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PUZZLES OLD AND NEW.
®
BY THE SAME AUTIIOB.
TRICKS WITH CARDS. Crown
8vo, cloth gilt.
CONJURER DICK. Large crown
8vo, cloth gilt, bevelled boards.
JSecp. 75,
PUZZLES
©lb ant) 1Rew>*
BY
PROFESSOR HOFFMANN,
Author of " Modern Magic? " Tricks with Cards? " Conjurer Dick? etc.
WITH ILLUSTRATIONS
" > ' ' o   ,  
oa o ae > 3 , * , o »>c «o a ■» 3* *.
r •
LONDON
FREDERICK WARNE AND
AND NEW YORK.
^p
„.»•*. *
Butler & Tanner,
The Selwood Printing Works,
Frome, and London.
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PREFACE.
The Natural History of tlie Puzzle has yet to be written.
It is a plant of very ancient growth, as witness the riddle of
the Sphinx, solved by (Edipus, and the enigma wherewith
Samson confounded his Philistine adversaries. Homer is
said by Plutarch to have died of chagrin at being unable to
guess a riddle ; and folklore abounds in instances where
the winning of a princess, or the issue of some perilous
adventure, is made to depend upon success in solving some
puzzle, verbal or otherwise. In more modern times grave
mathematicians, like Cardan and Euler, have not disdained to
employ their leisure in the fabrication of " posers " for the
puzzlement of their less erudite compeers.
The chief difficulties I have found in compiling the present
collection have been nomenclature and classification. In view
of the varieties of taste, some preferring a mathematical,
some a mechanical problem — one person a trial of skill,
another an exercise of patience — it seemed desirable to have
as many categories as possible. On the other hand, the
more numerous the divisions, the more, difficult does it be
come to assign a given puzzle definitely to one or another.
In many instances the same item might with equal propriety
be classed under either of several categories.
Nomenclature presents even greater difficulties. The
same title is often applied, with more or less appropriateness,
to two or three different puzzles. For example, there are
some halfdozen " cross " puzzles, more or less unlike, yet all
having a fair claim to the title, and being scarcely distin
viii Preface.
guishable by any other. Again, a mechanical puzzle is fre
quently described in the pricelists of different dealers by
different names, the " Arabian Mystery " of one being, say,
the " Egyptian. Paradox" or the " Ashantee Difficulty " of
another. Others are of necessity nameless, it being impos
sible to devise any short title which shall give any idea of
the nature of the problem .
The very wide class of verbal puzzles, comprising conun
drums, enigmas, charades, etc., is here, from considerations of
space, omitted. With this exception, it has been my endea
vour to make this little book as complete as possible, and I
have to acknowledge a substantial debt of thanks to the many
friends (notably Messrs. Paul Perkins and Edward Montau
ban) who have lent helping hands to make it so. The field
is, however, very wide, and it is almost a matter of course
that many "good things" should have been, through igno
rance or inadvertence, omitted. Should any reader note
such omissions, or have private information as to items of
special merit, I shall be glad, with a view to future editions,
to be made acquainted with them.
. LOUIS HOFFMANN.
TABLE OF CONTENTS,
CHAPTER I,
Puzzle3 Dependent on Dexterity and Perseverance,
I. The " PickMeUp " Puzzle
II. The Planet Puzzle .
III. The i Dotting Puzzle .
IV. The Spider and the Flies
V. The Tower Bridge Puzzle
VI. In the Soup .
VII. The Matrimonial Chair .
VIII. The " Tire 'Em Out" Puzzle
IX. The Electric Ball .
X. The " Hang Him " Puzzle
XI. BouciBoula . . .
XII. The Switchback
XIII. The Five HorseShoes .
XIV. The Two HorseShoes .
XV. The Maze
XVI. The Pitfall Puzzle .
XVII. The Fifteen Pellets Puzzle
XVIII. The Cross Puzzle .
XIX. The Hand of Cards .
XX. The Pig Puzzle
XXI. The Four Colours Puzzle
XXII. The Amphitheatre Puzzle
XXIII. The Persian Shah .
XXIV. The Balance Puzzle
XXV. The Fish Puzzle .
XXVI. The Marksman
XXVII. The Snake and Bird
XXVIII. The Coin and Card Puzzle
XXIX. The Egg and Card Puzzle
ix
PAGE
1
2
3
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4
5
6
7
8
9
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Table of Contents.
CHAPTER II.
Mechanical Puzzles Dependent on some Trick
or Secret.
I. The Barrel and Ball „
II. The Dice Box .
III. The Churn .
IV. The Cannon and Ball
V. The Cage and Ball .
VI. The Castle MoneyBox
VII. The New Castle MoneyBo
VIII. The Brass MoneyBox
IX. The Captive Sixpence
X. The Cannon and Cord
XI. The Heart Puzzle
XII. The Alliance (or Victoria) Puzzle
XIII. The Two Balls .
XIV. The Ariel Puzzle
XV. The Pen and Wheel .
XVI. The Balls and Rings .
XVII. The Chinese Ladder .
XVIII. The Staff .
XIX. The Imperial Scale . .
XX. The Sceptre
XXI. The Balls and Cha'n .
XXII. The Four Keys .
XXIII. The Screw Box .
XXIV. The Ball and Three Strings
XXV. The Lighthouse .
XXVI. The Jubilee Puzzle Box
XXVII. The Jubilee Penny
XXVIII. The Invisible Gift
XXIX. The Arabi Gun .
XXX. The Psycho MatchBox
XXXI. The "Touchmenot" MatchBox and
XXXII. The Magic Drawer MatchBox .
XXXIII. The " Unique " MatchBox.
XXXIV. The Surprise MatchBox .
XXXV. The New Brass Puzzle MatchBox
XXXVI. The Ne Plus Ultra MatchBox .
XXXVII. The Sphinx MatchBox
XXXVIII. The Puzzle SnuffBox
Problem Sol
ution
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. 41
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Table of Contents.
XI
Puzzles
Depended on some Thick on SEC^contirtued.^
XXXIX. Tlie New Puzzle SnuffBox.
XL. The Puzzle Ball .
XLI. The Ebony Puzzle Ball
XLII. The Puzzle Purse
XLIII. The Puzzle Pocket Knife .
XLIV. The Automatic Knife .
XLV. The Double Barrel and Ring
XLVI. The WeddingKing Box
XLYII. The New Money Box .
XLVIII. The Zulu Box .
XLIX. The New Persian Puzzle .
L. The Magic Handcuff .
LT. The Key and Bing Puzzle .
LII. The New Egg of Columbus .
TAGB
PAGE
. 42
66
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. 43
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. 44
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. 47
72
CHAPTER III.
Dissected or Combination Puzzles.
The Dissected Puzzles of fifty years ago.The Eichter
Building Stones • • • * *
Hints for the readier solution of Dissected Puzzles
The Eichter Puzzles—
I. The Anchor Puzzle .
II. The Tormentor Puzzle
III. The Pythagoras Puzzle
IV. The Cross Puzzle.
V. The Circular Puzzle .
VI. The Star Puzzle .
VII. The Zigzag Square .
VIII. The Extended Square
IX. The Octagon Puzzle ■••_'*"
X. The Patchwork Square
XI. The Two Squares
XII. The Latin Cross Puzzle .
XIII. The Greek Cross Puzzle .
XIV. The Protean Puzzle .
XV. The Caricature Puzzle
XVI. The Chequers Puzzle .
XVII. The (( Spots " Puzzle .
XYIII. The Endless Chain .
74
111
77
it
80
115
81
117
83
118
85
120
87
122
91
124
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92
125
ii
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93
ii
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12G
94
ii
127
j>
9G
128
97
129
93
130
99
131
Xll
Table of Contents.
Dissected on Combination Puzzles — continued,
XIX. The Hexagon .
XX. Eight Squares in One .
XXI. The Five Squares
XXII. The Geometrical Square
XXIII. The Dissected Square
XXIV. The Twenty Triangles
XXV, The New Triangle Puzzle
XXVI. The Japanese Square .
XXVII. The Chinese Square .
XXVIII. The Yankee Square .
XXIX. Another Cross Puzzle
XXX. The Carpenter's Puzzle. No. 1
XXXI. The Carpenter's Puzzle. No. 2
XXXII. The Cabinet Mater's Puzzle
XXXIII. The Bonbon Nut Puzzle .
XXXIV. The Battle Puzzle
XXXV. The Cross Keys (or Three piece)
XXXVI. The Nut (or Sixpiece) Puzzle
XXXVII. The Fairy Teatable .
XXXVIII. The Mystery
XXXIX. The Diabolical Cube .
XL. The Chinese Zigzag .
XLI. The Man of Many Parts
Puzzle
Problem Solution
PAC1E TAGE
100 132
101
102
103
104
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105
106
133
131
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135
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137
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139
107 141
108
142
109
143
110
144
CHAPTER IV.
Aeithmetical Puzzles
Elementary Properties of Numbers <, .
I. The "Fortyfive" Puzzle.
II. A Singular Subtraction .
III. A Mysterious Multiplicand
IV, Counting the Pigs „
V.» Another Pig Problem
VI. A Little Miscalculation ,
VII. A Simple Magic Square .
VIII. The " Thirtyfour " Puzzle
IX. The " Sixtyfive " Puzzle
X. The " Twentysix" Puzzle
XI. An Unmanageable Legacy
XII. Many Figures, but a Small Result
XIII. Can you Name It ? .
1U
143
147
174
181
182
133
186
191
?>
192
Table of Contents.
xiii
Arithmetical Puzzles— continued.
XIV. Squares, Product, and Difference
XV. A Peculiar Number .
XVI. A Novel Century
XVII. Another Century
XVIII. Another Way to Make 100
XIX. The Lucky Number .
XX. The Two Ages .
XXI. The Graces and the Muses
XXII. The Graces and the Muses again
XXIII. Just One Over . . •
XXIV. Scarcely Explicit
XXV. Making Things Even
XXVI. A Rejected Proposal .
XXVII. The Marketwoman and her Stock
XXVIII. The Captives in the Tower
XXIX. Father and Son . _ •
XXX. A Complicated Transaction
XXXI. A Long Family
XXXII. A Curious Number .
XXXIII. The Shepherd and his Sheep
XXXIV. A Difficult Problem .
XXXV. Well Laid Out .
XXXVI. The Two Travellers .
XXXVII. Measuring the Garden .
XXXVIII. When Will They Get It ?
XXXIX. Passing the Gate. .
XL. A Novel Magic Square .
XLI. Another Magic Square .
XLII. The Set of Weights .
XLIII. What Did He Lose ?
XL IV. A Difficult Division.
XLV. The Hundred Bottles of Wine
XL VI. The Last of her Stock .
XL^II. The Walking Match
XLVI1I. A Feat of Divination
XLIX. A Peculiar Number .
L. Another Peculiar Number
LI. The Three Legacies
LII. Another Mysterious Multiplicand
LIU. How to Divide Twelve among Thirteen
LIV. Tenth Man Out
LV. Ninth Man Out ....
Problem Solution
PAGE PAGE
148
192
»>
193
119 191
150
15 L
152
153
154
155
156
195
196
197
193
200
201
202
203
204
205
208
207
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209
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ti
XIV
Table of Contents.
Arithmetical Puzzles — continued.
LVI. The Three Travellers .
LVII. The Wolf, the Goat, and the Cabbages
LYIII. The Three Jealous Husbands .
LIX. The Captain and his Company
LX. The Treasure Trove .
LXI. The Kow of Counters
LXII. A Loan and a Present
LXIII. Eleven Guests in Ten Beds .
LXIV. A Difficult Division .
LXV. The Three MarketWomen
LXVI. The Farmer and his Three Daughters
LXVII. How Many for a Penny ? .
LXVIII. The Magic Cards .
LXIX. The " Fifteen" or " Boss " Puzzle.
LXX. The Pegaway Puzzle
LXXI. The Overpolite Guests .
LXXIL The " Koyal Aquarium " Thirteen Puzzl
LXXIII. An Easy Creditor .
LXXIV. The Three Arabs .
LXXV. An Eccentric Testator
LXXVI. Another Eccentric Testator
LXXVII. An Aggravating Uncle
LXXVIIL Apples and Oranges
LXXIX. The Two Squares . .
LXXX. A Curious Division
LXXXI. A Curious Multiplication.
LXXXII. The Two Schoolmasters .
LXXXIII. Nothing Left . . . .
LXXXIV. The Three Generations .
LXXXV. The Two Brothers .
LXXXVI. The Two Sons
LXXXVII. The Two Nephews .
LXXXYIII. The Beversed Number .
LXXXIX. Another Keversed Number
XC. The Shepherd and his Sheep .
XCI. The Shepherdess and her Sheep
XCII. A Weighty Matter .
XCIII. The Three Topers .
XCIV. The False Scales
XCV. An Arithmetical Policeman
XCVI. The Flock of Geese .
. XCVII. The Divided Cord .
I'rublem Solution
TAOE
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1G7
1G8
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170
Table of Contents.
xv
Arithmetical Puzzles — continued.
XOVIII. The Divided Number
XCIX. The Two Numbers .
C. The Horse and Trap .
CI. The Two Workmen .
CII. Another Divided Number
CHI. The Three Reapers .
CIV. The Bag of Marbles .
CV. The Expunged Numerals. A,
CVI. The Expunged Numerals. B.
CVII. A Tradesman in a Difficulty
CVIII. Profit and Loss t
CIX. A Curious Fraction .
CX. The Menagerie .
CXI. The Market Woman and her Eggs
CXII. The Cook and his Assistants
Problem
Solution
TAGE
FAG IS
170
230
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231
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ij
233
171
172
173
231
>>
235
236
55
237
CHAPTEE V.
Word and Letter Puzzles
I. A Puzzling Inscription
11= An Easy One
III. Pied Proverbs .
IV. Scattered Sentiment .
V DroppedLetter Proverbs .
VI. DroppedLetter Nursery Bhymes
VII. Transformations
VIII. Beheaded Words
IX. Anagrams c «
X. Word Squares .
XL Word Diamonds
XII. A Cross of DiamondSc
XIII. Knight's Tour Letter Puzzles
XIV. Knight's Tour Word Puzzle
XV. Hidden Proverbs
XVI. The Five Arab Maxims .
OQ
253
210
5*
»?
251
211
i'.
242
255
213
s;
211
256
215
)»
217
258
218
259
250
232
251
263
252
CHAPTEE VI.
Puzzles with Counters;
I. With 11 Counters, to make 12 rows of 3
I la With 9 Counters, to make 10 rows of 3 „
III. With 27 Counters, to make 9 rows of 6 ,
mi
275
276
XVI
Table of Contents,
Puzzles with Counters — continued.
IV. With 10 Counters, to make 5 rows of 4 .
V. With 12 Counters, to make 6 rows of 4 .
VI. With 19 Counters, to make 9 rows of 5 .
VII. With 16 Counters, to make 10 rows of 4
VIII. With 12 Counters, to make 7 rows of 4.
IX. With 9 white and 9 red Counters, to mi
rows of 3 white and 8 rows of 3 red
X. The Blind Abbot and his Monks .
XI. With 10 Counters, to make 8 rows of 4 .
XII. With 13 Counters, to make 12 rows of 5
XIII. The Eightpointed Star Puzzle
XIV. The " Okto " Puzzle .
XV. With 21 Counters, following the lines of a
figure, to form 30 rows of 3 ,
XVI. The " Crowning " Puzz 7 e
XVII. The " Eight and Left " Puzzle
XVIII. The „ „ (Improved) .
XIX. The "Four and Four" Puzzle
XX. The " Five and Five " Puzzle
XXI. The "Six and Six " Puzzle .
XXIL The " Thirtysix " Puzzle .
XXIII. The " Five to Four " Puzzle. , ,
XXIV. No Two in a Eow .;'...••..
XXV. The " Simple " Puzzle ,
XXVI. The " English Sixteen " Puzzle ,
XXVII. The Twenty Counters l
ce 10
given
ProMem
PAGE
265
2G6
Solution
PACK
277
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273
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279
266
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267
280
263
281
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282
269
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284
270
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271
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285
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272
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286
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273
287
274
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CHAPTER VII.
Puzzles with Lucifer Matches,
I. Of Eleven Matches, to make Nine » . , 9
II. Of Nine Matches, to make Three Dozen .
III. Of Nine Matches, to make Three and a halfdozen .
IV. Of Three Matches, to make Four ....
V, Of Three Matches, to make Six ...
VI. The Bridge of Three Matches .
VII. The Bridge of Four Matches . . .
VIII. From Twentyfour Matches, forming Nine Squares,
to take Eight, and leave Two Squares only
IX. The Bridge of Two Matches .....
X. From Seventeen Matches, forming Six Squares, to
take Five, and leave Three Squares only , ,
2S8 293
239
294
295
>»
296
Table of Contents.
xvii
Pozzles with Lucifer Matches — continued.
XI. From Seventeen Matches, forming Six Squares, to
take Six, and leave Two Squares only
XII. Twelve Matches being so placed as to form Four
equal Squares, to remove and replace Four so
as to form Three Squares only ....
