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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 soberminded 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 illfavoured 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, TitBits, 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 POSTOFFICE 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 postoffice 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 twopencehalfpenny 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 fiftyfive 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 fourshilling piece —
that I could have in my pocket without being
able to give change for a halfsovereign ?
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 MONEYBOXES.
Four brothers — named John, William, Charles,
and Thomas — had each a moneybox. 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 twentyfive 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 newlaid 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 CHRISTMASBOXES.
Some years ago a man told me he had spent one
hundred English silver coins in Christmasboxes,
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 chingchangs, that eleven with
square holes are worth sixteen chingchangs, and
that eleven with triangular holes are worth
seventeen chingchangs, 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 chingchang.
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
halfyearly. 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. halfyearly, 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 thirtyfour cents.
The only money he had was a dollar, a threecent
piece, and a twocent piece. The tradesman had
only a halfdollar and a quarterdollar. But
another customer happened to be present, and
when asked to help produced two dimes, a five
cent piece, a twocent piece, and a onecent
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
weddingring 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
pennypieces that may be laid flat on the surface
of a halfcrown, so that no piece lies on another
or overlaps the surface of the halfcrown ? 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, onehalf in
domestic expenses, and oneninth in other ways.
He has a balance of £190 remaining in the
bank. I know this last, because he acciden
tally left out his passbook 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 Mudleycum
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 fourpennypiece 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 oneyard
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 fortyeight onepound parcels of sugar
while the draper divided a roll of fortyeight
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 cattledealer 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 agoin' 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'sworth."
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
soninlaw'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 today."
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 oneeleventh 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
twentyfirst 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 today 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 fortyfive 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
sixtythree 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 onequarter the age
8
AMUSEMENTS IN MATHEMATICS.
of mother, and she is now onethird 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.— NEXTDOOR 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 dinnerparty, " 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
inlaw, 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 fatherinlaw, i mother
inlaw, and I daughterinlaw. Twentythree
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
inlaw, my brother's fatherinlaw, and also my
fatherinlaw'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^ingmachine 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
tomorrow, 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 sixtyfive 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 fortyninth 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 STOPWATCH.
We have here a stopwatch 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 tomorrow is yesterday,
today will be as far from Sunday as today was
from Simday when the day before yesterday
was tomorrow."
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 motorcar
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 thirtyfive 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 ladylove, 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 SlocombonSea 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 wateringplace 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 quartermiles, noted the fol
lowing results : — ^The first threequarters were
run in six and threequarter minutes, the first
halfmile 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 motorcar
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
onefigure 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 wideawake man, and he took out
his pocketbook 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 NUMBERCHECKS 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 splitring 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 99, 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 fourfigure
y^^:^7g^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 THIRTYTHREE PEARLS.
" A MAN I know," said Teddy Nicholson at a
certain family party, " possesses a string of
thirtythree 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 today. 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 LAMPPOSTS.
Tim Murphy and Pat Donovan were engaged
by the local authorities to paint the lampposts
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 lampposts on each side of the street,
the simple question is : Which man painted the
more lampposts, and just how many more ?
104.— CATCHING THE THIEF.
" Now, constable," said the defendant's counsel
in crossexamination, " you say that the prisoner
was exactly twentyseven 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
buryintheMarsh there were twentythree can
didates for nine seats. Each voter was qualified
to vote for nine of these candidates or for any
less number. One of the electorswants 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, twothirds 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 LEAPYEAR LADIES.
Last leapyear 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 oneeighth were widows. In conse
quence, a number of men were to be married
of whom oneeleventh were widowers. Of the
proposals made to widowers, onefifth were de
clined. AU the widows were accepted. Thirty
five fortyfourths 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 sugarplums. 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 twothirds of the parcel of sugarplums.
Bob at once grabbed threeeighths of these, and
Charlie managed to seize threetenths 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 onethird, David got twosevenths,
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 threequarters of what
Bob and Andrew had last acquired. The two
latter, with the greatest difficulty, recovered
fiveeighths of it in equal shares, but the three
others each carried ofE onefifth of the same.
Every sugarplum 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 sugarplums 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. 735804).
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 onethird, one
fourth, and onefifth 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 weekday, 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 SEESAW PUZZLE.
Necessity is, indeed, the mother of invention.
