square ; and so on, until 
we find that a seven-rail fence multiplied by 
four gives a field of twenty-eight miles square. 
In the case of our present problem, if the field 
be made smaller, then the number of rails will 
exceed the number of acres ; while if the field 
be made larger, the number of rails wiU be less 
than the acres of the field. 

118.— CIRCLING THE SQUARES. 

Though this problem might strike the novice 
as being rather difficult, it is, as a matter of 
fact, quite easy, and is made still easier by in- 
serting fomr out of the ten numbers. 



SOLUTIONS. 



163 



First, it will be found that squares that are 
diametrically opposite have a common differ- 
ence. For example, the difference between 
the square of 14 and the square of 2, in the 
diagram, is 192 ; and the difference between 
the square of 16 and the square of 8 is also 192. 
This must be so in every case. Then it should 
be remembered that the difference between 
squares of two consecutive numbers is always 
twice the smaller number plus i, and that the 
difference between the squares of any two 
numbers can always be expressed as the differ- 
ence of the numbers multiplied by their sum. 
Thus the square of 5 (25) less the square of 4 (16) 
equals (2X4)-f-i, or 9; also, the square of 7 
(49) less the square of 3 (9) equals (7+ 3) X (7 - 3)i 
0140. 

Now, the number 192, referred to above, may 
be divided into five different pairs of even 
factors : 2 X 96, 4 X 48, 6 x 32, 8 x 24, and 12 x 16, 
and these divided by 2 give us, 1x48, 2X24, 
3x16, 4x12, and 6x8. The difference and 
sum respectively of each of these pairs in turn 
produce 47, 49 ; 22, 26 ; 13, 19 ; 8, 16 ; and 
2, 14. These are the required numbers, four of 
which are already placed. The six numbers 
that have to be added may be placed in just 
six different ways, one of which is as follows, 
reading round the circle clockwise: 16, 2, 49, 22, 
19, 8, 14, 47, 26, 13. 

I will just draw the reader's attention to one 
other little point. In all circles of this kind, 
the difference between diametrically opposite 
numbers increases by a certain ratio, the first 
numbers (with the exception of a circle of 6) 
being 4 and 6, and the others formed by dou- 
bling the next preceding but one. Thus, in 
the above case, the first difference is 2, and 
then the numbers increase by 4, 6, 8, and 12. 
Of course, an infinite number of solutions may 
be found if we admit fractions. The number of 
squares in a circle of this kind must, however, 
be of the form 4»+6; that is, it must be a 
number composed of 6 plus a miiltiple of 4. 

119.— RACKBRANE'S LITTLE LOSS. 

The professor must have started the game with 
thirteen shillings, Mr. Potts with four shillings, 
and Mrs. Potts with seven shillings. 

120.— THE FARMER AND HIS SHEEP. 

The farmer had one sheep only ! If he divided 
this sheep (which is best done by weight) into 
two parts, making one part two-thirds and the 
other part one-third, then the difference he- 
tween these two numbers is the same as the 
difference between their squares — that is, one- 
third. Any two fractions will do if the denomi- 
nator equals the sum of the two numerators. 

121.— HEADS OR TAILS. 

Crooks must have lost, and the longer he went 
on the more he would lose. In two tosses he 
would be left with three-quarters of his money, 
in four tosses with nine-sixteenths of his money, 



in six tosses with twenty-seven sixty-fourths of 
his money, and so on. The order of the wins 
and losses makes no difference, so long as their 
number is in the end equal. 

122.— THE SEE-SAW PUZZLE. 

The boy's weight must have been about 39.79 
lbs. A brick weighed 3 lbs. Therefore 16 
bricks weighed 48 lbs. and 11 bricks 33 lbs. 
Multiply 48 by 33 and take the square root. 

123.— A LEGAL DIFFICULTY. 

It was clearly the intention of the deceased to 
give the son twice as much as the daughter, 
or the daughter half as much as the mother. 
Therefore the most equitable division would 
be that the mother should take two-sevenths, 
the son four-sevenths, and the daughter one- 
seventh. 

124.— A QUESTION OF DEFINITION. 

There is, of course, no difference in area be- 
tween a mile square and a square mile. But 
there may be considerable difference in shape, 
A mile square can be no other shape than square; 
the expression describes a surface of a certain 
specific size and shape. A square mile may be 
of any shape ; the expression names a unit of 
area, but does not prescribe any particular 
shape. 

125.— THE MINERS' HOLIDAY. 

Bill Harris must have spent thirteen shillings 
and sixpence, which would be three shillings 
more than the average for the seven men — half 
a guinea. 

126.— SIMPLE MULTIPLICATION. 

The number required is 3,529,411,764,705,882, 
which may be multiplied by 3 and divided by 
2, by the simple expedient of removing the 3 
from one end of the row to the other. If you 
want a longer number, you can increase this one 
to any extent by repeating the sixteen figures 
in the same order. 

127.— SIMPLE DIVISION. 

Subtract every number in turn from every 
other number, and we get 358 (twice), 716, 
1,611, 1,253, and 895. Now, we see at a glance 
that, as 358 equals 2 X 179, the only number that 
can divide in every case without a remainder 
will be 179. On trial we find that this is such 
a divisor. Therefore, 179 is the divisor we 
want, which always leaves a remainder 164 in 
the case of the original numbers given. 

128.--A PROBLEM IN SQUARES. 

The sides of the three boards measure 31 in., 
41 in., and 49 in. The common difference of 
area is exactly five square feet. Three num- 
bers whose squares are in A. P., with a com- 
mon difference of 7, are Hf » H^> Hi ', and with 



164 



AMUSEMENTS IN MATHEMATICS. 



a common difference of 13 are ftlH> W^W"> 
and V//?^/- I^ t^^ ^^s® °^ whole square num- 
bers the common difference will always be 
divisible by 24, so it is obvious that our squares 
must be fractional. Readers should now try 
to solve the case where the common difference 
is 23. It is rather a hard nut. 

129.— THE BATTLE OF HASTINGS. 

Any number (not itself a square number) may 
be multiplied by a square that will give a pro- 
duct I less than another square. The given 
number must not itself be a square, because a 
square multiplied by a square produces a square, 
and no square plus i can be a square. My re- 
marks throughout must be understood to apply 
to whole numbers, because fractional soldiers 
are not of much use in war. 

Now, of all the numbers from 2 to 99 in- 
clusive, 61 happens to be the most awkward 
one to work, and the lowest possible answer to 
our puzzle is that Harold's army consisted of 
3,119,882,982,860,264,400 men. That is, there 
would be 51,145,622,669,840,400 men (the 
square of 226,153,980) in each of the sixty-one 
squares. Add one man (Harold), and they 
could then form one large square with 
1,766,319,049 men on every side. The general 
problem, of which this is a particular case, is 
known as the " Pellian Equation " — apparently 
because PeU neither first propotmded the ques- 
tion nor first solved it ! It was issued as a 
challenge by Fermat to the English mathe- 
maticians of his day. It is readily solved by 
the use of continued fractions. 

Next to 61, the most difl&cult number under 
ICO is 97, where 97x6,377,3522+i = a square. 

The reason why I assumed that there must 
be something wrong with the figures in the 
chronicle is that we can confidently say that 
Harold's army did not contain over three 
trillion men ! If this army (not to mention 
the Normans) had had the whole surface of 
the earth (sea included) on which to encamp, 
each man would have had slightly more than 
a quarter of a square inch of space in which 
to move about ! Put another way : Allowing 
one square foot of standing-room per man, 
each small square would have required all the 
space allowed by a globe three times the 
diameter of the earth. 

130.— THE SCULPTOR'S PROBLEM. 

A LITTLE thought will make it clear that the 
answer must be fractional, and that in one case 
the nimierator will be greater and in the other 
case less than the denominator. As a matter 
of fact, the height of the larger cube must be 
f ft., and of the smaller f ft., if we are to have 
the answer in the smallest possible figures. 
Here the lineal measurement is \^ ft. — that is, 
i^ ft. What are the cubic contents of the 
two cubes ? First fxf X^=|H» ^ind secondly 
^Xfx^=-j^3. Add these together and the 
resmt is \H, which reduces to ^, or if ft. We 
thus see that the answers in cubic feet and 
lineal feet are precisely the same. 



The germ of the idea is to be found in the works 
of Diophantus of Alexandria, who wrote about 
the beginning of the fourth century. These 
fractional numbers appear in triads, and are 
obtained from three generators, a, b, c, where 
a is the largest and c the smallest. 

Then ab + c^ = denominator, and a^ — c^, 
b^—c^, and a^ — b^ wiU be the three numerators. 
Thus, using the generators 3, 2, i, we get f, f, 
f and we can pair the first and second, as in 
the above solution, or the first and third for a 
second solution. The denominator must al- 
ways be a prime number of the form 6m -fi, or 
composed of such primes. Thus you can have 
13, 19, etc., as denominators, but not 25, 55, 
187, etc. 

When the principle is understood there is no 
difficulty in writing down the dimensions of 
as many sets of cubes as the most exacting col- 
lector may require. If the reader would like 
one, for example, with plenty of nines, perhaps 
the following would satisfy him : IHflHi ^^^ 

-~9S 



FS 



TT- 



131.— THE SPANISH MISER. 



There must have been 386 doubloons in one 
box, 8,450 in another, and 16,514 in the third, 
because 386 is the smallest nxmaber that can 
occur. If I had asked for the smallest aggre- 
gate number of coins, the answer would have 
been 482, 3,362, and 6,242. It wiU be found 
in either case that if the contents of any two of 
the three boxes be combined, they form a square 
number of coins. It is a curious coincidence 
(nothing more, for it will net always happen) 
that in the first solution the digits of the three 
numbers add to 17 in every case, and in the 
second solution to 14. It should be noted that 
the middle one of the three numbers will always 
be half a square. 

132.— THE NINE TREASURE BOXES. 
Here is the answer that fulfils the conditions : — 



A=4 

D=2,ii6 

G=9,409 



B=3,364 
E=5,476 
H= 12,769 



0=6,724 
F=8,836 
I =16,129 



Each of these is a square number, the roots, 
taken in alphabetical order, being 2, 58, 82, 46, 
74, 94, 97, 113, and 127, while the required 
difference between A and B, B and C, D and E, 
etc., is in every case 3,360. 

133.— THE FIVE BRIGANDS. 

The sum of 200 doubloons might have been 
held by the five brigands in any one of 6,627 
different ways. Alfonso may have held any 
number from i to 11. If he held i doubloon, 
there are 1,005 different ways of distributing 
the remainder ; if he held 2, there are 985 ways ; 
if 3, there are 977 ways ; if 4, there are 903 
ways ; if 5 doubloons, 832 ways ; if 6 doubloons, 
704 ways ; if 7 doubloons, 570 ways ; if 8 
doubloons, 388 ways ; if 9 doubloons, 200 ways ; 
if 10 doubloons, 60 ways ; and if Alfonso held 
II doubloons, the remainder could be distri- 



SOLUTIONS. 



165 



buted in 3 different ways. More than 11 doub- 
loons he . could not possibly have had. It will 
scarcely be expected that I shall give all these 
6,627 ways at length. What I propose to do 
is to enable the reader, if he should feel so dis- 
posed, to write out all the answers where 
Alfonso has one and the same amount. Let us 
take the cases where Alfonso has 6 doubloons, 
and see how we may obtain all the 704 different 
ways indicated above. Here are two tables 
that will serve as keys to all these answers : — 



Table I. 
A=6. 
B=n. 

C={6s — 5n)-\-rn. 
D=(i28 + 4m) — 4m. 
E=3-\-3m. 



Table II. 
A=6. 

B = M 

C=i + m. 

D=(376— i6n) — 4m. 
E=(i5w-i83) + 3W. 



In the first table we may substitute for n any 
whole number from i to 12 inclusive, and m 
may be nought or any whole number from 
I to (31 -fw) inclusive. In the second table 
n may have the value of any whole number 
from 13 to 23 inclusive, and m may be nought 
or any whole number from i to (93 — 4M) in- 
clusive. The first table thus gives (32-Hw) 
answers for every value of w ; and the second 
table gives (94 — 4«) answers for every value of 
n. The former, therefore, produces 462 and 
the latter 242 answers, which together make 
704, as already stated. 

Let us take Table I., and say « = 5 and 
tn—2 ; also in Table II. take «=i3 and m=o. 
Then we at once get these two answers : — 



A= 6 




A= 6 


B= 5 




B= 13 


C= 40 




C= I 


D=i40 




D=i68 


E=_9 




E= 12 


200 


doubloons 


200 doubloons 



These will be found to work correctly. All the 
rest of the 704 answers, where Alfonso always 
holds six doubloons, may be obtained in this 
way from the two tables by substituting the 
different numbers for the letters m and w. 

Put in another way, for every holding of 
Alfonso the number of answers is the smn of 
two arithmetical progressions, the common 
difference in one case being i and in the other 
— 4. Thus in the case where Alfonso holds 6 
doubloons one progression is 33 + 34-}- 35 + 36 

+ + 43+44, and the other 42 + 38 + 34 

+ 30+ + 6 + 2. The sum of the first 

series is 462, and of the second 242 — results 
which again agree with the figures already given. 
The problem may be said to consist in finding 
the first and last terms of these progressions. I 
should remark that where Alfonso holds 9, 10, 
on I there is only one progression, of the second 
form. 

134.— THE BANKER'S PUZZLE. 

In order that a number of sixpences may not 
be divisible into a nimiber of equal piles, it is 
necessary that the number should be a prime. 



If the banker can bring about a prime number, 
he will win ; and I will show how he can always 
do this, whatever the customer may put in the 
box, and that therefore the banker will win to 
a certainty. The banker must first deposit 
forty sixpences, and then, no matter how many 
the customer may add, he will desire the latter 
to transfer from the counter the square of the 
number next below what the customer put in. 
Thus, banker puts 40, customer, we wiU say, 
adds 6, then transfers from the counter 25 (the 
square of 5), which leaves 71 in all, a prime 
number. Try again. Banker puts 40, cus- 
tomer adds 12, then transfers 121 (the square 
of 11), as desired, which leaves 173, a prime 
number. The key to the puzzle is the curious 
fact that any number up to 39, if added to its 
square and the sum increased by 41, makes a 
prime number. This was first discovered by 
Euler, the great mathematician. It has been 
suggested that the banker might desire the 
customer to transfer sufficient to raise the 
contents of the box to a given number ; but 
this would not only make the thing an ab- 
surdity, but breaks the rule that neither knows 
what the other puts in. 

135.— THE STONEMASON'S PROBLEM. 

The puzzle amounts to this. Find the smallest 
square number that may be expressed as the 
sum of more than three consecutive cubes, the 
cube I being barred. As more than three heaps 
were to be supplied, this condition shuts out the 
otherwise smallest answer, 233+24^+25^=2042. 
But it admits the answer, 253+263+27^+283 
+ 293=3152. The correct answer, however, re- 
quires more heaps, but a smaller aggregate num- 
ber of blocks. Here it is: 143+153+ 

up to 253 inclusive, or twelve heaps in all, 
which, added together, make 97,344 blocks of 
stone that may be laid out to form a square 312 
X 312. I will just remark that one key to the 
solution lies in what are called triangular 
numbers. (See pp. 13, 25, and 166.) 

136.— THE SULTAN'S ARMY. 

The smallest primes of the form 4»+i are 5, 13, 
17, 29, and 37, and the smallest of the form 
4«— I are 3, 7, 11, 19, and 23. Now, primes 
of the first form can always be expressed as 
the sum of two squares, and in only one way. 
Thus, 5=4 + 1; 13=9 + 4; 17=16 + 1; 29= 
25 + 4; 37=36+1. But primes of the second 
form can never he expressed as the sum of two 
squares in any way whatever. 

In order that a number may be expressed as 
the sum of two squares in several different ways, 
it is necessary that it shaU be a composite 
number containing a certain number of primes 
of our first form. Thus, 5 or 13 alone can only 
be so expressed in one way ; but 65, (5x13), 
can be expressed in two ways, 1,105, (5X^3 
X17), in four ways, 32,045, (5x13x17x29), 
in eight ways. We thus get double as many 
ways for every new factor of this form that 
we introduce. Note, however, that I say new 



i66 



AMUSEMENTS IN MATHEMATICS. 



factor, for the repetition of factors is subject 
to another law. We cannot express 25, (5x5), 
in two ways, but only in one ; yet 125, (5 X 5 X 5), 
can be given in two ways, and so can 625, (5 X 
5x5x5) ; while if we take in yet another 5 
we can express the number as the sum of two 
squares in three diSerent ways. 

If a prime of the second form gets into your 
composite number, then that number cannot be 
the sum of two squares. Thus 15, (3x5), will 
not work, nor will 135, (3x3x3x5); but if 
we take in an even number of 3's it will work, 
because these 3's will themselves form a square 
number, but you will only get one solution. 
Thus, 45, (3 X 3 X 5, or 9 X 5) = 36 + 9- Similarly, 
the factor 2 may always occur, or any power of 
2, such as 4, 8, 16, 32 ; but its introduction or 
omission will never affect the number of your 
solutions, except in such a case as 50, where it 
doubles a square and therefore gives you the 
two answers, 49 + 1 and 25 + 25. 

Now, directly a number is decomposed into 
its prime factors, it is possible to teU at a glance 
whether or not it can be split into two squares ; 
and if it can be, the process of discovery in how 
many ways is so simple that it can be done in 
the head without any effort. The number I 
gave was 130. I at once saw that this was 
2x5x13, and consequently that, as 65 can be 
expressed in two ways (64 + 1 and 49 + 16), 130 
can also be expressed in two ways, the factor 2 
not affecting the question. 

The smallest number that can be expressed 
as the sum of two squares in twelve different 
ways is 160,225, and this is therefore the 
smallest army that would answer the Sultan's 
pmT)ose. The number is composed of the fac- 
tors 5x5x13x17x29, each of which is of the 
required form. If they were all different factors, 
there would be sixteen ways ; but as one of the 
factors is repeated, there are just twelve ways. 
Here are the sides of the twelve pairs of squares : 
(400 and 15), (399 and 32), (393 and 76), (392 and 
81), (384 and 113), (375 and 140), (360 and 175), 
(356 and 183), (337 and 216), (329 and 228), 
(311 and 252), (265 and 300). Square the two 
numbers in each pair, add them together, and 
their sum will in every case be 160,225. 

137.— A STUDY IN THRIFT. 

Mrs. Sandy McAllister wiU have to save a 
tremendous sum out of her housekeeping allow- 
ance if she is to win that sixth present that her 
canny husband promised her. And the allow- 
ance must be a very liberal one if it is to 
admit of such savings. The problem required 
that we shoiild find five numbers higher than 36 
the imits of which may be displayed so as to 
form a square, a triangle, two triangles, and 
three triangles, using the complete number in 
every one of the four cases. 

Every triangular number is such that if we 
multiply it by 8 and add i the result is an odd 
square number. For example, multiply i, 3, 6, 
10, 15 respectively by 8 and add i, and we get 
9> 25i 49> 81, 121, which are the squares of the 
odd numbers 3, 5, 7, 9, 11. Therefore in every 



case where 8;t2+i=a square number, x^ is also 
a triangular. This point is dealt with in our 
puzzle, " The Battle of Hastings." I will now 
merely show again how, when the first solution 
is found, the others may be discovered with- 
out any difficulty. First of aU, here are the 
figures : — 



8 X 


l2 + I = 32 


8 X 


62 + I = 172 


8 X 


352 + I = 992 


8 X 


2048 + I = 5772 


8 X 


11892 + I = 33632 


8 X 


69302 + I = 196012 


8 X 


403912 + I = 1142432 


The successive pairs of numbers sire found in 


this way : — 




(1x3)+ (3 


Xi)= 6 (8xi)+ (3x3)= 17 


(1X17)+ (3 


X6)= 35 (8x6) + (3Xi7)= 99 


{iX99) + (3X 


35) = 204 (8x35) + (3x99) = 577 



and so on. Look for the numbers in the table 
above, and the method will explain itself. 

Thus we find that the numbers 36, 1225, 
41616, 1413721, 48024900, and i6fi43288i will 
form squares with sides of 6, 35, 204, 1189, 
6930, and 40391 ; and they wiU also form single 
triangles with sides of 8, 49, 288, 1681, 9800, 
and 57121. These numbers may be obtained 
from the last column in the first table above in 
this way : simply divide the numbers by 2 and 
reject the remainder. Thus the integral halves 
of 17, 99, and 577 are 8, 49, and 288. 

All the numbers we have found will form 
either two or three triangles at will. The fol- 
lowing little diagram wiU show you graphically 
at a glance that every square number must 
necessarily be the sum of two triangulars, and 
that the side of one triangle will be the same as 
the side of the corresponding square, while the 
other will be just i less. 




Thus a square may always be divided easily 
into two triangles, and the sum of two consecu- 
tive triangulars will always make a square. 
In numbers it is equally clear, for if we examine 
the first triangulars — i, 3, 6, 10, 15, 21, 28 — 
we find that by adding all the consecutive pairs 
in turn we get the series of square numbers — 
4, 9, 16, 25, 36, 49, etc. 

The method of forming three triangles from 
our numbers is equally direct, and not at all a 
matter of trial. But I must content myself with 
giving actual figures, and just stating that every 
triangular higher than 6 will form three tri- 
angulars. I give the sides of the triangles, and 
readers will know from my remarks when stat- 



SOLUTIONS. 



167 



ing the puzzle how to find from these sides the 
number of counters or coins in each, and so 
check the results if they so wish. 



writing out the series until we come to a square 
number does not appeal to the mathematical 
mind, but it serves to show how the answer to 



Number. 


Side of 
Square. 


Side of 
Triangle. 


Sides of Two 
Triangles. 


Sides of Three 
Triangles. 


36 

1225 

41616 

141 3721 

48024900 

1631432881 


6 

35 

204 

1189 

6930 

40391 


8 

49 

288 

1681 

9800 

57121 


6 + 5 

36 + 34 

204 + 203 

1189 + 1188 

6930+6929 

40391 + 40390 


5 + 5 + 3 
33 + 32 + 16 

192 + 192 + 95 
1121 + 1120 + 560 
6533 + 6533 + 3267 
38081 + 38080 + 19040 



I should perhaps explain that the arrange- 
ments given in the last two columns are not the 
only ways of forming two and three triangles. 
There are others, but one set of figures will 
fully serve our purpose. We thus see that 
before Mrs, McAllister can claim her sixth £5 
present she must save the respectable sum of 
£1,631,432,881. 



138. 



-THE ARTILLERYMEN'S 

DILEMMA. 



We were required to find the smallest number 
of cannon balls that we could lay on the ^oimd 
to form a perfect square, and could pile into a 
square pyramid. I will try to make the matter 
clear to the merest novice. 



X 


2 


3 


4 


5 


6 


7 


I 


3 


6 


10 


15 


21 


28 


I 


4 


10 


20 


35 


56 


84 


I 


5 


14 


30 


55 


91 


140 



Here in the first row we place in regular order 
the natural numbers. Each number in the 
second row represents the sum of the numbers 
in the row above, from the beginning to the 
number just over it. Thus i, 2, 3, 4, added 
together, make 10. The third row is formed in 
exactly the same way as the second. In the 
fourth row every number is formed by adding 
together the number just above it and the 
preceding number. Thus 4 and 10 make 14, 
20 and 35 make 55. Now, all the numbers in 
the second row are triangiilar numbers, which 
means that these numbers of cannon balls may 
be laid out on the ground so as to form equi- 
lateral triangles. The numbers in the third 
row will aU form our triangular pyramids, while 
the numbers in the fomrth row will all form 
square pyramids. 

Thus the very process of forming the above 
numbers shows us that every square pyramid 
is the sum of two triangular pyramids, one of 
which has the same number of balls in the side 
at the base, and the other one ball fewer. If we 
continue the above table to twenty-four places, 
we shall reach the number 4,900 in the fourth 
row. As this number is the square of 70, we 
can lay out the balls in a square, and can form 
a square pyramid with them. This manner of 



the particular puzzle may be easily arrived at 
by anybody. As a matter of fact, I confess 
my failure to discover any number other than 
4,900 that fulfils the conditions, nor have I 
found any rigid proof that this is the only 
answer. The problem is a difiicult one, and 
the second answer, if it exists (which I do not 
believe), certainly runs into big figures. 

For the benefit of more advanced mathe- 
maticians I will add that the general expression 
for square pyramid numbers is an^+sn^+n. 

6 
For this expression to be also a square number 
(the special case of i excepted) it is necessary 
that n=p^-i = 6t^, where 2p^-i=q^ (the 
" Pellian Equation "). In the case of our solu- 
tion above, w=24, p=5, t=2, q=7. 

139.— THE DUTCHMEN'S WIVES. 

The money pjaid in every case was a square 
number of shillings, because they bought i at 
IS., 2 at 2S., 3 at 3s., and so on. But every 
husband pays altogether 63s. more than his 
wife, so we have to find in how many ways 63 
may be the difierence between two square 
numbers. These are the three only possible 
ways : the square of 8 less the square of i, 
the square of 12 less the square of 9, and the 
square of 32 less the square of 31. Here i, 9, 
and 31 represent the number of pigs bought 
and the number of shillings per pig paid by 
each woman, and 8, 12, and 32 the same in the 
case of their respective husbands. From the 
further information given as to their purchases, 
we can now pair them off as follows : Cornelius 
and Gmrtrun bought 8 and i ; Elas and Katrhn 
bought 12 and 9 ; Hendrick and Anna bought 
32 and 31. And these pairs represent correctly 
the three married couples. 

The reader may here desire to know how we 
may determine the maximum number of ways 
in which a number may be expressed as the 
difierence between two squares, and how we are 
to find the actual squares. Any integer except 
I, 4, and twice any odd number, may be ex- 
pressed as the difference of two integral squares 
in as many ways as it can be split up into 
pairs of factors, counting i as a factor. Sup- 
pose the number to be 5,940. The factors are 



i68 



AMUSEMENTS IN MATHEMATICS. 



22.33.5.1 1. Here the exponents are 2, 3, i, i. 
Always deduct i from the exponents of 2 and 
add I to all the other exponents ; then we get 
I, 4, 2, 2, and half the product of these four 
numbers will be the required number of ways 
in which 5,940 may be the difference of two 
squares — that is, 8. To find these eight squares, 
as it is an even number, we first divide by 4 and 
get 1485, the eight pairs of factors of which are 
IX 1485, 3x495, 5x297, 9x165, 11x135, 
15x99, 27x55, and 33X45. The sum and 
difference of any one of these pairs will give 
the required numbers. Thus, the square of 
1,486 less the square of 1,484 is 5,940, the square 
of 498 less the square of 492 is the same, and so 
on. In the case of 63 above, the number is 
odd : so we factorize at once, i x 63, 3 x 21, 7 x 9. 
Then we find that half the sum and difierence 
will give us the numbers 32 and 31, 12 and 9, and 
8 and i, as shown in the solution to the puzzle. 
The reverse problem, to find the factors of a 
number when you have expressed it as the 
difference of two squares, is obvious. For 
example, the sum and difference of any pair 
of numbers in the last sentence will give us 
the factors of 63. Every prime number (except 
I and 2) may be expressed as the difference of 
two squares in one way, and in one way only. 
If a number can be expressed as the difference 
of two squares in more than one way, it is 
composite ; and having so expressed it, we may 
at once obtain the factors, as we have seen. 
Fermat showed in a letter to Mersenne or 
Fr6nicle, in 1643, how we may discover whether 
a number may be expressed as the difference 
of two squares in more than one way, or proved 
to be a prime. But the method, when dealing 
with large numbers, is necessarily tedious, 
though in practice it may be considerably 
shortened. In many cases it is the shortest 
method known for factorizing large numbers, 
and I have always held the opinion that Fermat 
used it in performing a certain feat in factorizing 
that is historical and wrapped in mystery. 



140.— FIND ADA'S SURNAME. 

The girls' names were Ada Smith, Annie Brown, 
Emily Jones, Mary Robinson, and Bessie Evans. 

141. —SATURDAY MARKETING. 

As every person's purchase was of the value of 
an exact number of shillings, and as the party 
possessed when they started out forty shilling 
coins altogether, there was no necessity for any 
lady to have any smaller change, or any evi- 
dence that they actually had such change. 
This being so, the only answer possible is that 
the women were named respectively Anne Jones, 
Mary Robinson, Jane Smith, and Kate Brown. 
It will now be foimd that there would be exactly 
eight shillings left, which may be divided equally 
among the eight persons in coin without any 
change being required. 

142.— THE SILK PATCHWORK. 







G 












H 




F 




1 




ly 








K 







Our illustration will show how to cut the 
stitches of the patchwork so as to get the square 
F entire, and four equal pieces, G, H, I, K, 
that wiU form a perfect Greek cross. The 
reader will know how to assemble these four 
pieces from Fig. 13 in the article. 

143.— TWO CROSSES FROM ONE. 

It will be seen that one cross is cut out entire, 
as A in Fig. 1, while the four pieces marked 






w 








^ / 


/ ^"^ v^ 


—-J 




t 





Fio. I. 



Fig. 2. 



SOLUTIONS. 



169 



B, C, D and E form the second cross, as in 
Fig. 2, which will be of exactly the same size as 
the other. I will leave the reader the pleasant 
task of discovering for himself the best way of 
finding the direction of the cuts. Note that 
the Swastika again appears. 

The difficult question now presents itself: 
How are we to cut three Greek crosses from one 
in the fewest possible pieces ? As a matter of 
fact, this problem may be solved in as few as 
thirteen pieces; but as I know many of my 
readers, advanced geometricians, will be glad 
to have something to work on of which they 
are not shown the solution, I leave the mjrstery 
for the present undisclosed. 

I44._THE CROSS AND THE TRIANGLE. 

The line A B in the following diagram repre- 
sents the side of a square having the same area 
as the cross. I have shown elsewhere, as stated, 
how to make a square and equilateral triangle 
of equal area. I need not go, therefore, into the 
preliminary question of finding the dimensions 
of the triangle that is to equal our cross. We 
will assume that we have already found this, 
and the question then becomes, How are we to 
cut up one of these into pieces that will form 
the other ? 

First draw the line A B where A and B are 
midway between the extremities of the two 
side arms. Next make the lines D C and E F 
equal in length to half the side of the triangle. 
Now from E and F describe with the same radius 
the intersecting arcs at G and draw FG. Fi- 
nally make I K equal to H C and L B equal to 
A D. If we now draw I L, it should be parallel 
to FG, and all the six pieces are marked out. 
These fit together and form a perfect equilateral 
triangle, as shown in the second diagram. Or 
we might have first found the direction of the 
line M N in our triangle, then placed the point 
O over the point E in the cross and turned 
roimd the triangle over the cross until the line 




I have seen many attempts at a solution in- 
volving the assumption that the height of the 
triangle is exactly the same as the height of 
the cross. This is a fallacy : the cross will 
always be higher than the triangle of equal 
area. 

145.— THE FOLDED CROSS. 

First fold the cross along the dotted line A B 
in Fig. I. You then have it in the form shown 



y 




Fig. I. 



Fig. 2. 



in Fig. 2. Next fold it along the dotted line 
CD (where D is, of course, the centre of the 
cross), and you get the form shown in Fig. 3. 





Fig. 3. 



Fig. 4. 



Now take your scissors and cut from G to F, 
and the four pieces, all of the same size and 




M N was parallel to A B. The piece 5 can then 1 shape, wiU fit together and fonn a square, as 
be marked off and the other pieces in succession. | shown in Fig. 4. 



170 



AMUSEMENTS IN MATHEMATICS. 



M 



146.— AN EASY DISSECTION PUZZLE. 




The solution to this puzzle is shown in the illus- 
tration. Divide the figure up into twelve equal 
triangles, and it is easy to discover the directions 
of the cuts, as indicated by the dark lines. 

147.— AN EASY SQUARE PUZZLE. 




The diagram explains itself, one of the five 
pieces having been cut in two to form a square. 

148.— THE BUN PUZZLE. 

The secret of the bim puzzle lies in the fact 
that, with the relative dimensions of the circles 



and Edgar the two halves marked D and E, 
they will have their fair shares — one quarter 
of the confectionery each. Then if we place 
the small bim, H, on the top of the remaining 
one and trace its circumference in the manner 
shown, Fred's piece, F, will exactly equal 
Harry's small bun, H, with the addition of the 
piece marked G — half the rim of the other. 
Thus each boy gets an exactly equal share, 
and there are only five pieces necessary. 

149.--THE CHOCOLATE SQUARES. 









. 










c 




V 


---."" A \ 








D , 














D * 









Square A is left entire ; the two pieces marked 
B fit together and make a second square ; the 
two pieces C make a third square ; and the four 
pieces marked D will form the fourth square. 

150.— DISSECTING A MITRE. 

The diagram on the next page shows how to cut 
into five pieces to form a square. The dotted 
lines are intended to show how to find the points 
C and F— the only difficulty. A B is half B D, 
and A E is parallel to B H. With the point of 
the compasses at B describe the arc H E, and 
A E wiU be the distance of C from B. Then F G 
equals B C less A B. 




as given, the three diameters will form a right- 
angled triangle, as shown by A, B, C. It follows 
that the two smaller buns are exactly equal to 
the large bvm. Therefore, if we give David 



This puzzle — with the added condition that 
it shall be cut into four parts of the same size 
and shape — I have not been able to trace to 
an earlier date than 1835. Strictly speaking, 



SOLUTIONS. 



171 




4 


\ / 


\^ 


/z 


5 


\3 



it is, in that form, impossible of solution ; but 
I give the answer that is always presented, 
and that seems to satisfy most people. 

We are asked to assume that the two portions 
containing the same letter — ^AA, BB, CC, DD — 
are joined by " a mere hair," and are, therefore, 
only one piece. To the geometrician this is 





/ 


AN. 


/ ^ 


X ^ / 





C 3> 



And yet it is apt to perplex the novice a good 
deal if he wants to do it in the fewest possible 
pieces — three. All you have to do is to find the 
point A, midway between B and C, and then 
cut from A to D and from A to E. The three 
piec^ then form a square in the manner shown. 
Of course, the proportions of the original figure 
must be correct ; thus the triangle BEF is just 
a quarter of the square BCDF. Draw lines 
from B to D and from C to F, and this will be 
clear. 

I52._AN0THER JOINER'S PROBLEM. 

C 



absurd, and the four shares are not equal in 
area unless they consist of two pieces each. If 
you make them equal in area, they will not be 
exactly alike in shape. 

-THE JOINER'S PROBLEM. 





Nothing could be easier than the solution of 
this puzzle — ^when you know how to do it. 



The point was to find a general rule for forming 
a perfect square out of another square com- 
bined with a " right-angled isosceles triangle." 
The triangle to which geometricians give this 
high-sounding name is, of course, nothing more 
or less than half a square that has been divided 
from corner to corner. 

The precise relative proportions of the square 
and triangle are of no consequence whatever. 



172 



AMUSEMENTS IN MATHEMATICS. 



It is only necessary to cut the wood or material 
into five pieces. 

Suppose our original square to be A C L F in 
the above diagram and our triangle to be the 
shaded portion C E D. Now, we first find half 
the length of the long side of the triangle (C D) 
and measure ofE this length at A B. Then we 
place the triangle in its present position against 
the square and make two cuts — one from B to 
F, and the other from B to E. Strange as it 
may seem, that is all that is necessary ! If we 
now remove the pieces G, H, and M to their new 
places, as shown in the diagram, we get the 
perfect square B E K F. 

Take any two square pieces of paper, of 
different sizes but perfect squares, and cut the 
smaller one in half from comer to comer. Now 
proceed in the manner shown, and you will find 
that the two pieces may be combined to form a 
larger square by making these two simple cuts, 
and that no piece wiU be required to be turned 
over. 

The remark that the triangle might be " a 
little larger or a good deal smaller in propor- 
tion " was intended to bar cases where area of 
triangle is greater than area of square. In 
such cases six pieces are necessary, and if tri- 
angle and square are of equal area there is an 
obvious solution in three pieces, by simply 
cutting the square in half diagonally. 

153.— A CUTTING-OUT PUZZLE. 



4 t ■ ' - ■ ' I • < I ■ f . r I I ■ > I I II I I ■ I ■ ■ II 

■ ■ T^ I L l i . .. iij 1 I... ■ 




The illustration shows how to cut the four 
pieces and form with them a square. First find 
the side of the square (the mean proportional 
between the length and height of the rectangle), 
and the method is obvious. If om: strip is 
exactly in the proportions 9x1, or 16x1, or 
25 X I, we can clearly cut it in 3, 4, or 5 rec- 
tangular pieces respectively to form a square. 
Excluding these special cases, the general law 
is that for a strip in length more than n^ times 
the breadth, and not more than {n+i)^ times 
the breadth, it may be cut in n-h2 pieces to 
form a square, and there will be n — i rec- 
tangular pieces like piece 4 in the diagram. 
Thus, for example, with a strip 24x1, the 
length is more than 16 and less than 25 times 



the breadth. Therefore it can be done in 6 
pieces (n here being 4), 3 of which will be 
rectangular. In the case where n equals i, 
the rectangle disappears and we get a solution 
in three pieces. V/ithin these limits, of covurse, 
the sides need not be rational : the solution is 
purely geometrical 

154.— MRS. HOBSON'S HEARTHRUG. 




As I gave full measurements of the mutilated 
rug, it was quite an easy matter to find the 
precise dimensions for the square. The two 
pieces cut off would, if placed together, make 
an oblong piece 12x6, giving an area of 72 
(inches or yards, as we please), and as the 
original complete rug measured 36 x 27, it had 
an area of 972. If, therefore, we deduct the 
pieces that have been cut away, we find that 
om: new rug will contain 972 less 72, or 900; 
and as 900 is the square of 30, we know that 
the new rug must measure 30 x 30 to be a per- 
fect square. This is a great help towards the 
solution, because we may safely conclude that 
the two horizontal sides measuring 30 each 
may be left intact. 

There is a very easy way of solving the puzzle 
in four pieces, and also a way in three pieces 
that can scarcely be called difficult, but the 
correct answer is in only two pieces. 

It will be seen that if, after the cuts are made, 
we insert the teeth of the piece B one tooth 
lower down, the two portions will fit together 
and form a square. 

155.— THE PENTAGON AND SQUARE. 

A REGULAR pentagon may be cut into as few 
as six pieces that wiU fit together without any 
turning over and form a square, as I shall show 
below. Hitherto the best answer has been in 
seven pieces — the solution produced some years 
ago by a foreign mathematician, Paul Busschop. 
We first form a parallelogram, and from that 
the square. The process will be seen in the 
diagram on the next page. 

The pentagon is A BCD E. By the cut 
A C and the cut F M (F being the middle point 
between A and C, and M being the same distance 
from A as F) we get two pieces that may be 
placed in position at G H E A and form the 
parallelogram G H D C. We then find the 
mean proportional between the length HD 
and the height of the parallelogram. This dis- 
tance we mark off from C at K, then draw C K, 



SOLUTIONS. 



173 



and from G drop the line G L, perpendicular to 
K C. The rest is easy and rather obvious. It 
will be seen that the six pieces will form either 
the pentagon or the square. 

I have received what purported to be a solu- 
tion in five pieces, but the method was based 
on the rather subtle fallacy that half the diag- 
onal plus half the side of a pentagon equals 
the side of a square of the same area. I say 
subtle, because it is an extremely close approxi- 
mation that will deceive the eye, and is quite 
difficult to prove inexact. I am not aware 
that attention has before been drawn to this 
curious approximation. 




Another correspondent made the side of his 
square ij of the side of the pentagon. As a 
matter of fact, the ratio is irrationsd. I calcu- 
late that if the side of the pentagon is i — inch, 
foot, or anything else — the side of the square 
of equal area is 1.3117 nearly, or say roughly 
1^. So we can only hope to solve the puzzle 
by geometrical methods. 

156.— THE DISSECTED TRIANGLE. 

Diagram A is our original triangle. We will 
say it measures 5 inches (or 5 feet) on each side. 
If we take off a slice at the bottom of any 
equilateral triangle by a cut parallel with the 
base, the portion that remains will always be 
an equilateral triangle ; so we first cut off piece 
I and get a triangle 3 inches on every side. 
The manner of finding directions of the other 
cuts in A is obvious from the diagram. 

Now, if we want two triangles, i will be one 
of them, and 2, 3, 4, and 5 will fit together, as 
in B, to form the other. If we want three 
equilateral triangles, i will be one, 4 and 5 will 
form the second, as in C, and 2 and 3 will form 
the third, as in D. In B and C the piece 5 is 
turned over ; but there can be no objection to 
this, as it is not forbidden, and is in no way 
opposed to the nature of the puzzle. 




I57._THE TABLE-TOP AND STOOLS. 