XIII. From Fifteen Matches, forming Five Squares, to
remove Three, and leave Three Squares only .
XIV. With Five Matches, to form Two Equilateral Tri
angles
XV. With Six Matches, to form Four Triangles of equal
size .......
XVI. To Lift Three Matches with One .
XVII. To Lift Nine Matches with One
XVIII. The Magnetised Matches ....
XIX. The Fifteen Matches Puzzle .
rrulilera Solution
PAGE
PAGE
290 296
297
291
292
293
299
300
CHAPTEE VIII.
Wire Puzzles
I. The United Hearts .
. ^ .
309
II. The Triangle ....
III. The Snake and Eing .
IV. The Hieroglyph ....
V. The Five Triangles .
303
. 301
>»
310
! J
VI. The Double Bow and Eing
. 305
311
VII. The Egyptian Mystery
fill. The Ball and Spiral .
IX. The Unionist Puzzle .
1 5)
. 306
312
X. The Eastern Question
XI. The Handcuff Puzzle .
307
313
XII. The Stanley Puzzle .
308
JJ
CHAPTEE IX.
''Quibble" or "Catch" Puzzles.
I. A Eemarkable Division ....
II. Subtraction Extraordinary ....
HI. Two Halves Greater than the Whole .
IV. A Distinction and a Difference .
515
*23
jj
jj
XV111
Table of Contents.
"Quibble" on "Catcii" Puzzles — continued.
rroblem Solution
l'AGK PAGE
V. The Family Party ....
. 315
323
VI. A Sum in Subtraction . . «,
. 31G
324
VII. Another Sum in Subtraction
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VIII. Three times Six .
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IX. A New Way of Writing 100.
. ,J
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X. A Seeming Impossibility
55
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XI. Multiplication Extraordinary
• ' ;i
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XII. A Question in Notation
. ,,
>7
XIII. The Miraculous Herrings .
. ,,
55
XIV. Two Evens make an Odd .
>i
325
XV. Six made Three . . ,
. .017
55
XVI. A Singular Subtraction
. . ,,
55
XVII. A Sum in Addition ....
9 ' !J
.,
XVIII. The Flying Sixpence ....
= . ,)
55
XIX. The Last Thing Out ....
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5)
XX. The Three Gingerbread Nuts
c .,
55
XXI. The Mysterious Obstacle .
, 318
326
XXII. The Bewitched Eight Hand
C . ,j
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XXIII. The Invisible Candle .
. . .;
»
XXIV. The Draper's Puzzle .
. . ,,
?>
XXV. The Portrait
' * 53
53
XXVI. Tue Charmed Circle . . .
. 319
55
XXVIL The Egg and the Cannonball .
5?
327
XXVIH. k Curious Window ....
. . ,,
5?
XXIX. A Queer Calculation ....
,,
35
XXX. An Arithmetical Enigma .
• • 5)
53
XXXI. A Short Year . . . .
• 55
55
XXXII. The Mysterious Addition .
. 320
33
XXXIII. Arithmetical Enigma ....
• 35
35
XXXIV. A New Valuation . ...
. . ,,
55
XXXV. Easy, When You Know It .
• ' 55
328
XXXVI. Necessity the Mother of Invention
* M
55
XXXVII. A Singular Subtraction . .
• " 55
53
XXXVIII. A Vanishing Number .
c • J)
53
XXXIX. A Queer Query
. 321
35
XL. The Mouse . . . ...
>>
35
XLI. The Fasting Man ...
• jj
33
XLII. The Family Party
• • V
33
XLIII. A Eeversible Fraction ....
' • M
329
XLIV. The Three Counters ....
. 322
?»
XLV. Magic Made Easy . ' .
• • j>
>;
Table of Contents.
xix
CHAPTER X.
Miscellaneous Puzzles.
I. The John Bull Puzzle,
; II. The Pig in Sty .
III. Hide and Seek .
IV. The Brahmin's Puzzle
V. Cardan's Rings .
VI. The Knight's Tour .
VII. The Knotted Handkerchief
VIII. Crossette .
IX. Single Stroke Figuies
X. The Balanced Egg (Another
XI. Solitaire Problems
XII. Skihi ....
XIII. A Card Puzzle .
XIV. Another Card Puzzle .
XV. The Floating Corks .
XVI. The Obstinate Cork .
XVII. Fixing the Ring .
XVIII. The Treasure at Medinet
XIX. The Four Wine Glasses
XX. One Peg to Fit Three Holes
XXI. The Balanced Pencil .
XXII. To Balance an Egg on the
Stick
XXIII. The Ashantee Horseshoe
XXIV. A Feat of Dexterity .
XXV. The Divided Square .
XXVI. The "Oval " Problem .
XXVI f. The Floating Ball
XXVIII. The Cut PlayingCard
XXIX. The Mitre Puzzle
XXX. The Five Straws .
XXXI. The Three Fountains .
XXXII. The Two Dogs .
XXXIII. Water Bewitched.
XXXIV. The Balanced Halfpenny
XXXV. The Balanced Sixpence
XXXVI. Silken Fetters .
XXXVII. The Orcharl Puzzle .
method)
Point
of
a Walking
roblem Solution
PAGE
PAGE
330
357
331
358
332
350
333
3CL
334
361
335
367
337
371
M
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338
375
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339
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340
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312
378
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381
311
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317
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319
350
382
383
384
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385
386
387
388
389
390
XX
Table of Contents,
Miscellaneous
XXXVIII. The
XXXIX. The
XL. The
XLI. The
XLII. The
XLIII. The
XLIV. The
XLV. The
XLVI. The
XLVII. The
XLVIlL The
P cjzzles — continued.
Cook in a Difficulty
Devil's Bridge
Two Corks .
Divided Farm
Conjurer's Medal
Maze Medal .
Puzzle Key ring
Singular Shilling 
Entangled Scissors
Penetrative Penny
Packer's Secret .
Problem Solution
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PUZZLES OLD AND NEW.
CHAPTER I.
PUZZLES DEPENDENT ON DEXTERITY AND PERSEVERANCE.
In this first section we propose to describe a class of puzzles
which do not depend upon any secret, or intellectual process,
but upon some knack, only to be obtained by repeated per
sistent effort of the " try and try again " kind on the part of
the operator.* Of these a very good example is —
No. I.— The "PickMeUp" Puzzle.
This puzzle was brought out by the proprietors of the
wellknown seriocomic weekly of the same name, and en
Fig. 1.
joyed for some time widespread popularity. It consists of a
cylindrical pasteboard box, four inches in diameter by two
in depth, with a glass top, as illustrated in Fig. 1.
_ * There will accordingly be no Key to the present Chapter. Any
hint likely to assist the operator will be incorporated in the text. Be
yond this, he must work out the solution for himself.
B
Puzzles Old and New.
The base is occupied by a shallow cone in plaster of Paris,
with an inclined plane in the form of a spiral (but diversified
by sundry irregularities and depressions) leading to the top,
where there is a minute "crater." At the bottom lies a
marble, and the puzzle is to get such marble to the top and
down again. The marble mast not be jerked into the re
quired position (this being a comparatively easy feat), but
worked steadily up and down the inclined plane.*
To an unpractised hand the feat seems almost impossible ;
but we have seen an expert work the ball to the top and
down again with the greatest apparent ease.
No. II.— The Planet Puzzle.
This very pretty puzzle also emanated from the Pick
MeUp Office. It is a shallow tray five inches in dia
meter by threequarters of an inch in depth, and, like the
puzzle last described, closed at top with a fixed glass lid.
(See Fig 2.) On the bottom are traced concentric rings, re
Fig. 2.
presenting the orbits of various heavenly bodies, with the
sun as centre. At a given point in each such orbit is a
minute depression, marked with the name and a pictorial
representation of the corresponding planet, the list com
* There is a miniature form of the " Pick Me Up " Puzzle made in
brass, about the size of a crownpiece, a swanshot taking the place of
the marble.
Puzzles of Dexterity and Perseverance. 3
prising (in addition to the Sun~ and Earth) Mercury, Venus,
Mars, Jupiter, Saturn, Uranus, and Neptune. The heavenly
bodies themselves are represented by minute marbles, of
various sizes and colours, corresponding Avith the represent
ations of the various stars ; and the task of the experimenter
is to coax each marble into its proper position on the chart —
a by no means easy matter. Indeed, by the time he suc
ceeds, the aspirant should have acquired a very fair idea
of the sizes and positions of the more important members of
the Solar System.
No. III.— The 1 Dotting' Puzzle.
This is a shallow cardboard box, with glass top, in which
are enclosed five leaden pellets. The central space (see
Fig. 3) is surrounded by vertical cardboard partitions, the
only access to the centre being by means of sundry minute
Fig. 3.
archways, cut in the lower halves of such partitions. ^ In the
centre is printed, in "script" type, the word Indivisibility,
and the task of the experimenter is so to manipulate the five
pellets that they shall each rest on the dot of one of the i's,
each of which has a minute depression to receive it.
No. IV.— The Spider and the Flies.
This is a puzzle on the same principle, but with a differ
ence which renders it one of the prettiest of its kind. It is
a little wooden tray with glass top. (See Fig. 4.)
Puzzles Old and New.
On the piece of cardboard which covers the bottom is
delineated a spider's web, with a depression in the centre
for the reception of the " spider," which is represented by a
globule of mercury, A. At different points of the web are
spots of colour, two blue and two red, while the "flies" are
represented by four circular wads of felt, a quarter of an
inch in diameter, these also being two of each colour.
The puzzle is, by means of the globule of mercury, to coax
the four flies to their proper positions, red on red, and blue
blue; and finally to leav.e the "spider" in the centre
on
of the web.
No. V.— The Tower Bridge Puzzle.*
An unusual amount of imagination has been expended
over the surroundings of this puzzle. The bottom of the
box (see Fig. 5) represents a moat, professedly dividing
Tooley Street from Great Tower Street. An inclined plane
leads from the bottom of the moat to a gallery or quay on
the Tooley Street side. This side is connected on the gallery
level with a balcony on the opposite side, the intervening
space being crossed by a narrow bridge of a somewhat pe
culiar construction. The bridge is divided in the centre, each
half working on a very light springhinge, so that it rests
normally in a horizontal position, but the moment any weight
is allowed to rest upon it, it drops, and the unwary passenger
* Manufactured by Messrs. Feltham .& Co.
Ptizzles of Dexterity and Perseverance. 5
is precipitated into the moat below. The floor space of the
balcony is divided into five squares, coloured white, black,
bine, pink, and green respectively, and at the bottom of the
moat lie five marbles of corresponding colours.
Fig. 5.
The puzzle is to get the marbles one by one up tlie inclined
plane, across the bridge, and into the balcony, each, on its
proper colour. By a knack acquired by practice, the marbles
can be shot across the bridge so quickly that they have not
time to drop into the moat below. Beyond this, the only
key is patience and perseverance.
No. VI.— In the Soup.
The appliances for this puzzle (introduced by Messrs.
Perry & Co.) are a little red earthenware bowl, three inches
in diameter by two in height, a glass marble, and a "spoon,"
Fig. 6.
consisting of a bit of wire set in a wooden handle, and
slightly curved and flattened at the opposite end. (See Fig. 6.)
The bowl is placed on a table or other flat surface, and the
puzzle is to get the marble out with no other aid but that of
Puzzles Old and New.
the spoon. It must not be tossed or flirted out, but gently
coaxed up the side and over the edge. _ *
This is by no means a puzzle of the most difficult kind,
but demands a steady hand and a considerable amount of
patience, the ball having a provoking way of escaping from
the spoon and falling "in the soup " again, just at the very
moment when the neophyte thinks that he has at last suc
ceeded. With juveniles it is invariably popular, and affords
a cood deal of amusement even to older members of the
community.
No. VII.— The Matrimonial Chair.
This again depends upon the manipulation of a marble,
but after a different fashion. Within a cardboard box, in
Fig. 7.
this case not glasscovered, but open, and five inches square
by one and a half deep, is constructed a sort of miniature
circus. At one point of the circle is fixed a little chair of
sheet brass, — the " matrimonial chair," — and facing it (see
Fig. 7) an anglepiece of the same material, of the same
height as the seat of the chair, and with its apex about an
inch distant from it. Loose in the box are four small
marbles of different colours, representing four different
characters : a blue marble standing for the gay young
bachelor ; a red marble for the old maid ; a green marble
appropriately representing a bashful young man ; and a
Ptizzles of Dexterity and Perseverance. 7
yellow marble (not quite so appropriately) a giddy young
girl.
The experimenter is invited to select "whichever of these
he pleases, and, without touching it, so to manipulate it as to
seat it in the chair.
Both dexterity and perseverance will be needed to achieve
success, but there is a definite method to follow, apart from
which neither dexterity nor perseverance will be of much
avail. Having chosen a given marble, get it singly within
the anglepiece, and depress the " chair " end of the box
very slightly, so that the marble may run down into the
angle. Then, holding the box in the left hand, with the
chair facing you, give a little fillip with the right thumb
against the bottom of the box, just below the marble ; when,
if your degree of force and angle of inclination have been
discreetly calculated, the marble will jump over the screen
and rest on the seat of the chair.
No. VIII.— The "Tire 'Em Out" Puzzle (L'Enervant).
This is a puzzle of French origin. It consists of a little
circular salver about four and a half inches in diameter,
rising gently towards the centre, at which point there is a
slight depression. (See Fig. 8 )
Fig. 8.
A marble, half an inch in diameter, is laid upon the salver,
and the operator is required to work it into the pit in the
centre.*
* Our young readers can construct a homemade version of this
puzzle in a rough and ready way, by gluing a ring of stout cardboard
(say one inch in external, and half an inch in internal diameter) in the
centre of an ordinary cheeseplate..
s
Puzzles Old and New.
The easiest method of getting the ball into position is to
work it close up to the centre, and then drop the salver
with a slight jerk to a depth of half an inch or so. With
a little dexterity the ball may easily be caught and retained
in the centre. This method is, however, scarcely legitimate.
The ball should be rolled, and not jerked, into the desired
position.
No. IX.— The Electric Ball.
This also is a French importation. It is more of a game
than a puzzle, though it may be presented in either shape. _ It
consists (see Fig. 9) of a hollow elbow piece, ABB, to which
is attached a sorb of miniature gallows, G. From the
Fig. 9.
middle of this projects a ring, D, and suspended from its
upper arm swings a little piece of strongly magnetized iron
wire, E. A gilt cork ball, F t into which are thrust six little
iron pins with their heads projecting, completes the appar
atus.
The ball being placed on the open end of B B, the operator
brings A to his lips and blows through the tube, endeavour
ing to force the ball upwards through the ring, and bring
one or other of the pinheads into contact with the magnet
ized wire, E, when the ball will remain suspended.
Puzzles of Dexterity and Perseverance. 9
Some little skill is required in order to blow in the right
way. The force nmst be gentle, but continuous. A novice
generally endeavours to blow the ball into position by a suc
cession of quick puffs, a method which inevitably results in
failure.
No. X.— The "Hang Him" Puzzle (Le Pendu).
This is a variation of that last described, and is productive
of even more amusement. The apparatus is in this case of
wood instead of metal. The little magnet is replaced by a
circular loop of fine wire, about an inch in diameter, and the
ball (which is somewhat smaller than the " Electric ") is
traversed by a thin wire, not quite two inches in length, ter
minating at its upper extremity in a little hook. The
straight portion of the wire projects about an inch in the oppo.
site direction. This is placed on top of the vertical tube cor
responding to BB in Fig. 9, the straight portion of the wire
passing down the tube. The experimenter then blows
through the mouthpiece, just as in the last case. The little
ball hovers in the air for a while, hook upwards, the weight
of the straight part of the wire keeping it pretty nearly
erect, and if the neophyte is fortunate, ultimately hooks
itself on to the little loop, and remains suspended.
The blowing, as before, should be gentle, but sustained,
when the ball will nutter in the air, after the manner of
the larger ball sometimes seen on the top of a fountain, with
extremely pretty effect.
No. XI.— BouciBoula.*
This again is a puzzle of French origin. It consists (see
Fig. 10) of a double bulb of blown glass about three inches in
length, in which are contained some fifty minute pellets
Fig. 10.
(resembling the sweets known to children as " hundreds and
thousands ").
* Presumably an abbreviation of " Bouleci, Boulela" (a ball to this
side, a ball to that side).
io Puzzles Old and Neiv.
Half of these are red, and half white, and the puzzle is to
separate them, getting all the white pellets into the one
Imlb, and the red into the other.
At first sight, one would be disposed to declare the task
impossible, but the puzzle is really an easy one of its kind.
The first step is to tilt all the pellets into one bulb, then
shake them till three or four of one colour, say red, lie
together near the neck. . Tap the full bulb on the under side
with the fingernail till these are one by one worked into the
opposite bulb. Then give the puzzle another little shake,
till a few more reds lie next to the red side ; work these
over, and so on, till all the reds have crossed the bridge. If,
as will occasionally happen, a jDellet'of a wrong colour
should find its way into the opposite bulb, it must be worked
back again before proceeding further.
With a steady hand and light touch, five or six minutes
should suffice to complete the separation,
No. XII.— The Switchback,
This is a bent glass tube, about seven inches in length by
half an inch in diameter, closed at each end. (See Fig. 11.)