I was amused the other day in watching a boy
who wanted to play seesaw 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 threequarter brick and
threequarters 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 twothirds 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 twothirds of the estate should go
to the mother and onethird 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 coalminers 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 evermemorable 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 warhatchet 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 battlecries, ' 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 newlydiscovered 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 sixtyone 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
twentyfive." 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 onethird 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 thirtysix 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 cannonballs 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 173940, 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 today ? 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 twentythree
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 cuttingout 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 today 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 cuttingout 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 rightangled 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 " rightangled 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 cuttingout
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 halfsquare. 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 halfsquare 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, 3132, 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 halfsquare 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 twentyone 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 heilfsquare 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 selfevident. 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 cuttingout 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 tabletop, 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 CUTTINGOUT PUZZLE.
Here is a little cuttingout 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 fivesided figure, if they sud
denly required to do so. A regular hexagon, or
sixsided 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 TABLETOP 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 tabletop, for which he had no use, into
seats for two oval stools, each with a handhole
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 handholes 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 handhole 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 tabletop 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
handholes 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 Feuhhi." 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 twentyfive 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 fretsaw 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 (16261700) 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. Today 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 fivepointed 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 fivepointed 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 twentythree
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 exEmperor 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 AngloChinese 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, ' seveningeniousplan '
or ' ingeniouspuzzle 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 motherofpearl, 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
hidebound 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 doublebass, whose atti
tude is lifelike, though he does stand at an un
usual distance from his instrument, and the
drummerboy, with his imposing musicstand.
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 twentyfive 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 cushioncover. 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 BELLROPES.
Two men broke into a church tower one night
to steal the bellropes. 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
fellowthief 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 squareshaped 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 milkingstool 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
ringfence 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 bestknown 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
halfacre 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 onehvmdredth 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 KITEFLYING PUZZLE.
While accompanying my friend Professor High
flite during a scientific kiteflying 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 foursided
spaces, one exactly three times as large as the
other, and (2) so as to enclose two fivesided
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 SHEEPPENS.
A ' ' ' ■ k.'
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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 sheeppens
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 treeplanting 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 fiftyfive
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 fortynine
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 TWENTYONE TREES.
A GENTLEMAN wishcd to plant twentyone 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 MINCEPIES.
It will be seen in our illustration how twelve
mincepies 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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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 fortynine trees,
planted in the form of a square as shown in
the accompanying illustration, he wishes to
know how he may cut down twentyseven of
the trees so that the twentytwo 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 TitBits 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
thirtytwo 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 (Shegoat 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 ninemen'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 ninepegmorris 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 longlived 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.
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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 puzzlemaker is pecxiliarly a " snapperup 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 LODGINGHOUSE DIFFICULTY.
CABINd
Piano
1 2
3
bF WRDfioei
BWKCASE
DRAVvERS 1
4 1 5^
' 1
The Dobsons secured apartments at Slocomb
onSea. 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 engineyard 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 stoppingplaces, 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 stoppingplace 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 enginedriver
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, flyshunting, 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 MOTORGARAGE 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 stoppingplaces 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 righthand
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 sevenlettered
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
thirtyone 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 : 51 7)
1210, 2612, 2426 (1311, 1125), 911 (2624,
2410, 1012), 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 twentyfive 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 titbit 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 menofwar 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 twentyfour large
jampots in as many pigeonholes. 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 pigeonholes, 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<p9oH>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 thirtyseven 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 thirtyone 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 thirtyone passages, each a furlong in length,
he will have to travel just thirtyone 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
startingpoint 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 suchandsuch 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
•• ! — rHll/^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 halfdozen 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
fT
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 MOTORCAR 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
wellknown 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 : 5X3Fo — 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 : 130 = 13; 83  3 = 80 ;
59213=579 ; 482183 = 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 thirtysix cells, six by six, and strike out
all the cells in the long diagonal from the
bottom lefthand corner to the top righthand
corner, also the five cells in the diagonal next
above it and the cell in the bottom righthand
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 CARDPLAYERS.
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
thousandodd 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
Titbits 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 twentyone.
Many persons have since tried hard to beat
this number, but so far have not succeeded.