One object that I had in view when presenting 
this little puzzle was to point out the uncer- 
tainty of the meaning conveyed by the word 
" oval." Though originally derived from the 
Latin word ovum, an egg, yet what we under- 
stand as the egg-shape (with one end smaller 
than the other) is only one of many forms of 
the oval ; while some eggs are spherical in shape, 
and a sphere or circle is most certainly not an 
oval. If we speak of an ellipse — a conical 
ellipse — we are on safer groimd, but here we 
must be careful of error. I recollect a Liver- 
pool town councillor, many years ago, whose 
ignorance of the poultry-yard led him to sub- 
stitute the word "hen" for "fowl," remarking. 




" We must remember, gentlemen, that althougii 
every cock is a hen, every hen is not a cock ! " 
Similarly, we must always note that although 
every ellipse is an oval, every oval is not an 



174 



AMUSEMENTS IN MATHEMATICS. 



ellipse. It is correct to say that an oval is an 
oblong " curvilinear figure, having two unequal 




THE TWO STOOLS. 

diameters, and bounded by a curve line return- 
ing into itself ; and this includes the ellipse, but 
all other figures which in any way approach 
towards the form of an oval without necessarily 
having the properties above described are in- 
cluded in the term " oval." Thus the follow- 
ing solution that I give to our puzzle involves 
the pointed "oval," known among architects 
as the " vesica piscis." 

The dotted lines in the table 
are given for greater clear- 
ness, the cuts being made 
along the other lines. It will 
be seen that the eight pieces 
form two stools of exactly the 
same size and shape with simi- 
lar hand-holes. These holes are 
a trifle longer than those in the 
schoolmaster's stools, but they 
are much narrower and of con- 
siderably smaller area. Of 
course 5 and 6 can be cut out 
in one piece — also 7 and 8 — 
making only six pieces in all. 
But I wished to keep the same 
number as in the original 
story. 

When I first gave the above 
puzzle in a London newspaper, 
in competition, no correct solu- 
tion was received, but an in- 
genious and neatly executed 
attempt by a man lying in a 
London infirmary was accom- 
panied by the following note : 
" Having no compasses here, I 
was compelled to improvise a 
pair with the aid of a small 
penknife, a bit of firewood from 
a bvmdle, a piece of tin from 
a toy engine, a tin tack, and 
two portions of a hairpin, for points. They 
are a fairly serviceable pair of compasses, 
and I shall keep them as a memento of your 
puzzle." 



158.— THE GREAT MONAD. 

The areas of circles are to each other as the 
squares of their diameters. If you have a circle 
2 in. in diameter and another 4 in. in diameter, 
then one circle will be four times as great in 
area as the other, because the square of 4 is four 
times as great as the square of 2. Now, if 
we refer to Diagram i, we see how two equal 
squares may be cut into four pieces that will 
form one larger square; from which it is self- 
evident that any square has just half the area 
of the square of its diagonal. In Diagram 2 I 
have introduced a square as it often occurs in 
ancient drawings of the Monad; which was 
my reason for believing that the S5nnbol had 
mathematical meanings, since it will be found 
to demonstrate the fact that the area of the 
outer ring or annulus is exactly equal to the 
area of the inner circle. Compare Diagram 2 
with Diagram i, and you will see that as the 
square of the diameter C D is double the square 
of the diameter of the inner circle, or CE, 
therefore the area of the larger circle is double 
the area of the smaller one, and consequently 
the area of the annulus is exactly equal to that 
of the inner circle. This answers our first 
question. 

In Diagram 3 I show the simple solution to 
the second question. It is obviously correct, 
and may be proved by the cutting and super- 
position of parts. The dotted lines will also 







serve to make it evident. The third question 
is solved by the cut C D in Diagram 2, but it 
remains to be proved that the piece F is really 
one-half of the Yin or the Yan. This we will 



SOLUTIONS. 



175 



do in Diagram 4. The circle K has one-quarter 
the area of the circle containing Yin and Yan, 
because its diameter is just one-half the length. 
Also L in Diagram 3 is, we know, one-quarter 
the area. It is therefore evident that G is ex- 
actly equal to H, and therefore half G is equal 
to half H. So that what F loses from L it gains 
from K, and F must be half of Yin or Yan. 

159.— THE SQUARE OF VENEER. 







— r 
• 












r- 




i 
1 










1 
1 
















i 






... 


.-. 


--« 


,-- 




;.. 


... 


^-m 


_. 


.__ 


i.- 











1 
1 
1 


























1 
















i 

1 










''! 


... 


** * "" 


k-~ 




- -• 






• — 










1 
1 








B 








\ 


c 




... 




--t 






I- 




--i 






kr 










1 
1 
















1 


^m 


''"''^ 






i 






\ 




' 






t 

1 


A 








1 


... 




T 




-1 


... 


"" 


1 
* 





-.- 












1 


I 






V 










; 


_ 


JD 




-ir 




T 

1 


«. 





Any square number may be expressed as the 
sum of two squares in an infinite number of 
different ways. The solution of the present 
puzzle forms a simple demonstration of this 
rule. It is a condition that we give actual 
dimensions. 



wards determine. Divide the square as shown 
(where the dotted lines indicate the original 
markings) into 169 squares. As 169 is the 
sum of the two squares 144 and 25, we will 
proceed to divide the veneer into two squares, 
measuring respectively 12 X 12 and 5x5; and 
as we know that two squares may be formed 
from one square by dissection in four pieces, 
we seek a solution in this number. The dark 
lines in the diagram show where the cuts are 
to be made. The square 5 X 5 is cut out 
whole, and the larger square is formed from the 
remaining three pieces, B, C, and D, which the 
reader can easily fit together. 

Now, » is clearly -^ of an inch. Consequently 
our larger square must be f§ in. X Ji in., and 
our smaller square f| in, X ff in. The square 
of ff added to the square of f| is 25. The square 
is thus divided into as few as four pieces that 
form two squares of known dimensions, and all 
the sixteen nails are avoided. 

Here is a general formula for finding two 
squares whose sum shall equal a given square, 
say a^. In the case of the solution of our puzzle 
P=3, ?=2, anda=5. 



2pq(i ^„. Ja2(j>2+g2) 



r2\2. 



Here x'^-\-y^=a^. 



P'^^<i' 



{2Pqa)^ ^y 



160.— THE TWO HORSESHOES. 

The puzzle was to cut the two shoes (including 
the hoof contained within the outlines) into four 
pieces, two pieces each, that would fit together 
and form a perfect circle. It was also stipulated 
that all four pieces should be different in shape. 
As a matter of fact, it is a puzzle based on the 
principle contained in that curious Chinese 
symbol the Monad. (See No. 158.) 




In this puzzle I ignore the known dimensions 
of our square and work on the assumption that 
it is I3» by i3«. The value of n we can after- 



The above diagrams give the correct solution 
to the problem. It will be noticed that i and 2 
are cut into the required four pieces, all difier- 



176 



AMUSEMENTS IN MATHEMATICS. 



ent in shape, that fit together and form the 
perfect circle shown in Diagram 3. It will 
further be observed that the two pieces A and B 
of one shoe and the two pieces C and D of the 
other form two exactly similar halves of the 
circle — the Yin and the Yan of the great Monad. 
It will be seen that the shape of the horseshoe 
is more easily determined from the circle than 
the dimensions of the circle from the horseshoe, 
though the latter presents no difficulty when 
you know that the curve of the long side of the 
shoe is part of the circumference of your circle. 
The difference between B and D is instructive, 
and the idea is useful in all such cases where it 
is a condition that the pieces must be difierent 
in shape. In forming D we simply add on a 
symmetrical piece, a curvilinear square, to the 
piece B. Therefore, in giving either B or 
D a quarter turn before placing in the new 
position, a precisely similar efiect must be 
produced. 

161.— THE BETSY ROSS PUZZLE. 

Fold the circular piece of paper in half along 
the dotted line shown in Fig. i, and divide the 



162.— THE CARDBOARD CHAIN. 

The reader wiU probably feel rewarded for any 
care and patience that he may bestow on cutting 
out the cardboard chain. We wiU suppose that 
he has a piece of cardboard measuring 8 in. by 
2^ in., though the dimensions are of no import- 
ance. Yet if you want a long chain you must, 
of course, take a long strip of cardboard. First 
rule pencil lines B B and C C, half an inch from 
the edges, and also the short perpendicular lines 
half an inch apart. (See next page.) Rule lines 
on the other side in just the same way, and in 
order that they shall coincide it is well to prick 
through the card with a needle the points where 
the short lines end. Now take your penknife 
and split the card from A A down to B B, and 
from D D up to C C. Then cut right through the 
card along all the short perpendicular lines, and 
half through the card along the short portions 
of B B and C C that are not dotted. Next turn 
the card over and cut half through along the 
short lines on B B and C C at the places that are 
immediately beneath the dotted lines on the 
upper side. With a little careful separation of 
the parts with the penknife, the cardboard may 




upper half into five equal parts as indicated. 
Now fold the paper along the lines, and it will 
have the appearance shown in Fig. 2. If you 
want a star like Fig. 3, cut from A to B ; if you 
wish one like Fig. 4, cut from A to C. Thus, 
the nearer you cut to the point at the bottom 
the longer will be the points of the star, and the 
farther off from the point that you cut the 
shorter wiU be the points of the star. 



now be diviaed into two interlacing ladder-like 
portions, as shown in Fig. 2 ; and if you cut 
away all the shaded parts you will get the 
chain, cut solidly out of the cardboard, with- 
out any join, as shown in the illustrations on 
page 40. 

It is an interesting variant of the puzzle to cut 
out two keys on a ring — in the same manner 
without join. 



SOLUTIONS. 



177 




2 




164.— THE POTATO PUZZLE. 

As many as twenty-two pieces may be obtained 
by the six cuts. The illustration shows a pretty 
symmetrical solution. The rule in such cases 




is that every cut shall intersect every other 
cut and no two intersections coincide ; that is 
to say, every line passes through every other 



(1,926) 



line, but more than two lines do not cross at the 
same point anywhere. There are other ways of 
making the cuts, but this rule must always be 
observed if we are to get the full number of 
pieces. 

The general formula is that with n cuts we 

can always produce ^^** ^^ + 1 pieces. One of 
2 

the problems proposed by the late Sam Loyd 
was to produce the maximum number of pieces 
by n straight cuts through a solid cheese. Of 
course, again, the pieces cut off may not be 
moved or piled. Here we have to deal with 
the intersection of planes (instead of lines), and 
the general formula is that with n cuts we may 

produce (^^Hilj^^L+il-f-w+i pieces. It is ex- 
o 

tremely difficult to " see " the direction and 
effects of the successive cuts for more than a 
few of the lowest values of n. 



165.— THE SEVEN PIGS. 

The illustration shows the direction for placing 
the three fences so as to enclose every pig in a 
separate sty. The greatest number of spaces 
that can be enclosed with three straight lines 
in a square is seven, as shown in the last puzzle. 
Bearing this fact in mind, the puzzle must be 
solved by trial. 



12 



178 



AMUSEMENTS IN MATHEMATICS. 




THE SEVEN PIGS. 

166.-— THE LANDOWNER'S FENCES. 
Four fences only are necessary, as follows :- 



THE WIZARD S CATS. 



line may be completed in an unlimited number 
of ways (straight or crooked), provided it be 




167.— THE WIZARD'S CATS. 



The illustration requires no explanation. It 
shows clearly how the three circles may be 
drawn so that every cat has a separate enclo- 
sure, and cannot approach another cat without 
crossing a line. 

168.— THE CHRISTMAS PUDDING. 

The illustration shows how the pudding may 
be cut into two parts of exactly the same size 
and shape. The lines must necessarily pass 
through the points A, B, C, D, and E. But, 
subject to this condition, they may be varied 
in an infinite number of ways. For example, 
at a point midway between A and the edge, the 



exactly reflected from E to the opposite edge. 
And similar variations may be introduced at 
other places. 

169.— A TANGRAM PARADOX. 

The diagrams will show how the figures are 
constructed — each with the seven Tangrams. It 
will be noticed that in both cases the head, hat, 
and arm are precisely alike, and the width at 
the base of the body the same. But this body 
contains four pieces in the first case, and in the 
second design only three. The first is larger 
than the second by exactly that narrow strip 
indicated by the dotted line between A and B. 



SOLUTIONS. 



179 



This strip is therefore exactly equal in area to 
the piece forming the foot in the other design. 



Z\ 




though when thus distributed along the side 
of the body the increased dimension is not 
easily apparent to the eye. 



170.— THE CUSHION COVERS. 

^ 1 B 




000 

0^0 9 




6 O « 
« 



o o^ 




6 « 

» « 




9 .<» Q 



» ^ 
o 






o 

000 




« (» o 

O O 9 




o o e 

• 09 







• « 
5 



O o 

900 




The two pieces of brocade marked A will fit 
together and form one perfect square cushion 
top, and the two pieces marked B will form the 
other. 

171.— -THE BANNER PUZZLE. 

The illustration explains itself. Divide the 
bimting into 25 squares (because this number is 
the sum of two other squares — 16 and 9), and 
then cut along the thick lines. The two pieces 




marked A form one square, and the two pieces 
marked B form the other. 

172.— MRS. SMILEY'S CHRISTMAS 
PRESENT. 



. ' I ' 
I I I I ; 
r-'-- »-7-'-r- 
I • * i 



--rT-7- ,-;- - - --r7H-ri" 

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l8o 



AMUSEMENTS IN MATHEMATICS. 



The first step is to find six different square 
numbers that sum to 196. For example, i + 4 + 
25 + 36+49 + 81 = 196; 1 + 4+9+25 + 36 + 121 
= 196; 1 + 9 + 16 + 25 + 64+81 = 196. The rest 
calls for individual judgment Jind ingenuity, and 
no definite rules can be given for procedure. 
The annexed diagrams wiU show solutions for 
the first two cases stated. Of course the three 
pieces marked A and those marked B will fit 
together and form a square in each case. The 
assembling of the parts may be slightly varied, 
and the reader may be interested in finding a 
solution for the third set of squares I have given. 

173.— MRS. PERKINS'S QUILT. 

The following diagram shows how the quilt 
should be constructed. 















r 






8 






























6 










2 












11 






















5 


































4 
















10 






























9 










































1 








































3 
































, , 


























L_ 



























There is, I believe, practically only one solu- 
tion to this puzzle. The fewest separate squares 
must be eleven. The portions must be of the 
sizes given, the three largest pieces must be 
arranged as shown, and the remaining group of 
eight squares may be " reflected," but cannot 
be differently arranged. 

174.— THE SQUARES OF BROCADE. 



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63 




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So far as I have been able to discover, there is 
only one possible solution to fulfil the conditions. 



The pieces fit together as in Diagram i. Dia- 
grams 2 and 3 showing how the two original 

Z 



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squares are to be cut. It will be seen that the 
pieces A and C have each twenty chequers, and 
are therefore of equal area. Diagram 4 (built 



w 




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sr 




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F 


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up with the dissected square No. 5) solves the 
puzzle, except for the small condition contained 
in the words, " I cut the two squares in the 
manner desired." In this case the smaller 
square is preserved intact. StiU I give it as 



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4- 



an illustration of a feature of the puzzle. It is 
impossible in a problem of this kind to give a 
quarter-turn to any of the pieces if the pattern 
is to properly match, but (as in the case of F, 
in Diagram 4) we may give a symmetrical piece 
a half-turn — that is, turn it upside down. 
Whether or not a piece may be given a quarter- 
turn, a half-turn, or no turn at all in these 
chequered problems, depends on the character 
of the design, on the material employed, and 
also on the form of the piece itself. 

I75._AN0THER PATCHWORK PUZZLE. 

The lady need only unpick the stitches along 
the dark lines in the larger portion of patchwork. 



SOLUTIONS. 



i8i 



i=!iL;-::i^::: 

— ^" — — — ■"" 

— ::-x::::: 
::.^i::-±:: 

E::EE:::ib 

^ — i-J-.^ -J I i ■■ —J — I ^i 



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when the four pieces will fit together and form a 
square, as shown in our illustration. 

176.— LINOLEUM CUTTING. 

There is only one solution that will enable us 
to retain the larger of the two pieces with as 
little as possible cut from it. Fig. i in the 
following diagram shows how the smaller piece 
is to be cut, and Fig. 2 how we should dissect 



^ 



^ 



i 



1 



1 



7F, 



B 



1 



1 



2: 



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c 

m 






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D 



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1 




the larger piece, while in Fig. 3 we have the new 
square 10 x 10 formed by the four pieces with 
all the chequers properly matched. It will 
be seen that the piece D contains fifty-two 
chequers, and this is the largest piece that it is 
possible to preserve omder the conditions. 

177.— ANOTHER LINOLEUM 
PUZZLE. 

Cut along the thick lines, and the four pieces will 
fit together and form a perfect square in the 
manner shown in the smaller diagram. 



I I I I • ' i 
I • I , , ' ', I 



I I 1 

I -L..'. J 



;:rTT:x 



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ANOTHER LINOLEUM PUZZLE. 

178.— THE CARDBOARD BOX. 

The areas of the top and side multiplied to- 
gether and divided by the area of the end give 
the square of the length. Similarly, the pro- 
duct of top and end divided by side gives the 
square of the breadth ; and the product of side 
and end divided by the top gives the square of 
the depth. But we only need one of these 
operations. Let us take the first. Thus, 120 X 
96 divided by 80 equals 144, the square of 12. 
Therefore the length is 12 inches, from which 
we can, of course, at once get the breadth and 
depth — 10 in. and 8 in. respectively. 

i79._STEALING THE BELL- ROPES. 

Whenever we have one side (a) of a right- 
angled triangle, and know the difference between 
the second side and the hypotenuse (which 
difference we will call 6), then the length of the 

2 /i 
hypotenuse wiU be —-|--. In the case of our 

puzzle this will be ^ ^/^ ■\-x\ in. = 32 ft. i J in., 
which is the length of the rope. 

180.— THE FOUR SONS. 




The diagram shows the most equitable division 
of the land possible, " so that each son shall 
receive land of exactly the same area and ex- 
actly similar in shape," and so that each shall 
have access to the well in the centre without 
trespass on another's land. The conditions 



1 82 



AMUSEMENTS IN MATHEMATICS. 



do not require that each son's land shall be in 
one piece, but it is necessary that the two 
portions assigned to an individual should be 
kept apart, or two adjoining portions might be 
held to be one piece, in which case the con- 
dition as to shape would have to be broken. 
At present there is only one shape for each 



183.— DRAWING A SPIRAL. 

Make a fold in the paper, as shown by the 
-dotted line in the illustration. Then, taking 
any two points, as A and B, describe semi- 
circles on the line alternately from the centres 
B and A, being careful to make the ends join, 




piece of land — half a square divided diagonally. 
And A, B, C, and D can each reach their land 
from the outside, and have each equal access 
to the well in the centre. 



181.— THE THREE RAILWAY 
STATIONS. 

The three stations form a triangle, with sides 
13, 14, and 15 miles. Make the 14 side the 
base ; then the height of the triangle is 12 and 
the area 84. Multiply the three sides together 
and divide by four times the area. The result 
is eight miles and one-eighth, the distance 
required. 

182.— THE GARDEN PUZZLE. 

Half the sum of the four sides is 144. From 
this deduct in turn the four sides, and we get 
64, 99, 44, and 81. Multiply these together, 
and we have as the result the square of 4,752. 
Therefore the garden contained 4,752 square 
yards. Of course the tree being equidistant 
from the four comers shows that the garden 
is a quadrilateral that may be inscribed in a 
circle. 



and the thing is done. Of course this is not a 
true spiral, but the puzzle was to produce the 
particular spiral that was shown, and that was 
drawn in this simple manner. 

184.— HOW TO DRAW AN OVAL. 

If you place your sheet of paper round the 
surface of a cylindrical bottle or canister, the 
oval can be drawn with one sweep of the com- 
passes. 

185.— ST. GEORGE'S BANNER. 

As the flag measures 4 ft. by 3 ft., the length 
of the diagonal (from comer to comer) is 5 ft. 
All you need do is to deduct half the length of 
this diagonal (2^ ft.) from a quarter of the dis- 
tance all round the edge of the flag (3^ ft.) — a 
quarter of 14 ft. The difference (i ft.) is the 
required width of the arm of the red cross. 
The area of the cross will then be the same as 
that of the white ground. 

186.— THE CLOTHES LINE PUZZLE. 

Multiply together, and also add together, the 
heights of the two poles and divide one resulr 



SOLUTIONS. 



183 



by the other. That is, if the two heights are 

a and 6 respectively, then -^— =- will give the 

height of the intersection. In the particular 
case of our puzzle, the intersection was there- 
fore 2 ft. II in. from the ground. The distance 
that the poles are apart does not affect the 
answer. The reader who may have imagined 
that this was an accidental omission will per- 
haps be interested in discovering the reason 
why the distance between the poles may be 
ignored. 

187.— THE MILKMAID PUZZLE. 




RiVtR 



X^OO-R 



STOOL 



Draw a straight line, as shown in the diagram, 
from the milking-stool perpendicular to the 
near bank of the river, and continue it to the 
point A, which is the same distance from that 
bank as the stool. If you now draw the straight 
line from A to the door of the dairy, it will cut 
the river at B. Then the shortest route wiU 
be from the stool to B and thence to the door. 
Obviously the shortest distance from A to the 
door is the straight line, and as the distance 
from the stool to any point of the river is the 
same as from A to that point, the correctness 
of the solution will probably appeal to every 
reader without any acquaintance with geometry. 

188.— THE BALL PROBLEM. 

If a round ball is placed on the level ground, six 
similar baUs may be placed round it (all on the 
ground), so that they shall all touch the central 
ball. 

As for the second question, the ratio of the 
diameter of a circle to its circumference we call 
pi ; and though we cannot express this ratio in 
exact numbers, we can get sufficiently near to 
it for all practical purposes. However, in this 
case it is not necessary to know the value of pi 
at all. Because, to find the area of the surface 
of a sphere we multiply the square of the 
diameter by pi ; to find the volume of a sphere 



we multiply the cube of the diameter by one- 
sixth of pi. Therefore we may ignore pi, and 
have merely to seek a number whose square 
shall equal one-sixth of its cube. This number 
is obviously 6. Therefore the baU was 6 ft. 
in diameter, for the area of its surface will be 
36 times pi in square feet, and its volume also 
36 times pi in cubic feet. 



189.— THE YORKSHIRE ESTATES. 

The triangular piece of land that was not for 
sale contains exactly eleven acres. Of covurse 
it is not difficult to find the answer if we follow 
the eccentric and tricky tracks of intricate trigo- 
nometry ; or I might say that the application 
of a. well-known formula reduces the problem 
to finding one-quarter of the square root of 
(4x370x116) — (37o-f-ii6 — 74)2 — that is a 
quarter of the square root of 1936, which is one- 
quarter of 44, or II acres. But aU that the 
reader really requires to know is the Pytha- 
gorean law on which many puzzles have been 
built, that in any right-angled triangle the 
square of the hypotenuse is equal to the sum 
of the squares of the other two sides. I shaU 
dispense with all " surds " and similar ab- 
surdities, notwithstanding the fact that the 
sides of our triangle are clearly incommensurate, 
since we cannot exactly extract the square roots 
of the three square areas. 




In the above diagram ABC represents our 
triangle. A D B is a right-angled triangle, 
A D measuring 9 and B D measuring 17, be- 
cause the square of 9 added to the square of 17 
equals 370, the known area of the square on 
A B. Also A E C is a right-angled triangle, 
and the square of 5 added to the square of 7 
equals 74, the square estate on A C. Similarly, 
C F B is a right-angled triangle, for the square 
of 4 added to the square of 10 equals 116, the 
square estate on B C. Now, although the sides 
of our triangular estate are incommensurate, 
we have in this diagram aU the exact figures 
that we need to discover the area with pre- 
cision. 

The area of our triangle A D B is clearly half 
of 9 X 17, or 76I acres. The area of A E C is 
half of 5x7, or 17^ acres ; the area of C F B 
is half of 4 X 10, or 20 acres ; and the area of 
the oblong E D F C is obviously 4x7, or 28 
acres. Now, if we add together 17^, 20, and 



184 



AMUSEMENTS IN MATHEMATICS. 



28 =65 J, and deduct this sum from the area of 
the large triangle A D B (which we have found 
to be 76^ acres), what remains must clearly be 
the area of A B C. That is to say, the area we 
want must be 76^ — 65+= 11 acres exactly. 

190.— FARMER WURZEL'S ESTATE. 

The area of the complete estate is exactly one 
hundred acres. To find this answer I use the 



following little formula, ^ 4ab-{a+b-c)^ 



where a, b, c represent the three square areas, 
in any order. The expression gives the area of 
the triangle A. This will be found to be 9 acres. 
It can be easily proved that A, B, C, and D 
are all equal in area ; so the answer is 26 + 20 + 
18+9 + 9 + 9 + 9=100 acres. 

Here is the proof. If every little dotted square 
in the diagram represents an acre, this must be 




a correct plan of the estate, for the squares of 
5 and I together equal 26 ; the squares of 4 and 
2 equal 20 ; and the squares of 3 and 3 added 
together equal 18. Now we see at once that 
the area of the triangle E is 2^, F is?4j, and G is 
4. These added together make 11 acres, which 
we deduct from the area of the rectangle, 20 
acres, and we find that the field A contains 
exactly 9 acres. If you want to prove that B, 
C, and D are equal in size to A, divide them in 
two by a line from the middle of the longest 
side to the opposite angle, and you will find that 
the two pieces in every case, if cut out, wiU 
exactly fit together and form A. 

Or we can get our proof in a still easier way. 
The complete area of the squared diagram is 
12x12=144 acres, and the portions i, 2, 3, 4, 
not included in the estate, have the respective 
areas of 12^, 17J, gl, and 4J. These added 
together make 44, which, deducted from 144, 
leaves 100 as the required area of the complete 
estate. 



191.— THE CRESCENT PUZZLE. 

Referring to the original diagram, let A C be 
X, let C D be ;»; - 9, and let E C be ;c - 5. Then 
^ — 5 is a mean proportional between x — g and x, 
from which we find that x equals 25. Therefore 
the diameters are 50 in. and 41 in. respectively. 



THE PUZZLE WALL. 




The answer given in all the old books is that 
shown in Fig. i, where the curved waU shuts 
out the cottages from access to the lake. But 
in seeking the direction for the " shortest pos- 
sible " waU most readers to-day, remembering 
that the shortest distance between two points 
is a straight line, wiU adopt the method shown 
in Fig. 2. This is certainly an improvement, 
yet the correct answer is really that indicated 
in Fig. 3. A measurement of the lines will 
show that there is a considerable saving of 
length in this wall. 

193.— THE SHEEP-FOLD. 

This is the answer that is always given and 
accepted as correct : Two more hurcfies would 
be necessary, for the pen was twenty-four by 
one (as in Fig. A on next page), and by moving 
one of the sides and placing an extra hurdle 
at each end (as in Fig. B) the area would be 
doubled. The diagrams are not to scale. Now 
there is no condition in the puzzle that requires 
the sheep-fold to be of any particular form. 
But even if we accept the point that the pen 
was twenty-four by one, the answer utterly fails, 
for two extra hurdles are certainly not at all 
necessary. For example, I arrange the fifty 
hurdles as in Fig. C, and as the area is increased 
from twenty-four " square hurdles " to 156, 
there is now accommodation for 650 sheep. 
If it be held that the area must be exactly 
double that of the original pen, then I construct 
it (as in Fig. D) with twenty-eight hiu-dles only, 
and have twenty-two in hand for other purposes 
on the farm. Even if it were insisted that aU 
the original hurdles must be used, then I should 
construct it as in Fig. E, where I can get the 
area as exact as any farmer could possibly 
require, even if we have to allow for the fact 
that the sheep might not be able to graze at 
the extreme ends. Thus we see that, from any 



SOLUTIONS. 



185 



24 



24 



n^ 




point of view, the accepted answer to this 
ancient little puzzle breaks down. And yet 
attention has never before been drawn to the 
absurdity. 

194.— THE GARDEN WALLS. 

The puzzle was to divide the circular field into 
four equal parts by three walls, each wall being 
of exactly the same length. There are two 
essential difficulties in this problem. These 
are : (i) the thickness of the walls, and (2) the 
condition that these walls are three in number. 
As to the first point, since we are told that the 
walls are brick walls, we clearly cannot ignore 
their thickness, while we have to find a solution 
that wiU equally work, whether the walls be 
of a thickness of one, two, three, or more bricks. 
The second point requires a little more con- 
sideration. How are we to distinguish between 
a wall and walls ? A straight wall without any 
bend in it, no matter how long, cannot ever 
become " walls," if it is neither broken nor 
intersected in any way. Also our circular field 
is clearly enclosed by one wall. But if it had 
happened to be a square or a triangular en- 
closure, would there be respectively four and 
three walls or only one enclosing wall in each 
case ? It is true that we speak of " the four 
walls " of a square building or garden, but this 
is only a conventional way of saying " the four 
sides. If you were speaking of the actual brick- 
work, you would say, " I am going to enclose this 
square garden with a wall." Angles clearly do 
not affect the question, for we may have a zig- 
zag wall just as well as a straight one, and the 
Great Wadl of China is a good example of a wall 
with plenty of angles. Now, if you look at 
Diagrams i, 2, and 3, you may he puzzled to 



declare whether there are in each case two or 
four new walls ; but you cannot call them three, 
as required ta our puzzle. The intersection either 
affects the question or it does not affect it. 

If you tie two pieces of string firmly together, 
or splice them in a nautical manner, they 
become " one piece of string." If you simply 
7* let them lie across one another or overlap, they 
remain " two pieces of string." It is aU a 
question of joining and welding. It may simi- 
larly be held that if two walls be built into one 
another — I might almost say, if they be made 
homogeneous — they become one wall, in which 
case Diagrams i, 2, and 3 might each be said to 
show one wall or two, if it be indicated that 
the four ends only touch, and are not really 
c. built into, the outer circular wall. 

The objection to Diagram 4 is that although 
it shows the three required walls (assuming the 
ends are not built into the outer circular wall), 
yet it is only absolutely correct when we assume 
the walls to have no thickness. A brick has 
thickness, and therefore the fact throws the 
whole method out and renders it only approxi- 
mately correct. 

Diagram 5 shows, perhaps, the only correct 
and perfectly satisfactory solution. It wiU be 
noticed that, in addition to the circular wall, 
there are three new walls, which touch (and so 
enclose) but are not built into one another. 
This solution may be adapted to any desired 
thickness of wall, and its correctness as to 
area and length of wall space is so obvious 
that it is unnecessary to explain it. I will. 




however, just say that the semicircular piece 
of ground that each tenant gives to his neigh- 
bour is exactly equal to the semicircular piece 
that his neighbour gives to him, while any 
section of wall space found in one garden is 
precisely repeated in aU the others. Of course 
there is an infinite number of ways in which 
this solution may be correctly varied. 



I8& 



AMUSEMENTS IN MATHEMATICS. 



LADY BELINDA'S GARDEN. 




All that Lady Belinda need do was this : She 
should measure from A to^ B, fold her tape in 
four and mark off the point'" E, which is thus one 
quarter of the side. Then, in the same way, 
mark off the point F, one-fourth of the side AD. 
Now, if she makes EG equal to AF, and GH 
equal to EF, then AH is the required width 
for the path in order that the bed shall be ex- 
actly half the area of the garden. An exact 
numerical measurement can only be obtained 
when the sum of the squares of the two sides 
is a square number. Thus, if the garden meas- 
ured 12 poles by 5 poles (where the squares 
of 12 and 5, 144 and 25, sum to 169, the square 
of 13), then 12 added to 5, less 13, would equal 
four, and a quarter of this, i pole, would be 
the width of the path. 

196.— THE TETHERED GOAT. 




This problem is quite simple if properly at- 
tacked. Let us suppose the triangle ABC to 
represent our half-acre field, and the shaded 
portion to be the quarter-acre over which the 
goat will graze when tethered to the comer C. 
Now, as six equal equilateral triangles placed 
together wiU form a regular hexagon, as shown, 
it is evident that the shaded pasture is just 
one-sixth of the complete area of a circle. 
Therefore aU we require is the radius (CD) of 
a circle containing six quarter-acres or i^ acres, 
which is equal to 9,408,960 square inches. As 
we only want our answer " to the nearest inch," 
it is sufficiently exact for our purpose if we as- 
simae that as i is to 3.1416, so is the diameter of 
a circle to its circumference. If, therefore, we 
divide the last number I gave by 3. 141 6, and 
extract the square root, we find that 1,731 



inches, or 48 yards 3 inches, is the required 
length of the tether " to the nearest inch." 

197.— THE COMPASSES PUZZLE. 

Let AB in the following diagram be the given 
straight line. With the centres A and B and 
radius AB describe the two circles. Mark off DE 
and EF equal to AD. V/ith the centres A and 
F and radius DF describe arcs intersecting at 
G. With the centres A and B and distance B G 
describe arcs GHK and N. Make HK equal 
to AB and HL equal to HB. Then with centres 
K and L and radius A B describe arcs intersect- 
ing at I. Make BM equal to B I. Finally, with 




the centre M and radius MB cut the line in C, 
and the point C is the required middle of the 
line AB. For greater exactitude you can mark 
off R from A (as you did M from B), and from 
R describe another arc at C. This also solves 
the problem, to find a point midway between 
two given points without the straight line. 

I win put the young geometer in the way of a 
rigid proof. First prove that twice the square 
of the line AB equals the square of the distance 
BG, from which it foUows that HABN are the 
four comers of a square. To prove that I is 
the centre of this square, draw a line from H to 
P through QIB and continue the arc HK to P. 
Then, conceiving the necessary lines to be 
drawn, the angle HKP, being in a semicircle, 
is a right angle. Let fall the perpendicular KQ, 
and by similar triangles, and from the fact that 
HKI is an isosceles triangle by the construc- 
tion, it can be proved that HI is half of HB. 
We can similarly prove that C is the centre of 
the square of which AIB are three comers. 

I am aware that this is not the simplest pos 
sible solution. 

198.— THE EIGHT STICKS. 

The first diagram is the answer that nearly 
every one will give to this puzzle, and at first 
sight it seems quite satisfactory. But consider 
the conditions. We have to lay " every one 
of the sticks on the table." Now, if a ladder be 



SOLUTIONS. 



187 



placed against a wall with only one end on the 
ground, it can hardly be said that it is *' laid on 







the ground." And if we place the sticks in the 
above manner, it is only possible to make one 
end of two of them touch the table : to say that 
every one lies on the table would not be cor- 
rect. To obtain a solution it is only necessary 
to have our sticks of proper dimensions. Say 
the long sticks are each 2 ft. in length and the 
short ones i ft. Then the sticks must be 3 in, 
thick, when the three equal squares may be 
enclosed, as shown in the second diagram. If 
I had said " matches " instead of " sticks," the 
puzzle would be impossible, because an ordinary 
match is about twenty-one times as long as it 
is broad, and the enclosed rectangles would 
not be squares. 

199.— PAPA'S PUZZLE. 

I HAVE found that a large number of people 
imagine that the following is a correct solution 
of the problem. Using the letters in the dia- 
gram below, they argue that if you make the 
distance B A one-third of B C, and therefore the 
area of the rectangle ABE equal to that of 
the triangular remainder, the card must hang 




with the long side horizontal. Readers will 
remember the jest of Charles II., who induced 
the Royal Society to meet and discuss the 



reason why the water in a vessel wiU not rise 
if you put a live fish in it ; but in the middle of 
the proceedings one of the least distinguished 
among them quietly slipped out and made the 
experiment, when he found that the water did 
rise 1 If my correspondents had similarly made 
the experiment with a piece of cardboard, they 
would have found at once their error. Area is 
one thing, but gravitation is quite another. 
The fact of that triangle sticking its leg out to 
D has to be compensated for by additional area 
in the rectangle. As a matter of fact, the ratio 
of B A to AC is as i is to the square root of 3, 
which latter cannot be given in an exact numeri- 
cal measure, but is approximately 1.732. Now 
let us look at the correct general solution. There 
are many ways of arriving at the desired result, 
but the one I give is, I think, the simplest for 
beginners. 

Fix your card on a piece of paper and draw 
the equilateral triangle BCF, BF and CF beuig 
equal to BC. Also mark off the point G so 
that DG shall equal DC. Draw the line CG 
and produce it until it cuts the line B F in H. 
If we now make HA parallel to BE, then A is 
the point from which our cut must be made to 
the corner D, as indicated by the dotted line. 

A curious point in connection with this 
problem is the fact that the position of the 
point A is independent of the side CD. The 
reason for this is more obvious in the solution 
I have given than in any other method that I 
have seen, and (although the problem may be 
solved with all the working on the cardboard) 
that is partly why I have preferred it. It will 
be seen at once that however much you may 
reduce the width of the card by bringing E 
nearer to B and D nearer to C, the line CG, 
being the diagonal of a square, will always lie 
in the same direction, and wiU cut BF in H. 
Finally, if you wish to get an approximate 
measure for the distance BA, all you have to 
do is to multiply the length of the card by the 
decimal .366. Thus, if the card were 7 inches 
long, we get 7 x. 366=2. 562, or a little more 
than 2^ inches, for the distance from B to A. 

But the real joke of the puzzle is this : We 
have seen that the position of the point A is 
independent of the width of the card, and de- 
pends entirely on the length. Now, in the 
illustration it will be found that both cards 
have the same length ; consequently all the 
little maid had to do was to lay the clipped 
card on top of the other one and mark off 
the point A at precisely the same distance 
from the top left-hand comer ! So, after all, 
Pappus' puzzle, as he presented it to his little 
maid, was quite an infantile problem, when he 
was able to show her how to perform the feat 
without first introducing her tO the elements of 
statics and geometry. 

200.— A KITE-FLYING PUZZLE. 

Solvers of this little puzzle, I have generally 
found, may be roughly divided into two classes : 
those who get within a mile of the correct 
answer by means of more or less complex calcu- 



1 88 



AMUSEMENTS IN MATHEMATICS. 



lations, involving *' pi" and those whose arith- 
metical kites fly hundreds and thousands of 
miles away from the truth. The compara- 
tively easy method that I shall show does not 
involve any consideration of the ratio that the 
diameter of a circle bears to its circumference. 
I call it the " hat-box method," 




Supposing we place our ball of wire, A, in a 
cylindrical hat-box, B, that exactly fits it, so 
that it touches the side all round and exactly 
touches the top and bottom, as shown in the 
illustration. Then, by an invariable law that 
should be known by everybody, that box con- 
tains exactly half as much again as the ball. 
Therefore, as the ball is 24 in. in diameter, a 
hat-box of the same circumference but two- 
thirds of the height (that is, 16 in. high) will have 
exactly the same contents as the ball. 

Now let us consider that this reduced hat- 
box is a cylinder of metal made up of an im- 
mense number of little wire cylinders close to- 
gether like the hairs in a painter's brush. By 
the conditions of the puzzle we are allowed to 
consider that there are no spaces between the 
wires. How many of these cylinders one one- 
hundredth of an inch thick are equal to the 
large cylinder, which is 24 in. thick ? Circles 
are to one another as the squares of their dia- 
meters. The square of yj^ is tuJ^iti ^^^ the 
square of 24 is 576 ; therefore the large cylinder 
contains 5,760,000 of the little wire cylinders. 
But we have seen that each of these wires is 
16 in. long; hence 16x5,760,000=92,160,000 
inches as the complete length of the wire. 
Reduce this to miles, and we get 1,454 miles 
2,880 ft. as the length of the wire attached to 
the professor's kite. 

Whether a kite would fly at such a height, 
or support such a weight, are questions that do 
not enter into the problem. 

201.— HOW TO MAKE CISTERNS. 

Here is a general formula for solving this 
problem. Call the two sides of the rectangle 

a and b. Then «+^- Vg'' + 62-a6 ^^^^^ ^^^ 

6 
side of the little square pieces to cut away. The 
measurements given were 8 ft. by 3 ft., and the 
above rule gives 8 in. as the side of the square 
pieces that have to be cut away. Of course. 



it wiU not always come out exact, as in this 
case (on account of that square root), but you 
can get as near as you like with decimals. 

202.— THE CONE PUZZLE. 

The simple rule is that the cone must be cut at 
one-third of its altitude. 

203.— CONCERNING WHEELS. 

If you mark a point A on the circumference of a 
wheel that runs on the surface of a level road, 
like an ordinary cart-wheel, the curve described 
by that point will be a common cycloid, as in 
Fig. I. But if you mark a point B on the cir- 
cumference of the flange of a locomotive-wheel, 
the curve will be a curtate cycloid, as in Fig. 2, 
terminating in nodes. Now, if we consider one 
of these nodes or loops, we shall see that " at 
any given moment " certain points at the bot- 
tom of the loop must be moving in the opposite 
direction to the train. As there is an infinite 




number of such points oh the flange's circum- 
ference, there must be an infinite number of 
these loops being described while the train is 
in motion. In fact, at any given moment cer- 
tain points on the flanges are always moving 
in a direction opposite to that in which the train 
is going. 