Fig. 11.
At each of the points A, B, C is a little hollow or depres
sion. Enclosed within the tube are three swanshot, and
the puzzle is to get one of these into each of the three de
pressions.
Success depends mainly upon dexterity and perseverance ;
but the task of the operator becomes very much easier if he
sets about it in the proper fashion. The expert begins by
turning the tube upside down, so that A, B and G shall, for the
time being, be uppermost. There is now below each of the
three points named a much larger depression, into which it
is comparatively easy to get one of the three balls. He then
gradually turns the tube round until A, B and are again
undermost ; and if his hand is steady enough, the trick is
done, each pellet resting at its proper point.
Puzzles of Dexterity and Perseverance. 1 1
The series of puzzles i. t described is for the most part
only niade in a very sma. size, suitable for the waistcoat
pocket. They are usually of stamped sheet brass with fixed
glass top, enclosing one or more leaden pellets (taking the
place of the marble in the larger puzzles), which the operator
is required to work into certain specified positions. We will
begin with
No. XIII.— The Five HorseShoes.
Five horse shoes are stamped in relief on the bottom of
the box (see Fig. 12) and five pellets are enclosed, one of
which is to be worked into each of the horseshoes.
A very moderate degree of perseverance will in this case
ensure success ; but there is another and more difficult form
Fig. 12.
of the same puzzle, in which the space enclosed by each
horseshoe is differently coloured, say red, green, blue, black,
and white respectively. The five pellets are of correspond
ing colours, and each must be coaxed into the space bearing
its own colour.
No. XIV.— The Two HorseShoes.
Fig. 13.
This is very nearly identical with the last item, save that
there are in this case two horseshoes only (see Fig. 13), and
12
Puzzles Old and New.
that their points face inwards, instead of outwards, while the
space on either side is occupied by three raised partitions.
There are eight pellets, and the experimenter is required to
work four of them into each horse shoe.
No. XV.— The Maze.
(See Fig. 14.) Here five pellets are enclosed, and the
operator is required to work them all into the central com
Fig. 14.
partment of the "maze." This is a more difficult puzzle
than that of the horseshoes, but still well within the reach
of a patient operator.
No. XVI.— The Pitfall Puzzle.
The puzzle to which, for want of a better, we have given
the above name is another variation of the same idea.
(See Fig. 15.) It has a, circular enclosure in the centre,
Fig. 15.
with a single narrow opening. Half a dozen pellets are to
be worked into this enclosure : in itself no easy matter, but
the difficulty is enhanced by the fact that the surrounding
Puzzles of Dexterity and Perseverance. 1 3
space is indented with some twenty or thirty minute de
pressions or pitfalls, which seriously impede the pellets in
their jonrney to the centre.
No. XVIL— The Fifteen Pellets Puzzle.
This is the same idea reversed. In this case (see Fig. 16)
there are fifteen pellets and a like number of indentations to
receive them, but there are five larger depressions, not to
Fig. 16.
mention a circular channel or moat in the centre of the
puzzle, into which the balls have an awkward knack of
rolling, rather than to their intended, destinations.
No. XVIII.— The Cross Puzzle.
This is a puzzle very simple in appearance, but demand
ing considerable skill for its successful manipulation. It has
(see Fig. 17) four pairs of raised partitions, arranged in the
Tig. 17.
form of a cross, with their opening towards the centre.
There are four pellets, and these are to be worked one into
each compartment.
H
Puzzles Old and New.
No. XIX.— The Hand of Cards.
This is on the same principle, but the figure occupying
Fig. 18.
the central space (see Fig. 18) represents a Land of cards.
There are five pellets, one of which must occupy each angle.
No. XX.— The Pig Puzzle.
We have here (see Fig. 19) a pig in basrelief, with a
ring growing out of the middle of his back. There are three
Fig. 19.
pellets, which are to be so manipulated as to rest, one be
tween his forelegs, one between his hind legs, and one in the
ring aforesaid.
No. XXL— The Four Colours Puzzle.
The ground plan of the puzzle is in this case as shown in
Fig. 20. There are four spaces or stalls, each distinguished
Puzzles of Dexterity and Perseverance. 1 5
by a letter, E, W, B, M (in the actual puzzle stamped in
the space itself), standing for red, white, black, and mauve
B W B M
Fig. 20.
respectively. There are four pellets, one of each colour ; and
the aspirant is required to work each pellet into its proper
compartment.
No. XXII.— The Amphitheatre Puzzle.
This (see Fig. 21) is, after a fashion, the converse of the
" PickMeUp " Puzzle, described at page 1. It has a series
of grooved channels, with openings at various intervals, and
Fig. 21.
the experimenter is required to work the pellets, of which
there are seven, down the inclines into the central space.
It is one of the easiest of this class of puzzles.
No. XXIII.— The Persian Shah,
In this case a little cardboard box with glass top encloses
the pictorial representation of the head of a handsome cat,
16
Puzzles Old and New.
the "cliat" referred to (see Fig. 22). A couple of pellets,
representing the pupils of the eyes, are to be worked into
their proper positions, slight indentations being made to
receive them.
No. XXIV.— The Balance Puzzle.
This (see Fig. 23) is a puzzle of a rather novel kind, and,
of its class, of more than average merit. It consists of a
little brass dish, a couple of inches in diameter, with glass
top, enclosing two leaden pellets. Revolving freely on a
Fig. 23.
pivot in the centre is a tiny brass bar, stamped into a
minute cup or hollow at each end, and forming a balance.
The task of the experimenter is to get the two pellets to
rest one in each scale, a work of considerable difficulty,
inasmuch as the merest touch disturbs the equilibrium of
the bar.
No. XXV.— The Fish Puzzle.
This is a variation of the same idea. The balance is in
this case replaced by the representation of a stamped metal
Ptizzles of Dexterity and Perseverance. 1 7
fish, in tolerably high relief, with open mouth. This re
volves freely on a central pivot, like the balance in the
puzzle last described (see Fig. 24). Enclosed with the fish
are some twenty very minute pellets, and the experimenter
is required to get such pellets one by one into the fish's
Fig. 24.
mouth, till all have found a restingplace in his belly. The
task is not easy, for the mouth is only just large enough to
allow the passage of one pellet at a time, and any slight
accidental movement of the fish will cause him to spin round
on the pivot, and consequently to disgorge all those already
swallowed.
No. XXVI.— The Marksman.
The place of the " fish " is here taken by a liRle figure of
a kneeling soldier, with his gun to his shoulder, as if in the
act of firing (see Fig. 25). This figure is not, like those of
Fig. 25.
the pig and fish, pivoted in the centre, but slides about
freely between the glass top and metal bottom of the box.
There are four pellets, and the aspirant is required to get
Q
i8
Puzzles Old and New.
three of them on to the barrel of the gun, and one on the
bent knee of the marksman.
No. XXVII.— The Snake and Bird.
This may be regarded either in the light of a game or of
a puzzle, though there is an element of chance about it
which to some extent bars its claim to rank as a puzzle
proper.
It consists of a cardboard medallion or counter represent
ing a bird, and a snake, of many convolutions, of the same
material, chopped up into ten segments, more or less curved,
and in length ranging from three to six inches (see Fig.
26).
The '' bird'' is laid upon the table, and the tail segment
of the snake placed haphazard a couple of feet or so away
from it. The experimenter is then required, beginning from
Fig. 26.
the tail, to reconstruct the snake by adding segments one
by one, till the head is reached ; his object being so to place
the head at the finish that the " bird " shall be just within
the open jaws. His prospect of effecting this will, of course,
depend upon the arrangement of the successive segments,
the choice of a wrong segment, or the selection of an upward
instead of a downward curl (or vice versa) at a given point,
making just the difference between success and failure.
The various segments (which are coloured on both sides,
and may therefore be used with either side uppermost) must
be so placed as to butt fairly one against another, no over
lapping being permissible,
Puzzles of Dexterity and Perseverance. 19
No. XXVIII.— The Coin and Card Puzzle.
Balance a playingcard horizontally on the top of your
left forefinger, and on it lay a coin, say a halfcrown or a
penny, so that both shall be in equilibrium. You are now
required to remove the card without touching the coin.
Any one not in the secret usually endeavours to draw
away the card by slow degrees, when failure is the inevitable
result. The proper method is as follows : give the corner
of the coin a smart "fillip" with the second finger of the
right hand. If this be done exactly in the plane of the card,
the latter will be shot away with a sort of spinning motion,
the coin remaining undisturbed.
■o
No. XXIX.— The Egg and Card Puzzle.
This is the same puzzle in a slightly altered form. Fill a
wineglass half full of water, and over its mouth lay a play
ingcard. On the centre of the card place a weddingring
(or other fairly stout ring of similar dimensions), and with
the aid of this balance an egg, small end upwards, upright
on the card. You are now required to remove the card, and
let the egg fall into the water, without touching egg or
ring.
The modus operandi is the same as above described. The
ooift being neatly flicked away with the second finger, the
egg and ring will fall into the glass. The water prevents
any injury to the egg.
CHAPTER II.
MECHANICAL PUZZLES DEPENDENT ON SOME TRICK OE SECRET*
We will commence this section with a few examples of the
pretty extensive class of puzzles in which the crux pro
pounded is to extract a marble or other small object from
some outer receptacle which has no visible opening large
enough to allow of its passage, or which cannot be opened
without the knowledge or discovery of some secret.
Fig. 27.
No. I.— The Barrel and Ball.
This is a small barrel of boxwood. There is a hole in the
* There are but few London dealers who make a speciality of Puzzles.
The largest assortment is probably that of Messrs. Hamley, of 231, High
Holborn, whose stock is peculiarly varied and extensive. Good selections
will also be found at Bland's, 35, New Oxford Street ; Jaques & Son's,
102, Hatton Garden ; and Passmore's, 124, Cheapside. The above are the
only firms, within the writer's knowledge, that go in at all extensively
for this class of goods.
Where the publisher or manufacturer of a particular puzzle is known,
he has been duly credited with it. The name of the actual inventor is,
in most cases, unfortunately unascertainable.
20
Puzzles Dependent on some Trick or Secret. 2 1
upper end, wherein rests, quite loosely, a little ebony pestle.
(See Fig. 27.)
The barrel contains an ordinary marble, considerably
larger than the hole (the only visible opening), and the task
of the experimenter is to get the marble out.
No. II.— The Dice Box.
This also is in the shape of a miniature barrel (see Fig. 28),
Fig. 28.
the bung and spigot being represented by movable plugs of
ebony. The barrel contains three little dice, and the puzzle
is to get them out.
No. Ill— The " Churn."
This is another boxwood puzzle. It is in the shape of a
little churn (see Fig. 29). There is a small hole in the top,
Fig. 29.
from which projects the handle of the plunger. Closer in
spection, however, discloses the fact that the handle is a
22
Puzzles Old and New.
handle only, being a mere ping of ebony, as shown at a,
quite unconnected with anything in the interior.
The churn contains a bone or wooden counter, which the
experimenter is required to extract from it.
No. IV.— The Cannon and Ball Puzzle.
The receptacle in this case takes the form of a cannon (see
Fig. 30). It contains a small marble, which the aspirant is
Fig. 30.
required to extricate. The only visible opening is a small
oval hole at the muzzle.
The breach is movable, as shown in the diagram, being
screwed into the cannon, but the cavity revealed by its
removal is only just large enough to receive it, and has no
connection with the portion containing the ball.
No. V.— The Cage and Ball.
In this case the experimenter is again required to extricate
a ball, but under different conditions. The ball (a glass
Fig. 31.
marble) is enclosed in a little boxwood cage (see Fig. 31)
consisting of a cylindrical top and bottom, united by four
upright pillars, between which the ball is confined.
Puzzles Dependent on some Trick or Secret. 2,
No. VI.— The Castle Moneybox.
This is a neatly turned box in the form of a castle or
tower (see Fig. 32), with a slit across the top large enough
The coin actually in the box is, however,
to admit a shilling.
Fig. 32.
a penny (which is too large to pass through the slit), and
the box has no other visible opening.
The aspirant is told that the box contains money, which
he may keep for himself if he can get it out ; but he is
very unlikely to do so without a knowledge of the secret.
No. VII.— The New Castle Moneybox.
This (see Fig. 33) is very much like the last puzzle in
Fig. 33.
appearance, taking the shape of a little boxwood castle or
24
Puzzles Old and New.
tower, about two and a half inches in height. It differs
from it, however, in one important particular — viz., that it
has not any slit at the top, nor indeed any visible opening
whatever. A coin is placed in the box, and the puzzle is to
extricate it. To an outsider the task seems hopeless, but any
one in the secret can extract the coin with perfect ease.
No. VIIL— The Brass Moneybox.
This also is a moneybox, but of brass, and of more elabo
rate construction. It is somewhat smaller than the last
Fig. 31.
described, and in outward appearance as depicted in Fig. 34.
A coin is dropped in through the slit at top, and the puzzle
is to extract it.
No. IX.— The Captive Sixpence.
This is a puzzle of American origin. It is a neat little
nickelplated box, of very small size, being only an inch in
Fig. 35.
height, and in diameter just large enough to admit a sixpence
(see Fig. 35). On the top, shut in by an overlapping edge of
metal, is seen an actual sixpence. The top, with the six
Puzzles Dependent on some Trick or Secret. 25
pence upon it, is depressible to the .extent of a quarter of an
inch. Just below this point is a horizontal slit, a sixteenth
of an inch wide, and extending halfway round the circum
ference; but the opening is closed by an inner tube of metal,
forming one with the movable top.
The puzzle is to extricate the sixpence. The experimenter
can see at a glance that if he could depress the top below the
level of the horizontal slit, the slit would be left open, and
would afford an exit for the coin. But the top stops just
short of this point, and the coin remains imprisoned until
the " open sesame " is discovered.
There is a large class of puzzles in which the task to be
performed consists of the removal of a ball or ring from a
silken cord, passed backwards and forwards through open
ings in a piece of wood or bone in a more or less complicated
manner. One of the simplest examples of these is —
No. X.— The Cannon and Cord.
Fig. 36.
This is a miniature boxwood cannon, a couple of inches in
26 Puzzles Old and New,
length. There is a perpendicular slot, starting at the "touch
hole " and traversing the whole length of the gun. A loop
of cord, with a small ball on its opposite end, is threaded
through the touchhole. To prevent the cord being drawn
out through the slot abovementioned, the cannon is encircled
by a brass ring, and this is in turn kept from coming off by
a couple of brass pins, as shown in Fig. 36.
The puzzle is to detach the cord and ball from the gun.
No. XI.— The Heart Puzzle.
This is a very easy puzzle. It consists of a heartshaped
piece of boxwood, through which is threaded a silken cord
Fig. 37.
terminating in a glass ball, as shown in Fig. 37. .The
problem to be solved is the detaching of the cord and ball
from the heart.
No. XII.— The Alliance (otherwise known as the
Victoria) Puzzle.
This is composed of two pieces of boxwood, attached
together by a cord, on which is strung a bone counter with
two holes in it (see Fig. 38, giving a front and back view
of the puzzle).
Puzzles Dependent on some Trick or Secret. 27
JtJ.ll.'tHl/sC. g »ffi^t ro»
Fig. 38.
The problem is to remove the counter from the cord.
No. XIII.— The Two Balls.
Fig. 39.
This consists of a flat piece of ivory or bone, from "which,
by means of a silken cord, two glass balls are suspended,
one on each bight of the cord, as shown in Fig. 39.*
The puzzle is to get both balls on the same bight of the
cord.
* Fig. 39 represents the puzzle, as sometimes sold, with the two end
knots coming from the back to the front, in the same direction as the
central loop. This introduces a needless element of complexity. If the
loop comes from back to front, the cords at the ends should pass from
front to back. If the neophyte chances to come across the puzzle in
the shape shown in our illustration, he is recommended to untie the end
knots, and retie them with the cords passed through in the opposite
direction.
28 Puzzles Old and New.
No. XIV.— The Ariel Puzzle.
The puzzle to which this name is given is composed (see
Fig. 40) of a flat wooden base with two turned pillars rising
from its upper side. Through the head of each is a hole.
Through these holes are threaded a couple of silken cords,
drawn taut, passing through other holes at the four corners
Fia. 40.
and secured on the under side by glass beads, the upper part
forming a sort of bridge between the two pillars. Over this
bridge is looped a third cord, which passes down through a
hole in the base between the two pillars, thence through a
glass ball, and then up, through a second hole, again to the
bridge, above which the two ends are tied in a knot. The
ball, therefore, hangs suspended below the base, and the puzzle
is to take it off the cord, of course without untying the knot.
No. XV.The Pen and Wheel.
The Pen and Wheel (brought out by Messrs. Perry & Co.)
is another ingenious application of the same principle. It
consists of a little iron wheel, with six spokes, and a hole in
the centre (see Fig. 41). A double cord, about twenty inches
in total length, is looped over one of the spokes, then laced
in and out between the remaining spokes, and finally brought
Puzzles Dependent on some Trick or Secret. 29
through the hole in the centre. It is then passed through
a bit of stamped brass in the form of a pen, and secured
by a double knot on the opposite side.