More than thirtyfive combinations of the
fifteen letters cannot be formed within the con
ditions. Theoretically, there cannot possibly
be more than twentythree 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 twentysecond 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 thirtyfive
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 MOUSETRAP PUZZLE.
This is a modern version, with a difference, of
an old puzzle of the same name. Number
twentyone 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 twentyone mice, but this is not here
possible, and if it were it would merely require
twentyone 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 twentyone 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
twentyeight 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
doublefour 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 chimneypot 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 thirtysix 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 cuttingout 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
chessplayer, 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 sixtyfour 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 twentyseven other
squares ; and if you try to hide her in a corner,
she still attacks twentyone 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 nonsymmetrical,
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
ninetytwo arrangements, as I have said, and
not ninetysix, as would happen if all twelve
were nonsymmetrical. 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
sixtytwo 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
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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 thirtytwo 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
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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 fourletter
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 THIRTYSIX LETTER
BLOCKS.
A
B
C
D
E
F
A
B
C
D
E
F
A
B
c
J>
E
F
A
B
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The illustration represents a box containing
thirtysix letterblocks. 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 fiftyone 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 twentyfive 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 FORTYNINE COUNTERS.
@ (g)@(g)@ ©)
m (D3) 64^ (gs) ($6) ^7)
Can you rearrange the above fortynine 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
ninetyone 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 sixtyfour 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 sixtyfour 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 sixtyfour 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 eightyone 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 HATPEG PUZZLE.
Here is a fivequeen puzzle that I gave in a
fanciful dress in 1897. As the queens were
94
AMUSEMENTS IN MATHEMATICS.
there represented as hats on sixtyfour pegs,
I will keep to the title, " The HatPeg 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.— LIONHUNTING.
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
longstanding 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 KNIGHTGUARDS.
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
KnightGuards.")
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).
NONATTACKING 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
2M3 — 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_i2m8
(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 stoppingplace,
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 twentyone 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 LJ 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 twentytwo 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
• * j > * I « ■ I » » I » ■ 1 <•■ i > ) « 1 < »•
^ — J ^.....^ J. — .^ I. — .,
' i"""i' 1 1 it t L_L_
A French prisoner, for his sins (or other
people's), was confined in an underground dun
geon containing sixtyfour 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 fiftyfive 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 sixtyfour 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,
— r I i I 1 i la
mm I 1 $ ■ I » e n I t I I ■ •+* ••+* •O*^
Hh H^^* + 4 + + 
i^«»L» ■ I I •— I* •*• 4* •+ —t— •
— h 4" 4 4" + *4~ "I —
■« • j ■ ■ I  ■ii j ■ ■ I ■ ■ I • » [ ■ ■ I * •■
5v T t I T T T
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 turningpoint at the enclosed D square,
we can obtain a solution starting from and
A
B
C
a c;
c
B
A
B
C
E
H H
E
D
B
F
1
1
F
E
C
<k
H
\
J
J
I
H
a
G
H
\
J
1
H
a
C
E
F
1
1
F
E
c
B
D
E
H
H
E
D
B
A
B
C
<^
a
C
B
A
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 startingpoints. There is no Queen's Tour
in fourteen moves (remember a tour must be
reentrant) that may start from a G, I, or J. But
we can have a nonreentrant 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 landlubbers, hoist your baby
jibtopsails, break out your spinnakers, ease
off your balloon sheets, and get your headsails
set!
Our race consists in starting from the point
at which the yacht is lying in the illustration
and touching every one of the sixtyfour 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 flagbuoy, and because it is a reentrant
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 sixtyfour 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 FORTYNINE 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 fortynine 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 chessprotector ! "
One year he had a large square field divided
into fortynine 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 twentyfifth in the
patch marked 25, the thirtyseventh in the one
marked 37, and the last, or fortyninth 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 lefthand 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 righthand 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 sixtyfour 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
reentrant 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 reentrant " 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 reentrant. 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 reentrant)
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 reentrant or not, on one
of the parts. The best reentrant 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 reentrant
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 HiChumChop, and his beauti
ful daughter, PeekyBo, had innumerable ad
mirers. One of her most ardent lovers was
WinkyHi, and when he asked the old mandarin
for his consent to their marriage, HiChumChop
presented him with the following puzzle and
promised his consent if the youth brought him
the correct answer within a week. WinkyHi,
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 HiChumChop
passed WinkyHi 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
twentyfive squares, as shown in the diagram.