In the case of the two wheels, the wheel that 
runs round the stationary one makes two revo- 
lutions round its own centre. As both wheels 
are of the same size, it is obvious that if at the 
start we mark a point on the circumference of 
the upper wheel, at the very top, this point will 
be in contact with the lower wheel at its lowest 
part when half the journey has been made. 
Therefore this point is again at the top of the 
moving wheel, and one revolution has been 
made. Consequently there are two such revolu- 
tions in the complete journey. 

204.— A NEW MATCH PUZZLE. 

I. The easiest way is to arrange the eighteen 
matches as in Diagrams i and 2, making the 
length of the perpendicular A B equal to a match 
and a half. Then, if the matches are an inch in 



SOLUTIONS. 



189 



length, Fig. i contains two square inches and 
Fig. 2 contains six square inches — 4X1^. The 
second case (2) is a little more difficult to solve. 
The solution is given in Figs. 3 and 4. For the 

1 2 A 



rli\ 







purpose of construction, place matches tempo- 
rarily on the dotted lines. Then it will be seen 
that as 3 contains five equal equilateral tri- 
angles and 4 contains fifteen similar triangles, 
one figure is three times as large as the other, 
and exactly eighteen matches are used. 

205.— THE SIX SHEEP-PENS. 




Place the twelve matches in the manner shown 
in the illustration, and you will have six pens of 
equal size. 

206.— THE KING AND THE CASTLES. 

There are various ways of building the ten 
castles so that they shall form five rows with 
four castles in every row, but the arrangement 
in the next column is the only one that also 
provides that two castles (the greatest number 
possible) shall not be approachable from the 
outside. It will be seen that you must cross 
the walls to reach these two. 

207.— CHERRIES AND PLUMS. 

There are several ways in which this problem 
might be solved were it not for the condition 
that as few cherries and plums as possible shall 
be planted on the north and east sides of the 
orchard. The best possible arrangement is that 
shown in the diagram, where the cherries, plums, 




THE KING AND THE CASTLES. 

and apples are indicated respectively by the 
letters C, P, and A. The dotted lines connect 
the cherries, and the other lines the plums. It 
will be seen that the ten cherry trees and the 
ten plum trees are so planted that each fruit 
forms five lines with four trees of its kind in 



r A A A 




C' ""P A A A 
P A— C^ P A P-^P A 



line. This is the only arrangement that allows 
of so few as two cherries or plums being planted 
on the north and east outside rows. 

208.— A PLANTATION PUZZLE. 

The illustration shows the ten trees that must 
be left to form five rows with four trees in every 



190 



AMUSEMENTS IN MATHEMATICS. 



<i. 


d 


• 


• 


• 




'• 


* 












• 


\ 1 ' 


• 


« 


• 


• 


• 


; • 


% 


V 

« 


• 


• 


• 


• 




* 






• 


• 


\ 




\ 


\ / 


' * 


* 






• 


• 
• 


X 


• 




• 


• 




. t' 


» 






'* 






• j» 








« 




• 




• \ 


• 


• 


\ 


• 
* 


/ 






% 








t" 


....q.. 


-—•*-■ 


•'-%- 


»•*»- 


••*••- 


'-■§ 



row. The dots represent the positions of the 
trees that have been cut down. 



209.— THE TWENTY-ONE TREES. 

I GIVE two pleasing arrangements of the trees. 
In each case there are twelve straight rows with 
five trees in every row. 



As all the points and lines puzzles that I have 
given so far, excepting the last, are variations 
of the case of ten points arranged to form five 
lines of four, it will be well to consider this 
particular case generally. There are six funda- 
mental solutions, and no more, as shown in the 
six diagrams. These, for the sake of conveni- 
ence, I named some years ago the Star, the Dart, 



\v 



•Vv 



..^' 



/» 




* 



0— O- -0-.< 



•O 



the Compasses, the Funnel, the Scissors, and 
the Nail. (See next page.) Readers will under- 
stand that any one of these forms may be dis- 
torted in an infinite niunber of difierent ways 
without destroying its real character. 

In " The King and the Castles " we have the 
Star, and its solution gives the Compasses. 
In the " Cherries and Plums " solution we find 
that the Cherries represent the Funnel and the 
Plums the Dart. The solution of the " Plan- 




210.— THE TEN COINS. 

The answer is that there are just 2,400 different 
ways. Any three coins may be taken from one 
side to combine with one coin taken from the 
other side. I give four examples on this and 
the next page. We may thus select three from 
the top in ten ways and one from the bottom 
in five ways, making fifty. But we may also 
select three from the bottom and one from the 
top in fifty ways. We may thus select the 
four coins in one hundred ways, and the four 
removed may be arranged by permutation in 
twenty-four ways. Thus there are 24 X 100= 
2,400 different solutions. 



tation Puzzle " is an example of the Dart dis- 
torted. Any solution to the " Ten Coins " will 
represent the Scissors. Thus examples of aU 
have been given except the Nail. 

On a reduced chessboard, 7 by 7, we may place 
the ten pawns in just three different ways, but 
they must all represent the Dart. The " Plan- 
tation " shows one way, the Plums show a 
second way, and the reader may like to find 
the third way for himself. On an ordinary 
chessboard, 8 by 8, we can also get in a beauti- 
ful example of the Funnel — symmetrical in 
relation to the diagonal of the board. The 
smallest board that will take a Star is one 9 by 7. 
The Nail requires a board 11 by 7, the Scissors 



SOLUTIONS. 



191 



• • • p 

/ .^•. '•. 




Q • • • 

■ • \ > 

&■ 6- 0-- 




• •• 9 * .^ 

#■ 

y / 1 \ 

d"d''6"'b ♦ 



II by 9, and the Compasses 17 by 12. At least 
these are the best results recorded in my note- 
book. They may be beaten, but Ido not think so. 
If you divide a chessboard into two parts by 
a diagonal zigzag line, so that the larger part 
contains 36 squares and the smaller part 28 



tions, it is clearly necessary to find what is 
the smallest number of heads that could form 
sixteen lines with three heads in every line. 
Note that I say sixteen, and not thirty-two, 
because every line taken by a bullet may be 
also taken by another bullet fired in exactly 




STAR 



DART 



COMPA.SSES FUNNEL ScrSSORS 



squares, you can place three separate schemes 
on the larger part and one on the smaller part 
(all Darts) without their conflicting — that is, 
they occupy forty dififerent squares. They can 
be placed in other ways without a division of 
the board. The smallest square board that 
will contain six different schemes (not funda- 
mentally different), without any line of one 
scheme crossing the line of another, is 14 by 14 ; 
and the smallest board that will contain one 
scheme entirely enclosed within the lines of a 
second scheme, without any of the lines of the 
one, when drawn from point to point, crossing 
a line of the other, is 14 by 12. 

211.— THE TWELVE MINCE- PIES. 

If you ignore the four black pies in our illus- 
tration, the remaining twelve are in their origi- 
nal positions. Now remove the four detached 
pies to the places occupied by the black ones, 
and you will have your seven straight rows of 
foiir, as shown by the dotted lines. 

212.— THE BURMESE PLANTATION. 

The arrangement on the next page is the most 
symmetrical answer that can probably be found 
for twenty-one rows, which is, I believe, the 
greatest number of rows possible. There are 
several ways of doing it. 

213.— TURKS AND RUSSIANS. 

The main point is to discover the smallest pos- 
sible number of Russians that there could have 
been. As the enemy opened fire from all direc- 



the opposite direction. Now, as few as eleven 
points, or heads, may be arranged to form the 



THE TWELVE MINCE PIES. 

required sixteen lines of three, but the discovery 
of this arrangement is a hard nut. The diagram 



192 



AMUSEMENTS IN MATHEMATICS. 




THE BURMESE PLANTATION. 



at the foot of this page will show exactly how 
the thing is to be done. 

If, therefore, eleven Russians were in the 
positions shown by the stars, and the thirty- 
two Turks in the positions indicated by the 
black dots, it will be seen, by the lines shown, 
that each Turk may fire exactly over the 
heads of three Russians. But as each bullet 
kills a man, it is essential that every Turk 
shall shoot one of his comrades and be shot 
by him in turn ; otherwise we should have 
to provide extra Russians to be shot, which 
would be destructive of the correct solution 
of our problem. As the firing was simul- 
taneous, this point presents no difSculties. 
The answer we thus see is that there were 
at least eleven Russians amongst whom there 
was no casualty, and that all the thirty- 
two Tiurks were shot by one another. It was 
not stated whether the Russians fired any shots, 
but it will be evident that even if they did their 
firing could not have been effective : for if one 
of their bullets killed a Turk, then we have 
immediately to provide another man for one of 
the Turkish bullets to kill; and as the Turks 
were known to be thirty-two in number, this 




SOLUTIONS. 



193 



would necessitate our introducing another Rus- 
sian soldier and, of course, destroying the solu- 
tion. I repeat that the difficulty of the puzzle 
consists in finding how to arrange eleven points 
so that they shall form sixteen lines of three. I 
am told that the possibility of doing this was 
first discovered by the Rev. Mr. Wilkinson 
some twenty years ago. 

214.— THE SIX FROGS. 

Move the frogs in the following order : 2, 4, 6, 
5, 3, I (repeat these moves in the same order 
twice more), 2, 4, 6. This is a solution in 
twenty-one moves — the fewest possible. 

If n, the number of frogs, be even, we require 

— -!— moves, of which will be leaps and 

2 2 

n simple moves. If n be odd, we shaU need 

— i-^ 4 moves, of which !L_Zi? will be leaps 

2 2 

and 2n— 4 simple moves. 

In the even cases write, for the moves, all the 
even numbers in ascending order and the odd 
numbers in descending order. This series must 
be repeated Jm times and followed by the even 
numbers in ascending order once only. Thus 
the solution for 14 frogs will be (2, 4, 6, 8, 10, 12, 
14, 13, II, 9, 7, 5, 3, i) repeated 7 times and 
followed by 2, 4, 6, 8, 10, 12, 14=105 moves. 

In the odd cases, write the even numbers in 
ascending order and the odd numbers in de- 
scending order, repeat this series |(»— i) times, 
follow with the even numbers in ascending 
order (omitting n— i), the odd numbers in 
descending order (omitting i), and conclude 
with all the numbers (odd and even) in their 
natural order (omitting i and n). Thus for 11 
frogs : (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, i) repeated 
5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 
7, 8, 9, 10=73 moves. 

This complete general solution is published 
here for the first time. 



215.— THE GRASSHOPPER 
PUZZLE. 

Move the counters in the following order. The 
moves in brackets are to be made four times 
in succession. 12, i, 3, 2, 12, 11, i, 3, 2 (5, 
7, 9, 10, 8, 6, 4), 3, 2, 12, II, 2, I, 2. The 
grasshoppers will then be reversed in forty-four 
moves. 

The general solution of this problem is very 
difficult. Of course it can always be solved by 
the method given in the solution of the last 
puzzle, if we have no desire to use the fewest 
possible moves. But to employ a full economy 
of moves we have two main points to consider. 
There are always what I call a lower movement 
(L) and an upper movement (U). L consists 
in exchanging certain of the highest numbers, 
such as 12, II, 10 in our " Grasshopper Puzzle," 
with certain of the lower numbers, i, 2, 3 ; the 
former moving in a clockwise direction, the 
latter in a non-clockwise direction. U consists 
in reversing the intermediate counters. In the 
above solution for 12, it wiU be seen that 12, 11, 
and I, 2, 3 are engaged in the L movement, 
and 4, 5, 6, 7, 8, 9, 10 in the U movement. 
The L movement needs 16 moves and U 28, 
making together 44. We might also involve 
10 in the L movement, which would result in 
L 23, U 21, making also together 44 moves. 
These I call the first and second methods. But 
any other scheme will entail an increase of 
moves. You always get these two methods 
(of equal economy) for odd or even counters, 
Ijut the point is to determine just how many to 
involve in L and how many in U. Here is the 
solution in table form. But first note, in giving 
values to «, that 2, 3, and 4 counters are special 
cases, requiring respectively 3, 3, and 6 moves, 
and that 5 and 6 covmters do not give a mini- 
mum solution by the second method — only by 
the first. 



First Method. 



Total No. 

of 
Counters. 


L Movement. 


U Movement. 


Total No. 
of Moves. 


No. of 
Counters. 


No. of 
Moves. 


No. of 
Counters. 


No. of 
Moves. 


4W 
4M — 2 

4W + I 
4n— I 


« - 1 and n 
n- I „ » 
n „ n-l-i 

M-I „ M 


2(w-l)2 + 5M-7 

2(n-i)2-f 5M-7 

2w2+5n — 2 
2(M-l)2 + 5W-7 


2M-f I 
2«— I 
2M 
2« 


2n2-f 3n-|-i 
2(W— i)2-f3M — 2 

2^2+ 3M — 4 
2W2-f 3» — 4 


4(«2-|-«-i) 

4^2-5 

2(2w2 + 4«-3) 

4n2-l-4w-9 



(1,926) 



Second Method. 



Total No. 

of 
Counters. 


L Movement. 


U Movement. 


Total No. 
of Moves. 


No. of 
Counters. 


No. of 
Moves. 


No. of 
Counters. 


No. of 
Moves. 


4n 
4»— 2 

4M-f I 

4»— I 


n and n 
n—x „ w-i 
n „ « 
n „ M 


2n2 + 3n — 4 

2(W-l)2+3M-7 
2^2+ 3M — 4 

2»2-i-3» — 4 


2W 
2M 

2W-f-I 
2M— I 


2(W— l)2+5n— 2 
2(W— l)2-|-5M— 2 
2«2 + 5n — 2 
2(»-l)2 + 5«-7 


4(n2+n-i) 
4^2-5 

2(2n2-|-4n— 3) 
4w2-f 4»— 9 



13 



194 



AMUSEMENTS IN MATHEMATICS. 



More generally we may say that with ni 
counters, where m is even and greater than 4, 

we require —^ moves ; and where m is 

odd and greater than 3, ^ +ow— 31 jQQygg. I 

4 
have thus shown the reader how to find the 
minimum number of moves for any case, and 
the character and direction of the moves. I 
will leave him to discover for himself how the 
actual order of moves is to be determined. 
This is a hard nut, and requires careful adjust- 
ment of the L and the U movements, so that 
they may be mutually accommodating. 

216.— THE EDUCATED FROGS. 

The following leaps solve the puzzle in ten 
moves : 2 to i, 5 to 2, 3 to 5, 6 to 3, 7 to 6, 4 to 7, 
I to 4, 3 to I, 6 to 3, 7 to 6. 

217.— THE TWICKENHAM PUZZLE. 

Play the coimters in the following order : K C E 
KWTCEHMKWTANCEHMIKCEH 
M T, and there you are, at Twickenham. The 
position itself will always determine whether you 
are to make a leap or a simple move. 

218.— THE VICTORIA CROSS PUZZLE. 

In solving this puzzle there were two things to 
be achieved : first, so to manipulate the coun- 
ters that the word VICTORIA should read 
roimd the cross in the same direction, only with 
the V on one of the dark arms ; and secondly, 
to perform the feat in the fewest possible moves. 
Now, as a matter of fact, it would be impos- 
sible to perform the first part in any way what- 
ever if all the letters of the word were different ; 
but as there are two I's, it can be done by mak- 
ing these letters change places — that is, the 
first I changes from the 2nd place to the 7th, and 
the second I from the 7th place to the 2nd. 
But the point I referred to, when introducing 
the puzzle, as a Little remarkable is this : that 
a solution in twenty-two moves is obtainable by 
moving the letters in the order of the follow- 
ing words : " A VICTOR ! A VICTOR ! A 
VICTOR I ! " 

There are, however, just six solutions in 
eighteen moves, and the following is one of 
them : I (i), V, A, I (2), R, O, T, I (i), I (2), 
A, V, I (2), I (i), C, I (2), V, A, I (i). The first 
and second I in the word are distinguished by 
the numbers i and 2. 

It will be noticed that in the first solution 
given above one of the I's never moves, though 
the movements of the other letters cause it to 
change its relative position. There is another 
peculiarity I may point out — that there is a 
solution in twenty-eight moves requiring no 
letter to move to the central division except 
the I's. I may also mention that, in each of 
the solutions in eighteen moves, the letters 
C, T, O, R move once only, while the second 
I always moves four times, the V always being 
transferred to the right arm of the cross. 



219.— THE LETTER BLOCK PUZZLE. 

This puzzle can be solved in 23 moves — the 
fewest possible. Move the blocks in the follow- 
ing order : A, B, F, E, C, A, B, F, E, C, A, B, 
D, H, G, A, B, D, H, G, D, E, F. , 

220.— A LODGING-HOUSE DIFFICULTY. ' 

The shortest possible way is to move the 
articles in the following order : Piano, book-, 
case, wardrobe, piano, cabinet, chest of drawers, 
piano, wardrobe, bookcase, cabinet, wardrobe, , 
piano, chest of drawers, wardrobe, cabinet, book- 
case, piano. Thus seventeen removals are neces- 
sary. The landlady could then move chest of 
drawers, wardrobe, and cabinet. Mr. Dobson 
did not mind the wardrobe and chest of drawers 
changing rooms so long as he secured the piano. 

221.— THE EIGHT ENGINES. 

The solution to the Eight Engines Puzzle is as 
follows : The engine that has had its fire drawn 
and therefore cannot move is No. 5. Move 
the other engines in the following order : 7, 6, , 
3, 7, 6, I, 2, 4, I, 3, 8, I, 3, 2, 4, 3, 2, seventeen 
moves in aU, leaving the eight engines in the 
required order. 

There are two other slightly dififerent solutions. 

222.— A RAILWAY PUZZLE. 

This little puzzle may be solved in as few as 
nine moves. Play the engines as follows : 
From 9 to 10, from 6 to 9, from 5 to 6, from 
3 to 5, from I to 2, from 7 to i, from 8 to 7, 
from 9 to 8, and from 10 to 9. You wiU then 
have engines A, B, and C on each of the three 
circles and on each of the three straight Lines. 
This is the shortest solution that is possible. 

223.— A RAILWAY MUDDLE. 




3 ^hm 



4f^f*im 



s^mm 



Only six reversals are necessary. The white 
train (from A to D) is divided into three sections, 
engine and 7 wagons, 8 wagons, and i wagon. 
The black train (D to A) never imcouples any- 
thing throughout. Fig. i is original position 



SOLUTIONS. 



195 



with 8 and i uncoupled. The black train pro- 
ceeds to position in Fig. 2 (no reversal). The 
engine and 7 proceed towards D, and black train 
backs, leaves 8 on loop, and takes up position 
in Fig. 3 (first reversal). Black train goes to 
position in Fig. 4 to fetch single wagon (second 
reversal). Black train pushes 8 ofi loop and 
leaves single wagon there, proceeding on its 
joiurney, as in Fig. 5 (third and fourth reversals). 
White train now backs on to loop to pick up 
single car and goes right away to D (fifth and 
sixth reversals). 

224.— THE MOTOR-GARAGE PUZZLE. 

The exchange of cars can be made in forty- 
three moves, as follows : 6 — G, 2 — B, i — E, 
3— H, 4—1, 3— L, 6— K, 4— G, I— I, 2— J, 
5— H, 4— A, 7— F, 8— E, 4— D, 8— C, 7— A, 
8— G, 5— C, 2— B, I— E, 8—1, I— G, 2— J, 
7— H, I— A, 7— G, 2— B, 6 — E, 3— H, 8— L, 
3—1, 7— K, 3— G, 6—1, 2— J, 5— H, 3— C, 
5 — G, 2 — B, 6 — E, 5 — I, 6 — J. Of course, 
"6 — G" means that the car numbered "6" 
moves to the point " G." There are other ways 
in forty-three moves. 

225.— THE TEN PRISONERS. 



\ 


* 


i 


Z' 


t 


^t 


\' 


\ 


k 


^, 




''<t 




~ "^-^ 


^- 


-^ 




Jf4 



It will be seen in the illustration how the 
prisoners may be arranged so as to produce as 
many as sixteen even rows. There are 4 such 
vertical rows, 4 horizontal rows, 5 diagonal 
rows in one direction, and 3 diagonal rows in 
the other direction. The arrows here show the 
movements of the four prisoners, and it will be 
seen that the infirm man in the bottom comer 
has not been moved. 

226.— ROUND THE COAST. 

In order to place words round the circle under 
the conditions, it is necessary to select words in 
which letters are repeated in certain relative 
positions. Thus, the word that solves o\ir 
puzzle is " Swansea," in which the first and 
fifth letters are the same, and the third and 
seventh the same. We make out jumps as 
follows, taking the letters of the word in their 
proper order : 2 — 5, 7 — 2, 4 — 7, i — 4, 6 — i, 
3 — 6, 8 — 3. Or we could place a word like 
" Tarapur " (in which the second and fourth 



letters, and the third and seventh, are alike) 
with these moves : 6 — i, 7 — ^4, 2 — 7, 5 — 2, 8 — 5, 
3 — 6, 8 — 3. But " Swansea " is the only word, 
apparently, that will fulfil the conditions of 
the puzzle. 

This puzzle should be compared with Sharp's 
Puzzle, referred to in my solution to No. 341, 
" The Four Frogs." The condition " touch 
and jump over two " is identical with " touch 
and move along a line." 

227.— CENTRAL SOLITAIRE. 

Here is a solution in nineteen moves ; the 
moves enclosed in brackets count as one move 
only: 19—17, 16—18, (29—17, 17—19), 30—18, 
27—25, (22 — 24, 24—26), 31 — 23, (4—16, 16 — 
28), 7 — 9, 10 — 8, 12 — 10, 3 — II, 18 — 6, (i — 3, 
3— II), (13—27, 27—25), (21—7, 7—9), (33— 
31, 31 — 23), (10 — 8, 8 — 22, 22 — 24, 24 — 26, 26 — 
12, 12 — 10), 5 — 17. All the counters are now 
removed except one, which is left in the central 
hole. The solution needs judgment, as one is 
tempted to make several jumps in one move, 
where it would be the reverse of good play. 
For example, after playing the first 3 — 11 
above, one is inclined to increase the length of 
the move by continuing with 11 — 25, 25 — 27, or 
with II — 9, 9 — 7. 

I do not think the number of moves can be 
reduced. 

228.— THE TEN APPLES. 

Number the plates (i, 2, 3, 4), (5, 6, 7, 8), (9, 10, 
II, 12), (13, 14, 15, 16) in successive rows from 
the top to the bottom. Then transfer the 
apple from 8 to 10 and play as follows, always 
removing the apple jumped over: 9 — 11, i — 9, 
13 — 5, i6— 8, 4 — 12, 12 — 10, 3—1, I — 9, 9— II. 

229.— THE NINE ALMONDS. 

This puzzle may be solved in as few as four 
moves, in the following manner : Move 5 over 

8, 9, 3, I. Move 7 over 4. Move 6 over 2 and 
7. Move 5 over 6, and all the counters are 
removed except 5, which is left in the central 
square that it originally occupied. 

230.— THE TWELVE PENNIES. 

Here is one of several solutions. Move 12 to 3, 
7 to 4, 10 to 6, 8 to I, 9 to 5, II to 2. 

231.— PLATES AND COINS. 

Number the plates from i to 12 in the order 
that the boy is seen to be going in the illustra- 
tion. Starting from i, proceed as follows, 
where " i to 4 " means that you take the coin 
from plate No. i and transfer it to plate No. 4 : 
I to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, II to 2, 
and complete the last revolution to i, making 
three revolutions in all. Or you can proceed 
this way : 4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 

9, 10 to I. It is easy to solve in four revolutions, 
but the solutions in three are more difficult to 
discover. 

This is " The Riddle of the Fishpond " (No. 
41, Canterbury Puzzles) in a different dress. 



196 



AMUSEMENTS IN MATHEMATICS. 



232.— CATCHING THE MICE. 

In order that the cat should eat every thir- 
teenth mouse, and the white mouse last of all, 
it is necessary that the count should begin at the 
seventh mouse (calling the white one the first) — 
that is, at the one nearest the tip of the cat's 
tail. In this case it is not at all necessary to 
try starting at all the mice in turn imtil you 
come to the right one, for you can just start 
anywhere and note how far distant the last one 
eaten is from the starting point. You will find 
it to be the eighth, and therefore must start at 
the eighth, counting backwards from the white 
mouse. This is the one I have indicated. 

In the case of the second puzzle, where you 
have to find the smallest number with which 
the cat may start at the white mouse and eat 
this one last of all, unless you have mastered 
the general solution of the problem, which is 
very difficult, there is no better course open to 
you than to try every number in succession 
imtil you come to one that works correctly. 
The smallest number is twenty-one. If you 
have to proceed by trial, you will shorten your 
labour a great deal by only counting out the 
remainders when the number is divided suc- 
cessively by 13, 12, II, 10, etc. Thus, in the 
case of 21, we have the remainders 8, 9, 10, i, 3, 
5, 7, 3, I, I, 3, I, I. Note that I do not give the 
remainders of 7, 3, and i as nought, but as 7, 3, 
and I. Now, count round each of these num- 
bers in turn, and you will find that the white 
mouse is killed last of aU. Of course, if we 
wanted simply any niunber, not the smallest, 
the solution is very easy, for we merely take 
the least common multiple of 13, 12, 11, 10, etc. 
down to 2. This is 360360, and you will find 
that the first count kills the thirteenth mouse, 
the next the twelfth, the next the eleventh, and 
so on down to the first. But the most arith- 
metically inclined cat could not be expected to 
take such a big number when a small one like 
twenty-one would equally serve its purpose. 

In the third case, the smallest number is 100. 
The number 1,000 would also do, and there are 
just seventy-two other numbers between these 
that the cat might employ with equal success. 

233.— THE ECCENTRIC CHEESEMONGER. 

To leave the three piles at the extreme ends of 
the rows, the cheeses may be moved as follows — 
the numbers refer to the cheeses and not to 
their positions in the row : 7 — 2, 8 — 7, 9 — 8, 
10 — 15, 6 — 10, 5 — 6, 14 — 16, 13 — 14, 12—13, 
3 — I, 4 — 3, II — 4. This is probably the easiest 
solution of all to find. To get three of the piles 
on cheeses 13, 14, and 15, play thus : 9 — 4, 10 — 
9, II — 10, 6—14, 5 — 6, 12 — 15, 8 — 12, 7 — 8, 
16 — 5, 3 — 13, 2 — 3, I — 2. To leave the piles 
on cheeses 3, 5, 12, and 14, play thus : 8 — 3, 
9—14, 16 — 12, I — 5, 10 — 9, 7 — 10, II — 8, 2 — I, 
4—16, 13—2, 6 — II, 15—4- 

234.— THE EXCHANGE PUZZLE. 

Make the following exchanges of pairs ; H — K, 
H— E, H— C, H— A, I— L, I— F, I— D, K— L, 



G— J, J— A, F— K, L— E, D— K, E— F, E— D, 
E— B, B— K. It wiU be found that, although 
the white counters can be moved to their proper 
places in II moves, if we omit aU consideration 
of exchanges, yet the black cannot be so moved 
in fewer than 17 moves. So we have to intro- 
duce waste moves with the white coimters to 
equal the minimum required by the black. 
Thus fewer than 17 moves must be impossible. 
Some of the moves are, of course, interchange- 
able. 



235.— TORPEDO PRACTICE. 




If the enemy's fleet be anchored in the formation 
shown in the illustration, it will be seen that as 
many as ten out of the sixteen ships may be 
blown up by discharging the torpedoes in the 
order indicated by the numbers and in the 
directions indicated by the arrows. As each 
torpedo in succession passes imder three ships 
and sinks the fourth, strike out each vessel 
with the pencil as it is simk. 

236.— THE HAT PUZZLE. 



1 


2 




3 

• 


4 






& 






e 



9 


10 




u 


12 


• 









9 





• 





• 










• 


• 








• 









• 










• 


m 















• 


• 










• 


• 

















• 































# 


• 


• 





I SUGGESTED that the reader should try this 
puzzle with counters, so I give my solution in 
that form. The silk hats are represented by 
black counters and the felt hats by white 
counters. The first row shows the hats in 
their original positions, and then each succes- 



SOLUTIONS. 



197 



sive row shows how they appear after one of 
the five manipulations. It will thus be seen 
that we first move hats 2 and 3, then 7 and 
8, then 4 and 5, then 10 and 11, and, finally, 
I and 2, leaving the four silk hats together, the 
four felt hats together, and the two vacant 
pegs at one end of the row. The first three 
pairs moved are dissimilar hats, the last two 
pairs being similar. There are other ways of 
solving the puzzle. 

237.— BOYS AND GIRLS. 

There are a good many different solutions to 
this puzzle. Any contiguous pair, except 7-8, 
may be moved first, and after the first move 
there are variations. The following solution 
shows the position from the start right through 
each successive move to the end : — 



. .12345678 
4312. .5678 
4312765- -8 



43127. .568 
4. .2713568 
48627135. • 



238.— ARRANGING THE JAM POTS. 

Two of the pots, 13 and 19, were in their proper 
places. As every interchange may result in a 
pot being put in its place, it is clear that twenty- 
two interchanges will get them all in order. 
But this number of moves is not the fewest 
possible, the correct answer being seventeen. Ex- 
change the following pairs : (3 — I, 2 — 3)» (15 — 4> 
16—15), (17—7, 20—17), (24—10, 11—24, 12— 
11), (8 — 5, 6 — 8, 21 — 6, 23 — 21, 22 — 23, 14 — 22, 
9 — 14, 18—9). When you have made the inter- 
changes within any pair of brackets, all numbers 
within those brackets are in their places. There 
are five pairs of brackets, and 5 from 22 gives 
the number of changes required — 17. 

239.— A JUVENILE PUZZLE. 

c ^ 



A 



r> 



conditions, since folding the paper is not actu- 
ally forbidden. Of course the lines are here left 
unjoined for the purpose of clearness. 

In the rubbing out form of the puzzle, first 
rub out A to B with a single finger in one stroke. 
Then rub out the line GH with one finger. 
Finally, rub out the remaining two vertical 
lines with two fingers at once ! That is the 
old trick. 

240.— THE UNION JACK. 
B 







yy 


> 




< 




y\ 





There are just sixteen points (all on the outside) 
where three roads may be said to join. These 
are called by mathematicians " odd nodes." 
There is a rule that tells us that in the case of a 
drawing like the present one, where there are 
sixteen odd nodes, it requires eight separate 
strokes or routes (that is, half as many as there 
are odd nodes) to complete it. As we have to 
produce as much as possible with only one of 
these eight strokes, it is clearly necessary to 
contrive that the seven strokes from odd node 
to odd node shall be as short as possible. 
Start at A and end at B, or go the reverse way. 

241.— THE DISSECTED CIRCLE. 



H 

As the conditions are generally understood, this 
puzzle is incapable of solution. This can be 
demonstrated quite easily. So we have to 
look for some catch or quibble in the statement 
of what we are asked to do. Now if you fold 
the paper and then push the point of your 
pencil down between the fold, you can with one 
stroke make the two lines C D and E F in our 
diagram. Then start at A, and describe the 
line ending at B. Finally put in the last line 
GH, and the thing is done strictly within the 




It can be done in twelve continuous strokes, 
thus : Start at A in the illustration, and eight 



198 



AMUSEMENTS IN MATHEMATICS. 



strokes, forming the star, will bring you back 
to A ; then one stroke round the circle to B, 
one stroke to C, one round the circle to D, and 
one final stroke to E — twelve in all. Of course, 
in practice the second circular stroke wiU be 
over the first one ; it is separated in the dia- 
gram, and the points of the star not joined to 
the circle, to make the solution clear to the eye. 

242.— THE TUBE INSPECTOR'S PUZZLE. 

The inspector need only travel nineteen miles 
if he starts at B and takes the following route : 
BADGDEFIFCBEHKLIHGJK. Thus the only 
portions of line travelled over twice are the two 
sections D to G and F to I. Of course, the route 
may be varied, but it cannot be shortened. 

243.— VISITING THE TOWNS. 

Note that there are six towns, from which only 
two roads issue. Thus i must lie between 9 
and 12 in the circular route. Mark these two 
roads as settled. Similarly mark 9, 5, 14, and 
4, 8, 14, and 10, 6, 15, and 10, 2, 13, and 3, 7, 13. 
All these roads must be taken. Then you will 
find that he must go from 4 to 15, as 13 is closed, 
and that he is compelled to take 3, 11, 16, and 
also 16, 12. Thus, there is only one route, as 
foUows : I, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 
II, 16, 12, I, or its reverse— reading the line 
the other way. Seven roads are not used. 

244.— THE FIFTEEN TURNINGS. 

O O 0^^3-^*0 O O 



O 7 Q O O Q 6 O 



O Oii Q O O Q Jo() O 



O C) d) O z 



O O O O O O 6 1 



o O ,(|-f 



o p-<h-<yo--eg o o 



o o o o o O g o 



® o o o 



It will be seen from the illustration (where the 
roads not used are omitted) that the traveller 
can go as far as seventy miles in fifteen turnings. 
The turnings are all numbered in the order in 
which they are taken. It wiU be seen that he 
never visits nineteen of the towns. He might 
visit them all in fifteen turnings, never entering 
any town twice, and end at the black town 
from which he starts (see "The Rook's Tour," 



No. 320), but such a tour would only take him 
sixty-four miles. 

245.— THE FLY ON THE OCTAHEDRON. 
Aj 




Though we cannot really see all the sides of the 
octahedron at once, we can make a projection 
of it that suits our pmrpose just as well. In the 
diagram the six points represent the six angles 
of the octahedron, and four lines proceed from 
every point under exactly the same conditions 
as the twelve edges of the solid. Therefore if 
we start at the point A and go over all the lines 
once, we must always end our route at A. And 
the number of different routes is just 1,488, 
counting the reverse way of any route as difier- 
ent. It would take too much space to show 
how I make the count. It can be done in 
about five minutes, but an explanation of the 
method is difficult. The reader is therefore 
asked to accept my answer as correct. 

246.— THE ICOSAHEDRON PUZZLE. 

Hi 




There are thirty edges, of which eighteen were 
visible in the original illustration, represented 



SOLUTIONS. 



199 



in the following diagram by the hexagon 
NAESGD. By this projection of the solid we 
get an imaginary view of the remaining twelve 
edges, and are able to see at once their direction 
and the twelve points at which all the edges 
meet. The difierence in the length of the lines 
is of no importance ; all we want is to present 
their direction in a graphic manner. But in 
case the novice should be puzzled at only find- 
ing nineteen triangles instead of the required 
twenty, I wiU point out that the apparently 
missing triangle is the outline HIK. 

In this case there are twelve odd nodes ; 
therefore six distinct and disconnected routes 
wiU be needful if we are not to go over any 
lines twice. Let us therefore find the greatest 
distance that we may so travel in one route. 

It wiU be noticed that I have struck out with 
little cross strokes five lines or edges in the 
diagram. These five lines may be struck out 
anywhere so long as they do not join one an- 
other, and so long as one of them does not 
connect with N, the North Pole, from which we 
are to start. It wiU be seen that the result of 
striking out these five lines is that aU the nodes 
are now even except N and S. Consequently 
if we begin at N and stop at S we may go over 
aU the lines, except the five crossed out, without 
traversing any line twice. There are many ways 
of doing this. Here is one route : N to H, I, K, 
S, I, E, S, G, K, D, H, A, N, B, A, E, F, B, C, 
G, D, N, C, F, S. By thus making five of the 
routes as short as is possible — simply from one 
node to the next — ^we are able to get the greatest 
possible length for our sixth line. A greater 
distance in one route, without going over the 
same ground twice, it is not possible to get. 

It is now readily seen that those five erased 
lines must be gone over twice, and they may 
be " picked up," so to speak, at any points of 
our route. Thus, whenever the traveller hap- 
pens to be at I he can run up to A and back 
before proceeding on his route, or he may wait 
until he is at A and then run down to I and 
back to A. And so with the other lines that 
have to be traced twice. It is, therefore, clear 
that he can go over 25 of the lines once only 
(25 X 10,000 miles = 250,000 miles) and 5 of the 
lines twice (5x20,000 miles= 100,000 miles), 
the total, 350,000 miles, being the length of his 
travels and the shortest distance that is possible 
in visiting the whole body. 

It will be noticed that I have made him end 
his travels at S, the South Pole, but this is not 
imperative. I might have made him finish at 
any of the other nodes, except the one from 
which he started. Suppose it had been required 
to bring him home again to N at the end of his 
travels. Then instead of suppressing the line 
A I we might leave that open and close IS. 
This would enable him to complete his 350,000 
miles tour at A, and another 10,000 miles would 
take him to his own fireside. There are a great 
many different routes, but as the lengths of 
the edges are all alike, one course is as good as 
another. To make the complete 350,000 miles 
tour from N to S absolutely clear to everybody, 
I will give it entire : N to H, I, A, I, K, H, K, 



S, I, E, S, G, F, G, K, D, C, D, H, A, N, B, E, B, 
A, E, F, B, C, G, D, N, C, F, S— that is, thirty- 
five lines of 10,000 miles each. 

247.— INSPECTING A MINE. 

Starting from A, the inspector need only travel 
36 furlongs if he takes the following route : 
A to B, G, H, C, D, I, H, M, N, I, J, O, N, S, R, 
M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, 
A, F, K, P, Q. He thus passes between A and 
B twice, between C and D twice, between F and 
K twice, between J and O twice, and between 
R and S twice — five repetitions. Therefore 31 
passages plus 5 repeated equal 36 furlongs. 
The little pitfall in this puzzle lies in the fact 
that we start from an even node. Otherwise 
we need only travel 35 furlongs. 

248.— THE CYCLIST'S TOUR. 

When Mr. Maggs replied, " No way, I'm sure," 
he was not saying that the thing was impossible, 
but was really giving the actual route by which 
the problem can be solved. Starting from the 
star, if you visit the towns in the order, NO 
WAY, I'M SURE, you will visit every town 
once, and only once, and end at E. So both 
men were correct. This was the little joke of 
the puzzle, which is not by any means difficult. 

249.— THE SAILOR'S PUZZLE. 





„„i .^..^paaa^g^ | 


\ 




g 



There are only four different routes (or eight, 
if we count the reverse ways) by which the sailor 
can start at the island marked A, visit all the 
islands once, and once only, and return again 
to A. Here they are : — 



200 



AMUSEMENTS IN MATHEMATICS. 



AIPTLOEHRQDCFUGNSKMBA 
AIPTSNGLOEUFCDKMBQRHA 
ABMKSNGLTPIOEUFCDQRHA 
AIPTLOEUGNSKMBQDCFRHA 

Now, if the sailor takes the first route he 
will make C his 12th island (counting A as i) ; 
by the second route he will make C his 13th 
island ; by the third route, his i6th island ; 
and by the fourth route, his 17th island. If 
he goes the reverse way, C will be respectively 
his loth, 9th, 6th, and 5th island. As these 
are the only possible routes, it is evident that 
if the sailor puts off his visit to C as long as 
possible, he must take the last route reading 
from left to right. This route I show by the 
dark lines in the diagram, and it is the correct 
answer to the puzzle. 

The map may be greatly simplified by the 
" buttons and string " method, explained in 
the solution to No. 341, " The Four Frogs." 

250.— THE GRAND TOUR. 

The first thing to do in trying to solve a puzzle 
like this is to attempt to simplify it. If you 
look at Fig. i, you will see that it is a simplified 
version of the map. Imagine the circular 
towns to be buttons and the railways to be 
connecting strings. (See solution to No. 341.) 
Then, it will be seen, we have simply " straight- 
ened out " the previous diagram without af- 
fecting the conditions. Now we can further 
simplify by converting Fig. i into Fig. 2, which 
is a portion of a chessboard. Here the direc- 
tions of the railways will resemble the moves 
of a rook in chess — that is, we may move in 
any direction parallel to the sides of the dia- 
gram, but not diagonally. Therefore the first 
town (or square) visited must be a black one ; 
the second must be a white ; the third must 
be a black ; and so on. Every odd square 
visited will thus be black and every even one 
white. Now, we have 23 squares to visit (an 




odd number), so the last square visited must 
be black. But Z happens to be white, so the 
puzzle would seem to be impossible of solution. 
As we were told that the man " succeeded " in 
carrying out his plan, we must try to find some 
loophole in the conditions. He was to " enter 
every town once and only once," and we find 
no prohibition against his entering once the 
town A after leaving it, especially as he has 
never left it since he was bom, and would thus 
be " entering " it for the first time in his life. 
But he must return at once from the first town 
he visits, and then he will have only 22 towns 
to visit, and as 22 is an even number, there is 
no reason why he should not end on the white 
square Z. A possible route for him is indicated 
by the dotted line from A to Z. This route is 
repeated by the dark lines in Fig. i, and the 
reader will now have no difficulty in applying 
it to the original map. We have thus proved 
that the puzzle can only be solved by a retium 
to A immediately after leaving it. 

251.— WATER, GAS, AND ELECTRICITY. 



/. 




*«. — »* 



According to the conditions, in the strict sense 
in which one at first understands them, there 




Fig. I. 



Fig. 2. 



SOLUTIONS. 