Fig. 41.
The puzzle is to detach the cord and pen from the wheel
■without unfastening the knot.
No. X VI.— The Balls and Rings.
We have here a flat mahogany ring, 3 inches in diameter,
and \ inch thick, in appearance not unlike a miniature life
buoy. In this are a number of holes, through which passes
Fig. 42.
a silken cord threaded through glass balls (usually six) on
the one side, and through a like number of bone rings on the
other. (See Fig. 42.) The puzzle is to reverse the balls and
rings— i.e., to bring the balls into the place of the rings, and
vice versa.
3°
Puzzles Old and New.
No. XVII.— The Chinese Ladder.
This is a puzzle of a different kind, the "loop" element
being in this case wanting. It is said to be a genuine
importation from China. It consists of a small wooden
ladder of four steps (see Fig. 43). Each step has two
Fig. 43.
boles in it. A silken cord, a yard in length, secured at
top with a knot and glass bead, is threaded through each
hole in succession (down one side and up the other). Be
tween each pair of holes it is also threaded through a hole
in a bone counter, so that there are two counters in each
compartment. To the free end of the cord is attached a
stout needle.
The puzzle is to bring the whole of the counters together
on the cord.
No. XVIILThe Staff.
This (see Fig. 44) is a miniature staff or truncheon — some
times shaped like that carried by a policeconstable; some
/
Puzzles Dependent on some Trick or Secret. 3 1
times of the plainer pattern shown in our illustration; — with
a hole running from end to end through the centre. Through
this is passed a double cord, knotted in four places, with a
Fig. 44.
ball at either end. Notwithstanding the knots, the cord
can be drawn backwards and forwards freely through the
staff.
The puzzle (a very easy one) is to remove the balls and
cord from the staff.
No. xix.— The Imperial Scale.
This is a puzzle of a more difficult character, though, like
many others, easy enough when you know it. It consists
(see Fig. 45) of a flat piece of boxwood, three inches square,
Fig. 45.
with eight holes in it, four in the centre and one at each
corner. Through each of the corner holes passes a silken
cord, about four inches in length, secured by a glass bead on
3 2
Puzzles Old and Nezv.
the under side, and united with the. rest in a knot at top.
A fifth cord, of rather more than double length, is passed
downwards through two of the centre holes, then up again
through the other two and through the loop formed by the
passage of the cord through the first pair of holes, the two
ends then being made to form part of the general knot at top.
Between the loop and the standing part of the central cord
is secured a ring, of bone or metal.
The problem is, without untying either of the knots, to
detach the ring from the cord, and again to restore it to its
position.
No. XX.— The Sceptre.
This (see Fig. 46) is a little wooden rod, a fanciful repre
sentation of a sceptre, with a couple of rings upon it, pre
Fig. 46,
vented from coming off: by an ornamental knob at each end.
Round each knob are a number of little bosses.
The puzzle is to remove the rings from the rod.
No. XXL— The Balls and Chain.
This is somewhat similar in general idea, but the prin
ciple is different. Two wooden balls (see Fig. 47) are here
Puzzles Dependent on some Trick or Secret, 35
united by a turned bar, on which they revolve freely. Round
the bar is an endless chain, too short to be passed over the
ball at either end.
The puzzle is to remove the chain.
No. xxil.The Four Keys.
This is a puzzle of a different variety. It consists of a
boxwood disc (2 inches in diameter by f inch in thickness)
with a crossshaped opening in the centre, as a in Fig. 48.
Fig. 18.
Through this disc (the "lock"), are passed four boxwood
"keys," each consisting of a narrow slip of wood uniting two
broad, flat ends (see b in the same figure). When all are in
serted, the effect is as shown in Fig. 49. The puzzle is to
Fio. 49.
disengage the keys from the lock ; no easy matter, for when
once inserted they instantly fall crosswise in various direc
tions, each blocking the others when you endeavour to
extricate them.
14
Puzzles Old and New.
No. XXIII.— The Screw Box.
This is a little box a couple of inches in height, and
shaped as shown in Fig. 50. The lid moves round and round
Fig. 50.
with the utmost freedom, but cannot be removed without a
knowledge of the secret.
No, XXIV.— The Ball and Three Strings.
Fig. 51.
This is a very pretty puzzle. It consists of a boxwood
ball, not quite two inches in diameter, with a cylindrical
opening down the centre (see Fig. 51). At equal distances
Puzzles Dependent on some Trick or Secret. 35
apart, round tlie circumference, are three smaller holes, con
verging* to the centre, and through these pass three loops of
cord, interlaced in the centre. The opposite ends of each pair
of cords are secured by a knot, and further prevented, by a
glass bead, from being pulled through the ball.
The puzzle is to detach the beads and string from the
central ball.
No. XXV.— The Lighthouse.
This is a miniature representation in boxwood of a light
house, with a ring of lignum vitce, b, loosely encircling its
centre (see Fig. 52). The " lantern " at top and the base at
bottom are both larger than the inner circumference of the
Fig. 52.
ring, which it is the object of the puzzle to remove. The solu
tion which naturally suggests itself is to screw off the lantern ;
but this is found to revolve freely at the point a, and the
only other movable portion of the puzzle is the little pinnacle
at top, which may be taken out at pleasure.
No. XXVI.— The Jubilee Puzzle Box.
This is a little cylindrical box, fashioned as shown in
Fig. 53, with a little movable pin at top. The experimenter
36
Puzzles Old and New.
is informed that it contains, somewhere or other, a portrait
of Her Majesty, and the puzzle is to find it.
Fig. 53.
No. XXVII.— The Jubilee Penny.
This is a French penny of the Second Empire, and bear
ing accordingly the eSigy of Napoleon III. The puzzle is
to find the Queen's head on it, a somewhat difficult thing to
do. And yet the likeness of Her Majesty, when found, is un
mistakable, and the ingenuity of the puzzle will deservedly
elicit high praise.
No. XXVIII.— The Invisible Gift.
This is a small box, very similar in appearance to that
depicted in Fig. 53. The Queen's head is in this case repre
sented by a concealed coin, a sixpenny or. threepenny piece,
and the puzzle is to extract it, the box being to all appear
ance empty.
No. XXIX.— The Arabi Gun.
This is a neatlyturned boxwood cannon, with the ball
peeping out at the muzzle (see Fig. 54). The ball is not a
fixture, but is held in position by the pressure of an internal
spring. It may be forced inward to a limited extent, but
Puzzles Dependent on some Trick or Secret,
the muzzle is too small to allow of its exit in that direction.
Some other means must, therefore, be found of effecting its
extraction, which is the problem required to be solved.
It should be mentioned that b and c are in one piece,
chase, a, is independent, revolving freely on c.
The
It is curious what a large amount of ingenuity has been
expended over the construction of puzzle matchboxes. . It
would almost seem as if human nature took a malignant
pleasure in offering a friend a match from a box which he
cannot open, and which, in some cases even does him a mild
personal injury by unexpectedly pricking his fingers. One
of the oldest, and at the same time one of the cleverest, of
puzzle matchboxes is now known as —
No. XXX.— The Psycho Matehbox.
This is an oblong box of wood or metal, of the shape in
dicated in Fig. 55. When made of wood it is usually about
3 inches in length by 1 \ inch in width, and f incli in depth ;
in metal, about half that size. In either case its principle
Fig. 55.
is the same. A portion (a) of the top works on a pivot, as
shown in our illustration, and any one taking up the box
for the first time naturally imagines that on pushing this
aside he will have access to the interior. But such is not
the case, nothing but a plain wooden surface (with or with
Puzzles Old and New.
out a striking plate for the matches) being visible below.
This under portion of the top — the true lid — is apparently a
fixture. The problem is to open it.
■No. XXXI.The "TouchMeNot" Matchbox.*
This is on a totally different principle. It is a nickel
plated box shaped as depicted in Fig. 56, and bearing on the
top the word " Precaution," The warning is justified by the
fact that any one endeavouring to open the box in the ordi
Fig. 56.
the "touchmenot" matchbox.
Fig. 57.
the " touchmenot " tobaccobox.
nary way (by pressing the little stud in front) produces no
result, save that a hidden needle in the stud pricks his finger,
and suggests the desirability of trying again in some other
quarter.
There is a puzzle tobaccobox, as illustrated in T?ig. 57,
made on the same principle.
No. XXXII.— The Magic Drawer Matchbox.
This is again a new departure. The box in this case takes
the form of a little drawer, working in an outer case (see
Fig. 58.
Jig. 58). There is not the least difficulty in pulling out
the drawer, but any one unacquainted with the secret finds
* This name is applied to boxes cf two or three different patterus
having as a common element the fingerpricking stud in front, this being
a rather favourite device of puzzle fuseebox makers.
Puzzles Dependent on some Trick or Secret. 39
it empty, though the owner may only the moment before
have shown it fall of matches, as it will again appear when
the secret is discovered.
No. XXXIII.— The " Unique" Matchbox,
This is a box of the upright pattern, with hinged lid (see
Fig. 59). It has the customary stud in front, but such stud
Fig. 59.
is in this case a mere makebelieve, the true " open sesame "
lying in another quarter, and being very ingeniously . con
cealed.
No. XXXIV.— The Surprise Matchbox,
This (see Fig. 60) is much like the lastmentioned in
appearance, though quite different in its working. It re
Fig. 60.
sembles the " TouchMeXot " in the fact that any one in
cautiously pressing the stud in front pricks his finger.
4°
Puzzles Old and New.
No. XXXV.— The New Brass Puzzle Matchbox.
This, sometimes described as the "Egyptian" Matchbox,
is fashioned like Fig. 61, being a ribbed cylinder of brass,
Fig. 61.
and having on the top a copper coin, with two holes in it.
This coin revolves in its centre, being rivetted to the top of
the box, which has no visible opening.
No. XXXVI.— The Ne Plus Ultra Matchbox.
This is, in our own opinion, .one of the best of puzzle
matchboxes. It is a flat box of plain metal, niokelled. It
has no visible hinges, and the lid, instead of overlapping',
Fig. 02.
is sunk within the box all round (see Fig. 62). The Ne Plus
Ultra is of the simplest possible construction, having neither
stud, spring, nor catch about it; but, save by accident, a
stranger will puzzle in vain to open it, though any one in
the secret can do so with almost magical ease.
Puzzles Dependent on some Trick or Secret. 41
No. XXXVII.— The Sphinx Matchbox.
We mention this next in order, the principle being very
nearly identical, though with some difference of detail. In
Fig. 63.
appearance the box is as shown in Fig. 63. It bears on the
lid the head of a sphinx, from which it takes its name.
No. XXXVIII.— The Puzzle Snuffbox.
This is a very old puzzle, but as good as it is old. It is a
neat little box, of the appearance shown in Fig. 64 ; air
tight, and in every respect a capital snuffbox. Its only
drawback to the uninitiated is that there is no apparent
way of openiug it. A keen eye can just trace what might
Fig. 64.
be the outline of a lid, but it sinks flush with, the surface of
the box, and the neophyte, after pulling and pressing in
every conceivable direction, is half inclined to believe that
he is the victim of a " sell," and that the box never was
intended to be opened at all. The owner, however, opens
it with perfect ease.
42 Puzzles Old and New.
No. XXXIX.— The New Puzzle Snuffbox.
This, though very similar in appearance, is a box of a
different kind. It is of the orthodox snuffbox shape, and
can be opened without difficulty, but no one can " take a
pinch," for the snuff lies below an inner slab, or false
bottom. There are two circular openings, each about the
size of a sixpence, through which the snuff is visible ; but
any one inserting his thumb and finger only pinches the
woodwork between the holes, and not the snuff.
The puzzle is, notwithstanding this obstacle, to take a
pinch of the snuff. This, of course, to be done in the legiti
mate manner, and not by emptying the snuff out of the
box.
No. XL.— The Puzzle Ball.
This is a boxwood ball, not quite two inches in diameter,
ornamented with six ebony bosses, with raised centres (see
Fig. 65.
Fig. 65). A threepennypiece or bonbon being inserted in
the ball, the experimenter is invited to extract it.
No. XLL— The Ebony Puzzle Ball.
This, sometimes known as the " Indian " puzzle ball, is
somewhat larger, of unpolished ebony, ornamented with rose
cutting in various directions (see Fig. 6Q). To the eye it
appears to be absolutely solid, the keenest eye failing to
detect any sign of an opening. And yet it can be opened
by any one having a knowledge of its secret, and will be
Puzzles Dependent on some Trick or Secret. 43
found to contain a receptacle for a dozen shillings or sove
reigns.
Fig. 66.
No. XLIL— The Puzzle Purse.
This (see Tig. 67) is a bag or purse of roan or morocco
leather, with two flaps, the inner one scored with longi
tudinal slits. What would ordinarily be the mouth of the
Fig. 67.
purse, aa, is stitched right across, so that it has no visible
opening. A coin is placed in the purse by some one in the
secret, and the puzzle is to extract it.
No. XLIIL— The Puzzle Pocketknife.
This is a very ingenious puzzle. It is a small single
bladed knife, with buckhorn handle, in appearance of the
simplest possible kind ; indeed, just such an article as a
schoolboy would buy for sixpence. Bat it has a peculiar
quality in the fact that no one, save with a knowledge of
its secret, can succeed in opening it. We have known hours
44
Puzzles Old and New.
expended, in vain, over the endeavour, though by the in
itiated it is as easily opened as any other pocketknife.
No. XLIV.— The Automatic Knife.
This is an elegant little knife, with oxidized silver handle,
and of the appearance shown in Yv*. 68. The puzzle is to
Fig. 63.
open it, the blades being unprovided with the customary
notch for the nail, or, apparently, any substitute for it.
No. XLV.— The Double Barrel and Ring*.
This (see Fig. 69) has a superficial resemblance to the
Balls and Chain (No. XXI.), but is wholly different in
principle. It consists of a substantial boxwood stem, with
a turned ball at each end, not revolving, but secured firmly
to it. On the stem slides an ebony ring, which it is the
object of the puzzle to remove.
No. XLVL— The Weddingring* Box.
Fig. 70.
This is a small box of hard wood, neatly turned and polished,
and fashioned as shown in Fig. 70, alike, top and bottom,
Ptizzles Dependent on some Trick or Secret. 45
Ifc contains a wedding ring, which can be heard to rattle
within, and which it is the object of the aspirant to extract.
No. XLVIL— The New Moneybox.
Fig. 71.
This is a box of different shape, being as depicted in Fig.
71. A coin is placed within, and the neophyte is invited to
extricate it, which he will find some difficulty in doing.
No. XLVIIL— The Zulu Box.
This has a considerable resemblance to the Screw Box
(No. XXIII.) . It is slightly different in shape, but has the
same peculiarity — viz., that the lid revolves freely on the
box, though without any nearer approach to the desired
consummation of removing it. The expedient which in
that case was found to be successful is here quite useless.
No. XLIX.— The New Persian Puzzle.
The puzzle to which this name is given is a little box
with a shaft running through the centre, and is professedly
intended to contain rings (see Fig. 72). The shaft termin
Fm. 72.
ates at each end in a little ornamental knob. The remainder
is in three separate parts — the cylinder in the middle and the
moulded portions at either end of it. All these revolve
freely on the central shaft, and the puzzle is to open the box.
4 6
Puzzles Old and New.
No. L.— The Magic Handcuff.
This is really more of a practical joke than a puzzle. It
consists of a woven tube, seven inches in length by half an
inch in diameter, made of some kind of very strong and
pliable grass. The victim is invited to insert his (or her)
little fingers one into each end of the tube (see Fig. 73).
He does so without suspicion, but when he again endeavours
Fig. 73.
to withdraw them, he finds it impossible to do so, the in
ternal friction holding the tube tight to the fingers, and
every additional "pull" causing the tube to contract more
and more in diameter, and so to grip the fingers the tighter.
In the case of a lady, the tube will probably be too
large for the little finger. If so, the second or third finger
may be inserted instead.
No. LI.— The Key and Ring Puzzle.
We have in this case a neatly finished brass key, of the
appearance shown in Fig. 74, with a wedding or other
fingerring upon its stem.
The aspirant is invited to remove the ring. The puzzle
Fig. 74.
may also be propounded in the converse manner — viz., the
key and ring may be handed separately to any person, and
he may be invited to pass the ring over the stem of the key
— an equally impossible task without a knowledge of the
secret.
Puzzles Dependent on some Trick or Secret. 47
]N T o. lil— The New Egg of Columbus.*
Every reader is, as a matter of course, acquainted with the
somewhat heroic expedient whereby Christopher Columbus
managed to make an egg stand upright. If the great
navigator were to try the same experiment with the new
egg that bears his name, ho might not succeed so easily, and
would probably have to appeal to some one in the secret to
show him how to do it.
The egg (see Fig. 75) is in this case of metal, with a little
raised disc or stud, a quarter of an inch in diameter, at its
base. The experimenter is required to make the ego* stand
upright on this. If the egg were empty, the task would be
easy enough, but some kind of weight is heard to roll about
within, and the continual shifting of this displaces the
centre of gravity. In spite of this, however, it is quite
possible to balance the egg, if the experimenter is acquaint
ed with the proper way to set about it.