On each of twentyfour 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 righthand 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 twentyfive 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.
" Chesseplay 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 thirtytwo 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.— THIRTYSIX MATES.
Place the remaining eight White pieces in
such a position that White shall have the choice
of thirtysix 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
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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
thirtytwo 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 fellowpassenger 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 fortythree 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), 15001559. 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 jigsaw 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 fivepint 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 fivequart and
a fourquart 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*
breakfasttables. 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 wineglass half fuU of wine, and another glass
twice the size onethird 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
tengaUon 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
halffuU, 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, halffull, 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 motorcar 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 twentyeight 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 lefthand 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 lefthand side
all add up 14, but the righthand 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 twentythree 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 offdrive, 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 HORSERACE 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 MOTORCAR 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 motorcar 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 : —
" Onethird of the cars in front of Goggle
smith added to threequarters 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 SlocombonSea. 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 squaretopped 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.
SEMINASlk
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
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^
Y
^
^
V
^
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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 SemiNasik, 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:
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«k
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^
^
9
«s
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rypi xn
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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 SemiNasik 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
SemiNasik (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 cuttingout 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 fortynine 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 fellowprisoners 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.
405CARD MAGIC SQUARES.
0^0
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4k
« ♦
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4r «
^ ^
^
r^ ^
\
r ■>
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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 TitBits
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
4S
IS
2/6
Si
4
4S
7^
2/6
IG
4
9
I OS
1296
Z
24
144
108
1
2
54
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 sixtyfour 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
twentyfive 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 today 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
chapterhouse 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 guardchamber
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 ChalonssiurMame, 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 " mizmazes " or " mize
mazes," it is not uncommon to find them locally
called " Troytowns," " 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 forthrights and meanders ! "
The Tempest, iii. 3.
There was such a maze at Comberton, in
Cambridgeshire, and another, locally called the
" mizmaze," 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 nonexist
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 Chartres 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 " Mizemaze." 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
]
[?.
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
1
' 1
I
1
1
1  ■■ ■■ ■■
1 — ■'
1
1 '
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 guidebook 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.
1
n
1
j
1
_J
"■"'""
1
crmn:
I ' —
imia
1
1
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 Ushaped 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 badlyconstructed 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 baldheaded 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
fyingglass, 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 ninetynine thousand nine hundred
and ninetynine 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 ninetynine thousand nine hundred
and ninetynine 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 hairsplitting, 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
billiardtable 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
visitingcard 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 billiardtable must, theoreti
cally, be spherical, just like a portion of the
orangepeel 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 twentysix miles from
France, and England is only twentyone 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 selfevident," 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 threefourths ? "
" By onefourth ! " 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 " onethird," although
Reginald tried to explain that three of anything,
if increased by onethird, becomes four.
" Uncle John, how do you pronoimce
' too ' ? " asked WiUie.
" • Too,' my boy."
" And how do you pronounce * two ' ? "
" 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 tablecloth; 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 PostOffice people were about to
erect a line of telegraphposts 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 PostOffice 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
halfmile."
" 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 nonmathematical 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 griefstricken 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 sixtyfour squares — eight by
eight. Now I draw a straight line from the top
lefthand corner, where the first and second
squares meet, to the bottom righthand 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 sixtythree squares, because seven
multiplied by nine makes sixtythree. 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 righthand 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 " tiringirons," and to be used by the
Norwegians today 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 tiringirons, 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 oneeighth 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 fortyfive stones, and there are
now only fortyone. 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 fortyfive rubies
originally arranged on the brooch ? The illus
tration shows exactly how the fortyone 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 HYMNBOARD 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 hymnboards, 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
puzzlesolving. 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 hymnboards, 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 hymnbook 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, iid. 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 hymnboards 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.— PHEASANTSHOOTING.
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 twentyfour
pheasants rose on the wing right in front of us.
I fired, and twothirds of them dropped dead 1
at my feet. Then the duke had a shot at what
were left, and brought down threetwenty
fomrths of them, wounded in the wing. Now,
out of those twentyfour 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 countryhouse,
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 weddingring. 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 POSTOFFICE PERPLEXITY.