20I 



is no possible solution to this puzzle. In such 
a dUemma one always has to look for some 
verbal quibble or trick. If the owner of house 
A will allow the water company to run their 
pipe for house C through his property (and we 
are not bound to assume that he would object), 
then the difi&culty is got over, as shown in our 
illustration. It wiU be seen that the dotted 
line from W to C passes through house A, but 
no pipe ever crosses another pipe. 

252.— A PUZZLE FOR MOTORISTS. 






® 



® 



■o 



S 



<b 



r© 



t 



1 



-© 



® 



® 



@>- 



(5)— I 



The routes taken by the eight drivers are shown 
in the illustration, where the dotted line roads 
are omitted to make the paths clearer to the 
eye. 

253.— A BANK HOLIDAY PUZZLE. 

The simplest way is to write in the number of 
routes to all the towns in this manner. Put a i 
on all the towns in the top row and in the first 
column. Then the number of routes to any 
town wiU be the sum of the routes to the town 
immediately above and to the town immedi- 
ately to the left. Thus the routes in the second 
row will be I, 2, 3, 4, 5, 6, etc., in the third row, 
I, 3, 6, 10, 15, 21, etc. ; and so on with the 



The general formula for the number of routes 
from one corner to the corner diagonally oppo- 
site on any such rectangular reticulated arrange- 
ment, under the conditions as to direction, is 
\ m+n I \m \n, where m is the number of 
towns on one side, less one, and n the number 
on the other side, less one. Our solution in- 
volves the case where there are 12 towns by 5. 
Therefore w=ii and n=4. Then the formula 
gives us the answer 1,365 as above. 

254.— THE MOTOR-CAR TOUR. 

First of all I will ask the reader to compare the 
original square diagram with the circular one 
shown in Figs, x, 2, and 3 below. If for the 
moment we ignore the shading (the purpose of 
which I shall proceed to explain), we find that 
the circular diagram in each case is merely a 
simplification of the original square one — that 
is, the roads from A lead to B, E, and M in both 
cases, the roads from L (London) lead to I, K, 
and S, and so on. The form below, being 
circular and symmetrical, answers my purpose 
better in applying a mechanical solution, and 
I therefore adopt it without altering in any way 
the conditions of the puzzle. If such a question 
as distances from town to town came into the 
problem, the new diagrams might require the 
addition of numbers to indicate these distances, 
or they might conceivably not be at all prac- 
ticable. 

Now, I draw the three circular diagrams, as 
shown, on a sheet of paper and then cut out 
three pieces of cardboard of the forms indicated 
by the shaded parts of these diagrams. It can 
be shown that every route, if marked out with 
a red pencil, will form one or other of the de- 
signs indicated by the edges of the cards, or a 
reflection thereof. Let us direct our attention 
to Fig. I. Here the card is so placed that the 
star is at the town T ; it therefore gives us (by 
following the edge of the card) one of the 
circular routes from London : L, S, R, T, M, 




other rows. It wiU then be seen that the only 
town to which there are exactly 1,365 different 
routes is the twelfth town in the fifth row — 
the one immediately over the letter E. This 
town was therefore the cyclist's destination. 



A, E, P, O, J, D, C, B, G, N, Q, K, H, F, I, L. 

If we went the other way, we should get L, I, 
F, H, K, Q, etc., but these reverse routes were 
not to be counted. When we have written out 
this first route we revolve the card imtil the 



202 



AMUSEMENTS IN MATHEMATICS. 



star is at M, when we get another different 
route, at A a third route, at £ a fourth route, 
and at P a fifth route. We have thus obtained 
five different routes by revolving the card as it 
lies. But it is evident that if we now take up 
the card and replace it with the other side upper- 
most, we shall in the same manner get five other 
routes by revolution. 

We therefore see how, by using the revolving 
card in Fig. i, we may, without any difficulty, 
at once write out ten routes. And if we employ 
the cards in Figs. 2 and 3, we similarly obtain 
in each case ten other routes. These thirty 
routes are all that are possible. I do not give 
the actual proof that the three cards exhaust 
all the possible cases, but leave the reader to 
reason that out for himself. If he works out 
any route at haphazard, he will certainly find 
that it falls into one or other of the three cate- 



gories. 



255.— THE LEVEL PUZZLE. 



Let us confine our attention to the L in the top 
left-hand comer. Suppose we go by way of the 
E on the right : we must then go straight on to 
the V, from which letter the word may be com- 
pleted in four ways, for there are four E's 
available through which we may reach an L. 
There are therefore four ways of reading through 
the right-hand E. It is also clear that there 
must be the same number of ways through the 
E that is inmiediately below our starting point. 
That makes eight. If, however, we take the 
third route through the E on the diagonal, we 
then have the option of any one of the three 
V's, by means of each of which we may complete 
the word in four ways. We can therefore spell 
LEVEL in twelve ways through the diagonal 
E. Twelve added to eight gives twenty read- 
ings, all emanating from the L in the top left- 
hand corner ; and as the four comers are equ^, 
the answer must be four times twenty, or 
eighty different ways. 

256.— THE DIAMOND PUZZLE. 

There are 252 different ways. The general 
formula is that, for words of n letters (not 
palindromes, as in the case of the next puzzle), 
when grouped in this manner, there are always 
2«+i — 4 different readings. This does not allow 
diagonal readings, such as you would get if you 
used instead such a word as DIGGING, where 
it would be possible to pass from one G to an- 
other G by a diagonal step. 

257.— THE DEIFIED PUZZLE. 

The correct answer is 1,992 different ways. 
Every F is either a comer F or a side F — stand- 
ing next to a comer in its own square of F's. 
Now, FIED may be read from a corner F in 
16 ways ; therefore DEIF may be read into a 
comer F also in 16 ways ; hence DEIFIED 
may be read through a corner F in i6x 16=256 
ways. Consequently, the four comer F's give 
4x256=1,024 wa)7s. Then FIED may be 
read from a side F in 11 ways, and DEIFIED 
therefore in 121 ways. But there are eight 



side F's ; consequently these give together ^ 
8x121 = 968 ways. Add 968 to 1,024 and we !' 
get the answer, 1,992. 

In this form the solution wJU depend on 
whether the number of letters in the palin- 
drome be odd or even. For example, if you 
apply the word NUN in precisely the same 
manner, you wiU get 64 different readings ; but 
if you use the word NOON, you wiU only get 
56, because you cannot use the same letter 
twice in immediate succession (since you must 
"always pass from one letter to another ") or 
diagonal readings, and every reading must in- 
volve the use of the central N. 

The reader may like to find for himself the 
general formula in this case, which is complex 
and difficult. I will merely add that for such 
a case as MADAM, dealt with in the same way 
as DEIFIED, the number of readings is 400. 

258.— THE VOTERS' PUZZLE. 

The number of readings here is 63,504, as in 
the case of "WAS IT A RAT I SAW" (No. 
30, Canterbury Puzzles). The general formula 
is that for palindromic sentences containing 
2«-hi letters there are [4(2«— 1)]2 readings. 

259.— HANNAH'S PUZZLE. 

Starting from any one of the N's, there are 17 
different readings of NAH, or 68 (4 times 17) 
for the 4 N's. Therefore there are also 68 ways 
of spelling HAN. If we were allowed to use 
the same N twice in a spelling, the answer would 
be 68 times 68, or 4,624 ways. But the con- 
ditions were, " always passing from one letter 
to another." Therefore, for every one of the 
17 ways of spelling HAN with a particular N, 
there would be 51 ways (3 times 17) of com- 
pleting the NAH, or 867 (17 times 51) ways 
for the complete word. Hence, as there are 
fomr N's to use in HAN, the correct solution of 
the puzzle is 3,468 (4 times 867) different ways. 

260.— THE HONEYCOMB PUZZLE. 

The required proverb is, " There is many a 
slip 'twixt the cup and the lip." Start at the 
T on the outside at the bottom right-hand 
comer, pass to the H above it, and the rest 
is easy. 

261.— THE MONK AND THE BRIDGES. 




The problem of the Bridges may be reduced to 
the simple diagram shown in illustration. The 



SOLUTIONS. 



203 



point M represents the Monk, the point I the 
Island, and the point Y the Monastery. Now 
the only direct ways from M to I are by the 
bridges a and h ; the only direct ways from I 
to Y are by the bridges c and d ; and there is a 
direct way from M to Y by the bridge e. Now, 
what we have to do is to coimt aU the routes 
that wiU lead from M to Y, passing over all the 
bridges, a, b, c, d, and e once and once only. 
With the simple diagram under the eye it is 
quite easy, without any elaborate rule, to count 
these routes methodically. Thus, starting 
from a, b, we find there are only two ways of 
completing the route ; with a, c, there are only 
two routes ; with a, d, only two routes ; and so 
on. It will be foimd that there are sixteen 
such routes in all, as in the following list : — 



a b e c d 


b c d a e 


a b e d c 


b c e a d 


a c d b e 


b d c a e 


a c e b d 


b d e a c 


a d c b e 


e c a b d 


a d e b c 


e c b a d 


b a e c d 


e d a b c 


b a e d c 


e d b a c 



If the reader will transfer the letters indicat- 
ing the bridges from the diagram to the cor- 
responding bridges in the original illustration, 
everything will be quite obvious. 

262.— THOSE FIFTEEN SHEEP. 

If we read the exact words of the writer in the 
cyclopaedia, we find that we are not told that 
the pens were all necessarily empty ! In fact, 
if the reader will refer back to the illustration, 
he wiU see that one sheep is already in one of 
the pens. It was just at this point that the 
wily farmer said to me, " Now I'm going to 
start placing the fifteen sheep." He thereupon 
proceeded to drive three from his flock into 
the already occupied pen, and then placed four 
sheep in each of the other three pens. " There," 
says he, " you have seen me place fifteen sheep 
in four pens so that there shall be the same 
number of sheep in every pen." I was, of 
coursej forced to admit that he was perfectly 
correct, according to the exact wording of the 
question. 

263.— KING ARTHUR'S KNIGHTS. 

On the second evening King Arthur arranged 
the knights and himself in the following order 
round the table : A, F, B, D, G, E, C. On the 
third evening they sat thus. A, E, B, G, C, F, 
D. He thus had B next but one to him on 
both occasions (the nearest possible), and G was 
the third from him at both sittings (the furthest 
position possible). No other way of sitting the 
knights would have been so satisfactory. 

264.— THE CITY LUNCHEONS. 

The men may be grouped as follows, where 
each line represents a day and each column a 
table : — 



AB 


CD 


EF 


GH 


IJ 


KL 


AE 


DL 


GK 


FI 


CB 


HJ 


AG 


LJ 


FH 


KC 


DE 


IB 


AF 


JB 


KI 


HD 


LG 


CE 


AK 


BE 


HC 


IL 


JF 


DG 


AH 


EG 


ID 


CJ 


BK 


LF 


AI 


GF 


CL 


DB 


EH 


JK 


AC 


FK 


DJ 


LE 


GI 


BH 


AD 


KH 


LB 


JG 


FC 


EI 


AL 


HI 


JE 


BF 


KD 


GC 


AJ 


IC 


BG 


EK 


HL 


FD 



Note that in every column (except in the case 
of the A's) all the letters descend cyclically in 
the same order, B, E, G, F, up to J, which is 
followed by B. 

265.— A PUZZLE FOR CARD-PLAYERS. 

In the following solution each of the eleven lines 
represents a sitting, each column a table, and 
each pair of letters a pair of partners. 



AB 


— IL 


E J 


— G K 


FH 


— CD 


AC 


— JB 


FK 


— HL 


GI 


— DE 


AD 


— KC 


GL 


— IB 


HJ 


— EF 


AE 


— LD 


HB 


— JC 


I K 


— FG 


AF 


— BE 


IC 


— KD 


JL 


— GH 


AG 


— CF 


k\ 


— L E 


KB 


— HI 


AH 


— DG 


— B F 


LC 


-IJ 


AI 


— EH 


LF 


— C G 


BD 


-JK 


AJ 


— FI 


BG 


— DH 


CE- 


— KL 


A K 


— GJ 


CH 


— EI 


D F 


— LB 


AL 


— H K 


DI 


-FJ 


EG 


— BC 



It will be seen that the letters B, C, D...L 
descend cyclically. The solution given above 
is absolutely perfect in aU respects. It will be 
found that every player has every other player 
once as his partner and twice as his opponent. 

266.— A TENNIS TOURNAMENT. 

Call the men A, B, D, E, and their wives a, b, 
d, e. Then they may play as foUows without 
any person ever plajnng twice with or against 
any other person : — 



ist Day 
2nd Day 
3rd Day 



First Court. 
A d against B e 
A e „ D b 
A b .. Ed 



Second Court. 
D a against E b 
E a „ B d 
B a „ D e 



It wiU be seen that no man ever plays with or 
against his own wife — an ideal arrangement. If 
the reader wants a hard puzzle, let him try to 
arrange eight married couples (in four courts on 
seven days) under exactly similar conditions. 
It can be done, but I leave the reader in this 
case the pleasure of seeking the answer and the 
general solution. 

267.— THE WRONG HATS. 

The nmnber of different ways in which eight 
persons, with eight hats, can each take the 
wrong hat, is 14,833. 

Here are the successive solutions for any 
number of persons from one to eight : — 



204 



AMUSEMENTS IN MATHEMATICS. 



I 


= 


2 


= X 


3 

4 
5 
6 

7 
8 


= 2 

= 9 

= 44 
= 265 

= 1,854 
= 14,833 



To get these numbers, multiply successively 
by 2, 3, 4, 5, etc. When the multiplier is even, 
add I ; when odd, deduct i. Thus, 3x1 — 1 = 
2; 4x2 + 1 = 9; 5x9-1=44; and so on. 
Or you can multiply the sum of the number of 
ways for n—i and « — 2 persons by w— i, and so 
get the solution for n persons. Thus, 4(2 + 9)= 
44 ; 5(9 +44) =265 ; and so on. 

268.— THE PEAL OF BELLS. 
The bells should be rung as follows : — 



X234 3124 2314 

2143 1342 3241 

2413 1432 3421 

4231 4123 4312 

4321 4213 4132 

3412 2431 X423 

3142 2341 1243 

1324 3214 2134 



I have constructed peals for five and six bells 
respectively, and a solution is possible for any 
number of beUs under the conditions previously 
stated. 

269.— THREE MEN IN A BOAT. 

If there were no conditions whatever, except 
that the men were all to go out together, in 
threes, they could row in an immense number 
of differ ent ways. If the reader wishes to 
know how many, the number is 455^. And 
with the condition that no two may ever be 
together more than once, there are no fewer 
than 15,567,552,000 different solutions — that is, 
difierent ways of arranging the men. With one 
solution before him, the reader will realize why 
this must be, for although, as an example, A 
must go out once with B and once with C, it 
does not necessarily follow that he must go out 
with C on the same occasion that he goes with 
B. He might take any other letter with him 
on that occasion, though the fact of his taking 
other than B would have its effect on the ar- 
rangement of the other triplets. 

Of course only a certain number of aU these 
arrangements are available when we have that 
other condition of using the smallest possible 
number of boats. As a matter of fact we need 
employ only ten different boats. Here is one of 
the arrangements : — 

istDay . (ABC) (DEF) (GHI) (JKL) MNO) 
8 6 7 9 10 

2nd Day . (ADG) (BKN) (COL) (JEI) (MHF) 
35412 

3rd Day . (AJM) (BEH) (CFI) (DKO) (GNL) 



76891 
4th Day . (AEK) (COM) (BOI) (DHL) (JNF) 

4 5 3 10 2 

5th Day . (AHN) (CD J) (BFL) (GEO) (MKI) 

6 7 8 10 I 

6th Day . (AFO) (BGJ) (CKH) (DNI) (MEL) 

54392 
7th Day . (AIL) (BDM) (CEN) (GKF) (JHO) 

It will be found that no two men ever go out 
twice together, and that no man ever goes out 
twice in the same boat. 

This is an extension of the well-known prob- 
lem of the " Fifteen Schoolgirls," by Kirkman. 
The original conditions were simply that fifteen 
girls walked out on seven days in triplets with- 
out any girl ever walking twice in a triplet with 
another girl. Attempts at a general solution of 
this puzzle had exercised the ingenuity of mathe- 
maticians since 1850, when the question was 
first propounded, until recently. In 1908 and 
the two following years I indicated (see Edu- 
cational Times Reprints, Vols. XIV., XV., and 
XVII.) that all our trouble had arisen from a 
failure to discover that 15 is a special case (too 
small to enter into the general law for all higher 
numbers of girls of the form 6 M + 3), and showed 
what that general law is and how the groups 
should be posed for any number of girls. I 
gave actual arrangements for numbers that had 
previously baffled all attempts to manipulate, 
and the problem may now be considered gener- 
ally solved. Readers will find an excellent full 
account of the puzzle in W. W. Rouse Ball's 
Mathematical Recreations, 5th edition. 

270.— THE GLASS BALLS. 

There are, in all, sixteen balls to be broken, 
or sixteen places in the order of breaking. 
Call the four strings A, B, C, and D — order is 
here of no importance. The breaking of the 
balls on A may occupy any 4 out of these 16 
places — that is, the combinations of 16 things, 

taken 4 together, wiU be 'yx'2 X3 xT^ ^''^'^^ 

ways for A. In every one of these cases B 
may occupy any 4 out of the remaining 12 

, . 9X 10X11X12 
places, making - ., ^^^ ^ ^ , =495 waj^. 



Thus 



1x2x3x4 

1920x495=950,400 different placings are open 
to A and B. But for every one of these cases 

C may occupy — — — =70 different places ; 

so that 950,400x70=66,528,000 different pla- 
cings are open to A, B, and C. In every one of 
these cases, D has no choice but to take the 
four places that remain. Therefore the correct 
answer is that the balls may be broken in 
66,528,000 different ways under the conditions. 
Readers should compare this problem with 
No. 345, " The Two Pawns," which they will 
then know how to solve for cases where there 
are three, four, or more pawns on the board. 



SOLUTIONS. 



205 



271.— FIFTEEN LETTER PUZZLE. 

The following will be found to comply with the 
conditions of grouping : — 



ALE 


MET 


MOP 


BLM 


BAG 


CAP 


YOU 


CLT 


IRE 


OIL 


LUG 


LNR 


NAY 


BIT 


BUN 


BPR 


AIM 


BEY 


RUM 


GMY 


OAR 


GIN 


PLY 


CGR 


PEG 


ICY 


TRY 


CMN 


CUE 


COB 


TAU 


PNT 


ONE 


GOT 


PIU 





The fifteen letters used are A, E, I, O, U, Y, 
and B, C, G, L, M, N, P, R, T. The number of 
words is 27, and these are aU shown in the first 
three columns. The last word, PIU, is a musi- 
cal term in common use; but although it has 
crept into some of our dictionaries, it is Italian, 
meaning " a little ; slightly." The remaining 
twenty-six are good words. Of course a TAU- 
cross is a T-shaped cross, also called the cross 
of St. Anthony, and borne on a badge in the 
Bishop's Palace at Exeter. It is also a name 
for the toad-fish. 

We thus have twenty-six good words and one 
doubtful, obtained under the required condi- 
tions, and I do not think it will be easy to 
improve on this answer. Of course we are not 
bound by dictionaries but by common usage. 
If we went by the dictionary only in a case of 
this kind, we should find ourselves involved in 
prefixes, contractions, and such absurdities as 
I.O.U., which NuttaU actually gives as a word. 

272.— THE NINE SCHOOLBOYS. 
The boys can walk out as follows : — 



ist Day. 


2nd Day. 


3rd Day. 


ABC 


B F H 


FAG 


D E F 


E I A 


I D B 


G H I 


C G D 


H C E 


4th Day. 


Sth Day. 


6th Day. 


A D H 


G B I 


D C A 


BEG 


C F D 


EH B 


F I C 


HA E 


IGF 



Every boy wiU then have walked by the side 
of every other boy once and once only. 

Dealing with the problem generally, 12 n+g 
boys may walk out in triplets under the condi- 
tions on 9 M-l-6 days, where n may be nought 
or any integer. Every possible pair will occmr 
once. Call the number of boys m. Then every 

fyt ^ I 
boy wiU pair m— i times, of which times he 



days, and 33 boys on 34 days. It is, perhaps, 
interesting to note that a school of 489 boys 
could thus walk out daily in one leap year, but 
it would take 731 girls (referred to in the solu- 
tion to No. 269) to perform their particular feat 
by a daily walk in a year of 365 days. 

273-— THE ROUND TABLE. 

The history of this problem will be found in 
The Canterbury Puzzles (No. 90). Since the 
publication of that book in 1907, so far as I 
know, nobody has succeeded in solving the 
case for that unlucky number of persons, 13, 
seated at a table on 66 occasions. A solution 
is possible for any number of persons, and I 
have recorded schedules for every number up 
to 25 persons inclusive and for 33. But as I 
know a good many mathematicians are still 
considering the case of 13, I will not at this 
stage rob them of the pleasure of solving it by 
showing the answer. But I will now display 
the solutions for all the cases up to 12 persons 
inclusive. Some of these solutions are now 
published for the first time, and they may afford 
useful clues to investigators. 

The solution for the case of 3 persons seated 
on I occasion needs no remark. 

A solution for the case of 4 persons on 3 
occasions is as follows : — 

1234 
1342 

1423 

Each line represents the order for a sitting, 
and the person represented by the last number 
in a line must, of course, be regarded as sitting 
next to the first person in the same line, when 
placed at the round table. 

The case of 5 persons on 6 occasions may be 
solved as follows : — 

12345 
12453 
12534 



will be in the middle of a triplet and 



m— I 



times 



on the outside. Thus, if we refer to the solution 
above, we find that every boy is in the middle 
twice (making 4 pairs) and four times on the 
outside (making the remaining 4 pairs of his 
8). The reader may now like to try his hand 
at solving the two next cases of 21 boys on 15 



13254 
14235 
15243 



The case for 6 persons on 10 occasions is 
solved thus : — 



124563 
135624 

146235 
152346 
163452 



It will now no longer be necessary to give the 
solutions in full, for reasons that I will explain. 
It will be seen in the examples above that 
the I (and, in the case of 5 persons, also the 2) 



206 



AMUSEMENTS IN MATHEMATICS. 



is repeated down the column. Such a number 
I caU a " repeater." The other numbers de- 
scend in cycUcal order. Thus, for 6 persons 
we get the cycle, 2, 3, 4, 5, 6, 2, and so on, 
in every column. So it is only necessary to 
give the two lines 123645 and 124563, and 
denote the cycle and repeaters, to enable any 
one to write out the full solution straight away. 
The reader may wonder why I do not start the 
last solution with the numbers in their natural 
order, 123456. If I did so the numbers in 
the descending cycle would not be in their 
natursd order, and it is more convenient to 
have a regular cycle than to consider the order 
in the first line. 

The difficult case of 7 persons on 15 occasions 
is solved as follows, and was given by me in 
The Canterbury Puzzles : — 



1234576 

1627534 
1352674 
1574362 



1527346 

In this case the i is a repeater, and there are 
two separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5. 
We thus get five groups of three lines each, for 
a fourth line in any group wiU merely repeat 
the first line. 

A solution for 8 persons on 21 occasions is as 
follows : — 

18634527 



18457236 

18273645 

The I is here a repeater, and the cycle 2, 3, 4, 5, 
6, 7, 8. Every one of the 3 groups will give 
7 lines. 

Here is my solution for 9 persons on 28 occa- 
sions : — 

219745638 



2 


9 


5 


I 


6 


8 


3 


4 


7 


2 


9 


3 


I 


8 


4 


7 


5 


6 



291564783 

There are here two repeaters, i and 2, and the 
cycle is 3, 4, 5, 6, 7, 8, 9, We thus get 4 groups 
of 7 lines each. 

The case of 10 persons on 36 occasions is 
solved as follows : — 



I 


10 


8 


3 


6 


5 


4 


7 


2 


9 


I 


10 


6 


5 


2 


9 


7 


4 


3 


8 


I 


10 


2 


9 


3 


8 


6 


5 


7 


4 



The repeater is i, and the cycle, 2, 3, 4, 5, 6, 7, 
8, 9, 10. We here have 4 groups of 9 lines each. 
My solution for 11 persons on 45 occasions 
is as follows : — 



2 


II 


9 


4 


7 


6 


5 


I 


8 


3 


10 


2 


I 


II 


7 


6 


3 


10 


8 


5 


4 


9 


2 


II 


10 


3 


9 


4 


8 


5 


I 


7 


6 


2 


II 


5 


8 


I 


3 


10 


6 


7 


9 


4 



2 II I 10 34 96758 

There are two repeaters, i and 2, and the cycle 
is, 3. 4, 5, • • • II- We thus get 5 groups of 9 
lines each. 

The case of 12 persons on 55 occasions is 
solved thus : — 



I 


2 


3 


12 


4 


II 


5 


10 


6 


9 


7 


8 


I 


2 


4 


II 


6 


9 


8 


7 


10 


5 


12 


3 


I 


2 


5 


10 


8 


7 


II 


4 


3 


12 


6 


9 


I 


2 


6 


9 


10 


5 


3 


12 


7 


8 


II 


4 



10 7 4 8 3 



956 



127 8 12 36 9 II 45 10 

Here i is a repeater, and the cycle is 2, 3, 4, 5, 
. . . 12. We thus get 5 groups of 11 lines each. 

274.— THE MOUSE-TRAP PUZZLE. 

If we interchange cards 6 and 13 and begin 
our count at 14, we may take up all the twenty- 
one cards — that is, make twenty-one " catches " 
— in the following order : 6, 8, 13, 2, 10, i, 11, 
4, 14, 3, 5, 7, 21, 12, 15, 20, 9, 16, 18, 17, 19. 
We may also exchange 10 and 14 and start at 
16, or exchange 6 and 8 and start at 19. 

275.— THE SIXTEEN SHEEP. 

The six diagrams on next page show solutions 
for the cases where we replace 2, 3, 4, 5, 6, and 
7 hurdles. The dark lines indicate the hurdles 
that have been replaced. There are, of course, 
other ways of making the removals. 

276.— THE EIGHT VILLAS. 

There are several ways of solving the puzzle, 
but there is very little difference between them. 
The solver should, however, first of all bear in 
mind that in making his calculations he need 
only consider the four villas that stand at the 
corners, because the intermediate villas can 
never vary when the corners are known. One 
way is to place the numbers nought to 9 one 
at a time in the top left-hand corner, and then 
consider each case in turn. 

Now, if we place 9 in the corner as shown in 
the Diagram A, two of the comers cannot be 
occupied, while the comer that is diagonally 
opposite may be filled by o, 1, 2, 3, 4, 5, 6, 7, 8, 
or 9 persons. We thus see that there sure lo 



SOLUTIONS. 



207 



oo|oo 
oojoo 
o|o.o^fo 
0000 



ooo|o 
00^0 



oloolo 
0000 



ofooo 
ojooo 



o|_goo 
ooolo 



ooloo 

00.I0 

000 



o 
o 
0000 




THE SIXTEEN SHEEP. 



solutions with a 9 in the corner. If, however, 
we substitute 8, the two corners in the same 
row and column may contain o, o, or i, i, or 
o, I, or I, o. In the case of B, ten different 
selections may be made for the fourth comer ; 
but in each of the cases C, D, and E, only nine 
selections are possible, because we cannot use 
the 9. Therefore with 8 in the top left-hand 
comer there are io+(3X9) = 37 different solu- 
tions. If we then try 7 in the comer, the result 
will be 10 + 27-1-40, or 'jy solutions. With 6 
we get 10 + 27-1-40+49=126 ; with 5, 10 + 27 + 
40+49 + 54=180 ; with 4, the same as with 5, 
-f 55 = 235 ; with 3, the same as with 4, +52= 
287; with 2, the same as with 3, +45 = 332 ; 
with I, the same as with 2, +34=366, and with 
nought in the top left-hand comer the number 

A B C 



9 









wa 




















8 









m 




X 







s 




1 




"///a 












of solutions will be found to be 10+27 + 40+ 
49 + 54 + 55 + 52+45 + 34 + 19=385. As there 
is no other number to be placed in the top left- 
hand comer, we have now only to add these 
totals together thus, 10 +37 +77 +126 + 180 + 
235 + 287+332 + 366 + 385=2,035. We there- 
fore find that the total number of ways in which 



tenants may occupy some or all of the eight 
villas so that there shaU be always nine persons 
living along each side of the square is 2,035. 
Of course, this method must obviously cover £dl 
the reversals and reflections, since each comer in 
turn is occupied by every number in all possible 
combinations with the other two comers that 
are in line with it. 
Here is a general formula for solving the 
(»2+3n+2)(M2+3n+3). 



puzzle : 



Whatever 



s 




1 




Wa 




1 







may be the stipulated number of residents along 
each of the sides (which number is represented 
by n), the total number of different arrangements 
may be thus ascertained. In our particular 
case the number of residents was nine. There- 
fore (81 + 27 + 2) X (81+27+3) and the product, 
divided by 6, gives 2,035. If the number 
of residents had been o, i, 2, 3, 4, 5, 6, 7, 
or 8, the total arrangements would be i, 
7, 26, 70, 155, 301, 532, 876, or 1,365 re- 
spectively. 

277.— COUNTER CROSSES. 



Let us first 
There are just 
numbers may 
Here they are 

12978 
34956 

23958 
14967 

12589 
34567 

14569 
23578 

15369 
24378 

24179 
35168 



deal with the Greek Cross, 
eighteen forms in which the 
be paired for the two arms. 



13968 
24957 
13769 
24758 

2 

I 



14958 
23967 

14759 
23768 

13579 
24568 



2o8 



AMUSEMENTS IN MATHEMATICS. 



Of course, the number in the middle is 
common to both arms. The first pair is the 
one I gave as an example. I will suppose that 
we have written out all these crosses, always 
placing the first row of a pair in the upright 
and the second row in the horizontal arm. 
Now, if we leave the central figure fixed, there 
are 24 ways in which the numbers in the up- 
right may be varied, for the four counters 
may be changed ^11x2x3x4=24 ways. 
And as the four in the horizontal may also be 
changed in 24 ways for every arrangement on 
the other arm, we find that there are 24 x 24 
= 576 variations for every form ; therefore, as 
there are 18 forms, we get 18 x 576 = 10,368 
ways. But this will include half the four re- 
versals and half the four reflections that we 
barred, so we must divide this by 4 to obtain 
the correct answer to the Greek Cross, which 
is thus 2,592 different ways. The division is by 
4 and not by 8, because we provided against 
half the reversals and reflections by always 
reserving one number for the upright and the 
other for the horizontal. 

In the case of the Latin Cross, it is obvious 
that we have to deal with the same 18 forms 
of pairing. The total number of different ways 
in this case is the full number, 18 x 576. 
Owing to the fact that the upper and lower 
arms are unequal in length, permutations will 
repeat by reflection, but not by reversal, for we 
cannot reverse. Therefore this fact only entails 
division by 2. But in every pair we may 
exchange the figures in the upright with those 
in the horizontal (which we could not do in the 
case of the Greek Cross, as the arms are there 
all alike) ; consequently we must multiply by 
2. This multiplication by 2 and division by 2 
cancel one another. Hence 10,368 is here the 
correct answer. 



278.— A DORMITORY PUZZLE. 



TnoN. 



Tves. 



1 


a 


1 


2 


m 


2 


1 


22 


1 



we.t>. 



1 


3 


1 


I 




I 


3 


19 


3 



1 


4» 


1 


1 


M 


1 


4 


16 


^ 



/tHURS^'. 



f=R|. 



SAT. 



2 



13 



2 


G 


2 


1 


W\ 


1 


7 


6 


r 




Arrange the nuns from day to day as shown 
in the six diagrams. The smallest possible 
number of nuns would be thirty-two, and the 
arrangements on the last three days admit of 
variation. 

279.— THE BARRELS OF BALSAM. 

This is quite easy to solve for any number of 
barrels — if you know how. This is the way to 



do it. There are five barrels in each row. 
Multiply the numbers i, 2, 3, 4, 5 together ; 
and also multiply 6, 7, 8, 9, 10 together. 
Divide one result by the other, and we get the 
number of different combinations or selections 
of ten things taken five at a time. This is here 
252. Now, if we divide this by 6 (i more than 
the number in the row) we get 42, which is the 
correct answer to the puzzle, for there are 42 
different ways of arranging the barrels. Try 
this method of solution in the case of six barrels, 
three in each row, and you will find the answer 
is 5 ways. If you check this by trial, you will 
discover the five arrangements with 123, 124, 
125, 134, 135 respectively in the top row, and 
you will find no others. 
The general solution to the problem is, in 



fact, this : 



'2n 

M-f I 



where 2« equals the number of 



barrels. The symbol C, of course, implies that 
we have to find how many combinations, or 
selections, we can make of 2« things, taken n at 
a time. 

280.— BUILDING THE TETRAHEDRON. 

Take your constructed pyramid and hold it so 
that one stick only lies on the table. Now, 
four sticks must branch off from it in different 
directions — two at each end. Any one of the 
five sticks may be left out of this connection ; 
therefore the four may be selected in 5 different 
ways. But these four matches may be placed 
in 24 different orders. And as any match may 
be joined at either of its ends, they may fur- 
ther be varied (after their situations are settled 
for any parti ciilar arrangement) in 16 different 
ways. In every arrangement the sixth stick 
may be added in 2 different ways. Now multi- 
ply these results together, and we get 5 x 24 X 
16x2=3,840 as the exact number of ways in 
which the pyramid may be constructed. This 
method excludes all possibility of error. 

A common cause of error is this. If you 
calculate your combinations by working up- 
wards from a basic triangle lying on the table, 
you will get half the correct niunber of ways, 
because you overlook the fact that an equal 
number of pyramids may be built on that tri- 
angle downwards, so to speak, through the table. 
They are, in fact, reflections of the others, and 
examples from the two sets of pyramids can- 
not be set up to resemble one another — except 
under fourth dimensional conditions ! 

281.— PAINTING A PYRAMID. 

It will be convenient to imagine that we are 
painting our pyramids on the flat cardboard, 
as in the diagrams, before folding up. Now, 
if we take any jour colours (say red, blue, green, 
and yellow), they may be applied in only 2 
distinctive ways, as shown in Figs, i and 2. 
Any other way will only result in one of these 
when the pyramids are folded up. If we take 
any three colours, they may be applied in the 3 
ways shown in Figs. 3, 4, and 5, If we take 
any two colours, they may be applied in the 3 



SOLUTIONS. 



209 



ways shown in Figs. 6, 7, and 8. If we take 
amy single colour, it may obviously be applied 
in only i way. But four colours may be 



1 


w 


V 


V 

3 


my 


V 


6 


\W 





2. 


w 


V V 




W 


\W 



selected in 35 ways out of seven ; three in 35 
ways ; two in 21 ways ; and one colour in 7 
ways. Therefore 35 applied in 2 ways = 70 ; 
35 in 3 ways = 105 ; 21 in 3 ways = 63 ; and 
7 in I way = 7. Consequently the pyramid 
may be painted in 245 different ways (70 + 105 + 
63 + 7), using the seven colours of the solar 
spectrum in accordance with the conditions of 
the puzzle. 

282.— THE ANTIQUARY'S CHAIN. 
A 




The number of ways in which nine things 
may be arranged in a row without any restric- 
tions is 1x2x3x4x5x6x7x8x9=362,880. 
But we are told that the two circular rings must 
never be together ; therefore we must deduct 
the number of times that this would occur. 
The number is 1x2x3x4x5x6x7x8= 
40,320x2=80,640, because if we consider the 
two circular links to be inseparably joined to- 
gether they become as one link, and eight links 
are capable of 40,320 arrangements ; but as 
these two links may always be put on in the 
orders A B or B A, we have to double this num- 
ber, it being a question of arrangement and not 
of design. The deduction required reduces our 
total to 282,240. Then one of our links is of a 
peculiar form, like an 8. We have therefore 
the option of joining on either one end or the 
other on every occasion, so we must double the 
last result. This brings up our total to 564,480. 
We now come to the point to which I directed 



the reader's attention — that every link may be 
put on in one of two ways. If we join the first 
finger and thumb of our left hand horizontally, 
and then link the first finger and thumb of the 
right hand, we see that the right thmnb may 
be either above or below. But in the case of 
our chain we must remember that although that 
8-shaped link has two independent ends it is like 
every other link in having only two sides — that 
is, you cannot turn over one end without turning 
the other at the same time. 

We wiU, for convenience, assume that each 
link has a black side and a side painted white. 
Now, if it were stipulated that (with the chain 
lying on the table, and every successive link 
falling over its predecessor in the same way, as 
in the diagram) only the white sides should be 
uppermost as in A, then the answer would be 
564,480, as above — ignoring for the present all 
reversals of the completed chain. If, however, 
the first link were allowed to be placed either 
side up, then we could have either A or B, and 
the answer would be 2X564,480=1,128,960; 
if two links might be placed either way up, the 
answer would be 4 X 564,480 ; if three links, 
then 8 x 564,480, and so on. Since, therefore, 
every link may be placed either side up, the 
number will be 564,480 multiplied by 2^, or by 
512. This raises our total to 289,013,760. 

But there is still one more point to be con- 
sidered. We have not yet allowed for the fact 
that with any given arrangement three of the 
other arrangements may be obtained by simply 
turning the chain over through its entire length 
and by reversing the ends. Thus C is really 
the same as A, and if we turn this page upside 
down, then A and C give two other arrangements 
that are still reaUy identical. Thus to get the 
correct answer to the puzzle we must divide 
our last total by 4, when we find that there are 
just 72,253,440 different ways in which the 
smith might have put those links together. 
In other words, if the nine links had originally 
formed a piece of chain, and it was known that 
the two circular Links were separated, then it 
would be 72,253,439 chances to i that the 
smith would not have put the links together 
again precisely as they were arranged before ! 

283.— THE FIFTEEN DOMINOES. 

The reader may have noticed that at each 
end of the line I give is a four, so that, if we 
like, we can form a ring instead of a line. It 
can easily be proved that this must always be 
so. Every line arrangement will make a cir- 
cular arrangement if we like to join the ends. 
Now, curious as it may at first appear, the 
following diagram exactly represents the con- 
ditions when we leave the doubles out of the 
question and devote our attention to forming 
circular arrangements. Each number, or half 
domino, is in line with every other number, so 
that if we start at any one of the five nmnbers 
and go over all the lines of the pentagon once 
and once only we shall come back to the starting 
place, and the order of our route will give us 
one of the circular arrangements for the ten 



(1,926) 



14 



2IO 



AMUSEMENTS IN MATHEMATICS. 



dominoes. Take your pencil and follow out the 
following route, starting at the 4 : 41 304210234. 
You have been over aU the lines once only, 
and by repeating aU these figures in this way, 
41 — 13 — 30 — 04 — 42 — 21 — 10 — 02 — 23 — 34, 
you get an arrangement of the dominoes (with- 
out the doubles) which will be perfectly clear. 
Take other routes and you will get other ar- 
rangements. If, therefore, we can ascertain 
just how many of these circular routes are 
obtainable from the pentagon, then the rest is 
very easy. 

Well, the number of different circular routes 
over the pentagon is 264. How I arrive at 
these figures I will not at present explain, 
because it would take a lot of space. The, 
dominoes may, therefore, be arranged in a 
circle in just 264 different ways, leaving out 
the doubles. Now, in any one of these circles 
the five doubles may be inserted in 2^=32 
different ways. Therefore when we include 
the doubles there are 264x32 = 8,448 differ- 
ent circular arrangements. But each of those 
circles may be broken (so as to form our straight 
line) in any one of 15 different places. Con- 
sequently, 8,448x15 gives 126,720 different 
ways as the correct answer to the puzzle. 

I purposely refrained from asking the reader 
to discover in just how many different ways 
the full set of twenty-eight dominoes may be 
arranged in a straight line in accordance with 
the ordinary rules of the game, left to right 



and right to left of any arrangement count- 
ing as different ways. It is an exceedingly 
difficult problem, but the correct answer ' is 
7,959>229,93i,520 ways. The method of solv- 
ing is very complex. 

284.— THE CROSS TARGET. 

Twenty-one different squares may be selected. 
Of these nine wiU be of the size shown by the four 





A's in the diagram, four of the size shown by 
the B's, four of the size shown by the C's, two 
of the size shown by the D's, and two of the 
size indicated by the upper single A, the upper 
single E, the lower single C, and the EB. It 
is an interesting fact that you cannot form any 
one of these twenty-one squares without using 
at least one of the six circles marked E. 

285.— THE FOUR POSTAGE STAMPS. 

Referring to the original diagram, the four 
stamps may be given in the shape i, 2, 3, 4, in 
three ways ; in the shape i, 2, 5, 6, in six ways ; 
in the shape i, 2, 3, 5, or i, 2, 3, 7, or i, 5, 6, 7, 
or 3, 5, 6, 7, in twenty-eight ways ; in shape i, 
2, 3, 6, or 2, 5, 6, 7, in fourteen ways ; in shape 
I, 2, 6, 7, or 2, 3, 5, 6, or i, 5, 6, 10, or 2, 5, 6, 9, 
in fourteen ways. Thus there are sixty-five 
ways in all. 