* Perry & Co.
KEY TO CHAPTER II.
MECHANICAL PUZZLES DEPENDENT ON SOME TRICK OR SECRET
No. I.— The Barrel and Ball. Solution.
For the explanation of the secret of this puzzle see Fig. 76,
representing a section of the barrel and its contents. It
will be observed that the lower end consists of a plug screwed
into the body of the barrel, the difficulty of the puzzle con
sisting, first, in discovering this, and, secondly, in getting
any hold on the movable part, which when in position is
sunk flush with or below the lower edge of the barrel.
Fig. 76.
To unscrew the end, hold the barrel with the second
finger on the top of the pestle, and the thumb on the bottom,
and press vigorously. This binds all together, and makes
the three (pestle, ball, and plug) practically one piece.
plow nip the barrel between the thumb and two first fingers
of the opposite hand, and screw from left to right. The
body of the barrel revolving round the movable portion,
while this latter remains stationary, the effect will be that
after a few turns the bottom is screwed out of the barrel,
and the ball is released.
Key to Trick or Secret Pzizzles. 49
No. II.— The Dice Box. Solution.
This is a puzzle of a much simpler character. The bottom
of the barrel is made slightly tapering, and retains its posi
tion by force of friction only. To release it, all that is
needful is to tap the barrel (holding it perfectly perpendi
cular) smartly on a table or other flat surface, and after a
few attempts the bottom will fall out and the dice with it.
The bung and spigot have nothing to do with the solution,
but are doubtless added in order to lead the experimenter in
a wrong direction, and so increase the difficulty of the puzzle.
The spigot, however, is not without its use, for it may now
and then happen that the bottom is driven in too tightly to
be got out in the orthodox manner. In such case it may be
forced out by thrusting a piece of stout wire, or some other
convenient implement, through the spigothole.
No. III.— The Churn. Solution.
On examination of the churn (Fig. 29), it will be noted
that there is in front, midway between top and bottom, a
black spot a quarter of an inch in diameter, being apparently
a piece of ebony let into the boxwood. A similar circle, a
shade less in diameter, is found on the opposite side. These
two black spots, though apparently designed for mere orna
ment, are in reality the key to the puzzle. The churn is
made in two parts, the one fitting tightly into the other, and
the point of division being just below the third hoop. The
two supposed black spots are in fact the opposite ends of a
tapering plug of ebony, which passes through corresponding
holes in the two parts of the churn, and so locks all together.
To force out the plug*, press with the end of the plunger
a on the smaller of the two black spots. The plug once
removed, the two portions of the churn may be pulled apart
without difficulty.
No. IV.— The Cannon and Ball. Solution.
The cannon is very ingeniously contrived. On closely
examining the muzzle, it will be seen that the small oval
opening is surrounded by an incised circle, apparently placed
there for mere ornament. This is, however, in reality the
dividing line of a circular plug screwed into the muzzle.
E
5<3 Puzzles Old and New.
The movable breech is in two portions, the one screwing
into the other (see Fig. 77). On the removal of the inner
portion, it will be found that its smaller end is fashioned
Fig. 77.
into a square stud, slightly tapering, just fitting the opening
in the muzzle, and serving as a key to screw out the plug
abovementioned.
No. V.— The Cage and Ball. Solution.
Of the four upright pillars, though seemingly all fixtures,
one is in reality movable. On turning such pillar partially
round on its own axis, it will be found, when it reaches a
certain point, to sink about a sixteenth of an inch deeper
into the socket at one end. This draws it in the same degree
out of the socket at the opposite end, and it may then be
withdrawn altogether.
No. VI.— The Castle Moneybox. Solution.
The secret lies in the fact that the top of the box is
screwed into the main portion. To extract the concealed
coin, take a shilling or halfpenny, and inserting it half
way into the slot at top, use it screw driver fashion to
screw out the movable portion.
No. VII.— The New Castle Moneybox. Solution.
It will be observed (see Fig. 33) that near the base of the
box is a black band. The opening of the box is midway in
this band, the lower part screwing into the upper (see Fig.
78). So far there would be no difficulty in opening it. The
secret lies in the fact that at one point of each portion is
bored a little cylindrical hole, that in the upper, a, being a
quarter of an inch deep, and that in the lower, b, an eighth
of an inch ; so that when the two holes are brought the one
over the other they have a joint length of threeeighths of
an inch. In this cavity is a little brass bolt, a quarter of an
inch in length, working freely up and down within it.
Key to Trick or Secret Ptizzles.
5i
When the lower part of the box is screwed into the other,
(which is done with the " castle " bottom upwards) the final
turn of the screw brings the two holes opposite one another.
The " castle " being then turned right end upwards, the
little bolt drops partially into the lower hole, thereby locking
the two portions together.
Fig. 78.
To open the box, invert it, and give it a rap on the table
or some hard substance. This drives the little bolt back
into the upper hole, and the bottom may then be unscrewed
without difficulty.
There is a mark on the black band, above and below,
which shows when the box has been screwed up tightly
enough to bring the two holes into apposition.
No. Till.— The Brass Moneybox. Solution.
Fig. 79.
It will be found on close examination of this box that
it consists externally (see Fig. 79) of two brass cylinders,
the upper one, a, being a shade larger in diameter than the
52 Puzzles Old and New.
lower, b. This is in fact a tube, working within the larger
tube, a, with an upward and downward play of nearly a
quarter of an inch, a is attached, by means of an inner
tube, <?, to the brass disc which forms the bottom. The
lower part of such inner tube is partially cut away, leaving
a slot, d. By pushing b upwards within a this slot is exposed,
as shown in the figure, and the coin contained in the box can
be let fall through it. A downward movement of b closes
the opening, and the box is again as shown in Fig. 34.
With a little practice it is quite possible to open and close
the moneybox with one hand, the coin being in the mean
while invisibly extracted. To do this the box should be held
with the first finger on the top and the third on the bottom
of the box, when the thumb and second finger will be in the
right position to grip 5, and move it upwards or downwards,
as may be necessary.
No. IX.— The Captive Sixpence. Solution.
To release the sixpence, turn the box upside down, holding
it as vertically as possible. A little ball, which normally pre
vents the depression of the movable portion beyond the
slot, will then drop into a central cavity provided for it;
and the movable portion, being no longer impeded by the
ball, may then be lowered beyond the slot, when the sixpence
will slide ont through the opening.
No. X.— The Cannon and Cord. Solution.
Draw the cord down the slot towards the muzzle, as far
as the brass ring will permit. Slide the ring back again.
Take hold of the loop now formed on the cord, draw it com
pletely out of the slot, pass it over the brass pin on either
side, and it may then be drawn away altogether.
To replace the ball, pass the loop of the cord through the
ring, from the breech towards the muzzle. Pass it over the
pin on either side, and into the slot. You can then draw it
back again into the breech portion.
No. XI —The Heart. Solution.
On closely examining this puzzle it will be seen that in
the centre the cord forms a loop, which passes round the
opposite portion. Draw the ball close up to the heart ; then
Key to Trick or Secret Puzzles. 53
by means of the " slack " thus gained draw out the loop as
far as possible. Pass it down through the centre hole, then
through the next two holes, and back again to the front
through the bottom hole. Finally slip the loop 07er the
ball.* Draw back the loop, which is now disengaged from
the rest of tlie cord, and all will come off together.
To replace the cord, pass the loop first from the front
through the bottom hole, up through, one of the adjoining
pair of holes, and back again through the other, then
through the next pair in like manner. You now have the
loop brought to the front. Pass it clown through the centre
hole and up through the bottom hole, then over the ball,
and draw it back again. Pall down the ball as far as the
cord permits, and all will be as at first.
No. XII.— The Alliance Puzzle. Solution.
It will be found on examination that the cord is secured
at each end by a loop passing round the other portion. To
remove the counter, draw up either loop, and pass it through
the hole nearest the counter, then over the counter and the
opposite piece of wood. The cord is now free, and may be
drawn away altogether, when there will no longer be any
obstacle to the removal of the counter.
To reinstate the cord and counter reverse the process.
No. XIII.— The Two Balls. Solution.
Hold the puzzle with the central loop passing through the
hole to the front. Draw down this loop to its full extent,
pass the right hand ball through it, and then pass the loop
(with due precaution against twisting) through the right
hand hole (from the side opposite to that on which the knot
is), over the knot, and draw it back again. Repeat at the
opposite side, and the loop will be free. Draw it through
the centre hole, and you will have the cord hanging in a
* In all puzzles of this class, wherever a loop has to be passed over a
knot, ball, or other obstacle, special care must be taken that the loop is
clear — i.e., that the two cords constituting it are not twisted. If the loop
be passed in a twisted condition over the ball, the result is " confusion
worse confounded," and the difficulty of the puzzle proportionately in
creased.
54 Puzzles Old and New.
single bight, with the two balls side by side upon it. Run
them along the cord to either end. Reform the loop in
centre, and pass it through the middle hole, then (from the
front) through one of the end holes, over the knot, and draw
back again. Repeat at the opposite end, and the trick is
done.
No. XIV.— The Ariel Puzzle. Solution.
To detach the ball, first draw up the " loop " end of the
cord, and pass it through the hole in the top of the pillar
nearest to it, then down one of the corner holes on the same
side, and over the bead ; back again inside the inclined cord,
down the other corner hole on the same side, over the bead,
and back again, then back through the hole in the pillar.
It will be found that the loop is now clear, and the ball can
be drawn off it without difficulty.
No. XV.— The Pen and Wheel. Solution.
To detach the pen, draw this latter close up, or nearly, to
the wheel, so as to make the "loop" as long as possible.
Pass this under and over the spokes, side by side with the
portion already wound, finally passing it through the hole
in the centre and over the pen, then draw it back again in
the same direction and the cord will be free from the wheel.
The caution given on page 53, as to not twisting the loop,
is especially necessary in the case of this puzzle, the succes
sive passages under and over the spokes rendering such
twisting a very likely occurrence. In such case the cord,
when drawn back again, will be found rather more entangled
than it was in the first instance.
To reinstate the cord and pen, reverse the process.
No. XVI.— The Balls and Rings. Solution.
It will be observed, on close examination of the larger
ring, that two of the holes are very close together. These
form the startingpoint of the solution. Taking the puzzle
in the left hand, with, say, the rings uppermost, and these
two holes next towards you, you will observe that in the case
of one of them the cord passes through a loop, and round
the wooden ring from the one side to the other. Draw the
loop up through the hole, and through it pass the nearest
Key to Trick or Secret Puzzles.
55
ring, which will then
fall below the wooden ring, though
still suspended by the cord. Draw up the second loop
through its hole, and pass the second ring through both this
and the first loop. Proceed in like manner with the re
maining rings (each having one additional loop to pass
through). When you have got thus far, the rings will hang
side by side on the same bight of the cord.
Turn the wooden ring over, and go through the same
process with the six balls, passing them one by one through
the six loops in succession. You will now have the balls
Fig. 80.
hanging side by side with the rings, as shown in Fig. 80.
Now pass the balls through the rings so that they shall lie
to the left, instead of the right.
Having done this, turn over the wooden ring again, and
work the rings back through the loops. All six rings must
pass through the first loop, five through the second, jour
through the third, three through the fourth, hvo through the
fifth, and one through the sixth.
Turn over the Avooden ring once more ; pass the balls
through the loops in like manner, and the deed is done, the
puzzle remaining ready for use again.
No. XVII.— The Chinese Ladder. Solution.
This, though at first sight it appears somewhat formid
able, is in reality a very simple puzzle. Take the ladder in
56 Puzzles Old and New.
the left hand, with the small bead and knot undermost at
the same side. Drawing the cord moderately taut, twist it
twice round the lower righthand end of the ladder. Then
pass the needle up through the lower hole on the same side,
through the first counter, and so on till you reach the top ;
then, in the same way, down the holes and through the
counters on the opposite side. Yon have now exactly re
versed the process by which the counters were threaded into
position, and if you were to release the hitch you made
round the lower end of the ladder, and pull on the cord in
its now doubled condition, it would be drawn clean oat, and
the counters would fall off it. You are, however, required
to keep the counters still on the cord. To effect this, you
must hitch the end, still drawn tightly, round the other foot
of the ladder, and then thread the remaining portion, with
the aid of the needle, through each counter in succession
(this time not passing through the holes in the ladder). This
done, unfasten your two hitches, and draw away the doubled
cord. It will now come clear away from the ladder, but the
counters will be left upon it, according to the conditions of
the puzzle.
No. XVIII.— The Staff. Solution.
The cord is in two parts, each terminating in a loop. Of
the four knots, one is a sham, being, in fact, merely the
junction of the two loops, which are interlaced, the one
Fig. 81.
asssa
Fig. 82
within the other, as shown in Fig. 81. Loosen this, and
draw the nearest ball through the loop. This frees this ball,
as shown in Fig. 82. The remaining portion of the cord
may then be drawn through the staff, and both will be free.
No. XIX.— The Imperial Scale. Solution.
To get the ring off, draw up the central loop a couple of
inches ; pass it through the hole a ; pass it under the bead
Key to Trick or Secret Ptizzles. 57
at that corner, and draw it back again outside the cord.
Repeat the same process at the corner, and the loop will
be outside both cords. Draw it up a little further, and pass
it over the central knot, then down through the holes c and d
in succession (over bead, and draw back under). The loop
will now no longer embrace the standing part of the central
cord, but will lie loose on the scale, and the ring will be
free.
Care must be taken not to twist the loop during either of
the foregoing operations.
To work the ring on again, draw the loop through the
ring, then down through the hole a, under the bead, and up
again outside the cord. Repeat at the corner b. You will
now have the loop outside two of the cords. Pass it over
the general knot at top, down through c, outside the cord,
over the bead, and up again inside. Repeat at d, and the
ring will be secure.
If the puzzle be homemade (and it is a very easy matter
to make it), it will be found an improvement to have the
centre cord of a different colour from the four others. By
adopting this plan there is much less risk of getting the
cords " mixed " in passing the loop through the corner
holes.
No. XX.— The Sceptre Puzzle. Solution.
It will be observed, on inspection of Fig. 46, that the two
knobs, with which the rod terminates, are surrounded with
small black studs, apparently for the purpose of ornament.
One of these studs is screwed in, its opposite end fitting into
a groove at the end of the rod, and thereby securing the knob.
(See Fig. 83.) On unscrewing this the rod and knob come
apart, and the rings may be removed.
5 8 Ptizzles Old and New.
No. XXI.— The Balls and Chain. Solution.
The secret lies in the fact that one of the balls is in two
portions, the smaller portion being attached to the central
rod or stem by an ordinary carpenter's screw, on which it
revolves, and the larger portion screwing on to this, as
shown in Fig. 84. The concentric grooves on the balls in a
great measure mask the join, but it will be discoverable on
minute inspection.
Fig. 81..
Having ascertained which is the " trick " ball, grasp the
rod firmly in the left hand, with the point of the thumb
and forefinger pressed, hard against the base of the ball.
The ball being thus prevented from revolving, the other
hand screws off the upper portion, when the state of things
will be as shown in the diagram, and the chain can be
removed.
No. XXII.— The Four Keys. Solution.
On careful examination of this puzzle it will be found that
the " web " (or broad portion) of one of the " keys " is a
shade narrower than those of the others. It will further be
discovered that one of the longer arms of the cross is of
slightly extra width, allowing just room for two of the keys
to lie in it, side by side.
To solve the puzzle, hold the " lock " horizontally in the
left hand, letting the keys hang perpendicularly. Arrange
the stem portions of two of them in the shorter arms of the
cross, and a third in the broad part of the longer arm above
mentioned. The narrow key may now be pressed out through
the longer arms of the cross, passing beside the key already
there. The removal of one key renders the removal of the
others a very easy matter.
Key to Trick or Secret Puzzles.
59
To replace the keys, reverse the process, beginning with
the three broader keys. Get these into their proper posi
tions, and then insert the narrow key.
No. XXIII.— The Screwbox. Solution.
The secret lies in the fact that the lid screws, not into
the visible box, but into an inner lining, prevented from
being withdrawn by an overlap of the upper edge of the
box (see Fig. 85). The lining fits loosely within the box,
Fia. 85.
and so in inexperienced hands moves round and round with
the lid, which cannot under such circumstances be unscrewed.
To open the box, pinch its sides smartly between the
thumb and finger while endeavouring to unscrew the lid.
The outer shell, which is very thin, is thereby pressed
against the inner lining, and holds it fast. The lid may
then be unscrewed without difficulty.*
No. XXIV.— The Ball and Three Strings. Solution.
The secret lies in the fact, not perceptible save to the
closest inspection, that one of the three " knots " which
apparently unite the ends of the three pairs of cords is in
reality tivo knots, one on each end of the cord, but made to
look like one by the bead which holds them together.
* There is another version of the Screwbox, known as the "Profane"
Box, from the fact that there is a hidden needlepoint in the lid, which,
pricking the unwary finger, is apt to elicit some more or less forcible
ejaculation.
6o
Puzzles Old and New.