The yotmg lady supplied 5 twopenny stamps,
30 penny stamps, and 8 twopencehalfpenny
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«fi), the next w(4n + i),
the next w(8n + 1), and so on to the first player,
who must have held fn(2'»^M4i).
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. 9d.
SOLUTIONS.
149
I4._SQUARE MONEY.
The answer is id. and 3d. Added together
they make 4id., and id. multiplied by 3 is
also 4d.
1 5. —POCKET MONEY.
The largest possible sum is 15s. gd., composed
of a crown and a halfcrown (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 fortyone 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 today. 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 moneybox two double florins
fSs.), WiUiam a halfsovereign 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 CHRISTMASBOXES.
The distribution took place " some years ago,"
when the fourpennypiece 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
pennypieces and one threepennypiece, and five
men each received five threepennypieces and
one fourpennypiece, 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. 5d., 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 chingchang is worth twopence and foUr
fifteenths of a chingchang, the remaining
elevenfifteenths of a chingchang must be worth
twopence. Therefore eleven chingchangs are
worth exactly thirty pence, or half a crown.
Now, the exchange must be made with seven
roimdholed coins and one squareholed coin.
Thus it will be seen that 7 roundholed coins
are worth sevenelevenths of 15 chingchangs,
and I squareholed coin is worth oneeleventh
of 16 chingchangs — that is, 'j'j rounds equal
105 chingchangs and 11 squares equal 16 ching
changs. Therefore "jy rounds added to 11
squares equal 121 chingchangs ; or 7 rounds
and I square equal 11 chingchangs, 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. halfyearly 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,
onehalf as against onequarter), 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 loopiece
must go to the tradesman, and it then follows
that the 50piece 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 locent 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 halfcrown, is one. A
second threepennypiece 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 threefourths 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 threepennypieces are admitted, there are
2n3+i5n2f33» , , ^, .
^g — hi ways. If sixpences are also
used there are ^^+22n3+i59n=^f4i4nf2i6
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 . . .
Threepennypiece in
Sixpence in
Shilling in .
Florin in . . .
Halfcrown in .
Double florin in .
Crown in . . .
Halfsovereign 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 halfminute delay, spent 24 min. 30 sec.
over the task ; but the draper had only to make
fortyseven cuts to divide the roll of cloth,
containing fortyeight 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 fortyeight cuts to
divide the roU into fortyeight 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 thirdpenny 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\.vQsixths 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 fivesixths 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 twentyone
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 weddingday, and her husband fiftyfour.
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 twentynine
years and twofifths, and Mary's nineteen years
and threefifths. When Marmaduke was aged
nineteen and threefifths, Mary was only nine
and fourfifths ; 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 twentyfive. 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 threefifths. Ann's age was sixteen and
fourfifths, the mother's thirtyeight and two
fifths, and the father's fifty and twofifths.
49.— NEXTDOOR 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 thirtyfive nuts. As 35 is con
tained in 770 twentytwo 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 brotherinlaw,
because my father married your sister Kate ;
you are my brother's fatherinlaw, because my
brother Alfred married your daughter Mary ;
and you are my fatherinlaw'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.
Twentysix 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
fortyninth 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 wellnigh 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 thirtysix 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
— («li) natural nimibers. In the case of the
puzzle «=3 ; therefore 12 — (3fi) = 8 and 1 +
2 f3 1415+61718 = 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 thirtysix 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 sixtysix 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
midday. 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 71 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 STOPWATCH.
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 twentyfour
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 tomorrow (Tuesday) is " yesterday,"
" today " will be Wednesday ; and when the
day before yesterday (Friday) was " to
morrow," " today " 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 rightangled tri
angles with a twelvemile 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
seventwentyfourths of a mile per minute and
the wind travelled fivetwentyfourths of a mile
per minute. Thus, going, the wind would help,
and the machine would do twelvetwenty
fourths, or half a mile a minute, and returning
only twotwentyfourths, or onetwelfth 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 twosevenths
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 fifteensixteenths — 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 motorcar 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 twentytwo 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 twentytwo 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 fivefigure 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
twofigure 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,
onehalf, onethird, onefourth, onefifth, one
sixth, oneseventh, oneeighth, and oneninth.