286.— PAINTING THE DIE. 

The 1 can be marked on any one of six different 
sides. For every side occupied by i we have a 
selection of four sides for the 2. For every 
situation of the 2 we have two places for the 3. 
(The 6, 5, and 4 need not be considered, as 
their positions are determined by the i, 2, and 
3.) Therefore 6, 4, and 2 multiplied together 
make 48 different ways — the correct answer. 

287.— AN ACROSTIC PUZZLE. 

There are twenty-six letters in the alphabet, 
giving 325 different pairs. Every one of these 
pairs may be reversed, making 650 ways. But 
every initial letter may be repeated as the final, 
producing 26 other ways. The total is there- 
fore 676 different pairs. In other words, the 
answer is the square of the number of letters 
in the alphabet. 

288.— CHEQUERED BOARD DIVISIONS. 

There are 255 different ways of cutting the 
board into two pieces of exactly the same size 



SOLUTIONS. 



an 




and shape. Every way must involve one of the 
five cuts shown in Diagrams A, B, C, D, and E. 
To avoid repetitions by reversal and reflection, 
we need only consider cuts that enter at the 
points a, b, and c. But the exit must always 
be at a point in a straight line from the entry 
through the centre. This is the most important 
condition to remember. In case B you cannot 
enter at a, or you will get the cut provided for 
in E. Similarly in C or D, you must not enter 



the key-line in the same direction as itself, or 
you will get A or B. If you are working on 
A or C and entering at a, you must consider 
joins at one end only of the key-line, or you 
will get repetitions. In other cases you must 
consider joins at both ends of the key ; but 
after leaving a in case D, turn always either 
to right or left — use one direction only. Figs. 
I and 2 are examples under A ; 3 and 4 are 
examples under B ; 5 and 6 come imder C ; 



212 



AMUSEMENTS IN MATHEMATICS. 



and 7 is a pretty example of D. Of course, E 
is a peculiar type, and obviously admits of 
only one way of cutting, for you clearly cannot 
enter at b or c. 

Here is a table of the results : — 







a 




b 




c 


Ways 


A 


= 


8 


+ 


17 


+ 


21 


= 46 


B 


= 


o 


+ 


17 


-f 


21 


= 38 


C 


= 


15 


+ 


31 


•+ 


39 


= 85 


U 


= 


17 


+ 


29 


+ 


39 


= 85 


E 


^^^ 


I 
41 


+ 


o 
94 


+ 


o 

I20 


= I 

255 



I have not attempted the task of enumerating 
the ways of dividing a board 8x8 — that is, 
an ordinary chessboard. Whatever the method 
adopted, the solution would entail considerable 
labour. 

289.— LIONS AND CROWNS. 




Here is the solution. It will be seen that each 
of the four pieces (after making the cuts along 
the thick lines) is of exactly the same size and 
shape, and that each piece contains a lion and a 
crown. Two of the pieces are shaded so as to 
make the solution quite clear to the eye. 

290.— BOARDS WITH AN ODD NUMBER 
OF SQUARES. 

There are fifteen different ways of cutting the 
5x5 board (with the central square removed) 
into two pieces of the same size and shape. 
Limitations of space will not allow me to give 
diagrams of all these, but I will enable the 
reader to draw them all out for himself with- 
out the slightest difficulty. At whatever point 
on the edge your cut enters, it must always 
end at a point on the edge, exactly opposite 
in a line through the centre of the square. 
Thus, if you enter at point i (see Fig. i) at the 



top, you must leave at point i at the bottom. 
Now, I and 2 are the only two really different 
points of entry ; if we use any others they 
will simply produce similar solutions. The 
directions of the cuts in the following fifteen 




Fig 2. 



solutions are indicated by the numbers on the 
diagram. The duplication of the numbers can 
lead to no confusion, since every successive 
number is contiguous to the previous one. 
But whichever direction you take from the 
top downwards you must repeat from the 
bottom upwards, one direction being an exact 
reflection of the other. 

I, 4, 8. 

I, 4, 3, 7, 8. 

1, 4, 3, 7, 10, 9. 

I, 4, 3, 7, 10, 6, 5, 9. 

TC, 4, 5, 9- 

I, 4, 5, 6, 10, 9. 

1, 4, 5, 6, 10, 7, 8. 

2, 3, 4, 8. 

2, 3, 4, 5, 9- 

2, 3, 4, 5, 6, 10, 9. 

2, 3, 4, 5, 6, 10, 7, 8. 

2, 3, 7, 8. 

2, 3, 7, 10, 9. 

2, 3, 7, 10, 6, 5, 9- 

2, 3, 7, 10, 6, 5, 4, 8. 

It will be seen that the fourth direction (t, 4, 3, 
7, 10, 6, 5, 9) produces the solution shown in 
Fig. 2. The thirteenth produces the solution 
given in propounding the puzzle, where the 
cut entered at the side instead of at the top. 
The pieces, however, will be of the same shape 
if turned over, which, as it was stated in 
the conditions, would not constitute a different 
solution. 

291.— THE GRAND LAMA'S PROBLEM. 

The method of dividing the chessboard so that 
each of the four parts shall be of exactly the 
same size and shape, and contain one of the 
gems, is shown in the diagram. The method of 
shading the squares is adopted to make the 
shape of the pieces clear to the eye. Two of 
the pieces are shaded and two left white. 

The reader may find it interesting to compare 
this puzzle with that of the " Weaver " (No. 14, 
Canterbury Puzzles), 



SOLUTIONS. 



213 




THE GRAND LAMA'S PROBLEM. 

292.— THE ABBOT'S WINDOW. 

The man who was " learned in strange mys- 
teries " pointed out to Father John that the 
orders of the Lord Abbot of St. Edmondsbury 
might be easily carried out by blocking up 
twelve of the lights in the window as shown by 
the dark squares in the following sketch : — • 



S" 



IHiTir 



K 



□[ 



¥ 



%^ 



F 



b 



'^ 



Ni^; 



D 



□ 



?5^ 



-^ 



111, i 


H 


n 


_ 


T= 


T 


j^- 






1 


1,. 








•4 


"^ 


ti" 


■ 






|li!|!'| 



Father John held that the four comers 
should also be darkened, but the sage explained 
that it was desired to obstruct no more light 
than was absolutely necessary, and he said, 
anticipating Lord Dundreary, " A single pane 
can no more be in a line with itself than one 
bird can go into a comer and flock in solitude. 
The Abbot's condition was that no diagonal 
lines should contain an odd number of lights." 

Now, when the holy man saw what had been 
done he was well pleased, and said, " Truly, 
Father John, thou art a man of deep wisdom, 
in that thou hast done that which seemed im- 
possible, and yet withal adorned our window 



with a device of the cross of St. Andrew, whose 
name I received from my godfathers and god- 
mothers." Thereafter he slept well and arose 
refreshed. The window might be seen intact 
to-day in the monastery of St. Edmondsbury, 
if it existed, which, alas ! the window does not. 

293.— THE CHINESE CHESSBOARD. 





Eighteen is the maximum ntimber of pieces. 
I give two solutions. The numbered diagram is 
so cut that the eighteenth piece has the largest 
area — eight squares — that is possible under the 
conditions. The second diagram was prepared 
under the added condition that no piece should 
contain more than five squares. 

No. 74 in The Canterbury Puzzles shows how 
to cut the board into twelve pieces, all differ- 



214 



AMUSEMENTS IN MATHEMATICS. 



ent, each containing five squares, with one 
square piece of four squares. 

294._THE CHESSBOARD SENTENCE. 

'"r"'r'"iT^ 




The pieces may be fitted together, as shown in 
the illustration, to form a perfect chessboard. 

295.— THE EIGHT ROOKS. 

Obviously there must be a rook in every row 
and every column. Starting with the top row, 
it is clear that we may put our first rook on 
any one of eight different squares. Wherever 
it is placed, we have the option of seven squares 
for the second rook in the second row. Then 
we have six squares from which to select the 
third row, five in the fourth, and so on. There- 
fore the number of our difierent ways must be 
8x7x6x5x4x3x2x1=40,320 (that is |8), 
which is the correct answer. 

How many ways there are if mere reversals 
and reflections are not counted as different has 
not yet been determined ; it is a difficult prob- 
lem. But this point, on a smaller square, is 
considered in the next puzzle. 

296.— THE FOUR LIONS. 

There are only seven different ways tmder the 
conditions. They are as follows : i 2 3 4, i 2 4 3, 
1324, 1342, 1432, 2143, 2413. Taking 
the last example, this notation means that we 
place a lion in the second square of first row, 
fourth square of second row, first square of 
third row, and third square of fourth row. The 
first example is, of course, the one we gave 
when setting the puzzle. 

297.— BISHOPS— UNGUARDED. 

This cannot be done with fewer bishops than 
eight, and the simplest solution is to place the 
bishops in Une along the fourth or fifth row of 




the board (see diagram). But it will be noticed 
that no bishop is here guarded by another, so 
we consider that point in the next puzzle. 

298.— BISHOPS— GUARDED. 

This puzzle is quite easy if you first of all give 
it a httle thought. You need only consider 
squares of one colour, for whatever can be 
done in the case of the white squares can always 
be repeated on the black, and they are here 
quite independent of one another. This equal- 
ity, of course, is in consequence of the fact that 
the number of squares on an ordinary chess- 
board, sixty-four, is an even number. If a 
square chequered board has an odd number of 
squares, then there will always be one more 
square of one colour than of the other. 




Ten bishops are necessary in order that every 
square shall be attacked and every bishop 
guarded by another bishop. I give one way 
of arranging them in the diagram. It will be 
noticed that the two central bishops in the group 



SOLUTIONS. 



215 



of six on the left-hand side of the hoard serve 
no purpose, except to protect those bishops that 
are on adjoining squares. Another solution 
would therefore be obtained by simply raising 
the upper one of these one square and placing 
the other a square lower down. 

299.— BISHOPS IN CONVOCATION. 

The fourteen bishops may be placed in 256 
different ways. But every bishop must always 
be placed on one of the sides of the board — 
that is, somewhere on a row or file on the ex- 
treme edge. The puzzle, therefore, consists in 
counting the number of different ways that we 
can arrange the fourteen round the edge of the 
board without attack. This is not a difficult 



are in a straight line in any oblique direction. 
This is the only arrangement out of the twelve 
fundamentally different ways of placing eight 
queens without attack that fulfils the last con- 
dition. 

301.— THE EIGHT STARS. 

The solution of this puzzle is shown in the first 
diagram. It is the only possible solution 
within the conditions stated. But if one of 
the eight stars had not already been placed as 
shown, there would then have been eight ways 
of arranging the stars according to this scheme, 
if we count reversals and reflections as different. 
If you tiurn this page round so that each side 
is in turn at the bottom, you will get the four 
reversals ; and if you reflect each of these in a 




matter. On a chessboard of n^ squares 2n — 2 
bishops (the maximum number) may always 
be placed in 2'* ways without attacking. On 
an ordinary chessboard n would be 8 ; therefore 
14 bishops may be placed in 256 different ways. 
It is rather curious that the general result should 
come out in so simple a form. 

300.— THE EIGHT QUEENS. 




The solution to this puzzle is shown in the 
diagram. It will be found that no queen 
attacks another, and also that no three queens 



mirror, you will get the four reflections. These 
are, therefore, merely eight aspects of one 
" fundamental solution." But without that 
first star being so placed, there is another fun- 
damental solution, as shown in the second dia- 
gram. But this arrangement being in a way 
symmetrical, only produces four different as- 
pects by reversal and reflection. 

302.— A PROBLEM IN MOSAICS. 



V 


Y 


R 


G 





w 


P 


B 


R 





B 


Y 


P 


V 


G 


w 


B 


w 


M. 





G 


1 


R 


V 


P 


G 


V 


W 


R 


B 





Y 


W 


B 





P 


Y 


G 


V 


R 


G 


R 


Y 


V 


B 


P 


\N 





Y 


V 


G 


R 


W 





B 


P 





P 


v^ 


B 


V 


R 


Y 


G 



The diagram shows how the tiles may be re- 
arranged. As before, one yellow and one 



2l6 



AMUSEMENTS IN MATHEMATICS. 



purple tile are dispensed with. I will here 
point out that in the previous arrangement the 
yellow and purple tiles in the seventh row 
might have changed places, but no other 
arrangement was possible. 

303.— UNDER THE VEIL. 

Some schemes give more diagonal readings of 
four letters than others, and we are at first 
tempted to favour these ; but this is a false 
scent, because what you appear to gain in this 
direction you lose in others. Of course it im- 
mediately occurs to the solver that every 
LIVE or EVIL is worth twice as much as 
any other word, since it reads both ways and 
always counts as 2. This is an important con- 
sideration, though sometimes those arrange- 
ments that contain most readings of these two 
words are fruitless in other words, and we lose 
in the general count. 



^ 






1 


V 


E 


L 








E 


V 


L 






1 




i 




L 






1 






V 


E 




1 




V 


E 








L 






E 






L 


V 




1 






L 






1 




E 


V 


/ 


V 




E 


L 






1 








1 






V 


E 


L 


% 



The above diagram is in accordance with the 
conditions requiring no letter to be in line with 
another similar letter, and it gives twenty read- 
ings of the five words — six horizontally, six 
vertically, four in the diagonals indicated by 
the arrows on the left, and four in the diagonals 
indicated by the arrows on the right. This is 
the maximum. 

Four sets of eight letters may be placed on the 
board of sixty-four squares in as many as 604 
different ways, without any letter ever being 
in line with a similar one. This does not count 
reversals and reflections as different, and it 
does not take into consideration the actual 
permutations of the letters among themselves ; 
that is, for example, making the L's change 
places with the E's. Now it is a singular fact 
that not only do the twenty word-readings 
that I have given prove to be the real maximum, 
but there is actually only that one arrangement 
from which this maximum may be obtained. 
But if you make the V's change places with the 
I*s, and the L's with the E's, in the solution 
given, you still get twenty readings — the same 
number as before in every direction. Therefore 
there are two ways of getting the maximum 



from the same arrangement. The miniilium 
number of readings is zero — that is, the letters 
can be so arranged that no word can be read in 
any of the directions. 

304.— BACHET'S SQUARE.' 





1 


L 




A 


K 


9 


J 


Q 


J 


A 


K 


J 


Q 


K 


A 


K 


A 


J 


Q 



AD 


KS 


QH 


JC 


(?C 


JH 


AS 


KD 


JS 


QO 


KC 


AH 


KH 


AC 


JD 


QS 



D 


s 


H 


c 


C 


H 


s 


D 


s 


D 


c 


H 


H 


C 


D 


S 


4 


S 


H 


C 


D 


D 


C 


H 


S 


H 


S 


D 


C 


C 


D 


s 


H 



Let us use the letters A, K, Q, J, to denote ace, 
king, queen, jack; and D, S, H, C, to denote 
diamonds, spades, hearts, clubs. In Diagrams 
I and 2 we have the two available ways of 
arranging either group of letters so that no 
two similar letters shall be in line — though a 
quarter-turn of i will give us the arrangement 
in 2. If we superimpose or combine these two 
squares, we get the arrangement of Diagram 3, 
which is one solution. But in each square we 
may put the letters in the top line in twenty- 
four different ways without altering the scheme 
of arrangement. Thus, in Diagram 4 the S's 
are similarly placed to the D's in 2, the H's to 
the S's, the C's to the H's, and the D's to the C's. 
It clearly follows that there must be 24X24= 
576 ways of combining the two primitive ar- 
rangements. But the error that Labosne fell into 
was that of assuming that the A, K, Q, J must 
be arranged in the form i, and the D, S, H, C 
in the form 2. He thus included reflections 
and half-turns, but not quarter-turns. They 
may obviously be interchanged. So that the 
correct answer is 2x576=1,152, coimting re- 
flections and reversals as different. Put in 
another manner, the pairs in the top row may 
be written in 16x9X4X1 = 576 different ways, 
and the square then completed in 2 ways, mak- 
ing 1,152 ways in all. 



305.- 



-THE THIRTY-SIX LETTER 
BLOCKS. 



I POINTED out that it was impossible to get all 
the letters into the box under the conditions, 
but the puzzle was to place as many as possible. 



SOLUTIONS. 



217 



This requires a little judgment and careful 
investigation, or we are liable to jump at the 
hasty conclusion that the proper way to solve 
the puzzle must be to first place all six of one 
letter, then all six of another letter, and so on. 
As there is only one scheme (with its reversals) 
for placing six similar letters so that no two 
shall be in a line in any direction, the reader 
will find that after he has placed four different 
kinds of letters, six times each, every place is 
occupied except those twelve that form the 
two long diagonals. He is, therefore, unable 
to place more than two each of his last two 
letters, and there are eight blanks left. I give 
such an arrangement in Diagram i. 







1 












* 


2. 






A 


B 


C 


D 


E 


F 




A 


B 


C 


D 


E 


F 


D 


F 


E 


B 


A 


C 


D 


E 


A 


F 


B 


C 


E 


C 






D 


B 


F 


C 






D 


A 


B 


D 






C 


E 


B 


D 






C 


E 


C 




B 


E 




B 


C 


A 


E 


B 


F 


D 




E 


D 


C 


B 




E 


P 


D 


C 


A 


B 



The secret, however, consists in not trying 
thus to place all six of each letter. It will be 
found that if we content ourselves with placing 
only five of each letter, this number (thirty in 
all) may be got into the box, and there will be 
only six blanks. But the correct solution is 
to place six of each of two letters and five of 
each of the remaining four. An examination 
of Diagram 2 will show that there are six each 
of C and D, and five each of A, B, E, and F. 
There are, therefore, only four blanks left, and 
no letter is in line with a similar letter in any 
direction. 

306.— THE CROWDED CHESSBOARD. 




Here is the solution. Only 8 queens or 8 
rooks can be placed on the board without attack, 
while the greatest number of bishops is 14, and 
of knights 32. But as all these knights must 
be placed on squares of the same colour, while 
the queens occupy four of each colour and the 
bishops 7 of each colour, it follows that only 21 
knights can be placed on the same colour in 
this puzzle. More than 21 knights can be placed 
alone on the board if we use both colours, but 
I have not succeeded in placing more than 21 
on the " crowded chessboard." I believe the 
above solution contains the maximum number 
of pieces, but possibly some ingenious reader 
may succeed in getting in another knight. 

307.— THE COLOURED COUNTERS. 

The counters may be arranged in this order : — 

Ri, B2, Y3, O4, G5. 

Y4, O5, Gi, R2, B3. 

G2, R3, B4, Y5, Oi. 

B5, Yi, O2, G3, R4. 

O3, G4, R5, Bi, Y2. 

308.— THE GENTLE ART OF STAMP- 
LICKING. 

The following arrangement shows how sixteen 
stamps may be stuck on the card, under the 
conditions, of a total value of fifty pence, or 
4s. 2d. : — 



4 3 5 2 

5 2 14 
14 3 5 
3 5 2 1 



If, after placing the four 5d. stamps, the 
reader is tempted to place four 4d. stamps also, 
he can afterwards only place two of each of 
the three other denominations, thus losing two 
spaces and counting no more than forty-eight 
pence, or 4s. This is the pitfall that was 
hinted at. (Compare with No. 43, Canterbury 
Puzzles.) 

309.— THE FORTY-NINE COUNTERS. 

The counters may be arranged in this order : — 

Ai, B2, C3, D4, E5, F6, G7. 

F4, G5, A6, B7, Ci, D2, E3. 

D7, Ei, F2, G3, A4, B5, C6. 

B3, C4, D5, E6, F7, Gi, A2 

G6, A7, Bi, C2, D3, E4, 

E2, F3, G4, A5, B6, C7, 

C5, D6, E7, Fi, G2, A3, 



F5. 
Di. 
B4. 



310.— THE THREE SHEEP. 



The number of different ways in which the 
three sheep may be placed so that every pen 



2l8 



AMUSEMENTS IN MATHEMATICS. 



shall always be either occupied or in line with 
at least one sheep is forty-seven. 

The following table, if used with the key in 
Diagram i, will enable the reader to place them 
in all these ways : — 



Two Sheep. 


Third Sheep. 


No. of 
Ways. 


A and B 


C, E, G, K, L, N, or P 


7 


A and C 


1, J, K, or 


4 


A andD 


M, N, or J 


3 


A and F 


J, K, L, or P 


4 


A and G 


H, J, K, N, 0, or P 


6 


AandH 


K, L, N, or 


4 


A and 


Kor L 


2 


B andC 


N 


I 


B and E 


F, H, K, or L 


4 


B and F 


G, J, N, or 


4 


B andG 


K, L, or N 


3 


B andH 


JorN 


2 


B and J 


Kor L 


2 


FandG 


J 


I 

47 



This, of course, means that if you place sheep 
in the pens marked A and B, then there are 



seven different pens in which you may place 
the third sheep, giving seven different solutions. 
It was understood that reversals and reflections 
do not count as different. 

If one pen at least is to be not in line with a 
sheep, there would be thirty solutions to that 
problem. If we counted all the reversals and 
reflections of these 47 and 30 cases respectively 
as different, their total would be 560, which is 
the number of different ways in which the sheep 
may be placed in three pens without any con- 
ditions. I will remark that there are three 
ways in which two sheep may be placed so that 
every pen is occupied or in line, as in Diagrams 
2, 3, and 4, but in every case each sheep is in 
line with its companion. There are only two 
ways in which three sheep may be so placed 
that every pen shall be occupied or in hne, but 
no sheep in line with another. These I show 
in Diagrams 5 and 6. Finally, there is only one 
way in which three sheep may be placed so that 
at least one pen shall not be in line with a 
sheep and yet no sheep in line with another. 
Place the sheep in C, E, L. This is practi- 
cally all there is to be said on this pleasant 
pastoral subject. 

311.— THE FIVE DOGS PUZZLE. 

The diagrams show four fundamentally differ- 
ent solutions. In the case of A we can reverse 



,A 


B 


C 


D 


E 


F 


& 


H 


1 


J 


K 


L 


M 


N 





P 


4 
































M- 







































/.:->, 











'^ 
























6 




^ 






















^i 




^ 







SOLUTIONS. 



219 



the order, so that the single dog is in the bottom 
row and the other four shifted up two squares. 
Also we may use the next column to the right 
and both of the two central horizontal rows. 
Thus A gives 8 solutions. Then B may be 









A 
















B 














• 
























































'• 










































• 


















• 














• 
















• 
















• 














• 


















• 










































• 




















reversed and placed in either diagonal, giving 
4 solutions. Similarly C will give 4 solutions. 
The line in D being symmetrical, its reversal 
will not be different, but it may be disposed in 
4 different directions. We thus have in all 20 
different solutions. 



312. 



-THE FIVE CRESCENTS OF 
BYZANTIUM. 



If that ancient architect had arranged his five 
crescent tiles in the manner shown in the follow- 
ing diagram, every tile would have been watched 




over by, or in a line with, at least one crescent, 
and space would have been reserved for a per- 
fectly square carpet equal in area to exactly 
half of the pavement. It is a very curious fact 
that, although there are two or three solutions 
allowing a carpet to be laid down within the 
conditions so as to cover an area of nearly 
twenty-nine of the tiles, this is the only possible 



solution giving exactly half the area of the pave- 
ment, which is the largest space obtainable. 

313.— QUEENS AND BISHOP PUZZLE. 

I. 




Fig. I. 

The bishop is on the square originally occupied 
by the rook, and the four queens are so placed 
that every square is either occupied or attacked 
by a piece. (Fig. i.) 

I pointed out in 1899 that if four queens are 
placed as shown in the diagram (Fig. 2), then 




Fig. 2. 

the fifth queen may be placed on any one of the 
twelve squares marked a, b, c, d, and e ; or a 
rook on the two squares, c ; or a bishop on 
the eight squares, a, b, and e ; or a pawn on 



220 



AMUSEMENTS IN MATHEMATICS. 



the square b ; or a king on the four squares, 
b, c, and e. The only known arrangement 
for four queens and a knight is that given by 
Mr, J. Wallis in The Strand Magazine for 
August 1908, here reproduced. (Fig. 3.) 




Fig. 3. 

I have recorded a large number of solutions 
with four queens and a rook, or bishop, but the 
only arrangement, I believe, with three queens 
and two rooks in which all the pieces are guarded 
is that of which I give an illustration (Fig. 4), 



Fig. 4- 

first published by Dr. C. Planck. But I have 
since found the accompanying solution with 
three queens, a rook, and a bishop, though the 
pieces do not protect one smother. (Fig, 5.) 





Fig. 5. 

314.— THE SOUTHERN CROSS. 

My readers have been so familiarized with the 
fact that it requires at least five planets to 
attack every one of a square arrangement of 
sixty-four stars that many of them have, 
perhaps, got to believe that a larger square 
arrangement of stars must need an increase of 
planets. It was to correct this possible error 
of reasoning, and so warn readers against 
another of those numerous little pitfalls in 
the world of puzzledom, that I devised this 
new stellar problem. Let me then state at 
once that, in the case of a square arrangement 
of eighty one stars, there are several ways of 
placing five planets so that every star shall be 
in line with at least one planet vertically, 
horizontally, or diagonally. Here is the solu- 
tion to the " Southern Cross " : — , 

^ tt ti ^ i^ A.,"^ ^ <f 

^ si? i^ jV'Vt^ rft-^ <^ 
^ ^ X^ tf? J^ i2i sir ft 

:-& ti^t!? •<^ -Ci ^ <=« 



SOLUTIONS. 



221 



I It will be remembered that I said that the 
' five planets in tbeir new positions " Will, of 
course, obscure five other stars in place of 
those at present covered." This was to exclude 
an easier solution in which only four planets 
need be moved. 

315.— THE HAT-PEG PUZZLE. 

The moves will be made quite clear by a refer- 
ence to the diagrams, which show the position 
on the board after each of the four moves. The 



queen attacks any other. In the case of the 
last move the queen in the top row might also 
have been moved one square farther to the left. 
This is, I believe, the only solution to the 
puzzle. 

316.— THE AMAZONS. 

It will be seen that only three queens have been 
removed from their positions on the edge of 
the board, and that, as a consequence, eleven 
squares (indicated by the black dots) are left 
unattacked by any queen. I will hazard the 




darts indicate the successive removals that 
have been made. It will be seen that at every 
stage all the squares are either attacked or 
occupied, and that after the fourth move no 



statement that eight queens cannot be placed 
on the chessboard so as to leave more than 
eleven squares unattacked. It is true that we 
have no rigid proof of this yet, but I have 



222 



AMUSEMENTS IN MATHEMATICS. 




entirely convinced myself of the truth of the 
statement. There are at least five different 
ways of arranging the queens so as to leave 
eleven squares unattacked. 

317.— A PUZZLE WITH PAWNS. 



77^ 



^ 



^ 



^ 




^ 






'4. 



':^ 



^y7:<!^> 



^ 



"m 



m 



c^ 



^ 



man may be placed on the same path, the re- 
sult must be the number of ways in which they 
will not be on the same path. The nmnber of 
ways in which they may be in hne is found with- 
out much diflBculty to be 816. Consequently, 
6,480—816=5,664, the required answer. 

The general solution is this : \n{n — i) 
(3«2— «-f 2). This is, of course, equivalent to 
sa3dng that if we call the number of squares on 
the side of a " chessboard " n, then the formula 
shows the nimaber of ways in which two 
bishops may be placed without attacking one 
another. Only in this case we must divide by 
two, because the two bishops have no distinct 
individuality, and cannot produce a different 
solution by mere exchange of places. 

319.— THE KNIGHT-GUARDS. 



W 



^ 




m 



Va 



'^z 



/^^ 
^ 



^ 



h 







1 



e 




^ 



^ 



^ 



Sixteen pawns may be placed so that no three 
shall be in a straight line in any possible direc- 
tion, as in the diagram. We regard, as the 
conditions required, the pawns as mere points 
on a plane. 

3 1 8.— LION-HUNTING. 

There are 6,480 ways of placing the man and 
the lion, if there are no restrictions whatever 
except that they must be on different spots. 
This is obvious, because the man may be 
placed on any one of the 81 spots, and in every 
case there are 80 spots remaining for the lion ; 
therefore 81 x 80=6,480. Now, if we deduct 
the number of ways in which the lion and the 



Diagram i. 

The smallest possible number of knights with 
which this puzzle can be solved is foiirteen. 




Diagram 2. 



SOLUTIONS. 



223 



It has sometimes been assumed that there are 
a great many different solutions. As a matter 
of fact, there are only three arrangements — 
not counting mere reversals and reflections as 




&:^- 



^W 



vm. 



i 






1 






W^ 



m 



^ 



% 



S 



y. 



^ 



W 




Diagram 3. 

different. Curiously enough, nobody seems 
ever to have hit on the following simple proof, 
or to have thought of dealing with the black 
and the white squares separately. 

Seven knights can be placed on the board 
on white squares so as to attack every black 
square in two ways only. These are shown in 
Diagrams i and 2. Note that three knights 
occupy the same position in both arrangements. 
It is therefore clear that if we turn the board so 
that a black square shall be in the top left-hand 



^ 



^ 



^ 



m 





A^ 



^ 



%. 



T. 



^ 



"^ 
^ 



% 



^ 



52 




% 



Diagram 4. 

comer instead of a white, and place the knights 
in exactly the same positions, we shall have 



two similar ways of attacking all the white 
squares. I will assume the reader has made 
the two last described diagrams on transparent 
paper, and marked them la and 2a. Now, by 
placing the transparent Diagram la over i you 
will be able to obtain the solution in Diagram 3, 
by placing 2a over 2 you will get Diagram 4, 
and by placing 2a over i you will get Diagram 5. 




Diagram 5. 

You may now try all possible combinations of 
those two pairs of diagrams, but you will only 
get the three arrangements I have given, or 
their reversals and reflections. Therefore these 
three solutions are all that exist. 

320.— THE ROOK'S TOUR. 



-- 


... 






— 


— 




""1 




r 


— 


— 


- - - 





-1 


1 ■■■■ 




• 


r ' 





— 


~T 












n 


"7 




' 








I 


1 


























b 










-- 


- - 




-- 


-- 


-- 


t ■ 

1 


-- 











The only possible minimum solutions are 
shown in the two diagrams, where it will be 
seen that only sixteen moves are required to 
perform the feat. Most people find it difficult 
to reduce the number of moves below seven- 
teen. 



224 



AMUSEMENTS IN MATHEMATICS. 



,-• 








. . - 











,- 















1 






„ „ 


• •. • 


_ „ 


. • >• 




-1 
























9fR 






-, 




# 












1^^ 





" 1 
















1 
















-f 














'— 


— 








-- 


•-? 


1 


— 





— 











1 



THE rook's tour. 

321.— THE ROOK'S JOURNEY. 











- 


3^ 


r* * 










"- •-" ^ 


■■ r ■ 




















1 




— 










1 




r- 











-1 I 1 




• 


1 













[ 


— , 








'— . 


- 1 






i- . J ' 


u . 


21 


1 


10 


-- 


-J 



I SHOW the route in the diagram. It will be 
seen that the tenth move lands us at the square 
marked " lo," and that the last move, the 
twenty-first, brings us to a halt on square " 21." 

322.— THE LANGUISHING MAIDEN. 

The dotted line shows the route in twenty-two 
straight paths by which the knight may rescue 
the maiden. It is necessary, after entering the 
first cell, immediately to return before entering 
another. Otherwise a solution would not be 
possible. (See "The Grand Tour," p. 200.) 

323.— A DUNGEON PUZZLE. 

If the prisoner takes the route shown in the 
diagram — where for clearness the doorways are 
omitted — he will succeed in visiting every cell 
once, and only once, in as many as fifty-seven 
straight lines. No rook's path over the chess- 
board can exceed this number of moves. 



t^ 



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THE LANGUISHING MAIDEN. 









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A DUNGEON PUZZLE. 

324.— THE LION AND THE MAN. 

First of all, the fewest possible straight lines in 
each case are twenty-two, and in order that no 
cell may be visited twice it is absolutely neces- 
sary that each should pass into one cell and then 
immediately " visit " the one from which he 
started, afterwards proceeding by way of the 
second available cell. In the following diagram 
the man's route is indicated by the unbroken 
lines, and the lion's by the dotted lines. It 
will be found, if the two routes are followed 
cell by cell with two pencil points, that the 
lion and the man never meet. But there was 
one little point that ought not to be overlooked 
— " they occasionally got glimpses of one an- 
other." Now, if we take one route for the 



SOLUTIONS. 



225 



lan and merely reverse it for the lion, we in- 
ariably find that, going at the same speed, 
hey never get a glimpse of one another. But in 
ur diagram it will be found that the man and 
he lion are in the cells marked A at the same 
Qoment, and may see one another through the 



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open doorways ; while the same happens when 
they are in the two cells marked B, the upper 
letters indicating the man and the lower the 
Hon. In the first case the lion goes straight 
for the man, while the man appears to attempt 
to get in the rear of the lion ; in the second 
case it looks suspiciously like running away 
from one another ! 

325.— AN EPISCOPAL VISITATION. 




In the diagram I show how the bishop may be 
made to visit every one of his white parishes 
in seventeen moves. It is obvious that we 
must start from one comer square and end at 
the one that is diagonally opposite to it. The 
puzzle cannot be solved in fewer than seven- 
teen moves. 

(1,926) 15 



326.— A NEW COUNTER PUZZLE. 

Play as follows : 2 — 3, 9 — 4, 10 — 7, 3 — 8, 4 — 2, 
7—5, 8—6, 5—10, 6—9, 2—5, 1—6, 6—4, 5—3, 
10 — 8, 4 — 7, 3 — 2, 8 — I, 7 — 10. The white 
counters have now changed places with the red 
ones, in eighteen moves, without breaking the 
conditions. 

327.— A NEW BISHOP'S PUZZLE. 



1 


2 


3 


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5 


6 


7 


8 


9 


10 


11 


12 


13 


14? 


15 


16 


17 


18 


19 


20 




A :a 

Play as follows, using the notation indicated 
by the numbered squares in Diagram A : — 

White. Black. White. [ Black. 

1. 18 — 15 I. 3 — 6 10. 20 — 10 10. I — II 

2. 17 — 8 2. 4 — 13 II. 3 — 9 II. 18 — 12 

3. 19 — 14 3. 2 — 7 12. 10 — 13 12. II — 8 
4- 15 — 5 4- 6 — 16 13. 19 — 16 13. 2 — 5 

5. 8 — 3 5. 13 — 18 14. 16 — I 14. 5 — 20 

6. 14 — 9 6. 7 — 12 15. 9 — 6 15. 12 — 15 

7. 5 — 10 7. 16 — II 16. 13 — 7 16. 8 — 14 

8. 9 — 19 8. 12 — 2 17. 6 — 3 17. 15 — 18 

9. 10 — 4 9. II — 17 18. 7— 2 18. 14 — 19 

Diagram B shows the position after the ninth 
move. Bishops at i and 20 have not yet moved, 
but 2 and 19 have sallied forth and returned. 
In the end, i and 19, 2 and 20, 3 and 17, and 
4 and 18 will have exchanged places. Note the 
position after the thirteenth move. 





328- 


-THE QUEEN'S TOUR. 




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226 



AMUSEMENTS IN MATHEMATICS. 



The annexed diagram shows a second way of 
performing the Queen's Tour. If you break 
the Hne at the point J and erase the shorter 
portion of that Une, you will have the required 
path solution for any J square. If you break 
the line at I, you will have a non-re-entrant 
solution starting from any I square. And if 
you break the line at G, you will have a solution 
for any G square. The Queen's Tour previ- 
ously given may be similarly broken at three 
different places, but I seized the opportvmity 
of exhibiting a second tour. 

329.— THE STAR PUZZLE. 



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The illustration explains itself. The stars are 
all struck out in fourteen straight strokes, 
starting and ending at a white star. 

330.— THE YACHT RACE. 




The diagram explains itself. The numbers will 
show the direction of the lines in their proper 
order, and it will be seen that the seventh 
course ends at the flag-buoy, as stipulated. 



331.— THE SCIENTIFIC SKATER. 

In this case we go beyond the boundary of th< 
square. Apart from that, the moves are al 
queen moves. There are three or four way 
in which it can be done. 

Here is one way of performing the feat : — 



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It will be seen that the skater strikes out all 
the stars in one continuous journey of fourteen 
straight lines, retmning to the point fron 
which he started. To follow the skater's 
course in the diagram it is necessary always- 
to go as far as we can in a straight line be' 
fore turning. 

332.— THE FORTY-NINE STARS. 




The illustration shows how all the stars may be 
struck out in twelve straight strokes, beginning 
and ending at a black star. 



SOLUTIONS. 
333.— THE QUEEN'S JOURNEY. 



227 















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The correct solution to this puzzle is shown in 
the diagram by the dark line. The five moves 
indicated will take the queen the greatest dis- 
tance that it is possible for her to go in five 
moves, within the conditions. The dotted line 
shows the route that most people suggest, but 
it is not quite so long as the other. Let us 
assume that the distance from the centre of any 
square to the centre of the next in the same 
horizontal or vertical line is 2 inches, and that 
the queen travels from the centre of her original 
square to the centre of the one at which she 
rests. Then the first route will be found to 
exceed 67.9 inches, while the dotted route is 
less than 67,8 inches. The difference is small, 
but it is sufficient to settle the point as to the 
longer route. All other routes are shorter still 
than these two. 

334.— ST. GEORGE AND THE DRAGON. 

We select for the solution of this puzzle one of 
the prettiest designs that can be formed by 
representing the moves of the knight by lines 
from square to square. The chequering of the 
squares is omitted to give greater clearness. 
St. George thus slays the Dragon in strict ac- 
cordance with the conditions and in the elegant 
manner we should expect of him. 

335.— FARMER LAWRENCE'S CORN- 
FIELDS. 

There are numerous solutions to this little 
agricultural problem. The version I give in the 
next column is rather curious on account of the 
long parallel straight lines formed by some of 
the moves. 

336.— THE GREYHOUND PUZZLE. 

There are several interesting points involved in 
this question. In the first place, if we had made 





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ST. GEORGE AND THE DRAGON. 



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FARMER Lawrence's cornfields. 

no stipulation as to the positions of the two 
ends of the string, it is quite impossible to form 
any such string unless we begin and end in 
the top and bottom row of kennels. We may 
begin in the top row and end in the bottom (or, 
of course, the reverse), or we may begin in one 
of these rows and end in the same. But we 
can never begin or end in one of the two central 
rows. Our places of starting and ending, how- 
ever, were fixed for us. Yet the first half of 
our route must be confined entirely to those 
squares that are distinguished in the following 
diagram by circles, and the second half will 
therefore be confined to the squares that are 
not circled. The squares reserved for the two 
half-strings will be seen to be sjonmetrical and 
similar. 

The next point is that the first half-string 
must end in one of the central rows, and the 



228 



AMUSEMENTS IN MATHEMATICS. 



second half-string must begin in one of these 
rows. This is now obvious, because they have 
to link together to form the complete string, and 
every square on an outside row is connected by 
a knight's move with similar squares only — 
that is, circled or non-circled as the case may 
be. The half-strings can, therefore, only be 
linked in the two central rows. 




Now, there are just eight different first half- 
strings, and consequently also eight second 
half-strings. We shall see that these combine 
to form twelve complete strings, which is the 
total nimiber that exist and the correct solution 
of our puzzle. I do not propose to give all the 
routes at length, but I will so far indicate them 
that if the reader has dropped any he will be 
able to discover which they are and work them 
out for himself without any difficulty. The 
following numbers apply to those in the above 
diagram. 

The eight first half-strings are : i to 6 (2 
routes) ; i to 8 (i route) ; i to 10 (3 routes) ; 
I to 12 (i route) ; and i to 14 (i route). The 
eight second half -strings are : 7 to 20 (i route) ; 
9 to 20 (x route) ; 11 to 20 (3 routes) ; 13 to 20 
(i route) ; and 15 to 20 (2 routes). Every 
different way in which you can link one half- 
string to another gives a different solution. 
These linkings will be found to be as follows : 
6 to 13 (2 cases) ; 10 to 13 (3 cases) ; 8 to 11 
(3 cases) ; 8 to 15 (2 cases) ; 12 to 9 (i case) ; 
and 14 to 7 (x case). There are, therefore, 
twelve different linkings and twelve different 
answers to the puzzle. The route given in 
the illustration with the greyhound will be 
found to consist of one of the three half-strings 
I to xo, linked to the half-string X3 to 20. It 
should be noted that ten of the solutions are 
produced by five distinctive routes and their 
reversals — that is, if you indicate these five 
routes by lines and then turn the diagrams 
upside down you will get the five other routes. 
The remaining two solutions are symmetrical 
(these are the cases where 12 to 9 and X4 to 7 
are the links), and consequently they do not 
produce new solutions by reversal. 



337.— THE FOUR KANGAROOS. 