Having discovered the two knots, pull out the bead to
which they belong as far as possible. This will bring out
the two loops of the other strings. Then push back the
bead beyond these loops, as shown in Fig. 86, and the cord
thus set free may then be drawn out through the loops,
thereby releasing the other two. It is best not to pull out
these latter completely, as it will be found a rather trouble
some matter to thread them through the holes again.*
No. XXV.— The Lighthouse. Solution.
On close examination of the little lighthouse it will be
found that there are four holes on its under side, and on
looking through these holes and at the same time turning
round the lantern portion, it will be seen that an inner core
revolves with it. At a particular point of its revolution a
small cavity, hollowed out in this core, is brought into
juxtaposition with one of the four holes. By removing the
little pinnacle at the top of the puzzle, and thrusting it
into this hole, the core is made a fixture, and the top may
then be unscrewed at the point a, and the ring released.
This is best done with a piece of bent wire.
Key to Trick or Secret Puzzles,
61
No. XXV /.—The Jubilee Puzzle Box. Solution.
The Queen's head is hidden in the lid, just under the
raised portion in centre. To find it, unscrew the lid, and
remove the little ornamental knob and top. Then take the
lower portion of the box, turn it upside down, and press it
down over the raised portion of the lid (see Fig. 53), which
it will be found to fit pretty tightly. This is then used as a
key to screw out this raised portion, when a photograph of
Her Majesty will be revealed.*
No. XXVII.— The Jubilee Penny. Solution.
Most people, on making the acquaintance of this very
clever puzzle for the first time, are apt to seek for some
fanciful resemblance to Her Majesty among the various
lines on the coin. In this endeavour they are necessarily
disappointed, for the secret of the coin is mechanical, as will
be seen by an inspection of Fig. 87. The centre of the coin
is cut away, and a photographic portrait of Her Majesty
Fig. 87.
pasted in the centre. The central portion of a second coin is
then fitted to the cavity, and held in position by an ingeni
ously constructed pinhinge, a. Pressure on this hinge,
from the " head " side of the coin, lifts the cover. The fact
that the penny, with the amount of skilled workmanship
that must have been expended upon it, can be sold (as it is)
for a shilling, is by no means the least surprising part of the
"puzzle."
.* There is another box on the same piinciple, known as the " Diplo
macy " puzzle. The bos. differs slightly in shape, and the portrait of her
Majesty (on a farthing) is visible through a glass disc in the top of the
lid. In other respects the two are identical.
62 P tizzies Old and New.
No. XXVIII. —The Invisible Gift. Solution.
It will be noticed, on examining the interior of the box,
that both box and lid are scored with one or more lightly cut
circles, apparently for the purpose of mere ornament. In
one of these, however, sometimes box, sometimes lid, lies the
secret of the puzzle. A small circular space has been ex
cavated in the centre, and then filled up again with a little
wooden disc, beneath which lies the hidden coin. To ex
tract it, tap the portion in question with the inside
downwards smartly on a table or other hard flat surface.
After you have tapped for a minute or two, the wooden disc
will begin to work itself out, and a little later will drop out
altogether, releasing the coin.
No. XXIX.— The Aratai Gun. Solution.
This is on the same principle as the "Lighthouse " puzzle
described on pp. 35, 60 — viz., the fixing of the revolving por
tion of the gun (a in Fig. 54), so that it can be unscrewed
and taken apart.
The first step is to unscrew the knob at the breech.
This knob is in reality in two portions, the one, e, screwed
Fig. 88.
into the other, d (see Fig. 88). Having separated them,
remove the trunnion /, and through the hole thus left
thrust the pin e, moving a round until the pin finds a deeper
hole (in an inner core), into which it sinks. The core
being now held fast, the portion a may be unscrewed from c,
and the ball extracted.
No. XXX.— The Psycho Matchbox. Solution.
As we have already indicated, the box has an upper and a
lower lid, the latter being the true one. The former, as we
have seen, can be moved aside at pleasure, but the lower
Key to Trick or Secret Puzzles. 63
remains a fixture, save when the box is turned upside down,
when an internal " bolt " drops out of position, and releases
it.
To open the box turn it upside down, then push aside the
movable portion a, as shown in Fig. 89. You will find that
Fio. 89.
you can now push the lower lid, 6, forward a quarter of an
inch in the direction of the pivot at the opposite end, on
which it moves. In this new position it just clears the
tongue or stop, c, which previously held it fast. Again turn
the box right side uppermost, and turn o to right or left,
when the box will be open.
No, XXXI.— The " TouchMeNot " Matchbox.,
Solution.
To open this box, press strongly with thumb and finger on
its two ends simultaneously, when the box will fly open.
The metal yields to the pressure and makes the box
"bulge" to a minute extent in front, thereby releasing the
catch.
The same method opens the "TouchMeNot" Tobaccobox.
No. XXXII. —The Magic Drawer Matchbox.
Solution.
The secret here lies in the fact that the " drawer " portion
is double, the true or inner drawer, which contains the
matches, working within a shell or dummy drawer. In the
normal condition of the box the inner drawer is held back
64 Puzzles Old and New.
by a spring, and only the "shell " or empty drawer comes
out. By pressure on a particular part of the bottom of the
box the catch is lifted ; and if the box be then opened, the
true drawer comes out with the dummy, and the matches
are exposed.
No. XXXIIL— The "Unique" Matchbox. Solution.
The opening is in this case effected by means of the
striking plate seen at the bottom of the box. This, being
drawn forward with the fingernail, releases the catch, and
the box flies open.
To open the box without disclosing its secret, take it
between the forefinger and thumb of the left hand. Then
make believe to press the little stud in front with the thumb
of the right hand, when the second finger of the same hand
will be brought underneath the box in just the right posi
tion to 'draw the striking plate forward, and so release the
spring.
No. XXXIV.— The Surprise Matchbox. Solution.
To open this box, take it in the right hand, with the little
projecting stud resting against the lower joint of the fore
finger. Then with the thumbnail press the nearer end of
the supposed " hinge," marked a in Fig. 60, and the box will
open at that point. The real hinge is at the opposite side,
immediately above the little stud.
No. XXXV.— The New Brass Puzzle Matehbox.
Solution.
The secret here is of the simplest, but very few persons
discover it ; the experimenter being usually led astray by
the revolving coin on top, with its two. holes, which he
naturally assumes to play some important part in the solu
tion.
The coin in question is, however, simply intended to mis
lead, the actual fact being that the box unscrews at about
onefourth of the way down; but the screw is a lefthanded
instead of a righthanded one, so that any one attempting to
unscrew it in the ordinary way only fixes it the tighter. If,
on the other hand, he makes the usual movement of screw
ing, the lid comes off without difficulty.
Key to Trick or Secret Puzzles.
65
No. XXXVI.— The Ne Plus Ultra Matchbox.
Solution.
It will be observed that one side of the box bears a
striking plate. This indicates the front, which would be
otherwise undiscoverable, there being no visible hinge. To
open the box, press down the opposite edge of the lid in the
Fig. CO.
direction shown by the arrow in Fig. 90. The secret lies
in the fact that the lid works on a pin hinge, pivoted be
tween the two ends of the box, though the pins do not come
through, and are not perceptible from the ontsicle. The lid
therefore forms, in fact, a lever, pressure on whose shorter
arm raises the opposite end, and thereby opens the box.
No. XXXVII.— The Sphinx Matchhox. Solution.
This is the "Ne Plus Ultra " over again, with an addition—
viz., that the striking plate, which is in this case at the back.
is movable, being hinged to the bottom of the box, with the
striking surface on its inner side. Open this, as shown in
Fig. 91, and then press down the edge of the lid, thereby
F
66
Ptizztes Old and New.
left exposed, and the box will open as depicted in our
illustration.
When the striking plate is closed the lid is "locked," and
the box cannot be opened.
No. XXXVIII. — The Puzzle Snuffbox. Solution.
The decorative markings of the box seiwe to conceal the
fact that the portion of the front marked A (see Fig. 92) is
movable, sliding up and down between the adjacent portions.
Fig. 92.
On pushing this up with the thumb, the lid is forced up
with it, and the box may then be opened without difficulty.
The sliding piece should be quietly pressed down again
before inviting any one to take a pinch, when there will be
nothing to disclose the secret.
No. XXXIX.— The New Puzzle Snuffbox. Solution.
There are in point of fact two lids to this box, opening in
opposite directions (see Fig. 93). The smaller is the
Fig. [)3.
normal lid, and is the one first opened. This done, the
second can be opened, as shown in the figure, and the snuf£
is then accessible.
Key to Trick or Secret Puzzles.
$7
No. XL.— The Puzzle Ball. Solution.
The secret here lies in the faet that one of the supposed orna
mental bosses is continued, in a tapering shape, right through
Fig. 94.
the ball, its smaller end forming the centre of the boss on
the opposite side (see Fig. 94). By tapping this centre,
which must be discovered by experiment, the boss is forced
out and the ball opened.
Fie. £5.
Fig. 96.
No. XLI.The Ebony Puzzle Ball. Solution.
On minute inspection of the ball, it will be found that
68 Puzzles Old and New.
close beside one 'of the little bosses is a minute depression,
scarce bigger than a pin's head. This serves to indicate the
position of the opening, of which the next boss (looking from
the boss abovementioned across the indicating mark) forms
the centre. Rap the side of the ball diametrically opposite
to this on the table, and the broad end of a tapering ping
will be forced out, as shown in Fig. 95. This plug, shown
in detail in Fig. 96, is hollow, with a screwlid, and forms a
receptacle for coins or any other small object.
The mechanism of the ball is so perfect and the opening
so cleverly masked by the rosecutting that its secret is
hardly ever discovered.
No. XLII.— The Puzzle Purse. Solution,
The back of the purse, like the front, is in two portions,
and is in like manner divided across the line a a by a line of
stitching. The lower portion of the back forms part of the
same piece of leather which is seen scored in front, and by
nipping this lower portion with the forefinger and thumb of
the right band, at the same time holding the hinder flap with
the left hand, the slit portion may be pulled down till it comes
below the stitched line a a at back, and access may then
be obtained through the slits to the interior. The stitches
are of such a size as just to allow the passage of the strips
of leather.
To close the purse again, take hold of its lower edge, and
pull up the slit portion, when all will be as at first.
No. XLIII.The Puzzle Pocket Knife. Solution.
Fig. 97.
The reader is probably acquainted with the ordinary
dagger knife, in which the blade is kept from closing by a
Key to Trick or Secret Puzzles. 69
catcli on the spring, which catch drops into a notch at the
back of the blade, and so holds it fast until, by the raising of
the spring, the catch is again withdrawn. The puzzle pocket
knife is on the same principle, but the manner of its appli
cation is here concealed. There is, in this case, one notch
to receive the catch when the knife is closed and a second to
receive it when open. One of the buckhorn sides is
attached to the free end of the spring. By drawing this back
with the left thumb, as shown in Fig. 97, the catch is
withdrawn, and the knife can be opened without difficulty.
When it is opened the blade is again held fast, and the
operation must be repeated to close it.
No. XLIV.— The Automatic Knife. Solution.
On minute examination it will be observed that there is
at either end of the knife (see Fig. 68) a little projecting
tongue of white metal, flat on one side and convex on the
other. By pi^essure on the flat side of either of these
tongues, the corresponding blade is made to fly open to a
small extent, assuming an angle of about 30° to the handle.
From this point it may be fully opened without difficulty.
The knife is closed in the ordinary manner.
Xo. XLV — The Double Barrel and Ring*. Solution.
This is somewhat on the principle of the Puzzle Ball (No
XL.). It will be observed that the ball at each end is cut
in circles, parallel to the central stem and growing smaller
and smaller as they pass higher or lower from the central
circle ; the final circle being a little boss, in pretty strong
relief. By rapping one of these (the right one must be
ascertained by experiment) on a table or other hard surface
a little plug wmicli binds the ball to the . stem is forced out
of position. • The ball may then be unscrewed from the stem
and the rins; released.
No. XLVL— The Weddingring Box. Solution.
This is a very clever puzzle, for the uninitiated invariably
take it for granted that the box must unscrew or otherwise
open at the mark round its centre, whereas the actual open 
7o
Puzzles Old and Neiv.
ing must be Bought in quite another direction. To open
the box, press firmly on one of the central bosses, and a
circular wooden piece, a, will be forced out below, as shown
in Fig. 98.
The ring lies around the central plug.
No. XLVIL— The New Moneybox. Solution.
This is on the same principle. The circular groove round
the larger circumference of the box is a mere blind, but by
pressing the central portion of the top (see Fig. 09) an inner
Fig. 99.
receptacle (after the manner of that in the Ebony Puzzle
Ball, No. XLI.) is forced out. The top, a, of this lifts off,
and in the lower part, Z>, is found the concealed coin,
Key to Trick or Secret Puzzles. 71
No. XLVIIL— The Zulu Box. Solution,
The revolving loose lid is, in this ease, in two portions,
the one screwing into the other. To open the box, take it in
the left hand, as shown in Fig. 100, and with the thumb and
Fig. 10Q.
fingers press the lower portion of the lid vigorously upwards,
This will make it, for the time being, a fixture, and the upper
part may then be unscrewed at the point a, as shown.
No. XLIX— The New Persian Puzzle, Solution,
The secret here lies in the fact that the central rod or shaft
is in two portions, the one screwed into the other. To open
the box, take hold of the little knob at each end simultaneously
and screw them in opposite directions, when the one knob
will come off, and the box will come apart in three portions.
No. L.— The Magic Handcuff. Solution.
To obtain your release, bring the imprisoned fingers as
near together as possible, thereby enlarging the diameter of
the tube. Then, with the thumb and forefinger of the right
hand take hold of the opposite end of the tube and draw it
off the finger, which in this manner you can do without
difficulty. f . „
Nq. LI.— The Key and Ring Puzzle. Solution.
The secret here lies in the fact that the point of the key,
which, it will be observed, projects some threequarters of
an inch beyond the " web," can be unscrewed, as shown in
Fig. 101. With the key in this condition the ring can be
passed gver the web, and on to the stem, without difficultv.
72 Puzzles Old and Neiv.
This puzzle is sometimes introduced in the guise of a con
juring trick. For use in this case, a little thumbscrew,
shown at a in Fig. 101, is supplied with the key. By the
aid of this the movable point of the key may be screwed
up so tightly that no one, save a Japanese dentist, could
possibly hope to remove it with the unassisted fingers,
though, the wizard, armed with the "fake," can do so at
pleasure. It is hardly necessary to remark that, for puzzle
purposes, the use of such an appliance is illegitimate, it
being an understood thing that every " fair " puzzle con
tains within itself all the elements necessary to its complete
solution.
jSTo. LIT.— The New Egg of Columbus. Solution.
The secret of this excellent puzzle lies in the fact that the
base of the egg is occupied by a hollow cone, with an open
ing on one side of its apex, and a shallow groove leading up
to such opening. The weight heard rolling about within the
ee 0, is a leaden bullet. When this is once fairly within the
cone, it naturally settles down in the centre, and the egg can
be made to stand upright without difficulty; but so long as the
bullet is outside the cone, its weight throws the egg to the
one side or the other, and makes it absolutely impossible to
balance it. The difficulty, therefore, resolves itself into the
getting the ball inside the cone. To do this, it is necessary
to know on which side the opening lies, and the guide as to
this is a little mark, a mere dent, on one side of the smaller
end of the egg. Holding the egg smaller end upwards, and
with the mark facing you, turn the point of the egg over
towards your own body. This brings the bullet on the same
side as the opening. Then turn the egg over, very slowly
and steadily, till it lies horizontal, with its point away from
Key to Trick or Secret P tizzies. 73
you, and, if you are lucky, the bullet will roll along the groove
into the cone, and you will have no further difficulty.
The entrance to the cone is purposely made somewhat
narrow, in order to diminish the likelihood of an accidental
discovery of the secret. The neophyte must, therefore, not
be discouraged if he finds that he has to repeat the above
process once or twice before he succeeds in lodging the bullet
in the proper quarter,
CHAPTER III.
" DISSECTED " OR COMBINATION PUZZLES.
The number of these is legion. We all remember the " dis
sected maps " of our early youth, whereby we learnt geo
graphy (and very thoroughly as far as it went), while we
fondly imagined we were only amusing ourselves.* With
these, however, we have for our present pnrpose nothing to
'do. The present chapter will be primarily devoted to the
(large class of puzzles in which, a given geometrical figure
having been cut up into various segments, the experimenter
is required to rearrange such segments so as to form another
figure, or figures, of a different character.
Foremost among these may be named the series of puzzles
issued by Messrs. Richter & Co., the inventors and manu
facturers of the admirable " Anchor " Stone Buildingblocks,
the ne plus ultra of Toydom, and delight of every kindergarten
and schoolroom which is fortunate enough to possess a set
of them. The Anchor building blocks may indeed them
selves be regarded as a puzzle, and one of a very interesting
kind. They are a modern and scientific develoj)ment of the
oldfashioned " box of bricks." Dr. Richter's blocks are not
of wood, like their prototypes, but of a mineral composition
compressed to the hardness of stone. They are of three
different colours : creamcolour, to represent sandstone ; red,
* We are sony to note that in these later days the dissected map has
fallen into comparative disfavour, being replaced by dissections cf fairy
tales, scripture histories, and the like, which have no natural aptitude for
being so dealt with. In the case of maps, on the other hand, if the .
dissections are properly made— i.e., so as to follow the. outlines of the
provinces, counties, or other divisions of the region represented— the
shape and position of such divisions are impressed upon the memory in
a manner a'most unattainable by any other method. The course of
rivers, situation of chief cities, etc., are unconsciously taken note of,
and, once thoroughly learnt in this manner, are never forgotten.