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 onethird, onefourth, onesixth,
oneseventh, and oneninth 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 onehalf and onefifth, 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,
oneeighth, 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 carryover 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 onefifth 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 onefifth 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 onesixth, nor an even figure
in the fraction onefifth, 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. 9d., 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 longsought 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, 9149, 82118, 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 82ii8 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 twentyfive
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)345 + (6X7) + (8X9)
= 100 (12 . . 20)
i + (2X3) + (4X5)6 + 7 + (8X9)
= 100 (11 . . 21)
(1+334) (56789)=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 + 34 + 5 + 67 + 8 + 9 = 100
■1234567 + 89 = 100
123 + 45 + 6789 = 100
123 + 4567 + 89 = 100 .
1234567 + 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 thirtyfive
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 halfdiameter, 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 THIRTYTHREE 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 threequarters. One candle had one
sixteenth of its total length left and the other
foursixteenths.
103.— PAINTING THE LAMPPOSTS.
Pat must have painted six more posts than
Tim, no matter how many lampposts 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 ninetyseven
104.— CATCHING THE THIEF.
The constable took thirty steps. In the same
time the thief would take fortyeight, which,
added to his start of twentyseven, carried him
seventyfive 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 lastmentioned number.
107.— THE SUFFRAGISTS' MEETING.
Eighteen were present at the meeting and
eleven left. If twelve had gone, twothirds
would have retired. If only nine had gone, the
meeting would have lost half its members.
108.— THE LEAPYEAR 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, " onefifth 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 fiveeighths, the fraction last men
tioned " — that is, oneeighth 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 onethird ; 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 onethird to onefourth — that
is in the proportion of fourtwelfths to three
II
1 62
AMUSEMENTS IN MATHEMATICS.
twelfths, which is the same as four to three.
Therefore Alfred takes foursevenths of the
hundred acres and Benjamin threesevenths.
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 fiftytwo Sundays and with
fiftythree Svmdays, and a leapyear with fifty
two and fiftythree 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 twentyeight 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 onerail
fence the field is four miles square ; a tworail
fence gives eight miles square; a threerail
fence, twelve miles square ; and so on, until
we find that a sevenrail fence multiplied by
four gives a field of twentyeight 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)fi, 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 twothirds and the
other part onethird, 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 threequarters of his money,
in four tosses with ninesixteenths of his money,
in six tosses with twentyseven sixtyfourths 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 SEESAW 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 twosevenths,
the son foursevenths, 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 sixtyone
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 standingroom 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=(i5wi83) + 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 (32Hw)
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 twentyfour 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 " rightangled isosceles triangle."
The triangle to which geometricians give this
highsounding 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 CUTTINGOUT 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 nh2 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 TABLETOP 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 eggshape (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 poultryyard 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 handholes. 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
onehalf of the Yin or the Yan. This we will
SOLUTIONS.
175
do in Diagram 4. The circle K has onequarter
the area of the circle containing Yin and Yan,
because its diameter is just onehalf the length.
Also L in Diagram 3 is, we know, onequarter
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
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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 ladderlike
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 twentytwo 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+ilfw+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
rT7 ,;   r7Hri"
« • . ' I r • : I I
! : L M J
rJrr'
'•rr'ri
I • I
r I • I ,
. ' I . I • . I
t ' 1 I i, 1 I I
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.
i^
7f
G^
^
^
^
^
<i>
A
^
^
€>
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Q>
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&
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<<2
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cV3
&
6^
^
63
^
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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^
^
<?>
A
<?>
&
^
^
Q>
S
Q?
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(k
i?
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t
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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
?
$>
^
^
F
^
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^
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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
^
Q>
^
^
w
&
E
&
^
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t
^
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F
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3
^
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fi»
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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
quarterturn 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 halfturn — that is, turn it upside down.
Whether or not a piece may be given a quarter
turn, a halfturn, 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
^ — iJ.^ J I i ■■ —J — I ^i
' 1
'
^1
r
■  
f="
^
JL
"■ "C '
 " "^r;~
_ __
'V
::\^ T
ir— 
:±""
 i
 ^B 
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:
^
m
^
c
m
:i
^
^
i
1
w
D
^
1
^
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 fiftytwo
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
1
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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 oneeighth, 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^OOR
STOOL
Draw a straight line, as shown in the diagram,
from the milkingstool 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 onesixth 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. wellknown formula reduces the problem
to finding onequarter of the square root of
(4x370x116) — (37ofii6 — 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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+bc)^
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 today, 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 SHEEPFOLD.