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A PRETTY sjnnmetrical solution to this puzzle 
is shown in the diagram. Each of the fom 
kangaroos makes his little excursion and re- 
turns to his comer, without ever entering a 
square that has been visited by another kan^i 
garoo and without crossing the central line/ 
It will at once occur to the reader, as a possible 
improvement of the puzzle, to divide the board 
by a central vertical line and make the condi- 
tion that this also shall not be crossed. Thisi 
would mean that each kangaroo had to confine 
himself to a square 4 by 4, but it would be* 
quite impossible, as I shall explain in the next, 
two puzzles. 

338.— THE BOARD IN COMPARTMENTS. 




In attempting to solve this problem it is first 
necessary to take the two distinctive compart- 
ments of twenty and twelve squares respec- 
tively and analyse them with a view to deter- 



SOLUTIONS. 



229 



mining where the necessary points of entry 
and exit he. In the case of the larger com- 
partment it will be found that to complete a 
tour of it we must begin and end on two of 
the outside squares on the long sides. But 
though you may start at any one of these ten 
squares, you are restricted as to those at which 
you can end, or (which is the same thing) you 
may end at whichever of these you like, pro- 
vided you begin your tour at certain particular 
squares. In the case of the smaller compart- 
ment you are compelled to begin and end at 
one of the six squares lying at the two narrow 
ends of the compartments, but similar restric- 
tions apply as in the other instance. A very 
little thought will show that in the case of the 
two small compartments you must begin and 
finish at the ends that lie together, and it then 
follows that the tours in the larger compart- 
ments must also start and end on the contigu- 
ous sides. 

In the diagram given of one of the possible 
solutions it will be seen that there are eight 
places at which we may start this particular 
tour ; but there is only one route in each case, 
because we must complete the compartment 
iji which we find ourself before passing into 
Another. In any solution we shall find that 
the squares distinguished by stars must be 
entering or exit points, but the law of reversals 
leaves us the option of making the other con- 
nections either at the diamonds or at the circles. 
In the solution worked out the diamonds are 
used, but other variations occur in which the 
circle squares are employed instead. I think 
these remarks explain all the essential points 
in the puzzle, which is distinctly instructive 
and interesting. 

339.— THE FOUR KNIGHTS' TOURS. 




re-entrant knight's tour may be made on each 
portion. There is only one possible route for 
each knight and its reversal. 

340.— THE CUBIC KNIGHT'S TOUR. 




It will be seen in the illustration how a chess- 
board may be divided into four parts, each of 
the same size and shape, so that a complete 



If the reader should cut out the above diagram, 
fold it in the form of a cube, and stick it together 
by the strips left for that purpose at the edges, 
he would have an interesting little curiosity. 
Or he can make one on a larger scale for himself. 
It will be found that if we imagine the cube to 
have a complete chessboard on each of its sides, 
we may start with the knight on any one of 
the 384 squares, and make a complete tour of 
the cube, always returning to the starting- 
point. The method of passing from one side 
of the cube to another is easily understood, 
but, of course, the difficulty consisted in finding 
the proper points of entry and exit on each 
board, the order in which the different boards 
should be taken, and in getting arrangements 
that would comply with the required conditions. 

341.— THE FOUR FROGS. 

The fewest possible moves, counting every 
move separately, are sixteen. But the puzzle 
may be solved in seven plays, as follows, if any 
number of successive moves by one frog count 
as a single play. All the moves contained 
within a bracket are a single play; the num- 
bers refer to the toadstools : (i — 5), (3 — 7» 
7—1), (8—4, 4—3, 3—7), (6—2, 2—8, 8—4, 
4—3), (5—6, 6—2, 2—8), (1—5, 5— 6), (7— I); 
This is the familiar old puzzle by Guanm, 
propounded in 15 12, and I give it here in order 
to explain my " buttons and string " method 
of solving this class of moving-counter problem. 



230 



AMUSEMENTS IN MATHEMATICS. 




Diagram A shows the old way of presenting 
Guarini's puzzle, the point being to make the 
white knights change places with the black 
ones. In " The Four Frogs " presentation of 
the idea the possible directions of the moves are 
indicated by lines, to obviate the necessity of 
the reader's understanding the nature of the 
knight's move in chess. But it will at once 
be seen that the two problems are identical. 
The central square can, of course, be ignored, 
since no knight can ever enter it. Now, regard 
the toadstools as buttons and the connecting 
lines as strings, as in Diagram B. Then by 
disentangling these strings we can clearly pre- 
sent the diagram in the form shown in Diagram 
C, where the relationship between the buttons 
is precisely the same as in B. Any solution on 
C will be applicable to B, and to A. Place 
your white knights on i and 3 and your black 
knights on 6 and 8 in the C diagram, and the 
simpUcity of the solution will be very evident. 
You have simply to move the knights round 
the circle in one direction or the other. Play 
over the moves given above, and you will find 
that. every little dif&culty has disappeared. 

In Diagram D I give another famihar puzzle 
that first appeared in a book published in 
Brussels in 1789, Les P elites A ventures de 
Jerome Sharp. Place seven counters on seven 
of the eight points in the following manner. 
You must always touch a point that is vacant 
with a counter, and then move it along a 
straight line leading from that point to the next 
vacant point (in either direction), where you 
deposit the counter. You proceed in the same 
way until all the counters are placed. Re- 
member you always touch a vacant place and 
sUde the coimter from it to the next place, 
which must be also vacant. Now, by the 



" buttons and string " method of simplification 
we can transform the diagram into E. Then 
the solution becomes obvious. " Always move 
to the point that you last moved from." This 
is not, of course, the only way of placing the 
counters, but it is the simplest solution to 
carry in the mind. 

There are several puzzles in this book that 
the reader will find lend themselves readily to 
this method. 

342.— THE MANDARIN'S PUZZLE. 

The rather perplexing point that the solver 
has to decide for himself in attacking this 
puzzle is whether the shaded numbers (those 
that are shown in their right places) are mere 
dmnmies or not. Ninety-nine persons out of a 
hundred might form the opinion that there can 
be no advantage in moving any of them, but 
if so they would be wrong. 

The shortest solution without moving any 
shaded number is in thirty-two moves. But 
the puzzle can be solved in thirty moves. The 
trick lies in moving the 6, or the 15, on the 
second move and replacing it on the nineteenth 
move. Here is the solution : 2 — 6 — 13 — 4 — i — 
21 — 4 — I — 10 — 2 — 21 — 10 — 2 — 5 — 22 — 16 — I 
— 13 — 6 — 19 — II — 2 — 5 — 22 — 16 — 5 — 13 — 4 — 
10 — 21. Thirty moves. 

343.— EXERCISE FOR PRISONERS. 

There are eighty different arrangements of 
the numbers in the form of a perfect knight's 
path, but only forty of these can be reached 
without two men ever being in a cell at the same 
time. Two is the greatest number of men that 
can be given a complete rest, and though the 



I 



SOLUTIONS. 



231 



knight's path can be arranged so as to leave 
either 7 and 13, 8 and 13, 5 and 7, or 5 and 13 
in their original positions, the following four 
arrangements, in which 7 and 13 are unmoved, 
are the only ones that can be reached under the 
moving conditions. It therefore resolves itself 
into finding the fewest possible moves that will 
lead up to one of these positions. This is cer- 
tainly no easy matter, and no rigid rules can 
be laid down for arriving at the correct answer. 
It is largely a matter for individual judgment, 
patient experiment, and a sharp eye for revolu- 
tions and position. 





D 



— I — I r— 

6.11 4^15 

1 14<,'r,10 



As a matter of fact, the position C can be 
reached in as few as sixty-six moves in the follow- 
ing manner: 12, II, 15, 12, II, 8, 4, 3, 2, 6, 5, 1, 6, 
5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 
4, 12, II, 3, 2, 5, 10, 15, 6, I, 8, 4, 9, 8, I, 6, 4, 9, 

12, 2, 5, 10, 15, 4, 9, 12, 2, 5, 3, II, 14, 2, 5, 14, 
11 = 66 moves. Though this is the shortest 
that I know of, and I do not think it can be 
beaten, I cannot state positively that there is 
not a shorter way yet to be discovered. The 
most tempting arrangement is certainly A ; but 
things are not what they seem, and C is really 
the easiest to reach. 

If the bottom left-hand comer cell might be 
left vacant, the following is a solution in forty-five 
moves by Mr. R. Elrick : 15, 11, 10, 9, 13, 14, 
II, 10, 7, 8, 4, 3, 8, 6, 9, 7, 12, 4, 6, 9, 5, 13, 7, 5, 

13, I, 2, 13, 5, 7, I, 2, 13, 8, 3, 6, 9, 12, 7, II, 

14, I, II, 14, I. But every man has moved. 

344.— THE KENNEL PUZZLE. 

The first point is to make a choice of the most 
promising knight's string and then consider the 
question of reaching the arrangement in the 
fewest moves. I am strongly of opinion that 
the best string is the one represented in the 



following diagram, in which it will be seen that 
each successive number is a knight's move from 
the preceding one, and that five of the dogs 
(i, 5, 10, 15, and 20) never leave their original 
kennels. 



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This position may be arrived at in as few as 
forty-six moves, as follows: 16 — 21, 16 — 22, 16 — 
23,17 — 16, 12 — 17, 12 — 22,12 — 21,7 — 12, 7 — 17, 
7 — 22, II — 12, II — 17, 2 — 7, 2 — 12, 6 — II, 8 — 7, 
8 — 6, 13 — 8, 18 — 13, II — 18, 2 — 17, 18 — 12, 
18—7, 18—2, 13—7, 3—8, 3—13, 4—3, 4—8, 
9—4, 9—3, 14—9, 14—4, 19—14, 19 — 9, 3—14, 
3 — 19, 6 — 12, 6 — 13, 6 — 14, 17 — II, 12 — 16, 
2 — 12, 7 — 17, II — 13, 16 — 18=46 moves. I 
am, of course, not able to say positively that a 
solution cannot be discovered in fewer moves, 
but I beheve it will be found a very hard task 
to reduce the number. 

345.— THE TWO PAWNS. 

Call one pawn A and the other B. Now, owing 
to that optional first move, either pawn may 
make either 5 or 6 moves in reaching the eighth 
square. There are, therefore, four cases to be 
considered : (i) A 6 moves and B 6 moves ; (2) 
A 6 moves and B 5 moves ; (3) A 5 moves and B 
6 moves ; (4) A 5 moves and B 5 moves. In 
case (i) there are 12 moves, and we may select 
any 6 of these for A. Therefore 7X8X9X10X 
11x12 divided by 1X2x3x4x5x6 gives us 
the number of variations for this case — that is, 
924. Similarly for case (2), 6 selections out 
of II will be 462 ; in case (3), 5 selections out of 
II will also be 462 ; and in case (4), 5 selections 
out of 10 will be 252. Add these four numbers 
together and we get 2,100, which is the correct 
number of different ways in which the pawns 
may advance under the conditions. (See No. 
270, on p. 204.) 

346.— SETTING THE BOARD. 

The White pawns may be arranged in 40,320 
ways, the White rooks in 2 ways, the bishops 
in 2 ways, and the knights in 2 ways. Multiply 
these numbers together, and we find that the 
White pieces may be placed in 322,560 different 



232 



AMUSEMENTS IN MATHEMATICS. 



ways. The Black pieces may, of course, be 
placed in the same number of ways. Therefore 
the men may be set up in 322,560 X 322,560 = 
104,044,953,600 ways. But the point that 
nearly everybody overlooks is that the board 
may be placed in two different ways for every 
arrangement. Therefore the answer is doubled, 
and is 208,089,907,200 different ways. 

347.— COUNTING THE RECTANGLES. 

There are 1,296 different rectangles in all, 204 
of which are squares, coimting the square board 
itself as one, and 1,092 rectangles that are not 
squares. The general formula is that aboard oin^ 

contains ^ — ^^—^ rectangles, of which 
4 

!— I ! — are squares and 

o 

3»4 + 2m3 



squares 



3» 



2_ 



zn 



12 
are rectangles that are not squares. It is 
curious and interesting that the total number 
of rectangles is always the square of the tri- 
angular number whose side is n. 

348.— THE ROOKERY. 

The answer involves the little point that in the 
final position the numbered rooks must be in 
numerical order in the direction contrary to that 
in which they appear in the original diagram, 
otherwise it cannot be solved. Play the rooks 
in the following order of thfeir numbers. As 
there is never more than one square to which 
a rook can move (except on the final move), the 
notation is obvious— 5, 6, 7, 5, 6, 4, 3, 6, 4, 7, 5, 4, 
7, 3, 6, 7, 3, 5, 4, 3, i, 8, 3, 4, 5, 6, 7, i, 8, 2, i, 
and rook takes bishop, checkmate. These are 
the fewest possible moves — thirty-two. The 
Black king's moves are all forced, and need not 
be given. 

349.— STALEMATE. 




stein, W. H. Thompson, and myself. So the 
following may be accepted as the best solution 
possible to this curious problem : — 



Working independently, the same position was 
arrived at by Messrs. S. Loyd, E. N. Franken- 



White. 
P-Q4 

Q—Qs 
g— KKt 

Kt— Q2 
P— R4 
P— R3 
R— R3 
Q— R2 
R— KKt 

10. P— QB 4 

11. P— B 3 

12. P— Q 5 



I. 
2. 
3. 
4- 
5- 
6. 

7. 
8. 

9. 



I. 
2. 

3. 
4- 
5- 
6. 

7. 
8. 

9. 
10. 
II. 
12. 



Black. 
P— K4 

Q— R5 
B— Kt 5 ch 
P— QR4 
P— Q3 
B— K3 
P— KB4 
P— B4 
B— Kt6 
P— B5 
P— K5 
P— K6 



And White is stalemated. 

We give a diagram of the curious position 
arrived at. It will be seen that not one of 
White's pieces may be moved. 

350.— THE FORSAKEN KING. 



Play as follows : — 






White. 


Black. 


I. 


P to K 4th I. 


Any move 


2. 


Q to Kt 4th 2. 


Any move except on 

KB file (a) 
K moves to royal row 


3. 


Q to Kt 7th 3. 


4. 


B to Kt 5th 4. 


Anv move 


5. 


Mate in two moves 






If 3, 


K other than to royal 


4. 


P to Q 4th 4, 


Any move [row 


5. 


Mate in two moves 






(a) If 2, 


Any move on KB file 


3. 


Q to Q 7th 3, 


K moves to royal row 


4. 


P to Q Kt 3rd 4, 


Any move 


5. 


Mate in two moves 






"3, 


K other than to royal 


4. 


P to Q 4th 4, 


Any move [row 


5. 


Mate in two moves 





Of course, by " royal row " is meant the 
row on which the king originally stands at the 
beginning of a game. Though, if Black plays 
badly, he may, in certain positions, be mated 
in fewer moves, the above provides for every 
variation he can possibly bring about. 

351.— THE CRUSADER. 





White. 




Black. 


I. 


Kt to QB 3rd 


I. 


P to Q 4th 


2. 


Kt takes QP 


2. 


Kt to QB 3rd 


3. 


Kt takes KP 


3. 


P to KKt 4th 


4. 


Kt takes B 


4. 


Kt to KB 3rd 


s. 


Kt takes P 


.5. 


Kt to K 5th 


6. 


Kt takes Kt 


6. 


Kt to B 6th 


7. 


Kt takes Q 


7. 


R to KKt sq 


8. 


Kt takes BP 


8. 


R to KKt 3rd 


Q. 


Kt takes P 


9. 


R to K 3rd 


10. 


Kt takes P 


ID. 


Kt to Kt 8th 


II. 


Kt takes B 


II. 


R to R 6th 


12. 


Kt takes R 


12. 


P to Kt 4th 


13. 


Kt takes P (ch) 


13. 


K to B 2nd 



SOLUTIONS. 



233 



Q. 


Kt to R 4 


10. 


Kt to Kt 6 


11. 


Kt takes R 


12. 


Kt to Kt 6 


IS. 


Kt takes B 


14. 


Kt to Q 6 


15. 


Q to K sq 


16. 


Kt takes Q 



White. Black. 

14. Kt takes P 14. K to Kt 3rd 

15. Kt takes R 15. K to R 4th 
i6» Kt takes Kt 16. K to R 5th 

White now mates in three moves. 

17. P to Q 4th 17- K to R 4th 

18. Q to Q 3rd 18. K moves 

19. Q to KR 3rd 

; (mate) If 17, K to Kt 5th 

i8.\ P to K 4th (dis. 18, K moves 

ch) 
19. P to KKt 3rd 
(mate) 

The position after the sixteenth move, with the 
mate in three moves, was first given by S. Loyd 
in Chess Nuts. 

352.— IMMOVABLE PAWNS. 

1. Kt to KB 3 

2. Kt to KR 4 

3. Kt to Kt 6 

4. Kt takes R 

5. Kt to Kt 6 

6. Kt takes B 

7. K takes Kt 

8. Kt toQB 3 
17. K takes Kt, and the position is reached. 

Black plays precisely the same moves as 
White, and therefore we give one set of moves 
only. The above seventeen moves are the 
fewest possible. 

353.— THIRTY-SIX MATES. 

Place the remaining eight White pieces thus : 
K at KB 4th, Q at QKt 6th, R at Q 6th, R at 
KKt 7th, B at Q 5th, B at KR 8th, Kt at QR 
5th, and Kt at QB 5th. The following mates 
can then be given : — 

By discovery from Q . . . . 8 

By discovery from R at Q 6th . 13 

By discovery from B at R 8th . 11 

Given by Kt at R 5th . . . . 2 

Given by pawns 1 2 

Total 36 

Is it possible to construct a position in which 
more than thirty-six different mates on the 
move can be given ? So far as I know, nobody 
has yet beaten my arrangement. 

354.— AN AMAZING DILEMMA. 

Mr. Black left his king on his queen's knight's 
7th, and no matter what piece White chooses 
for his pawn. Black cannot be checkmated. 
As we said, the Black king takes no notice of 
checks and never moves. White may queen 
his pawn, capture the Black rook, and bring 
his three pieces up to the attack, but mate is 
quite impossible. The Black king cannot be 
left on any other square without a checkmate 
being possible. 

The late Sam Loyd first pointed out the 
peculiarity on which this puzzle is based. 



355.~CHECKMATE ! 

Remove the White pawn from B 6th to K 4th 
and place a Black pawn on Black's KB 2nd. 
Now, White plays P to K 5th, check, and Black 
must play P to B 4th. Then White plays P, 
takes P en passant, checkmate. This was 
therefore White's last move, and leaves the 
position given. It is the only possible solution. 

356.— QUEER CHESS. 




If you place the pieces as follows (where only 
a portion of the board is given, to save space), 
the Black king is in check, with no possible 
move open to him. The reader will now see 
why I avoided the term " checkmate," apart 
from the fact that there is no White king. The' 
position is impossible in the game of chess, 
because Black could not be given check by 
both rooks at the same time, nor could he have 
moved into check on his last move. 

I believe the position was first published by 
the late S. Loyd. 

357.— ANCIENT CHINESE PUZZLE. 
Play as follows : — 



1. R-Q 6 

2. K— R 7 

3. R (R 6)- 



-B 6 (mate). 



Black's moves are forced, so need not be given. 

358.— THE SIX PAWNS. 

The general formula for six pawns on all squares 
greater than 2^ is this : Six times the square of the 
number of combinations of n things taken three 
at a time, where n represents the number of 
squares on the side of the board. Of course, where 
n is even the unoccupied squares in the rows and 
columns will be even, and where n is odd the 
number of squares will be odd. Here n is 8, 
so the answer is 18,816 different ways. This 
is " The Dyer's Puzzle " {Canterbury Puzzles, 
No. 27) in another form. I repeat it here in 
order to explain a method of solving that will 
be readily grasped by the novice. First of all, 
it is evident that if we put a pawn on any line, 
we must put a second one in that line in order 
that the remainder may be even in number. 
We cannot put four or six in any row without 
making it impossible to get an even number 
in all the columns interfered with. We have, 
therefore, to put two pawns in each of three 
rows and in each of three columns. Now, 
there are just six schemes or arrangements 
that fulfil these conditions, and these are shown 
in Diagrams A to F, inclusive, on next page. 



234 



AMUSEMENTS IN MATHEMATICS. 



I will just remark in passing that A and B 
are the only distinctive arrangements, because, 
if you give A a quarter-turn, you get F ; and if 
you give B three quarter-turns in the direction 
that a clock hand moves, you will get successively 
C, D, and E. No matter how you may place 



(3, 






1 


-f-F- 






« 


_L 





yoiir six pawns, if you have complied with the 
conditions of the puzzle they will fall under 
one of these arrangements. Of course it will 
be understood that mere expansions do not 
destroy the essential character of the arrange- 
ments. Thus G is only an expansion of form 
A. The solution therefore consists in finding 
the number of these expansions. Supposing 
we confine our operations to the first three 
rows, as in G, then with the pairs a and b placed 
in the first and second columns the pair c may 
be disposed in any one of the remaining six 
colimm.s, and so give six solutions. Now slide 
pair b into the third column, and there are five 
possible positions for c. Slide b into the fourth 
column, and c may produce four new solutions. 
And so on, until (still leaving a in the first 
column) you have 6 in the seventh column, 
and there is only one place for c — ^in the eighth 
colunm. Then you may put a in the second 
column, 6 in the third, and c in the fourth, and 
start sliding c and 6 as before for another series 
of solutions. 

We find thus that, by using form A alone 
and confining our operations to the three top 
rows, we get as many answers as there are 
combinations of 8 things taken 3 at a time. 

This is ^21221^=56. And it will at once strike 

1x2x3 
the reader that if there are 56 different ways of 
selecting the columns, there must be for each 
of these ways just 56 ways of selecting the rows, 
for we may simultaneously work that " sliding " 
process downwards to the very bottom in ex- 
actly the same way as we have worked from 
left to right. Therefore the total number of 
ways in which form A may be applied is 56X 
56=3,136. But there are, as we have seen, 
six arrangements, and we have only dealt with 
one of these, A. We must, therefore, multiply 
this result by 6, which gives us 3,136x6= 
18,816, which is the total number of ways, as 
we have already stated. 

359.— COUNTER SOLITAIRE. 

Play as follows : 3 — 11, 9 — 10, i — 2, 7 — 15. 
8 — 16, 8—7, 5—13. 1—4, 8—5, 6—14, 3—8, 



6 — 3, 6 — 12, I — 6, I — 9, and all the counters 
will have been removed, with the exception of 
No. I, as required by the conditions. 

360.— CHESSBOARD SOLITAIRE. 

Play as follows : 7 — 15, 8 — 16, 8 — 7, 2 — 10, 
1—9, 1—2, 5—13, 3—4, 6—3, II— I, 14—8, 
6—12, 5—6, 5— II, 31—23, 32—24, 32—31, 
26 — 18, 25 — 17, 25 — 26, 22 — 32, 14 — 22, 29 — 
21, 14 — 29, 27 — 28, 30 — 27, 25—14, 30—20, 
25 — 30, 25 — 5. The two counters left on the 
board are 25 and 19 — both belonging to the 
same group, as stipulated — and 19 has never 
been moved from its original place. 

I do not think any solution is possible in 
which only one counter is left on the board. 

361.— THE MONSTROSITY. 



I. 
2. 
3- 

4. 
5. 
6. 

7- 
8. 

9. 
10. 
II. 
12. 

13. 
14. 

15. 
16. 

17. 
18. 
19. 
20. 
21. 
22. 

23. 
24. 

25. 
26. 

27. 
28. 
29. 
30. 
31. 
32. 
33. 
34. 
35. 
36. 
37. 
38. 

39- 
40. 
41. 
42. 
43- 



White. 
P to KB 4 
K to B 2 
K to K3 
P to B 5 
Q to K sq 
Q to Kt 3 
Q to Kt 8 
Kt to KB 3 

Kt to K 5 
Q takes B 
P takes R 
R to R 4 
R to Q 4 
P to QKt 4 
Kto B 4 
Q to K8 
Kt to B 3, ch 
B to R 3 
R to Kt sq 
R to Kt 2 
K to Kt 5 
QtoRs 
P to Kt 5 
P to Kt 6 
P takes R 
P to B 8 (R) 
B to Q 6 
K to Kt 6 
Rto R8 
P to R4 

P to R5 
P takes Q 
P takes Q 
K to B 7 
Kto K8 
PtoB6 

P to B 7 
P to B 8 (B) 
B to Kt 8 
K toQ 8 
P takes Kt (R) 
K to B 7 
Q to B 7, ch 



Black. 
P to QB 3 
Q to R4 
K to Q sq 
K to B 2 
K to Kt 3 
Kt to QR 3 
P to KR 4 
Rto R 3 
R to Kt 3 
R to Kt 6, ch 
K to Kt 4 
Pto B 3 
P takes Kt 
P takes R, ch 
P to R 5 
P to R6 
P takes Kt 
Pto R7 
P to R 8 (Q) 
P takes R 
Q to KKt 8 

Kto R 5 

R to B sq 

R to B 2 

P to Kt 8 (B) 

Q to B 2 

Kt to Kt 5 

KtoR6 

K to Kt 7 

Q (Kt 8) to Kt 3 

KtoB8 

KtoQ8 

Kto K8 

Kt to KR 3, ch 

B to R 7 

B to Kt sq 

K takes B 

Kt to Q 4 

Kt to B 3, ch 

Kt to K sq 

Kt to B 2, ch 

Kt to Q sq 

K to Kt 8 



And the position is reached. 



The order of the moves is immaterial, and 
this order may be greatly varied. But, al- 



SOLUTIONS. 



235 



though many attempts have been made, nobody 
has succeeded in reducing the number of my 
moves. 

362.— THE WASSAIL BOWL. 

The division of the twelve pints of ale can be 
made in eleven manipulations, as below. The 
six columns show at a glance the quantity of 
ale in the barrel, the five-pint jug, the three- 
pint jug, and the tramps X, Y, and Z respec- 
tively after each manipulation. 



Barrel. 

7 
7 
7 
7 
4 
o 
o 
o 
o 
o 
o 



5-pint. 

■ 5 
2 
O 

■ 3 
. 3 

■ 3 
. 5 
. 5 

2 
o 
o 



3-pint. 
O 

3 
3 
o 

3 
3 
z 
o 
3 
3 
o 



X. 

o 
o 

2 
2 
2 
2 
2 
2 
2 
4 
4 



And each man has received his four pints of ale. 

363.— THE DOCTOR'S QUERY. 

The mixture of spirits of wine and water is in 
the proportion of 40 to i, just as in the other 
bottle it was in the proportion of i to 40. 

364.— THE BARREL PUZZLE. 




All that is necessary is to tilt the barrel as in 
Fig. I, and if the edge of the surface of the water 
exactly touches the lip a at the same time that 
it touches the edge of the bottom h, it will be 
just half full. To be more exact, if the bottom 
is an inch or so from the ground, then we can 
allow for that, and the thickness of the bottom, 
at the top. If when the surface of the water 
reached the lip a it had risen to the point c in 
Fig. 2, then it would be more than half full. If, 
as in Fig. 3, some portion of the bottom were 
visible and the level of the water fell to the 
point d, then it would be less than half full. 



This method applies to all symmetrically con- 
structed vessels. 

365.— NEW MEASURING PUZZLE. 

The following solution in eleven manipulations 
shows the contents of every vessel at the start 
and after every manipulation : — 



quart. 


lo-quart. 


S-quart 4-quart 


ID 


10 


.. 


5 . 


10 


5 .. 


5 . 


10 


I .. 4 


9 


ID 


I .. 


9 


6 . 


. I .. 4 


9 


7 . 


. .. 4 


9 


7 . 


. 4 .. 


9 


3 


• 4 .. 4 


9 


3 . 


. 5.-3 


9 


8 


. .. 3 


4 


8 


. 5.-3 


4 


10 


. 3 .. 3 



366.— THE HONEST DAIRYMAN. 

Whatever the respective quantities of milk 
and water, the relative proportion sent to 
London would always be three parts of water 
to one of milk. But there are one or two 
points to be observed. There must originally 
be more water than milk, or there will be no 
water in A to double in the second transaction. 
And the water must not be more than three 
times the quantity of milk, or there will not 
be enough liquid in B to effect the second trans- 
action. The third transaction has no effect 
on A, as the relative proportions in it must be 
the same as after the second transaction. It 
was introduced to prevent a quibble if the 
quantity of milk and water were originally the 
same ; for though double " nothing " would 
be " nothing," yet the third transaction in 
such a case could not take place. 

367.— WINE AND WATER. 

The wine in small glass was one-sixth of the 
total liquid, and the wine in large glass two- 
ninths of total. Add these together, and we 
find that the wine was seven-eighteenths of 
total fluid, and therefore the water eleven- 
eighteenths. 

368.— THE KEG OF WINE. 

The capacity of the jug must have been a little 
less than three gallons. To be more exact, it 
was 2.93 gallons. 

369.— MIXING THE TEA. 

There are three ways of mixing the teas. 
Taking them in the order of quality, 2S. 6d., 
2S. 3d., IS. 9d,, mix 16 lbs., i lb., 3 lbs. ; or 14 lbs., 
4 lbs., 2 lbs. ; or 12 lbs., 7 lbs., 1 lb. In every 
case the twenty pounds mixture should be worth 
2S. 4^. per pound ; but the last case requires the 
smallest quantity of the best tea, therefore it 
is the correct answer. 



236 



AMUSEMENTS IN MATHEMATICS. 



370.— A PACKING PUZZLE. 

On the side of the box, 14 by 22^, we can 
arrange 13 rows containing alternately 7 and 
6 balls, or 85 in all. Above this we can place 
another layer consisting of 12 rows of 7 and 6 
alternately, or a total of 78. In the length of 
24^ inches 15 such layers may be packed, 
the alternate layers containing 85 and 78 balls. 
Thus 8 times 85 added to 7 times 78 gives us 
1,226 for the full contents of the box. 

371.— GOLD PACKING IN RUSSIA. 

The box should be 100 inches by 100 inches by 
II inches deep, internal dimensions. We can 
lay flat at the bottom a row of eight slabs, 
lengthways, end to end, which will just fill one 
side, and nine of these rows will dispose of 
seventy-two slabs (all on the bottom), with a 
space left over on the bottom measuring 100 
inches by i inch by i inch. Now make eleven 
depths of such seventy-two slabs, and we have 
packed 792, and have a space 100 inches by i 
inch by II inches deep. In this we may exactly 
pack the remaining eight slabs on edge, end to 
end. 

372.— THE BARRELS OF HONEY. 

The only way in which the barrels could be 
equally divided among the three brothers, so 
that each should receive his 3^ barrels of honey 
and his 7 barrels, is as follows : — 



A 
B 
C 



There is one other way in which the division 
could be made, were it not for the objection 
that all the brothers made to taking more than 
four barrels of the same description. Except 
for this difficulty, they might have given B his 
quantity in exactly the same way as A above, 
and then have left C one full barrel, five half- 
full barrels, and one empty barrel. It will 
thus be seen that in any case two brothers 
would have to receive their allowance in the 
same way. 

373.— CROSSING THE STREAM. 

First, the two sons cross, and one returns. 
Then the man crosses and the other son returns. 
Then both sons cross and one returns. Then 
the lady crosses and the other son returns. 
Then the two sons cross and one of them returns 
for the dog. Eleven crossings in all. 

It would appear that no general rule can be 
given for solving these river-crossing puzzles. 
A formula can be found for a particular case 
(say on No. 375 or 376) that would apply to any 
number of individuals under the restricted con- 
ditions ; but it is not of much use, for some little 
added stipulation will entirely upset it. As in 
the case of the measuring puzzles, we generally 
have to rely on individual ingenuity. 



Full. 


Half-full. 


Empty. 


3 


I 


3 


2 


3 


2 


2 


3 


2 



374.— CROSSING THE RIVER 
AXE. 

Here is the solution : — 



"5 .... 


. .. ( T \ . . . 


...G T8 3 
. . .G T 8 <i 


5 . . . . 


...Jg 3V. . 


T T 8 


53.. 

5 3.- 

J 5 


...r G V... 


.... J X 

.... T T 8 


... (t a... 


...G 8 
...G 8 
T ^ 


J 5 .... 


^1 3|--- 
- . . /g 8 ^ . . . 


G 8 


..•Ti «,!-... 


■ ••• 

T 9 


G 8 


y 5/ 

...-It t\ . . . 


eo 


JT8 




« 


JT8 




D 

< 


G T8 3 .. 


|G 3) 


« 


GTS 3 .. 


...{J 5)... 





G, J, and T stand for Giles, Jasper, and 
Timothy ; and 8, 5, 3, for £800, £500, and £300 
respectively. The two side columns represent 
the left bank and the right bank, and the middle 
column the river. Thirteen crossings are neces- 
sary, and each line shows the position when 
the boat is in mid-stream during a crossing, 
the point of the bracket indicating the direc- 
tion. 

It will be found that not only is no person left 
alone on the land or in the boat with more than 
his share of the spoil, but that also no two 
persons are left with more than their joint 
shares, though this last point was not insisted 
upon in the conditions. 

375.— FIVE JEALOUS HUSBANDS. 

It is obvious that there must be an odd number 
of crossings, and that if the five husbands had 
not been jealous of one another the party might 
have all got over in nine crossings. But no 
wife was to be in the company of a man or men 
tmless her husband was present. This entails 
two more crossings, eleven in all. 

The following shows how it might have been 
done. The capital letters stand for the hus- 
bands, and the small letters for their respective 
wives. The position of affairs is shown at the 
start, and after each crossing between the left 
bank and the right, and the boat is represented 
by the asterisk. So you can see at a glance that 
a, b, and c went over at the first crossing, that 
b and c returned at the second crossing, and 
so on. 



SOLUTIONS. 



237 



ABCDE abcde * 



I. 

2. 
3. 
4- 
5- 
6. 

7. 

8. 

9. 
10. 
II. 



ABCDE 
ABCDE 
ABCDE 
ABCDE 
DE 
CDE 



de 

bcde 

e 

de 

de 

cde 

cde 

bcde 

e 

be e 



* 

# 

* 
* 
* 



* ABC 
AB 

* ABCDE 
ABCDE 

* ABCDE 
ABCDE 

* ABCDE 



abc 
a 

abed 

abc 

abc 

ab 

ab 

a 

abed 

a d 

abcde 



There is a little subtlety concealed in the 
words " show the quickest way," 

Everybody correctly assumes that, as we are 
told nothing of the rowing capabilities of the 
party, we must take it that they all row equally 
well. But it is obvious that two such persons 
should row more quickly than one. 

Therefore in the second and third crossings 
two of the ladies should take back the boat to 
fetch d, not one of them only. This does not 
affect the number of landings, so no time is lost 
on that account. A similar opportunity occurs 
in crossings 10 and 11, where the party again 
had the option of sending over two ladies or 
one only. 

To those who think they have solved the 
puzzle in nine crossings I would say that in 
every case they will find that they are wrong. 
No such jealous husband would, in the circum- 
stances, send his wife over to the other bank 
to a man or men, even if she assured him that 
she was coming back next time in the boat. If 
readers will have this fact in mind, they will at 
once discover their errors. 

376.— THE FOUR ELOPEMENTS. 

If there had been only three couples, the island 
might have been dispensed with, but with four 
or more couples it is absolutely necessary in 
order to cross under the conditions laid down. 
It can be done in seventeen passages from land 
to land (though French mathematicians have 
declared in their books that in such circum- 
stances twenty- four are needed), and it cannot 
be done in fewer. I will give one way. A, B, 
C, and D are the young men, and a, b, c, and d 
are the girls to whom they are respectively en- 
gaged. The three columns show the positions 
of the different individuals on the lawn, the 
island, and the opposite shore before starting 
and after each passage, while the asterisk indi- 
cates the position of the boat on every occasion. 
Having found the fewest possible passages, 
we should consider two other points in deci(£ng 
on the " quickest method " : Which persons 
were the most expert in handling the oars, and 
which method entails the fewest possible delays 
in getting in and out of the boat ? We have 
no data upon which to decide the first point, 
though it is probable that, as the boat belonged 
to the girls' household, they would be capable 
oarswomen. The other point, however, is im- 



Lawn. 



ABCD abed 

ABCD cd 

ABCD 

ABCD 

ABCD 

CD 

BCD 

BCD 

BCD 

D 

D 

D 

B D 



bed 

d 

cd 

cd 

cd 



d 
d 
d 
d 
d 
d 
d 
d 
cd 



Island. 



be • 
b 
b 
b 

bed ♦ 
be 
be 
abc * 
b 
b 
b 
be * 



Shore. 



ab 

a 

a 

a 

a 

a 

a 

a 

a 



AB 

A 

A 

A 

ABC 

ABC 

ABC a e 

AC a e 

ABCD a c 

ABCD a 

ABCD abc 

ABCD ab 

ABCD abed 



portant, and in the solution I have given (where 
the girls do 8-i3ths of the rowing and A and D 
need not row at all) there are only sixteen 
gettings-in and sixteen gettings-out. A man 
and a girl are never in the boat together, and 
no man ever lands on the island. There are 
other methods that require several more ex- 
changes of places. 

377.— STEALING THE CASTLE 
TREASURE. 

Here is the best answer, in eleven manipula 
tions : — 

Treasure down. 

Boy down — treasure up. 

Youth down — boy up. 

Treasure down. 

Man down — youth and treasure up. 

Treasure down. 

Boy down — treasure up. 

Treasure down. 

Youth down — ^boy up. 

Boy down — treasure up. 

Treasure down. 

378.— DOMINOES IN PROGRESSION. 

There are twenty-three different ways. You 
may start with any domino, except the 4 — 4. 
and those that bear a 5 or 6, though only cer- 
tain initial dominoes may be played either way 
round. If you are given the common differ- 
ence and the first domino is played, you have 
no option as to the other dominoes. Therefore 
all I need do is to give the initial domino for all 
the twenty-three ways, and state the common 
difference. This I will do as follows : — 

With a common difference of i, the first 
domino may be either of these : o — o, o — i, 
I — o, o — 2, I — I, 2 — o, o — 3, I — 2, 2 — I, 3 — o, 
o — 4, I — 3, 2 — 2, 3 — I, I — 4, 2 — 3, 3 — 2, 2—4, 
3 — 3, 3 — 4. With a difference of 2, the first 
domino may be o — o, o — 2, or o — i. Take the 
last case of all as an example. Having played 
the o — I, and the difference being 2, we are 



238 



AMUSEMENTS IN MATHEMATICS. 



compelled to continue with i — 2, 2 — 3, 3 — 4, 
4 — 5, 5 — 6. There are three dominoes that 
can never be used at all. These are o — 5, o — 6, 
and X — 6. If we used a box of dominoes ex- 
tending to 9 — 9, there would be forty different 
ways. 

379.— THE FIVE DOMINOES. 

There are just ten different ways of arranging 
the dominoes. Here is one of them : — 

(2—0) (0—0) (o— I) (1—4) (4—0). 
I will leave my readers to find the remaining 
nine for themselves. 

380.— THE DOMINO FRAME PUZZLE. 



It 






THE 

DOMiNO FRAME 
PUZZLE 

SOLUTION 






The illustration is a solution. It will be found 
that all four sides of the frame add up 44. The 
sum of the pips on all the dominoes is 168, and 
if we wish to make the sides sum to 44, we must 
take care that the four comers sum to 8, be- 
cause these corners are counted twice, and 168 
added to 8 will equal 4 times 44, which is neces- 
sary. There are many different solutions. 
Even in the example given certain interchanges 
are possible to produce different arrangements. 
For example, on the left-hand side the string 
of dominoes from 2 — 2 down to 3 — 2 may be 
reversed, or from 2 — 6 to 3 — 2, or from 3 — o 
to 5 — 3. Also, on the right-hand side we may 
reverse from 4 — 3 to 1 — 4. These changes 
will not affect the correctness of the solution. 

381.— THE CARD FRAME PUZZLE. 

The sum of all the pips on the ten cards is 55. 
Suppose we are trying to get 14 pips on every 
side. Then 4 times 14 is 56. But each of the 
four comer cards is added in twice, so that 55 
deducted from 56, or i, must represent the sum 
of the four corner cards. This is clearly im- 
possible ; therefore 14 is also impossible. But 
suppose we came to tr5ang 18. Then 4 times 
18 is 72, and if we deduct 55 we get 17 as the 
sum of the comers. We need then only try 



different arrangements with the four corners 
always summing to 17, and we soon discover 
the following solution : — 







♦ ♦♦ 


♦ 




♦ 4 


♦ ♦ 


4 






♦ ♦ 



The final trials are very limited in number, 
and must with a little judgment either bring 
us to a correct solution or satisfy us that a 
solution is impossible under the conditions we 
are attempting. The two centre cards on the 
upright sides can, of course, always be inter- 
changed, but I do not call these different solu- 
tions. If you reflect in a mirror you get 
another arrangement, which also is not con- 
sidered different. In the answer given, how- 
ever, we may exchange the 5 with the 8 and 
the 4 with the i. This is a different solution. 
There are two solutions with 18, four with 19, 
two with 20, and two with 22 — ten arrange- 
ments in all. Readers may like to find all 
these for themselves. 