7*
Dissected'' or Combination Puzzles. 75
for bricks ; and slatecolour, for roofing material. The boxes
range in price from sixpence up to five guineas, and in
weight from fourteen ounces to nearly threequarters of a
hundredweight. Each block is shaped with mathematical
^lipssu.
Fig. 102.
accuracy; and this fact, combined with the extra weight of
the material, gives buildings constructed with them a degree
of stability, as well as elegance, in which the oldfashioned
woodbrick structures were wholly wanting.*
* So perfect are the Richter blocks in poii.t of accuracy and finish
that they are now largely used, by professional architects to form tem
porary models of intended structures. Our illustrations (Figs. 102 and
103) give a notion of what may be attempted with the higher priced
boxes. The handsome castle depicted in the Fron'ispiece represents a
still higher flight ; and it must be remembered that this is not a mere
fanciful design, but can really be constructed to the minutest detail by
a sufficiently skilful operator. The box used in this case is No. 23,
which contains 1549 stones, and eight books of designs. This is one of
the most expensive sets, costing £4 6s. Of course, the construction of.
76
Puzzles Old and New.
By an ingenious system of "supplements" a person
originally buying a box, of however small size, may by sub
sequent purchases, singly of small amount, so increase his
store of material as ultimately to bring it up to the level of
one of the largest sets, and for the same total outlay. Each
new acquisition enables the purchaser to attempt a further
and more ambitious series of designs, plans of which ac
company his purchase accordingly. The ample supply of
Fig." 103.
building designs is one of the most valuable features of the
boxes. The multiplicity of these designs may be imagined
from the fact that two skilled architects, with half a dozen
draughtsmen under them, are employed all the year round
producing new ones. Each stone of the elevation is clearly
shown, duly drawn to scale, a front and back view being
given where necessary ; where from the nature of the case
some portion of the stones are liot visible, or their arrange
ment offers any special difficulty, a horizontal section is
a building of this magnitude demands not merely more than average
lightness of finger, but a very considerable amount of intellectual effort.
The endeavour to build a very much less ambitious structure (using
a box costing, say, ten or twelve shillings) will be found amply to justify
our assertion that the Kichter building stones not only come within,
but hold a very high place in the province of Puzzledom.
("Dissected" or Combination Puzzles. yj
added, giving a ground plan of the particular " course " in
question.*
Passing from the Richter building blocks to the Richter
puzzles proper, we have, in the first place —
Ho. I.— The "Anchor" Puzzle.
This consists of a square slab of the "Anchor" building
block composition, divided into seven segments, of the
shapes shown in Fig. 104 (viz., two large triangles, a, a, a
smaller one, b, and two still smaller, c, c, a square, eZ, and a
rhomboid, e), neatly packed in a flat box, three inches square.
Fig. 101.
Even the replacement of the pieces in the box, when once
fairly mixed, will be found a matter of some little difficulty,
but this is the smallest of the problems presented by the
puzzle. The box contains, in addition to the seven " stones,"
a little book, containing over 150 different designs, which
may be formed by combining them in different ways. Space
only allows of our giving a few examples (see Figs. 105115.
It is to be noted that the whole seven stones must be employed
in the formation of each design ; and it is a curious fact that
some of those designs whichare simplest in appearance are
the most difficult to work out. The letters of the alphabet,
each one of whiqh can be formed (in more or less grotesque
shape) by means of the seven blocks, give far less trouble
* The building stones are manufactured at Budolstadt, in Thuringia,
but may be procured through any highclass toy dealer. The London
branch of the firm is at 12, Jewin Street, E.C., where any desired infor
mation can be obtained.
78
Puzzles Old and New.
than simpler forms. No more aggravating, and at the same
time fascinating, puzzles have come under our notice than
this series.
Fig. 105.
Fig. 106.
Fig. 1C7.
'" Dissected' 1 or 'Combination Puzzles. jg
Fig. 108.
FlGe 109.
Fig. 110.
Fig. 111.
Fig. 112.
Fig. 113.
Fig. 114.
Fig. 115.
8o Puzzles Old and Nezv.
No. II.— The Tormentor Puzzle.
This, the second of the series, consists of a square divided
into eight pieces, of the shapes depicted in Fig. 116. Here
again Ave have over 150 designs which can be formed by
Fig. 116.
combination of the several segments. Wc append (Figs.
117124) a few of the possible combinations.
Fig. 117.
Fig, 118.
Fig. 119.
"Dissected" or Combination Puzzles.
Fig. 120.
Fig. 121.
Fig, 122..
Fig. 123.
Fig. 124.
No. III.— The Pythagoras Puzzle.
Here we have again (Fig. 125) a square, divided into
seven segments, and a selection of 181 figures which may be
formed by their recombination. We append a few speci
mens (Figs. 126134).
G
Sc
Puzzles Old and New
Fig. 126.
Fig. 127.
Fig. 128,
Fig. 129.
Fig. 130.
''Dissected" or Combination P tizzies. 85
Fig. 131.
Fig. 132.
Fig. 133.
Fig. 134.
No. IV.— The Cross Puzzle.
This is so called from the fact that the first of the figures
which the experimenter is invited to form with the
Fig. 135.
several segments is a cross. The rectangular slab of which
the puzzle is composed is not in this case a perfect square,
J 2
Puzzles Old and New.
but slightly oblong, being 3J inches in length by 3 in width
(see Fig. 135). Over 130 designs are given, which may be
composed by the aid of the seven segments. Of these
examples will be found in Figs. 136144.
Fig. 136.
Fig. 137.
Fig. 138,
Pig. .139.
Fig. 140.
Fig. 141.
"Dissected* of Combination Puzzles. 85
Fig. 112.
Fig. 143.
Fig. 144.
No. V.— The Circular Puzzle.
This is a new departure, the slab out of which the several
segments are formed being circular, instead of rectangular.
The number of segments is in this case ten (see Fig. 145).
Fig. 145.
These facts lend greater plasticity to the resulting combina
tions, some of which, as may be gathered from an inspection
of the diagrams appended (Figs. 146154), are very artistic,
and, with the aid of a little imagination, may be taken to
S6
Puzzles Old and New.
represent familiar objects — as, Fig. 146, a crown; Fig. 149, a
fool's cap ; Fig. 150, a lobster ; and so on.
Fig. 146.
Fig. 147.
Fig. 148.
Fig. 149.
Fig. 150.
Fig. 151.
"Dissected" or Combination Puzzles. &J
Fig. 152.
Fig. 153.
Fig. 154.
Over a hundred such figures are given in the little book
which accompanies the puzzle, but this number might be
largely increased, as indeed is the case with the whole of the
series.
Ko. VI.— The "Star" Puzzle.
The Star Puzzle consists of 48 segments, 24 black and 24
grey, the grey forming in their original positions in the
box a star. (See Fig. 155). This, from the use of the two
colours, and the larger number of pieces used, is capable of
an extraordinary variety of effect, and will by most persons
be considered the prettiest puzzle of the series. Figs. 156
167 will give some faint idea of the almost kaleidoscopic
83
Puzzles Old and NeiO.
effects which can be produced by combination of these
simple elements.
Fig, 155.
Fig. 156.
Fig. 157.
"Disserted" or Combination Puzzles. 89
Fig. 158.
Fig. 159.
Fig. 160c
Fig. 1G1.
Fig. 162.
Fig. 163.
go
Puzzles Old and Neio.
Fig. 164.
Fig. 165.
Fig. 165.
Fig. 167.
The " Star" is the last issued of the " Anchor " series. ;
It should be mentioned that in addition to the various figures
resulting from the rearrangement of the segments of a single
puzzle of this series, there are many others which may be
produced by the combination of the segments of a given pair
of puzzles. For these, however, we must refer the reader to
the little book which accompanies each puzzle.
"Dissected" or Combination Puzzles. 91
No. VII.— The Zigzag Square.*
Given twelve pieces of paper or cardboard, four of eacli
shape given in Fig. 168.
Fig. 16S.
Required, so to combine these twelve pieces as to form a
perfect square.
No. VIII.— The Extended Square.
Required, so to cut a cardboard square, as a in the dia
gram (Fig. 169), into two portions, in such manner that by
Fig. 169.
successive shiftiugs of their relative positions, they may form
the parallelogram/ b, and the eccentric figure, c, in the same
diagram.
* There is another class of puzzle which goes by the same name — viz.,
a square of thin wood cu with a fretsaw iuto a number of segments of
curved outline, after the fashion of a dissected map.
92
Puzzles Old and New.
No. IX.— The Octagon Puzzle.
Given twelve pieces of paper or cardboard, five of each
shape represented in Fig. 170.
V
^
Fig. 170.
Required, so to combine them as to form an octagon.
No. X.— The Patchwork Square.
Given, eight pieces of paper or cardboard shaped as shown
in Fig. 171.
Fig. 171.
Required, by arrangement of these segments to form a
perfect square.
"Dissected" or Combination P tizzies, 93
No. XI.— The Two Squares.
Given, a piece of paper or cardboard of the shape depicted
in Fig. 172, being that of a small square in juxtaposition
with one four times its size.
/CsU&2^(s7
Fig. 172.
Required, by two cuts (each in a straight line) so to
divide the piece of cardboard that the resulting segments
shall, differently arranged, form one perfect square.
No. XII.— The Latin Cross Puzzle.
This has no affinity with the Cross Puzzle described at
page 83.
Given, five pieces of paper or cardboard, one as a, one as
b, and three as c (Fig. 173).
Fig. 173.
1
L J
Fig. 174.
Required, of these five segments to form a Latin cross, as
Fig. 174,
94
Puzzles Old and Nezv.
No. XIII.— The Greek Cross Puzzle.
Given, a piece of paper or cardboard in the form of a
Greek or equalarmed cross, as Fig. 175.
Fig. 175.
Required, by two straight cuts so to divide it that the
pieces when reunited shall form a square.
No. XIV.— The Protean Puzzle.
This is a puzzle of the same class as the Anchor, Tormen
tor, and Pythagoras (pp. 7783), but very much easier, from
the greater number of the component parts. It consists of
eleven pieces of cardboard, forming an oblong square, as
Fig. 176.
shown in Fig. 176. Figs. 177182 illustrate a few only of
the many shapes which can be constructed out of the above
elements.
"Dissected" or Combination Ptizzles. 95
Fig. 177.
Fig, 178.
J
LJ
Fig. 179.
Fig. ISO.
Fig. 131.
Fig. 182.
9 6
Puzzles Old and Neiv,
No. XV.— The Caricature Puzzle.
This puzzle consists of seven pieces of blackened card
board; a square, five rightangled triangles, and a rhomboid,
of the respective proportions shown in Fig. 183.*
With these the experimenter is required to construct
Fig. 183.
grotesque representations of the human figure, in various
positions. Space only allows of our giving half a dozen
examples (Figs. 184189) but the possible number of such
combinations is extraordinary, and many of them are most
comical in effect.
/
Fig. 181.
Fig. 185.
Fig. 186.
* It will/be \ observed by the acute reader that these segments are
identical in number and shape with those of the "Anchor" Puzzle,
described at page 77. The fact that such very different results are
here produced with the same elements illustrates the extraordinary
fertility of this class of puzzles. New combinations are constantly
offering themselves, the attempt to construct one figure suggesting some
other, totally different, but equally effective.
"Dissected" or Combination Puzzles. 97
Fig. 187.
Fig. 188.
Fig. 189.
No. XVI.— The Chequers Puzzle.*
This is a dissected puzzle of a very novel kind. It con
sists of a miniature chessboard, five inches square, divided
Fig. 190.
into fourteen pieces, each consisting of from three to five
squares, as illustrated in Fig. 190.
The experimenter is required, out of these fourteen seg
ments, to construct the chessboard anew. At first sight,
•• Published by Messrs. Feltham & Co.
II
9 8
Puzzles Old and New.
the task would seem to be one of the easiest possible, but
any such idea very soon vanishes when the matter is put
to the test of experiment. The pieces drop into position
with enticing facility till the board is about three parts
complete, but at that point the neophyte usually finds him
self with half a dozen segments still in hand, which abso
lutely decline to accommodate themselves to the spaces left
for them.
It may be mentioned, for the encouragement of the faint
hearted, that there are, according to the publishers, no less
than fifty ways in which the puzzle may be solved, and it
should therefore be merely a matter of time and persever
ance to discover one or other of them. *
No. XVIL— The " Spots" Puzzle.f
This puzzle also is very much more difficult than it looks.
It consists (see Fig. 191) of a wooden cube, not quite three
inches each way, cut into nine bars, of equal size. Each
Fig. 191.
of these is decorated with one or more " spots," half an inch
in diameter; and the experimenter is required to put together
these bars in such manner that the resulting cube shall
represent an enlarged model of the die familiar to the back
* Beaders who have found pleasure in the " Chequers " may be glad
to know of another very pretty series of puzzles of the same kind — tbe
" Peel " puzzles, issued by Messrs. Jaques & Son. They are three in
number, each consisting of nine pieces of cardboard, beariDg, some two,
some three, squares of different colours. With these a larger square or
other figure is to be formed, in such manner that the same colour shall
not occur more than once in any row, horizontal or vertical.
We regret that it is impossible, without the aid of coloured illustra
tions, to give a more detailed description.
t Manufactured by Messrs. Wolff & Son.
"Dissected" or Combination Ptczzles. 99
gammon player (see Fig. 192), with all its spots in proper
position. These, it may be mentioned for the benefit of the
uninitiated, are arranged as follows : — The " ace " point is
on the opposite side to the " six " ; the two on the opposite
side to the " five," and the "three " on the opposite side to the
" four " ; the total of each pair of opposite sides being always
Fig. 192.
seven. A die which does not answer these conditions is re
garded as fraudulent.
As an aid to the memory in this particular, the outside
of the pasteboard box or case in which the segments are
contained is itself an enlarged fac simile of a die, with the
spots in proper position.
No. XVIII.— The Endless Chain.
Fig. 193.
IOO
Ptizzles Old and New.
Yet another good dissected puzzle is the " Endless Chain."
A piece of cardboard, six inches square, and bearing the
representation (in gold, on a blue ground) of an endless
chain, is cut into eighteen pieces, of various sizes, and such
pieces are placed hapbazard, as shown in Fig. 193. The puzzle
is to rearrange them, within the limits of the square box
which contains them, so as to reform the endless chain,
with each link in proper connection.
No. XIX.— The Hexagon.
Given, five pieces of wood or cardboard, one of each shape
depicted in Fig. 194.
Fig. 194.
Required, with these to form a regular hexagon.
No. XX —Eight Squares in One.
Cut out eight squares of cardboard, of equal size, and
divide four of them diagonally from corner to corner. This
will give you twelve pieces, four square, and eight triangular.
Required, so to arrange them as to form a single perfect
square.
No. XXI— The Five Squares.
Given, five squares of paper of cardboard, alike in size.
Required, so to cut them that by re arrangement of the
pieces you can form one large square.
"Dissected" or Combination Puzzles. 101
No. XXII.— The Geometrical Square.
Given, six pieces of wood or cardboard, two as a, and four
as b (Fig. 195).
Fig. 195.
Required, so to combine them as to form one complete
square, or two smaller squares.
No. XXIII.— The Dissected Square.
We have in this case nine pieces, three as a, three as b,
and three as c (Fig. 196).
Required so to combine them as to form one perfect
square, or three smaller squares.
Fig. 196.
The formation of the three smaller squares will be found
easy enough, but that of the single square will give more
trouble, being, in fact, one of the most difficult of this class
of puzzles.
No, XXIV.— The Twenty Triangles.
We have Here twenty triangles, of size and shape as in
Fig, 197.
Fig. 197. The experimenter is required to form with them
a perfect square.
102
Puzzles Old and Neiv.
No. XXV.— The New Triangle Puzzle.
The puzzle which goes by this name is of more than
average merit. It has long ceased to be " new," though for
distinction's sake it retains the qualifying adjective. It
A J
Fig. 19S.
consists of seven pieces, one as a (Fig. 198), three as b, and
three as c. With these seven segments it is required to
form an equilateral triangle.
No. XXVI.— The Japanese Square.
Given, ten pieces of wood or cardboard, four as a, four as
b, and two as c (Fig. 199).
Fig. 199.
Required, to form a square with them.
No. XXVII.— The Chinese Square.
Given, sixteen pieces of wood or cardboard, four as a, four
as b, and eight as c (Fig, 200).
<X/
Z\ L
Fig. 200.
Required, to form a square with them.
"Dissected" or Combination Puzzles. 103
No. XXVIII.— The Yankee Square.