This is the answer that is always given and
accepted as correct : Two more hurcfies would
be necessary, for the pen was twentyfour 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 sheepfold to be of any particular form.
But even if we accept the point that the pen
was twentyfour 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 twentyfour " 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 twentyeight hiudles only,
and have twentytwo 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, onefourth 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 halfacre field, and the shaded
portion to be the quarteracre 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
onesixth of the complete area of a circle.
Therefore aU we require is the radius (CD) of
a circle containing six quarteracres 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 twentyone 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 onethird 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 lefthand 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 KITEFLYING 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 " hatbox method,"
Supposing we place our ball of wire, A, in a
cylindrical hatbox, 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
hatbox 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'' + 62a6 ^^^^^ ^^^
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
onethird 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 cartwheel, 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 locomotivewheel,
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 SHEEPPENS.
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.
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row. The dots represent the positions of the
trees that have been cut down.
209.— THE TWENTYONE 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
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0— O 0.<
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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
twentyfour 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
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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 thirtytwo,
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 twentyone 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 thirtytwo 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
twentyone 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 fortyfour
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 nonclockwise 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 „ nli
MI „ M
2(wl)2 + 5M7
2(ni)2f 5M7
2w2+5n — 2
2(Ml)2 + 5W7
2Mf I
2«— I
2M
2«
2n2f 3ni
2(W— i)2f3M — 2
2^2+ 3M — 4
2W2f 3» — 4
4(«2«i)
4^25
2(2w2 + 4«3)
4n2l4w9
(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
4Mf I
4»— I
n and n
n—x „ wi
n „ «
n „ M
2n2 + 3n — 4
2(Wl)2+3M7
2^2+ 3M — 4
2»2i3» — 4
2W
2M
2WfI
2M— I
2(W— l)2+5n— 2
2(W— l)25M— 2
2«2 + 5n — 2
2(»l)2 + 5«7
4(n2+ni)
4^25
2(2n24n— 3)
4w2f 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 twentytwo 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 twentyeight 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 LODGINGHOUSE 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 MOTORGARAGE 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 fortythree 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 twentyone. 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
twentyone 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 seventytwo 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 78,
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<yoeg 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
sixtyfour 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 MOTORCAR 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
lefthand 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 righthand 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 righthand
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 CARDPLAYERS.
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; 5x91=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 wellknown 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
twentysix are good words. Of course a TAU
cross is a Tshaped 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 toadfish.
We thus have twentysix 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 Ml6 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 MOUSETRAP 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 twentyone " 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 lefthand 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
oooo
oojoo
oo.o^fo
0000
oooo
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 lefthand
comer there are io+(3X9) = 37 different solu
tions. If we then try 7 in the comer, the result
will be 10 + 27140, or 'jy solutions. With 6
we get 10 + 27140+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 lefthand 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 thirtytwo, 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
Mf 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
8shaped 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 twentyeight 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.
Twentyone 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 twentyone 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 twentyeight 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 sixtyfive
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 twentysix 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 keyline 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 keyline, 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
today 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, sixtyfour, 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 lefthand 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 sixtyfour 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 wordreadings
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
quarterturn 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 halfturns, but not quarterturns. 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 THIRTYSIX 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 fortyeight
pence, or 4s. This is the pitfall that was
hinted at. (Compare with No. 43, Canterbury
Puzzles.)
309.— THE FORTYNINE 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 fortyseven.
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
twentynine 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
sixtyfour 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 HATPEG 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 KNIGHTGUARDS.
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.— LIONHUNTING.
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 lefthand
^
^
^
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.

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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.
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THE rook's tour.
321.— THE ROOK'S JOURNEY.

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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
twentyfirst, brings us to a halt on square " 21."
322.— THE LANGUISHING MAIDEN.
The dotted line shows the route in twentytwo
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 fiftyseven
straight lines. No rook's path over the chess
board can exceed this number of moves.
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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 twentytwo, 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
4?
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 nonreentrant
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 flagbuoy, 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 FORTYNINE 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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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
halfstrings will be seen to be sjonmetrical and
similar.