382.— THE CROSS OF CARDS. 

There are eighteen fundamental arrangements, 
as follows, where I only give the numbers in 
the horizontal bar, since the remainder must 
naturally fall into their places. 



6 8 
6 7 



It will be noticed that there must always be 
an odd number in the centre, that there are 
four ways each of adding up 23, 25, and 27, but 
only three ways each of summing to 24 and 
26. 



SOLUTIONS. 



239 



383.— THE "T" CARD PUZZLE. 

If we remove the ace, the remaining cards may 
be divided into two groups (each adding up 
alike) in four ways ; if we remove 3, there are 
three ways ; if 5, there are four ways ; if 7, 
there are three ways ; and if we remove 9, 
there are four ways of making two equal groups. 
There are thus eighteen different ways of group- 
ing, and if we take any one of these and keep 
the odd card (that I have called " removed ") 
at the head of the column, then one set of 
numbers can be varied in order in twenty-four 
ways in the column and the other four twenty- 
four ways in the horizontal, or together they may 
be varied in 24x24=576 ways. And as there 
are eighteen such cases, we multiply this num- 
ber by 18 and get 10,368, the correct number 
of ways of placing the cards. As this number 
includes the reflections, we must divide by 2, 
but we have also to remember that every hori- 
zontal row can change places with a vertical 
row, necessitating our multipl5nng by 2 ; so 
one operation cancels the other. 

384.— CARD TRIANGLES. 

The following arrangements of the cards show 
(i) the smallest possible sum, 17 ; and (2) the 
largest possible, 23. 



I 

9 6 

4 8 

3752 



7 

4 2 

3 6 

9518 



It will be seen that the two cards in the middle 
of any side may always be interchanged without 
affecting the conditions. Thus there are eight 
ways of presenting every fundamental arrange- 
ment. The number of fundamentals is eighteen, 
as follows : two summing to 17, four summing 
to 19, six summing to 20, four summing to 21, 
and two summing to 23. These eighteen funda- 
mentals, multiplied by eight (for the reason 
stated above), give 144 as the total number of 
different wa3rs of placing the cards. 

385.— "STRAND" PATIENCE. 

The reader may find a solution quite easy in 
a little over 200 moves, but, surprising as it 
may at first appear, not more than 62 moves 
are required. Here is the play : By " 4 C up " 
I mean a transfer of the 4 of clubs with all the 
cards that rest on it. i D on space, 2 S on 
space, 3 D on space, 2 S on 3 D, i H on 2 S, 2 C 
on space, i D on 2 C, 4 S on space, 3 H on 4 S 
(9 moves so far), 2 S up on 3 H (3 moves), 5 H 
and 5 D exchanged, and 4 C on 5 D (6 moves), 

3 D on 4 C (i), 6 S (with 5 H) on space (3), 

4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space 
(i), 6 C up on 7 D (3), 8 S on space (i), 7 H on 

8 S (i), 8 C on 9 D (i), 7 H on 8 C (i), 8 S on 

9 H (1), 7 H on 8 S (i), 7 D up on 8 C (5), 4 C up 
on 5 D (9), 6 S up on 7 H (3), 4 S up on 5 H (7) = 
62 moves in all. This is my record; perhaps 
the reader can beat it. 



386.— A TRICK WITH DICE. 

All you have to do is to deduct 250 from the 
result given, and the three figures in the answer 
will be the three points thrown with the dice. 
Thus, in the throw we gave, the number given 
would be 386 ; and when we deduct 250 we get 
136, from which we know that the throws were 
I, 3, and 6. 

The process merely consists in giving looa-H 
10&-I-C -1-250, where a, b, and c represent the 
three throws. The result is obvious. 



387.— THE VILLAGE CRICKET 
MATCH. 





N\T. JJiVTixKins >j> — * 


1 


I 


<=nii^ 






^H^^^- 




IVl-r. Liif+<y »-* 


1 




-^ " 


2 






___ — 



^..^IMr. Stru^^les 



The diagram No. i will show that as neither 
Mr. Podder nor Mr. Dumkins can ever have 
been within the crease opposite to that from 
which he started, Mr. Dumkins would score 
nothing by his performance. Diagram No. 2 
will, however, make it clear that since Mr. 
Luffey and Mr. Struggles have, notwithstanding 
their energetic but careless movements, con- 
trived to change places, the manoeuvre must 
increase Mr. Struggles's total by one run. 



388.— SLOW CRICKET. 

The captain must have been " not out " and 
scored 21. Thus : — 



2 

4 

I 

3 

_i 

II 



men (each Ibw) . 
men (each caught) . 
man (run out) 
men (each bowled) . 
man (captain — ^not out) . 



19 

17 

o 

9 
21 

66 



The captain thus scored exactly 15 more than 
the average of the team. The " others " who 
were bowled could only refer to three men, as 
the eleventh man would be " not out." The 
reader can discover for himself why the captain 
must have been that eleventh man. It would 
not necessarily follow with any figures. 



240 



AMUSEMENTS IN MATHEMATICS. 



389.— THE FOOTBALL PLAYERS. 

The smallest possible number of men is seven. 
They could be accounted for in three different 
ways : i. Two with both arms sound, one with 
broken right arm, and four with both arms 
broken. 2. One with both arms soxmd, one 
with broken left arm, two with broken right 
arm, and three with both arms broken, 3. Two 
with left arm broken, three with right arm 
broken, and two with both arms broken. But 
if evp«ry man was injured, the last case is the 
only one that would apply. 

300.— THE HORSE-RACE PUZZLE. 

Tm «inswer is: £12 on Acorn, £15 on Blue- 
bottle, £20 on Capsule. 

391.— THE MOTOR-CAR RACE. 

The first point is to appreciate the fact that, 
in a race round a circular track, there are the 
same number of cars behind one as there are 
before. All the others are both behind and 
before. There were thirteen cars in the race, 
including Gogglesmith's car. Then one-third 
of twelve added to three-quarters of twelve will 
give us thirteen — the correct answer. 

392.— THE PEBBLE GAME. 

In the case of fifteen pebbles, the first player 
wins if he first takes two. Then when he holds 
an odd number and leaves i, 8, or 9 he wins, 
and when he holds an even number and leaves 

4, 5, or 12 he also wins. He can always do 
one or other of these things until the end of 
the game, and so defeat his opponent. In the 
case of thirteen pebbles the first player must 
lose if his opponent plays correctly. In fact, 
the only numbers with which the first player 
ought to lose are 5 and multiples of 8 added to 

5, such as 13, 21, 29, etc. 

393— THE TWO ROOKS. 

The second player can always win, but to en- 
sure his doing so he must always place his rook, 
at the start and on every subsequent move, on 
the same diagonal as his opponent's rook. He 
can then force his opponent into a comer and 
win. Supposing the diagram to represent the 
positions of the rooks at the start, then, if 
Black played first. White might have placed 
his rook at A and won next move. Any square 
on that diagonal from A to H will win, but the 
best play is always to restrict the moves of the 
opposing rook as much as possible. If White 
played first, then Black should have placed his 
rook at B (F would not be so good, as it gives 
White more scope) ; then if White goes to C, 
Black moves to D ; White to E, Black to F ; 
White to G, Black to C ; White to H, Black to 
I ; and Black must win next move. If at any 
time Black had failed to move on to the same 
diagonal as White, then White could take 
Black's diagonal and wio. 




THE TWO ROOKS. 

394.— PUSS IN THE CORNER. 

No matter whether he plays first or second, the 
player A, who starts the game at 55, must win. 
Assuming that B adopts the very best lines of 
play in order to prolong as much as possible his 
existence, A, if he has first move, can always 
on his 12th move capture B ; and if he has 
the second move, A can always on his 14th 
move make the capture. His point is always 
to get diagonally in line with his opponent, 
and by going to 33, if he has first move, he 
prevents B getting diagonally in line with 
himself. Here are two good games. The nima- 
ber in front of the hyphen is always A's move ; 
that after the hyphen is B's : — 

33-8, 32-15, 31-22, 30-21, 29-14, 22-7, 15-6, 
14-2, 7-3, 6-4, II-, and A must capture on his 
next (12th) move, -13, 54-20, 53-27, 52-34, 
51-41, 50-34, 42-27, 35-20, 28-13, 21-6, 14-2, 
7-3, 6-4, II-, and A must capture on his next 
(14th) move. 

395.— A WAR PUZZLE GAME. 

The Britisher can always catch the enemy, no 
matter how clever and elusive that astute 
individual may be ; but curious though it may 
seem, the British general can only do so after 
he has paid a somewhat mysterious visit to 
the particular town marked " i " in the map, 
going in by 3 and leaving by 2, or entering by 
2 and leaving by 3. The three towns that are 
shaded and have no numbers do not really 
come into the question, as some may suppose, 
for the simple reason that the Britisher never 
needs to enter any one of them, while the 
enemy cannot be forced to go into them, and 
would be clearly ill-advised to do so voluntarily. 
We may therefore leave these out of considera- 
tion altogether. No matter what the enemy 
may do, the Britisher should make the follow- 



SOLUTIONS. 



241 



ing first nine moves : He should visit towns 24, 
20, 19, 15, II, 7, 3, I, 2. If the enemy takes it 
into his head also to go to town i, it will be 
found that he will have to beat a precipitate 
retreat the same way that he went in, or the 




Britisher will infallibly catch him in towns 2 or 
3, as the case may be. So the enemy will be 
wise to avoid that north-west comer of the 
map altogether. 

Now, when the British general has made the 
nine moves that I have given, the enemy will 
be, after his own ninth move, in one of the 
towns marked 5, 8, 11, 13, 14, 16, 19, 21, 24, 
or 27. Of course, if he imprudently goes to 3 
or 6 at this point he will be caught at once. 
Wherever he may happen to be, the Britisher 
" goes foT him," and has no longer any difficulty 
in catching him in eight more moves at most 
(seventeen in all) in one of the following ways. 
The Britisher will get to 8 when the enemy is 
at 5, and win next move ; or he will get to 19 
when the enemy is at 22, and win next move ; 
or he will get to 24 when the enemy is at 27, 
and so win next move. It will be foimd that 
he can be forced into one or other of these 
fatal positions. 

In short, the strategy really amounts to this : 
the Britisher plays the first nine moves that I 
have given, and although the enemy does his 
very best to escape, our general goes after his 
antagonist and always driving him away from 
that north-west comer ultimately closes in with 
him, and wins. As I have said, the Britisher 
never need make more than seventeen moves 
in all, and may win in fewer moves if the enemy 
plays badly. But after playing those first nine 
moves it does not matter even if the Britisher 
makes a few bad ones. He may lose time, but 
cannot lose his advantage so long as he now 
keeps the enemy from town i, and must eventu- 
ally catch him. 

This is a complete explanation of the puzzle. 

It may seem a little complex in print, but in 

practice the winning play will now be quite 

(1,926) 



easy to the reader. Make those nine moves, 
and there ought to be no difiiculty whatever in 
finding the concluding line of play. Indeed, 
it might almost be said that then it is difficult 
for the British general not to catch the enemy. 
It is a question of what in chess we call the 
" opposition," and the visit by the Britisher to 
town I " gives him the jump " on the enemy, 
as the man in the street would say. 

Here is an illustrative example in which the 
enemy avoids capture as long as it is possible 
for him to do so. The Britisher's moves are 
above the line and the enemy's below it. Play 
them alternately. 

24 20 19 15 II 7 3 I 2 6 10 14 18 19 20 24 



13 9 13 17 21 20 24 23 19 15 19 23 24 25 27 

The enemy must now go to 25 or B, in either 
of which towns he is immediately captured. 

396.— A MATCH MYSTERY. 

If you form the three heaps (and are therefore 
the second to draw), any one of the following 
thirteen groupings will give you a win if you 
play correctly : 15, 14, i ; 15, 13, 2 ; 15, 12, 3 ; 
15,11,4; 15,10,5; 15,9.6; 15,8,7; 14,13, 
3 ; 14, II, 5 ; 14, 9, 7 ; 13, ", 6 ; 13, 10, 7 ; 
12, II, 7- 

The beautiful general solution of this problem 
is as follows. Express the number in every 
heap in powers of 2, avoiding repetitions and 
remembering that 2®=!. Then if you so leave 
the matches to your opponent that there is 
an even number of every power, you can win. 
And if at the start you leave the powers even, 
you can always continue to do so throughout 
the game. Take, as example, the last group- 
ing given above — 12, 11, 7. Expressed in 
powers of 2 we have — 

12 = 8 4 — — 

II = 8 — 2 I 

7 = — 4 2 I 



As there are thus two of every power, you must 
win. Say your opponent takes 7 from the 12 
heap. He then leaves — 

5 = — 4 — I 

II = 8 — 2 X 

7 = — 4 2 1 



Here the powers are not all even in number, 
but by taking 9 from the 11 heap you immedi- 
ately restore your winning position, thus — 



5 = 
2 = 

7 = 



4 — I 
— 2 — 
421 



And so on to the end. This solution is quite 



16 



242 



AMUSEMENTS IN MATHEMATICS. 



general, and applies to any nximber of matches 
and any number of heaps. A correspondent 
informs me that this puzzle game was first 
propoimded by Mr. W, M. F. Mellor, but when 
or where it was published I have not been able 
to ascertain. 

397.— THE MONTENEGRIN DICE GAME. 

The players should select the pairs 5 and 9, 
and 13 and 15, if the chances of winning are to 
be quite equal. There are 216 different ways 
in which the three dice may fall. They may 
add up 5 in 6 different ways and 9 in 25 different 
ways, making 31 chances out of 216 for the 
player who selects these numbers. Also the 
dice may add up 13 in 21 different ways, and 
15 in 10 different ways, thus giving the other 
player also 31 chances in 216. 

398.— THE CIGAR PUZZLE. 

Not a single member of the club mastered this 
puzzle, and yet I shall show that it is so simple 
that the merest child can understand its solu- 
tion — when it is pointed out to him ! The 
large majority of my friends expressed their 
entire bewilderment. Many considered that 
" the theoretical result, in any case, is deter- 
mined by the relationship between the table 
and the cigars ; " others, regarding it as a 
problem in the theory of Probabilities, arrived 
at the conclusion that the chances are sHghtly 
in favour of the first or second player, as the 
case may be. One man took a table and a 
cigar of particular dimensions, divided the 
table into equal sections, and proceeded to 
make the two players fill up these sections so 
that the second player should win. But why 
should the first player be so accommodating ? 
At any stage he has only to throw down a cigar 
obhquely across severai of these sections en- 
tirely to upset Mr. 2's calculations ! We have 
to assume that each player plays the best 
possible ; not that one accommodates the other. 

The theories of some other friends would be 
quite sound if the shape of the cigar were that 
of a torpedo — ^perfectly symmetrical and pointed 
at both ends. 

I will show that the first player should in- 
falhbly win, if he always plays in the best 
possible manner. Examine carefully the fol- 
lowing diagram, No. i, and all will be clear. 





1 




v 


^ 


«^. 




.r^-* 


c 


'^^ 


.^ 


^ 




The first player must place his first cigar on 
end in the exact centre of the table, as indicated 



by the little circle. Now, whatever the second 
player may do throughout, the first player must 
always repeat it in an exactly diametrically 
opposite position. Thus, if the second player 
places a cigar at A, I put one at AA ; he places 
one at B, I put one at BB ; he places one at C, 
I put one at CC ; he places one at D, I put one 
at DD ; he places one at E, I put one at EE ; 
and so on until no more cigars can be placed 
without touching. As the cigars are supposed 
to be exactly aUke in every respect, it is per- 
fectly clear that for every move that the second 
player may choose to make, it is possible 
exactly to repeat it on a line drawn through 
the centre of the table. The second player 
can always duplicate the first player's move, 
no matter where he may place a cigar, or 
whether he places it on end or on its side. As 
the cigars are all ahke in every respect, one will 
obviously balance over the edge of the table 
at precisely the same point as another. Of 
coiurse, as each player is supposed to play in 
the best possible manner, it becomes a matter 
of theory. It is no valid objection to say that 
in actual practice one would not be sufficiently 
exact to be sure of winning. If as the first 
player you did not win, it would be in conse- 
quence of your not having played the best 
possible. 

The second diagram will serve to show why 
the first cigar must be placed on end. (And 
here I will say that the first cigar that I selected 
from a box I was able so to stand on end, and 
I am allowed to assume that all the other cigars 
would do the same.) If the first cigar were 
placed on its side, as at F, then the second 
player could place a cigar as at G — as near as 
possible, but not actually touching F. Now, 
in this position you cannot repeat his play on 
the opposite side, because the two ends of the 
cigar are not ahke. It will be seen that GG, 
when placed on the opposite side in the same 
relation to the centre, intersects, or Hes on 
top of, F, whereas the cigars are not allowed 
to touch. You must therefore put the cigar 
farther away from the centre, which would 
result in your having insufl&cient room between 
the centre and the bottom left-hand comer to 
repeat everything that the other player would 
do between G and the top right-hand comer. 
Therefore the result would not be a certain 
win for the first player. 

399.— THE TROUBLESOME EIGHT. 



4t 


8 


Zi 


3 


5 


T 


ri 


z 


51 



The conditions were to place a different number 
in each of the nine cells so that the three rows, 



SOLUTIONS. 



243 



three columns, and two diagonals should each 
add up 15. Probably the reader at first set 
himself an impossible task through reading 
into these conditions something which is not 
there — a common error in puzzle-solving. If 
I had said " a different figure," instead of " a 
different number," it would have been quite 
impossible with the 8 placed anywhere but in 
a comer. And it would have been equally 
impossible if I had said " a different whole 
number." But a number may, of course, be 
fractional, and therein lies the secret of the 
puzzle. The arrangement shown in the figure 
will be found to comply exactly with the condi- 
tions : all the numbers are different, and the 
square adds up 15 in all the required eight ways. 

400.— THE MAGIC STRIPS. 

There are of course six different places between 
the seven figures in which a cut may be made, 
and the secret lies in keeping one strip intact 
and cutting each of the other six in a different 
place. After the cuts have been made there 
are a large number of ways in which the thirteen 
pieces may be placed together so as to form a 
magic square. Here is one of them : — 



1 


2 


3 


4 


5 


6 


r 


3 


4 


5 


6 


7 


1 


2 


5 


& 


7 


1 


2 


3 


4 



71 1 1 2 3 4 5 6 1 



2 


3 


4 5 6 


7 1 


4 


5 


6 7 1 


2 3 


6 


7 


1 2 3 


4 5 



The arrangement has some rather interesting 
features. It will be seen that the uncut strip 
is at the top, but it will be found that if the 
bottom row of figures be placed at the top the 
numbers will still form a magic square, and that 
every successive removal from the bottom to 
the top (carrying the uncut strip stage by stage 
to the bottom) will produce the same result. If 
we imagine the numbers to be on seven complete 
perpendicular strips, it will be found that these 
columns could also be moved in succession 
from left to right or from right to left, each 
time producing a magic square. 

401.— EIGHT JOLLY GAOL-BIRDS. 

There are eight waj^ of forming the magic 
square — all merely different aspects of one 



fundamental arrangement. Thus, if you give 
our first square a quarter turn you will get the 
second square ; and as the four sides may be in 
turn brought to the top, there are four aspects. 
These four in turn reflected in a mirror produce 
the remaining four aspects. Now, of these eight 
arrangements only four can possibly be reached 
under the conditions, and only two of these 
four can be reached in the fewest possible moves, 
which is nineteen. These two arrangements 





are shown. Move the men in the following 
order : 5, 3, 2, 5, 7, 6, 4, i, 5, 7, 6, 4, i, 6, 4, 8, 
3, 2, 7, and you get the first square. Move them 
thus : 4, I, 2, 4, X, 6, 7, i, 5, 8, t, 5, 6, 7, 5, 6, 4, 
2, 7, and you have the arrangement in the second 
square. In the first case every man has moved, 
but in the second case the man numbered 3 has 
never left his cell. Therefore No. 3 must be the 
obstinate prisoner, and the second square must 
be the required arrangement. 

402.— NINE JOLLY GAOL BIRDS. 

There is a pitfall set for the unwary in this 
little puzzle. At the start one man is allowed 
to be placed on the shoulders of another, so as 
to give always one empty cell to enable the 
prisoners to move about without any two ever 
being in a cell together. The two imited 
prisoners are allowed to add their numbers 
together, and are, of course, permitted to remain 
together at the completion of the magic square. 
But they are obviously not compelled so to 
remain together, provided that one of the pair 
on his final move does not break the condition 
of entering a cell already occupied. After the 
acute solver has noticed this point, it is for him 




to determine which method is the better one — 
for the two to be together at the coimt or to 
separate. As a matter of fact, the puzzle can 
be solved in seventeen moves if the men are 
to remain together ; but if they separate at the 



16 a 



244 



AMUSEMENTS IN MATHEMATICS. 



end, they may actually save a move and per- 
form the feat in sixteen ! The trick consists 
in placing the man in the centre on the back 
of one of the corner men, and then working 
the pair into the centre before their finsJ 
separation. 

Here are the moves for getting the men into 
one or other of the above two positions. The 
numbers are those of the men in the order in 
which they move into the cell that is for the time 
being vacant. The pair is shown in brackets : — 

Place 5 on i. Then, 6, g, 8, 6, 4, (6), 2, 4, 9, 3, 

4, 9, (6), 7, 6, I. 
Place 5 on 9. Then, 4, i, 2, 4, 6, (14), 8, 6, i, 7, 

6, I, (14), 3, 4, 9. 
Place 5 on 3. Then, 6, (8), 2, 6, 4, 7, 8, 4, 7, i, 

6, 7, (8), 9, 4, 3. 
Place 5 on 7. Then, 4, (12), 8, 4, 6, 3, 2, 6, 3, 9, 

4, 3, (12), I, 6, 7. 

The first and second solutions produce Dia- 
gram A ; the second and third produce Diagram 
B. There are only sixteen moves in every case. 
Having found the fewest moves, we had to 
consider how we were to make the burdened 
man do as little work as possible. It will at 
once be seen that as the pair have to go into 
the centre before separating they must take at 
fewest two moves. The labour of the burdened 
man can only be reduced by adopting the other 
method of solution, which, however, forces us 
to take another move. 

403.— THE SPANISH DUNGEON. 



12 3 4 

5 6 Y S 
-^ « I * • I ■ I I » •— 

13} l^lgj 



10} 9} 7}4 

6 s: 11 s 
Li 



}h}h. 



This can best be solved by working backwards — 
that is to say, you must first catch your square, 
and then work back to the original position. 
We must first construct those squares which are 
foimd to require the least amount of readjust- 
ment of the numbers. Many of these we know 
cannot possibly be reached. When we have 
before us the most favomrable possible arrange- 
ments, it then becomes a question of careful 
analysis to discover which position can be 
reached in the fewest moves. I am afraid, 
however, it is only after considerable study 
and experience that the solver is able to get 
such a grasp of the various " areas of disturb- 
ance " and methods of circulation that his 
judgment is of much value to him. 

The second diagram is a most favourable 
magic square position. It will be seen that 
prisoners 4, 8, 13, and 14 are left in their 
original cells. This position may be reached 



in as few as thirty-seven moves. Here are the 
moves : 15, 14, 10, 6, 7, 3, 2, 7, 6, 11, 3, 3, 7, 6, 
II, 10, 14, 3, 2, II, 10, 9, 5, I, 6, 10, 9, 5, I, 6, 
10, 9, 5, 2, 12, 15, 3. This short solution will 
probably surprise many readers who may not 
find a way imder from sixty to a hundred 
moves. The clever prisoner was No. 6, who 
in the original illustration will be seen with 
his arms extended calUng out the moves. He 
and No. 10 did most of the work, each chan- 
ging his cell five times. No. 12, the man with 
the crooked leg, was lame, and therefore for- 
tunately had only to pass from his cell into the 
next one when his time came round. 

404.— THE SIBERIAN DUNGEONS. 



I A r^A J^A "A 



In attempting to solve this puzzle it is clearly 
necessary to seek such magic squares as seem 
the most favourable for our purpose, and then 
carefully examine and try them for " fewest 
moves." Of course it at once occurs to us that 
if we can adopt a square in which a certain 
number of men need not leave their original 
cells, we may save moves on the one hand, but 
we may obstruct our movements on the other. 
For example, a magic square may be formed 
with the 6, 7, 13, and 16 unmoved ; but in such 
case it is obvious that a solution is impossible, 
since cells 14 and 15 can neither be left nor 
entered without breaking the condition of no 
two men ever being in the same cell together. 

The following solution in fourteen moves was 
found by Mr. G. Wotherspoon : 8 — 17, 16 — 21, 
6 — 16, 14 — 8, 5 — 18, 4 — 14, 3 — 24, II — 20, 10 — 
19, 2 — 23, 13 — 22, 12 — 6, I — 5, 9 — 13. .As 
this solution is in what I consider the theoretical 
minimum number of moves, I am confident 
that it cannot be improved upon, and on this 
point Mr. Wotherspoon is of the same opinion. 

405.— CARD MAGIC SQUARES. 

Arrange the cards as follows for the three new 
squares : — 



3 


2 


4 


6 


5 


7 


9 8 10 


4 


3 


2 


7 


6 


5 


10 9 8 


2 


4 


3 


5 


7 


6 


8 10 9 



Three aces and one ten are not used. The 
summations of the four squares are thus : 9, 15, 
x8, and 27 — all different, as required. 



SOLUTIONS. 



245 



406.— THE EIGHTEEN DOMINOES. 



• ••,♦ 

...1 •• 


• • 
• 1 • 

• • 


• 


t 


• • • • 

• ••' « 


• • • 


• • 


« • • 


• • 

« 1 

• • 


t 1 


• 

• • • 


• • • • • 

• • •' • • 


1 * 


• • • '••• 


• 




1 


• • • • 

• 1 • 



The illustration explains itself. It will be found 
that the pips in every column, row, and long 
diagonal add up 18, as required. 

407.— TWO NEW MAGIC SQUARES. 

Here are two solutions that fulfil the condi- 
tions : — 



sue.TR/ 


ACTING 


f- 


DIVIDING 


n 


4 


»4 


13 




56 


8 


54 


27 


16 


7 


1 


2 


216 


IZ 


/ 


2. 


6 


5 


3 


\Z 


6 


3 


4 


7^ 


9 


10 


S 


15 


9 


18 


24- 


10^ 



The first, by subtracting, has a constant 8, 
and the associated pairs all have a difference 
of 4. The second square, by dividing, has a 
constant 9, and all the associated pairs produce 
3 by division. These are two remarkable and 
instructive squares. 

408.— MAGIC SQUARES OF TWO 
DEGREES. 

The following is the square that I constructed. 
As it stands the constant is 260. If for every 
number you substitute, in its allotted place, 
its square, then the constant will be 11,180. 
Readers can write out for themselves the second 
degree square. 

The main key to the solution is the pretty 
law that if eight numbers sum to 260 and their 
squares to 11,180, then the same will happen 
in the case of the eight numbers that are com- 
plementary to 65. Thus i-f-i8 + 23-|-26+3i-f- 



48-1-56-1-57=260, and the sum of their squares 
is 11,180. Therefore 644-47-1-424- 39 -f 34 -f- 
17 + 9 + 8 (obtained by subtracting each of the 
above numbers from 65) will sum to 260 and 
their squares to 11,180. Note that in every 



r 


53 


41 


27 


2 


52 


4S 


30 


12 


S& 


3S 


24 


la 


63 


35 


n 


5] 


1 


29 


47 


54 


8 


2& 


42 


64 


14 


1^ 


3fc 


57 


11 


23 


31 


25 


43 


55 


5 


32 


46 


50 


4 


22 


40 


GO 


10 


]9 


35 


61 


15 


45 


31 


3 


49 


44 


26 


6 


5fc 


34 


20 


ie> 


62 


39 


21 


9 


59 



one of the sixteen smaller squares the two 
diagonals sum to 65. There are four columns 
and four rows with their complementary 
columns and rows. Let us pick out the niun- 
bers fotmd in the 2nd, ist, 4th, and 3rd rows 
and arrange them thus : — 

1 8 28 29 42 47 51 54 

2 7 27 30 41 48 52 53 

3 6 26 31 44 45 49 56 

4 5 25 32 43 46 50 55 

Here each column contains four consecutive 
numbers cyclically arranged, four running in one 
direction and four in the other. The numbers in 
the 2nd, 5th, 3rd, and 8th columns of the square 
may be similarly grouped. The great dilB&culty 
lies in discovering the conditions governing 
these groups of numbers, the pairing of the 
complementaries in the squares of four and 
the formation of the diagonals. But when a 
correct solution is shown, as above, it discloses 
all the more important keys to the mystery. 
I am inclined to think this square of two 
degrees the most elegant thing that exists in 
magics. I beheve such a magic square cannot 
be constructed in the case of any order lower 
than 8. 

409.--THE BASKETS OF PLUMS. 

As the merchant told his man to distribute the 
contents of one of the baskets of plums " among 
some children," it would not be permissible to 
give the complete basketful to one child ; and 
as it was also directed that the man was to give 
" plums to every child, so that each should 
receive an equal number," it would also not be 
allowed to select just as many children as there 
were plums in a basket and give each child a 
single plum. Consequently, if the number o( 



246 AMUSEMENTS I]^ 

plums in every basket was a prime number, 
then the man would be correct in saying that 
the proposed distribution was quite impossible. 
Our puzzle, therefore, resolves itself into form- 
ing a magic square with nine different prime 
numbers. 

A 3 



r 


61 


43 


73 


37 


1 


31 


13 


67 



83 


29 


101 


89 


71 


53 


41 

_ _. J 


113 


59 



D 



103 


79 


37 


7 


73 


J 39 


109 


67 


43 



1669 


m 


1249 


619 


J039 


ms 


829 


m9 


409 



In Diagram A we have a magic square in 
prime numbers, and it is the one giving the 
smallest constant sum that is possible. As to 
the little trap I mentioned, it is clear that 
Diagram A is barred out by the words " every 
basket contained plums," for one plum is not 
plums. And as we were referred to the baskets, 
" as shown in the illustration," it is perfectly 
evident, without actually attempting to count 
the plums, that there are at any rate more than 
7 plums in every basket. Therefore C is also, 
strictly speaking, barred. Numbers over 20 and 
under, say, 250 would certainly come well within 
the range of possibility, and a large number of 
arrangements would come within these limits. 
Diagram B is one of them. Of comse we can 
allow for the false bottoms that are so fre- 
quently used in the baskets of fruitseUers to 
make the basket appear to contain more fruit 
than it really does. 

Several correspondents assumed (on what 
groimds I cannot think) that in the case of this 
problem the numbers cannot be in consecutive 
arithmetical progression, so I give Diagram D 
to show that they were mistaken. The num- 
bers are 199, 409, 619, 829, 1,039, 1,249, i,459, 
1,669, and 1,879 — all primes with a common 
difference of 210. 

410.— THE MANDARIN'S "T" PUZZLE. 

There are many different ways of arranging 
the numbers, and either the 2 or the 3 may be 
omitted from the "T" enclosure. The ar- 
rangement that I give is a " nasik " square. 
Out of the total of 28,800 nasik squares of the 
fifth order this is the only one (with its one 
reflection) that fulfils the " T " condition. 
This puzzle was suggested to me by Dr. C. 
Planck. ' 



MATHEMATICS. 




1 i 


19 


23 


11 


5 


7 [ 
13 


1 


10 


It 


24 


22 


14 


3 


6 


20 1 


S 


16 


25 


12 


4. 


15 


2 


9 


IS 


21 



THE MANDARIN S 



PUZZLE. 



411. 



-A MAGIC SQUARE OF COM- 
POSITES. 



The problem really amounts to finding the 
smallest prime such that the next higher prime 
shall exceed it by 10 at least. If we write out 
a little list of primes, we shall not need to exceed 
150 to discover what we require, for after 1131 
the next prime is 127. We can then form the 
square in the diagram, where every number is < 
composite. This is the solution in the smallest 
nmnbers. We thus see that the answer is arrived ' 
at quite easily, in a square of the third order, 
by trial. But I propose to show how we may 
get an answer (not, it is true, the one in smallest 
numbers) without any tables or trials, but in a 
very direct and rapid manner. 



m 


114 


119 


116 


118 


1^0 


117 


122 


115 



First write down any consecutive numbers, 
the smallest being greater than i — say, 2, 3, 4, 
5, 6, 7, 8, 9, 10. The only factors in these 
numbers are 2, 3, 5, and 7, We therefore mul- 



SOLUTIONS. 



247 



tiply these four numbers together and add the 
product, 210, to each of the nine numbers. The 
result is the nine consecutive composite num- 
bers, 212 to 220 inclusive, with which we can 
form the required square. Every number will 
necessarily be divisible by its difference from 
210. It will be very obvious that by this 
method we may find as many consecutive compo- 
sites as ever we please. Suppose, for example, 
we wish to form a magic square of sixteen 
such numbers ; then the numbers 2 to 17 
contain the factors 2, 3, 5, 7, 11, 13, and 17, 
which, multiplied together, make 5 105 10 to be 
added to produce the sixteen numbers 5 105 12 
to 510527 inclusive, all of which are composite 
as before. 

But, as I have said, these are not the answers 
in the smallest numbers: for if we add 523 to 
the numbers i to 16, we get sixteen consecutive 
composites ; and if we add 1,327 to the numbers 
I to 25, wa get twenty-five consecutive com- 
posites, in each case the smallest numbers 
possible. Yet if we required to form a magic 
square of a hundred such numbers, we should 
find it a big task by means of tables, though by 
the process I have shown it is quite a simple 
matter. Even to find thirty-six such numbers 
you will search the tables up to 10,000 without 
success, and the difficulty increases in an ac- 
celerating ratio with each square of a larger 
order. 

412.— THE MAGIC KNIGHT'S TOUR. 



46 



4^ 



5it!S2Q.hk 



sz 



31 



15 



55>^A 



ISJifT 



ir>^53M 



1649 



301 1 



19 



k 



50 



5S 



12 



33 



511429643926 



%1 



6 



59 



Zft 



61 



28 



36 



S 



11 



Z3 



3762 



35 



Here each successive number (in numerical 
order) is a knight's move from the preceding 
number, and as 64 is a knight's move from i, 
the tour is " re-entrant." All the columns and 
rows add up 260. Unfortunately, it is not a 
perfect magic square, because the diagonals are 
incorrect, one adding up 264 and the other 
256 — requiring only the transfer of 4 from 
one diagonal to the other. I think this is the 
best result that has ever been obtained (either 
re-entrant or not), and nobody can yet say 
whether a perfect solution is possible or im- 
possible. 



413.— A CHESSBOARD 


FALLACY. 








X 
















\ 














\ 
















\ 














\ 




B 












\, 




B 












V 














V, 


\, 














^ 










A 




\ 


s. 










A 




^ 


s.. 














\ 


■i. 














\ 
















\ 
















\ 
















N 

















The explanation of this little fallacy is as 
follows. The error lies in assuming that the 
little triangular piece, marked C, is exactly the 
same height as one of the little squares of the 
board. As a matter of fact, its height (if we 
make the sixty-four squares each a square inch) 
will be i^ in. Consequently the rectangle is 
really gf in. by 7 in., so that the area is sixty- 
four square inches in either case. Now, su- 
though the pieces do fit together exactly to 
form the perfect rectangle, yet the directions of 
the horizontal lines in the pieces will not co- 
incide. The new diagram above will make 
everything quite clear to the reader. 

414.— WHO WAS FIRST ? 

Biggs, who saw the smoke, would be first ; 
Carpenter, who saw the bullet strike the water, 
would be second ; and Anderson, who heard the 
report, would be last of all. 

415.— A WONDERFUL VILLAGE. 

When the sun is in the horizon of any place 
(whether in Japan or elsewhere), he is the length 
of half the earth's diameter more distant from 
that place than in his meridian at noon. As the 
earth's semi-diameter is nearly 4,000 miles, the 
sun must be considerably more than 3,000 miles 
nearer at noon than at his rising, there being no 
valley even the hundredth part of 1,000 miles 
deep. 

416.— A CALENDAR PUZZLE. 

The first day of a century can never fail on a 
Sunday ; nor on a Wednesday or a Friday. 

417.— THE TIRING-IRONS. 

I WILL give my complete working of the solution, 
so that readers may see how easy it is when 
you know how to proceed. And first of all, as 
there is an even number of rings, I will say that 
they may all be taken off in one-third of 
(2»-f-i— 2) moves ; and since n in our case is 14, 
all the rings may be taken off in 10,922 moves. 
Then I say 10,922 — 9,999=923, and proceed to 
find the position when only 923 out of the 
10,922 moves remain to be made. Here is the 
curious method of doing this. It is based on 



248 



AMUSEMENTS IN MATHEMATICS. 



the binary scale method used by Monsieur L. 
Gros, for an accoimt of which see W. W. Rouse 
Ball's Mathematical Recreations. 

Divide 923 by 2, and we get 461 and the re- 
mainder I ; divide 461 by 2, and we get 230 and 
the remainder i ; divide 230 by 2, and we get 
115 and the remainder nought , Keep on dividing 
by 2 in this way as long as possible, and all the 
remainders will be found to be i, i, i, o, o, i, 1, 
o, I, I, the last remainder being to the left and 
the first remainder to the right. As there are 
foiurteen rings and only ten figures, we place 
the difference, in the form of four noughts, in 
brackets to the left, and bracket all those 
figures that repeat a figure on their left. Then 
we get the following arrangement : (0000)1 
(i i) o (o) I (i) o I (i). This is the correct 
answer to the puzzle, for if we now place rings 
below the line to represent the figures in 
brackets and rings on the line for the other 
figures, we get the solution in the required 
form, as below : — 

00 



0000 



00 



This is the exact position of the rings after 
the 9,999th move has been made, and the 
reader will find that the method shown will 
solve any similar question, no matter how many 
rings are on the tiring-irons. But in working 
the inverse process, where you are required to 
ascertain the number of moves necessary in 
order to reach a given position of the rings, the 
rule will require a little modification, because 
it does not necessarily follow that the position 
is one that is actually reached in course of 
taking off all the rings on the irons, as the reader 
will presently see. I will here state that where 
the total number of rings is odd the number of 
moves required to take them all off is one- 
third of (2»+i-i). 

With n rings (where n is odd) there are 2" 
positions counting all on and all off. In ^ 
(2"+^ + 2) positions they are all removed. The 
number of positions not used is ^ (2"— 2). 

With n icings (where n is even) there are 2" 
positions counting all on and all off. In ^ 
(2"+^ + i) positions they are all removed. The 
number of positions not used is here ^ (2"— i). 

It will be convenient to tabulate a few 
cases. 



No. of 


Total 


Positions 


Positions 


Rings. 


Positions. 


used. 


not used. 


I 


2 


2 





3 


8 


6 


2 


5 


32 


22 


10 


7 


128 


86 


42 


9 


512 


342 


170 


2 


4 


3 


I 


4 


16 


II 


5 


6 


64 


43 


21 


8 


256 


171 


85 


10 

I 


1024 


683 


341 



Note first that the number of positions used i 
is one more than the number of moves required I 
to take all the rings off, because we are including k 
" all on " which is a position but not a move. 
Then note that the number of positions not useii 
is the same as the number of moves used to 
take off a set that has one ring fewer. For r 
example, it takes 85 moves to remove 7 rings/;', 
and the 42 positions not used are exactly the f 
number of moves required to take off a set ofi 
6 rings. The fact is that if there are 7 rings 
and you take off the first 6, and then wish to 
remove the 7th ring, there is no course open 
to you but to reverse all those 42 moves that 
never ought to have been made. In other . 
words, you must replace all the 7 rings on the* 
loop and start afresh ! You ought first to haveii 
taken off 5 rings, to do which you should have% 
taken off 3 rings, and previously to that i ring.! 
To take off 6 you first remove 2 and then 4 rings. / 

418.— SUCH A GETTING UPSTAIRS. i 

Number the risers in regular order upwards, , 
I to 8. Then proceed as follows : i (step back ( 
to floor), I, 2, 3 (2), 3, 4, 5 (4), 5, 6, 7 (6), 7, 8, 
landing (8), landing. The steps in brackets,! 
are taken in a backward direction. It will thus 1 
be seen that by returning to the floor after thet 
first step, and then always going three steps i 
forward for one step backward, we perform the I 
required feat in nineteen steps. 



419.— THE FIVE PENNIES. 




First lay three of the pennies in the way shown 
in Fig. I. Now hold the remaining two pennies 
in the position shown in Fig, 2, so that they 
touch one another at the top, and at the base 
are in contact with the three horizontally 
placed coins. Then the five pennies will be 
equidistant, for every penny will touch every 
other penny. 

420.— THE INDUSTRIOUS BOOKWORM. 

The hasty reader will assume that the book- 
worm, in boring from the first to the last page 



SOLUTIONS. 



249 



of a book in three volumes, standing in their 
proper order on the shelves, has to go through 
all three volumes and four covers. This, in our 
case, would mean a distance of gi in., which 
is a long way from the correct answer. You 
will find, on examining any three consecutive 
volumes on your shelves, that the first page of 
Vol. I. and the last page of Vol. III. are actually 
the pages that are nearest to Vol, II., so that 
the worm would only have to penetrate four 
covers (together, ^ in.) and the leaves in the 
second volume (3 in.), or a distance of si inches, 
in order to tunnel from the first page to the 
last. 

421.— A CHAIN PUZZLE. 

To open and rejoin a link costs threepence. 
Therefore to join the nine pieces into an end- 
less chain would cost 2S. 3d., whereas a new 
chain would cost 2s. 2d. But if we break up 
the piece of eight links, these eight will join 
together the remaining eight pieces at a cost 
of 2s. But there is a subtle way of even im- 
proving on this. Break up the two pieces con- 
taining three and four links respectively, and 
these seven will join together thel remaining 
seven pieces at a cost of only is. gd. 

422.— THE SABBATH PUZZLE. 

The way the author of the old poser proposed 
to solve the difficulty was as follows : From the 
Jew's abode let the Christian and the Turk set 
out on a tour round the globe, the Christian 
going due east and the Turk due west. Readers 
of Edgar Allan Poe's story. Three Sundays in a 
Week, or of Jules Verne's Round the World in 
Eighty Days, will know that such a proceeding 
will result in the Christian's gaining a day and 
in the Turk's losing a day, so that when they 
meet again at the house of the Jew their 
reckoning will agree with his, and all three may 
keep their Sabbath on the same day. The cor- 
rectness of this answer, of comrse, depends on 
the popular notion as to the definition of a day 
— the average duration between successive sun- 
rises. It is an old quibble, and quite sound 
enough for puzzle purposes. Strictly speaking, 
the two travellers ought to change their reckon- 
ings on passing the i8oth meridian ; otherwise 
we have to admit that at the North or South 
Pole there would only be one Sabbath ip seven 
years. 

423.— THE RUBY BROOCH. 

In this case we were shown a sketch of the 
brooch exactly as it appeared after the four 
rubies had been stolen from it. The reader 
was asked to show the positions from which 
the stones " may have been taken ; " for it is 
not possible to show precisely how the gems 
were originally placed, because there are many 
such ways. But an important point was the 
statement by Lady Littlewood's brother : " I 
know the brooch well. It originally contained 
forty-five stones, and there are now only forty- 
one. Somebody has stolen four rubies, and then 
reset as small a number as possible in such a 



way that there shall always be eight stones in 
any of the directions you have mentioned." 




The diagram shows the arrangement before 
the robbery. It will be seen that it was only 
necessary to reset one ruby — the one in the 
centre. Any solution involving the resetting 
of more than one stone is not in accordance 
with the brother's statement, and must there- 
fore be wrong. The original arrangement was, 
of course, a little imsymmetrical, and for thv 
reason the brooch was described as " rather 
eccentric," 

424.— THE DOVETAILED BLOCK. 




The mystery is made clear by the illustration. 
It will be seen at once how the two pieces slide 
together in a diagonal direction. 

425.— JACK AND THE BEANSTALK. 

The serious blunder that the artist made in 
this drawing was in depicting the tendrils ^f 



IBiO 



AMUSEMENTS IN MATHEMATICS. 




the bean climbing spirally as at A above, 
whereas the French bean, or scarlet runner, 
the variety clearly selected by the artist in 
the absence of any authoritative information 
on the point, always climbs as shown at B. 
Very few seem to be aware of this curious 
little fact. Though the bean always insists 
on a sinistrorsal growth, as B, the hop prefers 
to chmb in a dextrorsal manner, as A. Why, 
is one of the mysteries that Nature has not 
yet unfolded. 

426.— THE HYMN-BOARD POSER. 

This puzzle is not nearly so easy as it looks at 
first sight. It was required to find the smallest 
possible number of plates that would be neces- 
sary to form a set for three hjonn-boards, each 
of which would show the five hymns sxmg at 
any particular service, and then to discover 
the lowest possible cost for the same. The 
hymn-book contains 700 hymns, and therefore 
no higher number than 700 could possibly be 
needed. 

Now, as we are required to use every legiti- 
mate and practical method of economy, it 
should at once occur to us that the plates 
must be painted on both sides ; indeed, this 
is such a common practice in cases of this 
kind that it would readily occur to most solvers. 
We shoTild also remember that some of the 
figures may possibly be reversed to form other 
figures ; but as we were given a sketch of the 
actual shapes of these figures when painted on 
the plates, it would be seen that though the 6's 
may be turned upside down to make 9's, none 
of the other figures can be so treated. 

It will be found that in the case of the figures 
I, 2, 3, 4, and 5, thirty-three of each will be 
required in order to provide for every possible 
emergency ; in the case of 7, 8, and o, we can 
only need thirty of each ; while in the case of 
the figure 6 (which may be reversed for the 



figure 9) it is necessary to provide exactly forty- 
two. 

It is therefore clear that the total number of 
figures necessary is 297 ; but as the figures are 
painted on both sides of the plates, only 149 
such plates are required. At first it would 
appear as if one of the plates need only have a 
number on one side, the other side being left 
blank. But here we come to a rather subtle point 
in the problem. 

Readers may have remarked that in real life 
it is sometimes cheaper when making a purchase 
to buy more articles than we require, on the 
principle of a reduction on taking a quantity : 
we get more articles and we pay less. Thus, if 
we want to buy ten apples, and the price asked 
is a penny each if bought singly, or ninepence a 
dozen, we should both save a penny and get 
two apples more than we wanted by buying the 
full twelve. In the same way, since there is a 
regular scale of reduction for plates painted 
alike, we actually save by having two figures 
painted on that odd plate. Supposing, for 
example, that we have thirty plates painted 
alike with 5 on one side and 6 on the other. 
The rate wo\ild be 4|d., and the cost irs. lo^d. 
But if the odd plate with, say, only a 5 on one 
side of it have a 6 painted on the other side, 
we get thirty-one plates at the reduced rate of 
4^d., thus saving a farthing on each of the 
previous thirty, and reducing the cost of the 
last one from is. to 4id. 

But even after these points are all seen there 
comes in a new difficulty : for although it will be 
found that all the 8's may be on the backs of 
the 7's, we cannot have all the 2's on the backs 
of the I's, nor all the 4's on the backs of the 3's, 
etc. There is a great danger, in our attempts 
to get as many as possible painted aUke, of our 
so adjusting the figures that some particular 
combination of hynms cannot be represented. 

Here is the solution of the difficulty that was 
sent to the vicar of Chumpley St. Winifred. 
Where the sign X is placed between two figmres, 
it implies that one of these figures is on one side 
of the plate and the other on the other side. 



31 plates painted 5 



30 

21 

21 

12 

12 

12 

8 

I 

I 




149 plates @ 6d. each 



: 3 14 6 
£7 19 I 



Of course, if we could increase the number 
of plates, we might get the painting done for 
nothing, but such a contingency is prevented 
by the condition that the fewest possible plates 
must be provided. 



SOLUTIONS. 



251 



This puzzle appeared in Tit-Bits, and the 
following remarks, made by me in the issue for 
nth December 1897, may be of interest. 

The " Hymn-Board Poser " seems to have 
created extraordinary interest. The immense 
nimiber of attempts at its solution sent to me 
from all parts of the United Kingdom and from 
several Continental countries show a very kind 
disposition amongst our readers to help the 
worthy vicar of Chumpley St. Winifred over 
Ms parochial difficulty. Every conceivable esti- 
mate, from a few shillings up to as high a sum 
as £1,347, 1 OS., seems to have come to hand. But 
the astonishing part of it is that, after going 
carefully through the tremendous pile of corre- 
spondence, I find that only one competitor has 
succeeded in maintaining the reputation of the 
Tit-Bits solvers for their capacity to solve any- 
thing, and his solution is substantially the same 
as the one given above, the cost being identical. 
Some of his figures are differently combined, 
but his grouping of the plates, as shown in the 
first coliimn, is exactly the same. Though a 
large majority of competitors clearly hit upon 
all the essential points of the puzzle, they com- 
pletely collapsed in the actual arrangement of 
the figures. According to their methods, some 
possible selection of h5nnns, such as in, 112, 
121, 122,211, cannot be set up. A few corre- 
spondents suggested that it might be possible 
so to paint the 7's that upside down they would 
appear as 2's or 4's ; but this would, of course, 
be barred out by the fact that a representation 
of the actual figures to be used was given. 

427.— PHEASANT-SHOOTING. 

The arithmetic of this puzzle Is very easy indeed. 
There were clearly 24 pheasants at the start. 
Of these 16 were shot dead, i was wounded in 
the wing, and 7 got away. The reader may have 
concluded that the answer is, therefore, that 
*' seven remained." But as they flew away it 
is clearly absurd to say that they " remained." 
Had they done so they would certainly have 
been killed. Must we then conclude that the 
17 that were shot remained, because the others 
flew away ? No ; because the question was 
not " how many remained ? " but " how many 
^ still remained ? " Now the poor bird that was 
wounded in the wing, though unable to fly, 
was very active in its painful struggles to nm 
away. The answer is, therefore, that the 16 
birds that were shot dead " still remained," or 
" remained still." 



428. 



-THE GARDENER AND THE 
COOK. 



Nobody succeeded in solving the puzzle, so I 
had to let the cat out of the bag — an operation 
that was dimly foreshadowed by the puss in 
the original illustration. But I first reminded 
the reader that this puzzle appeared on April i, 
a day on which none of us ever resents being 
made an "April Fool;" though, as I practically 
" gave the thing away " by specially drawing 
attention to the fact that it was All Fools' Day, 



it was quite remarkable that my correspond- 
ents, without a single exception, fell into the 
trap. 

One large body of correspondents held that 
what the cook loses in stride is exactly made up 
in greater speed ; consequently both advance at 
the same rate, and the result must be a tie. 
But another considerable section saw that, 
though this might be so in a race 200 ft. 
straight away, it could not really be, because 
they each go a stated distance at " every 
bound," and as 100 is not an exact multiple of 
3, the gardener at his thirty-fourth bound will 
go 2 ft. beyond the mark. The gardener will, 
therefore, run to a point 102 ft. straight away 
and return (204 ft. in all), and so lose by 4 ft. 
This point certainly comes into the puzzle. 
But the most important fact of all is this, that 
it so happens that the gardener was a pupil 
from the Horticultural College for Lady Gar- 
deners at, if I remember aright, Swanley ; 
while the cook was a very accomphshed French 
chef of the hemale persuasion ! Therefore 
" she (the gardener) made three boimds to 
his (the cook's) two." It will now be found 
that while the gardener is running her 204 ft. 
in 68 bounds of 3 ft., the somewhat infirm old 
cook can only make 45^ of his 2 ft. bounds, 
which equals 90 ft. 8 in. The result is that the 
lady gardener wins the race by 109 ft. 4 in. at 
a moment when the cook is in the air, one-third 
through his 46th boimd. 

The moral of this puzzle is twofold : (i) Never 
take things for granted in attempting to solve 
puzzles ; (2) always remember All Fools' Day 
when it comes round. I was not writing of 
any gardener and cook, but of a particular couple, 
in " a race that I witnessed." The statement 
of the eye-witness must therefore be accepted : 
as the reader was not there, he cannot contra- 
dict it. Of course the information supplied was 
insufficient, but the correct reply was : Assum- 
ing the gardener to be the ' he,' the cook wins 
by 4 ft. ; but if the gardener is the ' she,* then 
the gardener wins by 109 ft. 4 in." This would 
have won the prize. Cmiously enough, one soli- 
tary competitor got on to the tight track, but 
failed to follow it up. He said : "Is this a 
regular April i catch, meaning that they only 
ran 6 ft. each, and consequently the race was 
unfinished ? If not, I think the following must 
be the solution, supposing the gardener to be 
the * he ' and the cook the ' she.* " Though 
his solution was wrong even in the case he sup- 
posed, yet he was the only person who suspected 
the question of sex. 

429.— PLACING HALFPENNIES. 

Thirteen coins may be placed as shown on 
page 252. 

430.— FIND THE MAN'S WIFE. 

There is no guessing required in this puzzle. 
It is all a question of ehmination. If we can 
pair off any five of the ladies with their re- 
spective husbands, other than husband No. 10, 
then the remainini? lady must be No. lo's wife. 



252 



AMUSEMENTS IN MATHEMATICS. 





3 




9 


(^'} 


( 1 


J''"""""^ 


( ^ 


a 


3 j 






c-> 


S 


A ^ y 



PLACING HALFPENNIES. 

I will show how this may be done. No. 8 is 
seen carrying a lady's parasol in the same hand 
with his walking-stick. But every lady is pro- 
vided with a parasol, except No. 3 ; therefore 
No. 3 may be safely said to be the wife of No. 8. 
Then No. 12 is holding a bicycle, and the dress- 
guard and make disclose the fact that it is a 
lady's bicycle. The only lady in a cycling skirt 



is No. 5 ; therefore we conclude that No. 5 is 
No. 12's wife. Next, the man No. 6 has a dog, 
and lady No. 11 is seen carrying a dog chain, 
So we may safely pair No. 6 with No. 11. Then 
we see that man No. 2 is paying a newsboy for 
a paper. But we do not pay for newspapers ia 
this way before receiving them, and the gentle- 
man has apparently not taken one from the 
boy. But lady No. 9 is seen reading a paper. 
The inference is obvious — that she has sent the 
boy to her husband for a penny. We therefore 
pair No. 2 with No. 9. We have now disposed 
of all the ladies except Nos. i and 7, and of all 
the men except Nos. 4 and 10. On looking at 
No. 4 we find that he is carr3dng a coat over 
his arm, and that the buttons are on the left 
side — ^not on the right, as a man wears them. 
So it is a lady's coat. But the coat clearly 
does not belong to No. 1, as she is seen to be 
wearing a coat already, while No. 7 lady is very 
lightly clad. We therefore pair No. 7 lady with 
man No. 4. Now the only lady left is No. i, 
and we are consequently forced to the conclusion 
that she is the wife of No. 10. This is therefore 
the correct answer. 



INDEX. 



Abbot's Puzzle, The, 20, 161. 

Window, The, 87, 213. 

Academic Courtesies, 18, 160. 
Acrostic Puzzle, An, 84, 210. 
Adam and Eve and the Apples, 18. 
Aeroplanes, The Two, 2, 148. 
Age and Kinship Puzzles, 6. 

Concerning Tommy's, 7, 153. 

Mamma's, 7, 152. 

Mrs. Timpkins's, 7, 152. 

Rover's, 7, 152. 

Ages, The Family, 7, 152. 

Their, 7, 152- 

Alcuin, Abbot, 20, 112. 
Almonds, The Nine, 64, 193. 
Amazons, The, 94, 221. 
Andrews, W. S., 125. 
Apples, A Deal in, 3, 149. 

Buying, 6, 151. 

The Ten, 64, 195. 

Approximations in Dissection, 28. 
Arithmetical and Algebraical Problems, i. 

Various, 17. 

Arthur's Knights, King, 'j'jy 203. 
Artillerymen's Dilemma, 26, 167. 
Asparagus, Bundles of, 140. 
Aspects all due South, 137. 
Associated Magic Squares, 120. 
Axiom, A Puzzling, 138. 

Bachet de M^ziriac, 90, 109, 112. 
Bachet's Square, 90, 216. 
Ball Problem, The, 51, 183. 
Ball, W. W. Rouse, 109, 204, 248. 
Balls, The Glass, 78, 204. 
Banker's Puzzle, The, 25, 165. 
Bank Holiday Puzzle, A, 73, 201. 
Banner Puzzle, The, 46, 179. 

St. George's, 50, 182. 

Barrel Puzzle, The, 109, 235. 
Barrels of Balsam, The, 82, 208. 
Beanfeast Puzzle, A, 2, 148. 
Beef and Sausages, 3, 149. 
Beer, The Barrel of, 13, 155. 
Bell-ropes, Stealing the, 49, 181. 
Bells, The Peal of, 78, 204. 
Bergholt, E., 116, 119, 125. 
Betsy Ross Puzzle, The, 40, 176. 
Bicycle Thief, The, 6, 152. 
Bishops — Guarded, 88, 214. 

in Convocation, 89, 215. 

Puzzle, A New, 98, 225. 



Bishops — Unguarded, 88, 214. 
Board, The Chess-, 85. 

in Compartments, The, 102, 228. 

Setting the, 105, 231. 



Boards with Odd Number of Squares, 86, 212. 
Boat, Three Men in a, 78, 204. 
Bookworm, The Industrious, 143, 248. 
Boothby, Guy, 154. 
Box, The Cardboard, 49, 181. 
The Paper, 40. 



Boys and Girls, 67, 197. 

Bridges, The Monk and the, 75, 202. 

Brigands, The Five, 25, 164. 

Brocade, The Squares of, 47, 180. 

Bun Puzzle, The, 35, 170. 

Busschop, Paul, 172. 

Buttons and String Method, 230. 

Cab Numbers, The, 15, 157. 

Calendar Puzzle, A, 142, 247. 

Canterbury Puzzles, The, 14, 28, 58, 117, 121, 

195, 202, 205, 206, 212, 213, 217, 233. 
Card Frame Puzzle, The, 114, 238. 
— — Magic Squares, 123, 244. 

Players, A Puzzle for, 78, 203. 

Puzzle, The " T," 115, 239. 

Triangles, 115, 239. 



Cards, The Cross of, 115, 238. 
Cardan, 142. 
Carroll, Lewis, 43. 

Castle Treasure, Steahng the, 113, 237. 
Cats, the Wizard's, 42, 178. 
Cattle, Judkins's, 6, 151. 
Market, At a, i, 148. 



Census Puzzle, A, 7, 152. 
Century Puzzle, The, 16, 158. 
The Digital, 16, 159. 



Chain Puzzle, A, 144, 249. 

The Antiquary's, 83. 209. 

The Cardboard, 40, 176. 



Change, Giving, 4, 150. 
Ways of giving, 151. 



Changing Places, 10, 154. 
Channel Island, 138. 
Charitable Bequest, A, 2, 148. 
Charity, Indiscriminate, 2, 148. 
Checkmate, 107, 233. 
Cheesemonger, The Eccentric, 66, 196. 
Chequered Board Divisions, 85, 210. 
Cherries and Plums, 56, 189. 
Chess Puzzles, Dynamical, 96. 
Statical, 88. 



254 



AMUSEMENTS IN MATHEMATICS. 



Chess Puzzles, Various, 105. 

Queer, 107, 233. 

Chessboard, The, 85. 

Fallacy, A, 141, 247. 

Guarded, 95. 

Non-attacking Arrangements, 96. 

Problems, 84. 

Sentence, The, 87, 214. 

SoUtaire, 108, 234. 

The Chinese, 87, 213. 

The Crowded, 91, 217. 

Chestnuts, Buying, 6, 152. 
Chinese Money, 4, 150. 

Puzzle, Ancient, 107, 233. 

The Fashionable, 43. 

Christmas Boxes, The, 4, 150. 

Present, Mrs. Snailey's, 46, 179. 

Pudding, The, 43, 178. 

Cigar Puzzle, The, 119, 242. 
Circle, The Dissected, 69, 197. 
Cisterns, How to Make, 54, 188. 
Civil Service " Howler," 154. 
Clare, John, 58. 
Clock Formulae, 154. 

Puzzles, 9. 

The Club, 10, 154. 

The Railway Station, 11, 155. 

Clocks, The Three, 11, 154. 
Clothes Line Puzzle, The, 50, 182. 
Coast, Roimd the, 63, 195. 
Coincidence, A Queer, 2, 148. 
Coins, The Broken, 5, 150. 

The Ten, 57, 190. 

Two Ancient, 140. 

Combination and Group Problems, 76. 
Compasses Puzzle, The, 53, 186. 
Composite Magic Squares, 127, 246. 
Cone Puzzle, The, 55, 188. 
Com, Reaping the, 20, 161. 
Cornfields, Farmer Lawrence's, loi, 227. 
Gostermonger's Puzzle, The, 6, 152. 
Coimter Problems, Moving, 58. 

Puzzle, A New, 98, 225. 

Solitaire, 107, 234. 

Coimters, The Coloured, 91, 217. 

The Forty-nine, 92, 217. 

The Nine, 14, 156. 

The Ten, 15, 156. 

Crescent Puzzle, The, 52, 184. 

Crescents of Byzantium, The Five, 92, 219. 

Cricket Match, The Village, 116, 239. 

Slow, 116, 239, 

Cross and Triangle, 35, 169. 

of Cards, 115, 238, 

The Folded, 35, 169. 

The Southern, 93, 220, 

Crosses, Coimter, 81, 207. 

from One, Two, 35, 168. 

Three, 169. 

Crossing River Problems, 112. 
Crusader, The, 106, 232. 
Cubes, Sums of, 165. 
Cushion Covers, The, 46, 179. 
Cutting-out Puzzle, A, 37, 172. 
Cyclists' Feast, The, 2, 148. 

Dairyman, The Honest, no, 235. 
Definition, A Question of, 23, 163. 



De Fonteney, 112. 

Deified Puzzle, The, 74, 203. 

Delannoy, 112. 

De Morgan, A., 27. 

De Tudor, Sir Edwyn, 12, 155. 

Diabohque Magic Squares, 120. 

Diamond Puzzle, The, 74, 202. 

Dice, A Trick with, 116, 239. 

Game, The Montenegrin, 119, 242, 

Numbers, The, 17, 160. 



Die, Painting the, 84, 210. 
Digital Analysis, 157, 158. 

Division, 16, 158. 

MultipUcation, 15, 156. 

Puzzles, 13. 



Digits, Adding the, 16, 158. 

and Squares, 14, 155. 

Odd and Even, 14, 156. 



Dilemma, An Amazing, 106, 233. 
Diophantine Problem, 164. 
Dissection Puzzle, An Easy, 35, 170. 

Puzzles, 27. 

Various, 35. 



Dividing Magic Squares, 124. 
Division, Digital, 16, 158. 
Simple, 23, 163. 



Doctor's Query, The, 109, 235. 
Dogs Puzzle, The Five, 92, 218. 
Domestic Economy, 5, 151. 
Domino Frame Puzzle, The, 114, 238. 
Dominoes in Progression, 114, 237. 

The Eighteen, 123, 245. 

The Fifteen, 83, 209. 

The Five, 114, 238. 



Donkey Riding, 13, 155. 
Dormitory Puzzle, A, 81, 208. 
Dovetailed Block, The, 145, 249. 
Drayton's Polyolbion, 58. 
Dimgeon Puzzle, A, 97, 224. 
Dungeons, The Siberian, 123, 244. 
The Spanish, 122, 244. 



Dutchmen's Wives, The, 26, 167. 
D3mamical Chess Puzzles, 96. 

t 
Earth's Girdle, The, 139. 
Educational Times Reprints, 204. 
Eggs, A Deal in, 3, 149. 
Obtaining the, 140. 



Election, The Muddletown, 19, 161. 
The Parish Council, 19, 161. 



Eleven, The Mystic, 16, 159. 

Elopements, The Four, 113, 237. 

Elrick, E., 231. 

Engines, The Eight, 61, 194. 

Episcopal Visitation, An, 98, 225. 

Estate, Farmer Wurzel's, 51, 184. 

Estates, The Yorkshire, 51, 183. 

EucUd, 31, 138. 

Euler, L., 165. 

Exchange Puzzle, The, 66, 196. 

Fallacy, A Chessboard, 141, 247. 
Family Party, A, 8, 153, 
Fare, The Passenger's, 13, 155. 
Fajmer and his Sheep, The, 22, 163. 
Fence Problem, A, 21, 162. 
Fences, The Landowner's, 42, 178. 
Fermat, 164, 168. 



INDEX. 



253 



Find the Man's Wife, 147, 251. 
Fly on the Octahedron, The, 70, 198. 
Fog, Mr. Gubbins in a, 18, 161. 
Football Players, The, 116, 240. 
Fraction, A Puzzling, 138. 
Fractions, More Mixed, 16, 159. 
Frame Puzzle, The Card, 114, 238. 

The Domino, 114, 238. 

Frankenstein, E. N., 232. 
Fr6nicle, B., 119, 168. 
Frogs, The Educated, 59, 194. 

The Four, 103, 229. 

The Six, 59, 193. 

Frost, A. H., 120. 

Games, Puzzle, 117. 

Problems concerning, 114. 

Garden, Lady Belinda's, 52, 186. 

Puzzle, The, 49, 182. 

Gardener and the Cook, The, 146, 251, 
Geometrical Problems, 27. 

Puzzles, Various, 49. 

George and the Dragon, St., loi, 227. 
Getting Upstairs, Such a, 143, 248. 
Girdle, the Earth's, 139. 
Goat, The Tethered, 53, 186. 
Grand Lama's Problem, The, 86, 212. 
Grasshopper Puzzle, The, 59, 193. 
Greek Cross Piizzles, 28. 

Three from One, 169. 

Greyhound Puzzle, The, loi, 227. 
Grocer and Draper, The, 5, 151. 
Gros, L., 248. 

Group Problems, Combination and, 76. 
Groups, The Three, 14, 156. 
Guarini, 229. 

Hairdresser's Puzzle, The, 137. 
Halfpennies, Placing, 147, 251. 
Hampton Court Maze solved, 133. 
Hannah's Puzzle, 75, 202. 
Hastings, The Battle of, 23, 164. 
Hatfield Maze solved, 136. 
Hat Puzzle, The, 67, 196. 
Hat-peg Puzzle, The, 93, 221. 
Hats, The Wrong, 78, 203. 
Hay, The Trusses of, 18, 161. 
Heads or Tails, 22, 163. 
Hearthrug, Mrs. Hobson's, 37, 172. 
Helmholtz, Von, 41. 
Honey, The Barrels of, iii, 236. 
Honeycomb Puzzle, The, 75, 202. 
Horse Race Puzzle, The, 117, 240. 
Horseshoes, The Two, 40, 175. 
Houdin, 68. 

Hydroplane Question, The, 12, 155. 
Hymn-board Poser, The, 145, 250. 

Icosahedron Puzzle, The, 70, 198. 

Jack and the Beanstalk, 145, 249. 
Jackson, John, 56. 
Jaenisch, C. F. de, 92. 
Jampots, Arranging the, 68, 197. 
Jealous Husbands, Five, 113, 236. 
Joiner's Problem, The, 36, 171. 

Another, 37, 171. 

Jolly Gaol-Birds, Eight, 122, 243. 



Jolly Gaol-Birds, Nine, 122, 243. 
Journey, The Queen's, 100, 227. 
The Rook's, 96, 224. 



Junior Clerks' Puzzle, The, 4, 150, 
Juvenile Puzzle, A, 68, 197. 

Kangaroos, The Four, 102, 228. 
Kelvin, Lord, 41. 
Kennel Puzzle, The, 105, 231. 
King and the Castles, The, 56, 189. 
The Forsaken, 106, 232. 



Kite-flying Puzzle, A, 54, 187. 
Knight-guards, The, 95, 222. 
Knights, King Arthur's, 77, 203. 

Tour, Magic, 127, 247. 

The Cubic, 103, 229. 

The Four, 103, 229. 



Labosne, a., 25, 90, 216. 

Labourer's Puzzle, The, 18, 160. 

Ladies' Diary, 26. 

Lagrange, J. L., 9. 

Laisant, C. A., 76. 

Lamp-posts, Painting the, 19, 161. 

Leap Year, 155. 

Ladies, The, 19, i6r,. 



Legacy, A Puzzling, 20, 161. 
Legal Difficulty, A, 23, 163. 
Le Plongeon, Dr., 29. 
Letter Block Puzzle, The, 60, 194. 

Blocks, The Thirty-six, 91, 216. 

Puzzle, The Fifteen, 79, 205. 



Level Puzzle, The, 74, 202. 
Linoleum Cutting, 48, 181. 

Puzzle, Another, 49, 181. 



Lion and the Man, The, 97, 224. 
Hunting, 94, 222. 



Lions and Crowns, 85, 212. 
The Four, 88, 214. 



Lockers Puzzle, The, 14, 156. 
Locomotion and Speed Puzzles, 11. 
Lodging-house Difficulty, A, 61, 194. 
London and Wise, 131. 
Loyd, Sam, 8, 43, 44, 98, 144, 232, 233. 
Lucas, Edouard, 16, 76, 112, 121. 
Luncheons, The City, yj, 203. 

MacMahon, Major, 109. 
Magic Knight's Tour, 127, 247. 

Square Problems, 119. 

Card, 123, 244. 

of Composites, 127, 246. 

of Primes, 125. 

of Two Degrees, 125, 245. 

Two New, 125, 245. 



Strips, 121, 243. 

Magics, Subtracting, Multiplying, and Dividing, 

124. 
Maiden, The Languishing, 97, 224. 
Mandarin's Puzzle, The, 103, 230. 

" T " Puzzle, The, 126, 246. 



Marketing, Saturday, 27, 168. 
Market Women, The, 3, 149. 
Mary and Marmaduke, 7, 152. 
Mary, How Old was, 8, 153. 
Massacre of Innocents, 139. 
Match Mystery, A, 118, 241. 
Puzzle, A New, 55, 188. 



256 



AMUSEMENTS IN MATHEMATICS. 



Mates, Thirty-six, io6, 233. 

Mazes and how to thread Them, 127. 

Measuring, Weighing, and Packing Puzzles, 109. 

Puzzle, New, no, 235. 

Meeting, The Suffragists', 19, 161, 
Mellor, W. M. F., 242. 
Menages, Probleme de, 76. 
Mersenne, M., 168. 
Mice, Catching the, 65, 196. 
Milkmaid Puzzle, The, 50, 183. 
Millionaire's Perplexity, The, 3, 149. 
Mince Pies, The Twelve, 57, 191. 
Mine, Inspecting a, 71, 199. 
Miners' Holiday, The, 23, 163. 
Miser, The Converted, 21, 162. 
Mitre, Dissecting a, 35, 170. 
Monad, The Great, 39, 174. 
Money, A Queer Thing in, 2, 148. 

Boxes, The Puzzhng, 3, 149. 

, Pocket, 3, 149. 

Puzzles, I. 

Puzzle, A New, 2, 148. 

, Square, 3, 149. 

Monist, The, 125. 

Monk and the Bridges, The, 75, 202. 

Monstrosity, The, 108, 234. 

Montenegrin Dice Game, The, 119, 242. 

Moreau, 76. 

Morris, Nine Men's, 58. 

Mosaics, A Problem in, 90, 215. 

Mother and Daughter, 7, 152. 

Motor-car Race, The, 117, 240. 

Tour, The, 74, 201. 

Garage Puzzle, The, 62, 195. 

Motorists, A Puzzle for, 73, 201. 
Mouse-trap Puzzle, The, 80, 206. 
Moving Coimter Problems, 58. 
Multiplication, Digital, 15, 156. 

Queer, 15, 157. 

Simple, 23, 163. 

Multiplying Magic Squares, 124. 
Muncey, J. N., 125. 
Murray, Sir James, 44. 

Napoleon, 43, 44. 
Nasik Magic Squares, 120. 
Neighbours, Next-Door, 8, 153. 
Newton, Sir Isaac, 56. 
Nine Men's Morris, 58. 
Notation, Scales of, 149. 
Noughts and Crosses, 58, 117. 
Nouvelles Annates de Mathematiques, 14. 
Number Checks Puzzle, The, 16, 158, 
Numbers, Curious, 20, 162. 
Nuts, The Bag of, 8, 153. 

Observation, Defective, 4, 150. 
Octahedron, The Fly on the, 70, 198. 
Oval, How to draw an, 50, 182. 
Ovid's Game, 58. 

Packing in Russia, Gold, iii, 236. 

Puzzles, Measuring, Weighing, and, 109. 

Puzzle, A, III, 236. 

Pandiagonal Magic Squares, 120. 
Papa's Puzzle, 53, 187. 
Pappus, 53. 
Paradox Party, The, 137. 



Party, A Family, 8, 153. 
Patchwork Puzzles, 46. 

Puzzle, Another, 48, 180. 

The Silk, 34, 168. 



Patience, Strand, 116, 239. 
Pawns, A Puzzle with, 94, 222. 

Immovable, 106, 233. 

The Six, 107, 233. 

The Two, 105, 231. 



Pearls, The Thirty- three, 18, 160. 
Pebble Game, The, 117, 240. 
Pedigree, A Mixed, 8, 153. 
PeUian Equation, 164, 167. 
Pennies, The Five, 143, 248. 
The Twelve, 65, 195. 



Pension, Drawing her, 12, 155. 
Pentagon and Square, The, 37, 172. 
Drawing a, 37. 



Pfeffermann, M,, 125. 
Pheasant-Shooting, 146, 251. 
Philadelphia Maze solved, 137. 
Pierrot's Puzzle, The, 15, 156. 
Pigs, The Seven, 41, 177. 
Planck, C, 220, 246. 
Plane Paradox, 138. 
Plantation Puzzle, A, 57, 189. 
The Burmese, 58, 191. 



Plates and Coins, 65, 195. 
Plums, The Baskets of, 126, 245. 
Poe, E. A., 249. 
Points and Lines Problems, 56. 
Postage Stamps, The Four, 84, 210. 
Post-0£&ce Perplexity, A, i, 148. 
Potato Puzzle, The, 41, 177. 
Potatoes, The Basket of, 13, 155. 
Precocious Baby, The, 139. 
Presents, Buying, 2, 148. 
Prime Magic Squares, 125. 
Printer's Error, A, 20, 162. 
Prisoners, Exercise for, 104, 230. 
The Ten, 62, 195. 



Probabilities, Two Questions in, 5, 150. 
Problems concerning Games, 114. 
Puss in the Comer, 118, 240. 
Puzzle Games, 117, 
Pyramid, Painting a, 83, 208. 
Pyramids, Square and Triangular, 167. 
Pythagoras, 31. 

" Queen, The," 120, 
Queens and Bishop Puzzle, 93, 219. 
The Eight, 89, 215. 



Queen's Journey, The, 100, 227. 
Tour, The, 98, 225. 



Quilt, Mrs. Perkins's, 47, 180. 

Race Puzzle, The Horse-, 117, 240. 
The Motor-car, 117, 240. 



Rackbrane's Little Loss, 21, 163. 
Railway Muddle, A, 62, 194. 

Puzzle, A, 61, 194. 

Stations, The Three, 49, 182. 



Rational Amusement for Winter Evenings, 56. 

Rectangles, Counting the, 105, 232. 

Reiss, M., 58. 

Relationships, Queer, 8, 153. 

Reversals, A Puzzle in, 5, 151. 

River Axe, Crossing the, 112, 236. 



INDEX. 



257 



River Problems, Crossing, 112. 
Rookery, The, 105, 232. 
Rook's Journey, The, 96, 224. 

Tour, The, 96, 223. 

Rooks, The Eight, 88, 214. 

The Two, 117, 240. 

Round Table, The, 80, 205. 

Route Problems, Unicursal and, 68. 

Ruby Brooch, The, 144, 249. 

Sabbath Puzzle, The, 144, 249. 
Sailor's Puzzle, The, 71, 199. 
Sayles, H. A., 125. 
Schoolboys, The Nine, 80, 205. 
Schoolgirls, The Fifteen, 80, 204. 
Scramble, The Great, 19, 161. 
Sculptor's Problem, The, 23, 164. 
Second Day of Week, 139. 
See-Saw Puzzle, The, 22, 163. 
Semi-Nasik Magic Squares, 120. 
Senior and Junior, 140. 
Sevens, The Four, 17, 160. 
Sharp's Puzzle, 230. 
Sheepfold, The, 52, 184. 
Sheep Pens, The Six, 55, 189. 

The Sixteen, 80, 206. 

The Three, 92, 217. 

Those Fifteen, 77, 203, 

Shopping Perplexity, A, 4, 150. 
Shuldham, C. D., 125, 126. 
Siberian Dungeons, The, 123, 244. 
Simpleton, The Village, 11, 155. 
Skater, The Scientific, 100, 226. 
Skeat, Professor, 127. 
Solitaire, Central, 63, 195. 

Chessboard, 108, 234. 

Counter, 107, 234. 

Sons, The Four, 49, 181. 
Spanish Dungeons, The, 122, 244. 

Miser, The, 24, 164. 

Speed and Locomotion Puzzles, 11. 

Average, 11, 155. 

Spiral, Drawing a, 50, 182. 
Spot on the Table, The, 17, 160. 
Square Numbers, Check for, 13. 

Digital, 16, 159. 

of Veneer, The, 39, 175. 

Puzzle, An Easy, 35, 170. 

Squares, A Problem in, 23, 163. 

Circling the, 21, 162. 

Difference of Two, 167. 

Magic, 119. 

Sum of Two, 165, 175. 

The Chocolate, 35, 170. 

Stalemate, 106, 232. 

Stamp-licking, The Gentle Art of, 91, 217. 

Star Puzzle, The, 99, 226. 

Stars, The Eight, 89, 215. 

The Forty-nine, 100, 226. 

Statical Chess Puzzles, 88. 
Sticks, The Eight, 53, 186. 
Stonemason's Problem, The, 25, 165. 
Stop-watch, The, 11, 154. 
Strand Magazine, The, 44, 116, 220. 
Strand Patience, 116, 239. 
Stream, Crossing the, 112, 236. 
Strutt, Joseph, 59. 
Subtracting Magic Squares, 124. 
(1,926) 



Sultan's Army, The, 25, 165, 
Suppers, The New Year's Eve, 3, 149. 
Surname, Find Ada's, 27, 168. 
Swastika, The, 29, 31, 169. 

" T " Card Puzzle, The, 115, 239. 
Table, The Round, 80, 205. 
Table-top and Stools, The, 38, I73. 
Tangram Paradox, A, 43, 178. 
Target, The Cross, 84, 210. 
Tarry, 112. 

Tartaglia, 25, 109, 112. 
Tea, Mixing the, in, 235. 
Telegraph Posts, The, 139. 
Tennis Tournament, A, 78, 203. 
Tetrahedron, Building the, 82, 208. 
Thief, Catching the, 19, 161. 
Thrift, A Study in, 25, 166. 
Thompson, W. H., 232. 
Ticket Puzzle, The Excursion, 5» i5i. 
Time Puzzle, A, 10, 153. 

What was the, 10, 153. 



Tiring Irons, The, 142, 247. 
Tit-Bits, 58, 79, 124, 251. 
Tom Number, The, 20, 162. 
Torpedo Practice, 67, 196. 
Tour, The Cyclists', 71, 199. 

The Grand, 72, 200. 

The Queen's, 98, 225. 

The Rook's, 96, 223. 



Towns, Visiting the, 70, 198. 
Trains, The Two, 11, 155. 
Treasure Boxes, The Nine, 24, 164. 
Trees, The Twenty-one, 57, i90. 
Tr6maux, M., 133, 135. 
Triangle, The Dissected, 38, 173. 
Triangular Numbers, 13, 25, 166. 
Check for, 13. 



Troublesome Eight, The, 121, 242. 
Tube Inspector's Puzzle, The, 69, 198. 
Railway, Heard on the, 8, 153. 



Turks and Russians, 58, 191. 
Turnings, The Fifteen, 70, 198. 
Twickenham Puzzle, The, 60, 194. 
Two Pieces Problem, The, 96. 

Unclassified Puzzles, 142. 
Unicursal and Route Problems, 68. 
Union Jack, The, 50, 69, 197. 

Vandermonde, a., 58, 103. 

Veil, Under the, 90, 216. 

Verne, Jules, 249. 

Victoria Cross Puzzle, The, 60, 194. 

Village, A Wonderful, 142, 247. 

Villages, The Three, 12, 155. 

Villas, The Eight, 80, 206. 

Vortex Rings, 40. 

Voter's Puzzle, The, 75> 202. 

Wall, The Puzzle, 52, 184. 
Wallis, J., 142. 

(Another), 220. 



Walls, The Garden, 52, 183. 
Wapshaw's Wharf Mystery, The, 10, 153. 
War Puzzle Game, The, 118, 240. 
Wassail Bowl, The, 109, 235. 
Watch, A Puzzling, 10, 153. 



17 



258 



AMUSEMENTS IN MATHEMATICS. 



Water, Gas, and Electricity, 73, 200. 
Weekly Dispatch, The, 28, 124, 125, 146, 148, 
Weighing Puzzles, Measuring, Packing, and, 

109. 
Wheels, Concerning, 55, 188. 
Who was First ? 142, 247. 
Whvte, W. T., 147. 
Widow's Legacy, The, 2, 148. 
Wife, Find the Man's, 147, 251. 
Wilkinson, Rev. Mr., 193. 



Wilson, Professor, 29. 
Wilson's Poser, 9, 153. 
Wine and Water, no, 235. 
The Keg of, no, 235. 



Wotherspoon, G., 244. 

Yacht Race, The, 99, 226. 
Youthful Precocity, i, 148. 

ZSKO, 139. 



THE END. 



PRINTED IN GREAT BRITAIN. 



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