The next point is that the first halfstring
must end in one of the central rows, and the
228
AMUSEMENTS IN MATHEMATICS.
second halfstring 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 noncircled as the case may
be. The halfstrings can, therefore, only be
linked in the two central rows.
Now, there are just eight different first half
strings, and consequently also eight second
halfstrings. 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 halfstrings 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 halfstrings
I to xo, linked to the halfstring 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.
reentrant 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 movingcounter 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. Ninetynine 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 thirtytwo 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 sixtysix 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 lefthand comer cell might be
left vacant, the following is a solution in fortyfive
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
fortysix 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 — thirtytwo. 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.
PQ4
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.— THIRTYSIX 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 thirtysix 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. RQ 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 quarterturn, you get F ; and if
you give B three quarterturns in the direction
that a clock hand moves, you will get successively
C, D, and E. No matter how you may place
(3,
1
fF
«
_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 fivepint 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
5pint.
■ 5
2
O
■ 3
. 3
■ 3
. 5
. 5
2
o
o
3pint.
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.
loquart.
Squart 4quart
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 onesixth of the
total liquid, and the wine in large glass two
ninths of total. Add these together, and we
find that the wine was seveneighteenths 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
seventytwo 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 seventytwo 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 rivercrossing 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.
Halffull.
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 midstream 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 8i3ths of the rowing and A and D
need not row at all) there are only sixteen
gettingsin and sixteen gettingsout. 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 twentythree 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 twentythree 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 lefthand 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 righthand 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 twentyfour
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 looaH
10&IC 1250, 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^^^
IVlr. 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 HORSERACE PUZZLE.
Tm «inswer is: £12 on Acorn, £15 on Blue
bottle, £20 on Capsule.
391.— THE MOTORCAR 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 onethird
of twelve added to threequarters 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 : —
338, 3215, 3122, 3021, 2914, 227, 156,
142, 73, 64, II, and A must capture on his
next (12th) move, 13, 5420, 5327, 5234,
5141, 5034, 4227, 3520, 2813, 216, 142,
73, 64, 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 illadvised 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 northwest 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 northwest 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 lefthand comer to
repeat everything that the other player would
do between G and the top righthand 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 puzzlesolving. 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 GAOLBIRDS.
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 thirtyseven 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 ifi8 + 2326+3if
48156157=260, and the sum of their squares
is 11,180. Therefore 644471424 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 twentyfive 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 thirtysix 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 " reentrant." 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
reentrant 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 sixtyfour 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 semidiameter 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 TIRINGIRONS.
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 onethird of
(2»fi— 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 tiringirons. 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»+ii).
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
fortyfive 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 HYMNBOARD 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 hjonnboards, 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
hymnbook 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, thirtythree 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 4d., 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 thirtyone 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 TitBits, and the
following remarks, made by me in the issue for
nth December 1897, may be of interest.
The " HymnBoard 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
TitBits 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.— PHEASANTSHOOTING.
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 thirtyfourth 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, onethird
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 eyewitness 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 walkingstick. 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.
Bellropes, 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.
Nonattacking 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 Fortynine, 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.
Cuttingout 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.
Hatpeg 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.
Hymnboard 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 GaolBirds, Eight, 122, 243.
Jolly GaolBirds, 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.
Kiteflying Puzzle, A, 54, 187.
Knightguards, 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.
Lampposts, 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 Thirtysix, 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.
Lodginghouse 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, Thirtysix, 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.
Motorcar Race, The, 117, 240.
Tour, The, 74, 201.
Garage Puzzle, The, 62, 195.
Motorists, A Puzzle for, 73, 201.
Mousetrap 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, NextDoor, 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.
PheasantShooting, 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.
Post0£&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 Motorcar, 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.
SeeSaw Puzzle, The, 22, 163.
SemiNasik 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.
Stamplicking, The Gentle Art of, 91, 217.
Star Puzzle, The, 99, 226.
Stars, The Eight, 89, 215.
The Fortynine, 100, 226.
Statical Chess Puzzles, 88.
Sticks, The Eight, 53, 186.
Stonemason's Problem, The, 25, 165.
Stopwatch, 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.
Tabletop 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.
TitBits, 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 Twentyone, 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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