Given, eleven pieces of wood or cardboard, two as a, one
as b, three as c, two as d, two as e, and one as/ (Fig. 201)
Fig. 201.
Required, to form a square with them.
No. XXIX.— Another Cross Puzzle.
Given, five pieces of wood or cardboard, three shaped as a
(Fig. 202), and two as b.
U
Fig. 202.
Required, so to arrange them as to form a Latin cross.
No. XXX.— The Carpenter's Puzzle. No. l.
Given, a slip of w r ood, 15 inches long by 3 wide.
How is it possible to cut it so that the pieces when re
arranged shall form a perfect square ?
No. XXXI.— Carpenter's Puzzle. No. 2.
A carpenter's apprentice has a board 3 feet in length by 1
in width. With this he is required to fill up a space 2 feet in
length by 18 inches in width, but he is not to cut the board
into more than tw r o pieces.
How can he manage it ?
104
Puzzles Old and New.
No. XXXII.— The Cabinet Maker's Puzzle.
A cabinetmaker lias a circular piece of veneer, with
which he desires to veneer the tops of two officestools,
leaving a handhole in the centre of each. The circular
piece is exactly sufficient for the job, provided that it be cut
without waste.
How can he cut it to the best advantage ?
CLUSTER PUZZLES.
There is a large class of puzzles generically known as
" Cluster " Puzzles. These consist of a number of pieces of
wood, so interlocked together as to form some more or less
fanciful shape; not flat, like the "dissected" form of puzzle,
but solid. The experimenter is required to take the puzzle
to pieces, and reconstruct it as at first. Many of these are
extremely clever, but for Our present purpose they labour
under the disadvantage that it is almost impossible, even
with the fullest aid from diagrams, to describe the modus
operandi in writing. We shall, however, make the attempt
with regard to a few of them ; bespeaking the reader's in
dulgence for any unavoidable obscurity in our explanations.
No. XXXIII.— The Bonbon Nut Puzzle.
This has a very complex appearance (see Fig. 203), but it
in reality consists of six pieces of wood only, generally box
or willow. There is a small space in the centre, which
encloses a carraway comfit, offered as prize to any one
  ' Fig. 203.
who may succeed in extracting it. As usually happens,
however, in these cases, the taking of the puzzle apart is a
comparatively straightforward matter, the real difficulty
being found in the endeavour to put it together again.
"Dissected" or Combination Puzzles. 105
No. XXXIV.— The Rattle Puzzle.
This (see Fig. 204) is another puzzle of the same kind, but
differing in form, being shaped like a child's rattle. It con
sists of twelve short pieces — fashioned exactly like those of
the " Bonbon Nut " — two long flat pieces, with a square head
at each end, — and a central piece forming the handle. There
Fig. 204.
is a narrow space down the centre, in which is inserted a
bonbon, or leaden pellet, the shaking about of this producing
the " rattle."
The experimenter is required to take the rattle to pieces,
and put it together again.
106 Puzzles Old and New.
No. XXXV.— The CrossKeys op Threepiece
Puzzle.
This (see Fig. 205) is a very ingenious puzzle of its kind.
It is one of the simplest, in one sense, being composed of
only three pieces of wood, but they are interlocked with
extreme ingenuity, and the endeavour to separate them
Fig. 205.
will give a good deal of trouble. Indeed, one's first im
pression, on a casual inspection, is that the whole must have
been carved out of a single piece.
Contrary to the usual rule, the reconstruction of the
puzzle will here be found easier than its separation.
No. XXXVI.— The Nut (or Sixpiece) Puzzle.
This puzzle is (as its second name implies) composed of
six pieces of wood, so cut as to fit together as shown in
Fig. 206.
Fig. 206.
The experimenter is required to take the pieces apart, and
put them together again.
"Dissected" or Combination Puzzles. 107
No. XXXVII.— The Fairy Teatable.
This is a pretty little table (see Fig. 207), of various
coloured woods. It is composed of sixteen different pieces,
even the pillar which supports it beiDg in four different
sections.
Fig. 207.
The neophyte is required to take the table to pieces, and
reconstruct it, as shown in the figure. The former is a very
simple operation.
No. XXXVUl.The Mystery.
Fig. 208.
The puzzle to which this somewhat highsounding name
ioS
Puzzles Old and New.
is given is as depicted in Fig. 208. It is an elegantlooking
affair, being made of fancy woods in three different colours.
It consists of ten different sections, which the aspirant is
required to take apart, and put together again.
No. XXXIX.— The Diabolical Cube.
This is a puzzle of a much simpler character, but it will,
nevertheless, give some trouble to any one attempting it for
the first time. It consists of six pieces, shaped as a, b, c, cl,
e, and/, respectively, in Fig. 209.
Fig. 209.
Of these six segments the experimenter is required to
form a cube.
KEY TO CHAPTER III.
" DISSECTED " OR COMBINATION PUZZLES,
.No. I.— The Anchor Puzzle, Solutions.
The first step towards the solution of puzzles of this class
is to study the relative proportions of the various segments,
and note what results can be obtained from the combination
of a given pair. These will, of course, vary according to the
particular manner in which the two parts are brought into
juxtaposition, and a slight alteration in this respect will often
supply the solution of an apparently hopeless puzzle.
Examining, from this point of view, the seven segments of
the Anchor Puzzle, we find (see Fig. 104) that we have two
Fig. 212.
Fig. 213.
Fig. 214.
Fig. 215.
rightangled triangles, a a. each equal to onefourth of the
complete square; a third, 6, equal to oneeighth, and two
smaller, c c, each equal to onesixteenth of the complete
square. The three last, "joined together as Fig. 212, are
exactly equal to one of the larger triangles, a a. Besides
these, we have a square, d ; and a rhomboid, e ; each exactly
double the area of one of the triangles c c, and therefore
also together equal to one of the larger triangles. It should
also be noted that the two larger triangles, a a, together
constitute onehalf the square, and the remaining five seg
ments the other half. The apprehension of these facts will
often help the experimenter out of a diificulty, by enabling
in
I 12
Puzzles Old and New.
him to substitute segments. differently shaped but geometric
ally equivalent, for others found unsuited to the required
figure.
Further, as to the change of shape produced by change of
relative position, it is to be noted that the juxtaposition of
two rightangled triangles of equal size, as c c, with their
hypotenuses together, produces a square (Fig. 213). With
two of their shorter sides in contact, they appear as
Fig. 214, or Fig. 215. Reverse one of them, and again
unite, and the resulting figure is a rhomboid, as Fig.
216. A]3ply one of the shorter sides of each of the triangles
Fig. 216.
Fig. 217.
Fig. 218.
c c to the square d, and you have again a rhomboid, though
of different proportions, as in Fig. 217. Apply one of the
shorter sides of each of the triangles c c to one of the
shorter sides of the rhomboid e, and you have once more a
rhomboid (Fig. 218), though much narrower in proportion
to its length. Apply the hypotenuse of each of the triangles
c o to one of the longer sides of the rhomboid e, and the
e " /
&
Cs
Fig. 2iy.
Fig. 220. Fig. 221. Fig. 222.
resultant figure is a rectangle, as in Fig. 219. Apply the
hypotenuse of one only of the smaller triangles c c to one
of the longer sides of e, and we have the trapezoid depicted
in Fig. 220. Apply one of the shorter sides of the other
triangle c to one side of the square d, and we have a trape
zoid (Fig. 221) of the same size and shape, though produced
in a different manner. Place these two trapezoids in juxta
Key to " Dissected" or Combination Ptizzles. 113
position, and insert the triangle b in the concave space
between them, and we have a square (half the size of the
original), as in Fig. 222.
Having thus made himself acquainted with the elementary
capabilities of his materials, the neophyte will be the better
prepared to attempt the formation of the more elaborate
patterns depicted on pp. 78, 79.
Beginning with Fig. 105, we note in the first place that it
is a rectangular parallelogram, consisting of two equal
squares. We know that the two large triangles a a con
stitute one half of our available material, and that they can
be made to form a square (i.e., half the required figure). We
have then only to form a second square, as shown in Fig.
222, side by side with this, and our rectangle is complete.
(See Fig, 223.)
a/
\ ^
I
/
CL>
Cs \
d,
Fig. 223.
Fig. 224.
To form Fig. 106, we retain the square last formed by
way of centre, remove the outer large triangle from the left
hand side, and transfer it, the reverse way up, to the opposite
side. (See Fig. 224.) Again removing the triangle last placed
in position, and transferring it (see Fig. 225) to the top of the
square, in such manner that its hypotenuse shall be in line
with the hypotenuse of the triangle on the left, we have
Fig. 107.
For the composition of the remaining figures we must refer
I
114
Puzzles Old and New,
Fig. 225.
the reader to the diagrams following (Figs. 226233.) It
will be found that all the designs are constructed on the
same principle, and, by means of a little intelligent analysis,
can readily be resolved into their component factors.
Fig. 226.
Fig. 227,
Fig. 228.
Fig. 220,
Fig. 230.
Key to " Dissected" or Combination Puzzles. 115
Fig. 231.
Fig. 232.
Fig. 233,
No. II.— The Tormentor Puzzle. Solutions,
See Figs. 234241 below.
Fig. 234.
Fig. 235.
n6
Puzzles Old and New.
Fig. 236,
Fig. 237 a
Fig. 238.
Fig. 239.
Fig. 240.
Fig. 241.
Key to "Dissected" or Combination Puzzles. 117
No. III.— The Pythagoras Puzzle. Solutions.
See Figs. 242250 below.
Fig. 242.
Fig. 243,
Fig. 244.
Fig. 245.
Fig. 246.
n8 Puzzles Old and New.
Fig. 247.
Fig. 248.
Fig. 249.
Fig. 250.
'Uo. IV.— The Gross Puzzle. Solutions.
See Figs. 251259 below.
Fig. 251.
Fig, 252.
Key to "Dissected" or Combination Puzzles.
119
Fig. 253.
Fig. 255.
Fig. 254.
Fig. 256,
Fig. 259.
120
Puzzles Old and Ncu).
Na v.— The Circular Puzzle. Solutions.
See Figs. 260268 below.
Fig. 260.
Fig. 261.
Fig. 262.
Fig. 263.
Fig. 264.
Fig. 265.
Key to "Dissected" or Combination Puzzles. 121
Fig. 266.
Fig. 267.
Fig. 268.
122
Puzzles Old and New.
No. VI.— The Star Puzzle. Solutions.
See Figs. 269280 below. Comparison with the problems
will indicate which of the segments should be of the darker,
and which of the lighter material.
Fig. 2CD.
Fig. 270.
Fig. 271.
Fig. 272.
Key to "Dissected" or Combination Puzzles. 123
Fm, 273.
Fig. 274.
Fig. 275.
K 7
Fig. 276,
Fig, 277,
Fig. 278,
124
Puzzles Old and Nezv.
Fig. 279.
Fig, 280.
No. VII.— The Zigzag Square. Solution,,
Arrange the various segments as shownin Fig, 281.
WI/M/l/ii
umm
1
«iil
1
'Huiiil
www
§
Wllill/lii
■■
'mm//
iiiiiiiiiii
Mm
Fig. 281.
No. VIII.— The Extended Square. Solution.
ah c
Fig. 282.
See Fig. 282. The card is cut as indicated in a. The
Key to "Dissected" or Combination Puzzles. 125
upper part is then shifted backward one step, to form the
shape b, or two steps, to form the shape c.
No. IX.— The Octagon Puzzle. Solution.
Fig. 283.
The segments are arranged as shown in Fig. 283.
No. X.— The Patchwork Square. Solution,
The segments are arranged as shown in Fig. 284.
Fig, 284.
No. XI.— The Two Squares. Solution.
See Figs. 285, 286.
First divide the larger square by pencil lines from a to b,
and c to d, then cut from e to c, and from c to/ (Fig. 285).
126
Puzzles Old and New.
The card will now be in three pieces, which, duly re
arranged, will form a square, as shown in Fig. 286.
Fig. 286.
No. XII.— The Latin Cross Puzzle, Solution.
The segments are arranged as shown in Fig. 287.
Fig. 2.87,
No. XIII.— The Greek Cross. Solution.
Divide the cross as shown by the dotted lines in Fig. 288,
and rearrange as shown in Fig 289.
Fig. 288.
Fig. 289.
Key to "Dissected" or Combination Ptizzles. 127
No. xiv.— The Protean Puzzle. Solutions.
See Figs, 290295.
/J Y i N
Fig, 290.
Fig. 291.
Fig, 292.
Fig. 294,
Fig. 295,
128
Puzzles Old and New.
No. XV.— The Caricature Puzzle. Solutions,
See Figs. 296301. A host of other figures, equally
comical, may be formed after a similar fashion.
Fig. 296.
Fig. 297.
Fig. 298,
Fig. 299.
FiG, 300.
Fig. 301.
Key to "Dissected" or Combination Puzzles. 129
No. XVI.— The Chequers Puzzle. Solution.
We have not undertaken to verify the fifty ways in which
it is said that this puzzle may be solved, but we append
Fig. 302.
nWiff*
SSS^
w
■
.
^^3
•
_
Mm
— *
in
i'
Ji
i
'mi
/
~
111
i
_J
Fig. 303.
two (see Figs, 302, 303), leaving the remaining fortyeight
to the ingenuity of our readers.
K
130
Puzzles Old and New.
It will be seen that the two solutions given are radically
different.
No. XVIL— The " Spots" Puzzle. Solution.
It will be found a great assistance towards the solution
of this puzzle to close the box with the lid in the proper
position (in which condition, as we have mentioned, it is an
exact representation of a die), and work from it by way of
model. We will assume, for the purpose of our explanation,
that it is placed as shown in Fig. 192, the six being to the
front, and, consequently, the one to the rear ; the two at top,
consequently the five at bottom ; and the four to the left,
consequently the three to the right, on the side concealed
from view.
The lower stratum will consist of three bars as under.
(See Fig. 304.)
Fig. 304.
The Hinder bar. — Two spots on under side, and one on the
end to the left.
Middle. — One spot in centre of under side. Otherwise
blank.
Front. — Two spots on under side, two in front, and one
on each end.
We next come to the middle layer ; and here most people
give themselves a good deal of unnecessary trouble by taking
it for granted that all the nine bars must lie in the same
direction. As a matter of fact, the three at top and three at
bottom should lie parallel, but the three in the middle at
Key to "Dissected" or Combination Puzzle si 131
right angles to them. This middle layer is formed as
follows : — ■
Lefthand. — Bar with one spot on end towards front.
Otherwise blank.
Middle. — Bar with spot oh hinder end. Otherwise blank.
Bighthand. — Bar with spot on forward end, and one in
centre of righthand side.
(These, as we have said, are to be laid across the three
bars already placed.)
Upper layer. — Back. Spot on each end, and one on
upper side to the right. Further side blank.
Middle. — Blank throughout.
Front. — Two spots on front, one on the end to the left,
and one on the top, at the same end.
These last three bars being laid on the top, parallel to the
bottom section, the die will be complete, having the appear
ance shown in the figure.
No. XVIII.— The Endless Chain. Solution.
Fig. 305.
See Fig. 305, which shows the proper arrangement of the
various segments.
132 Ptizzles Old and New.
No. XIX.— The Hexagon. Solution.
Fig. 300.
The segments are arranged as shown in Fig. 306.
No. XX.— Eight Squares in One. Solution,
Fig. 307.
Place the four squares together to form the centre, and
arrange the smaller pieces round them as shown in Fig. 307.
No. XXI.— The Five Squares. Solution.
Find the centre of either side of a given square, and cut
the card in a straight line from that point to one of the
opposite corners, as shown in Fig. 308. Treat four of the
five squares in this manner. Rearrange the eight segments
Key to "Dissected" or Combination Puzzles. 133
thus made with the uncut square in the centre, as shown in
Fig. 309, and you will have a single perfect square.
Fig. 308.
Fig. 309.
No. XXII.— The Geometrical Square. Solution.
Fig. 310.
Fig. 311.
See Fig. 310, representing the larger square, and Fig. 311,
showing the arrangement of the two smaller squares.
134
Puzzles Old and New,
No. XXIII.— The Dissected Square. Solution.
Fig. 312.
\ ^ /
Cs
S\h
<xA
\ °"
/b
e
Fig. 313.
Fig. 314.
The three smaller squares are formed as sliown in Fig. 312.
The larger square either as Fig. 313 or Fig. 314.
No. XXIV.— The Twenty Triangles. Solution.
Place ten of the triangles alternately, so as to form a
Fig. 315.
rhomboid, and complete the square with the remaining ten,
as shown in Fig. 315.
Key to "Dissected" or Combination Puzzles. 135
No. XXV.— The New Triangle Puzzle. Solution.
Fig. 316.
The seven segments are arranged as shown in Fig. 316.
No. XXVI.— The Japanese Square. Solution.
Fig. 317.
The segments are arranged as shown in Fig. 317.
No. XXVII.— The Chinese Square. Solution.
>v ^ / \ ^ s
C X
/ h \
/ Cs
\a//
<^Jh
fc^>
e /
\ ~b /
X ^
/ & \ / & \.
Fig. 318.
The segments are arranged as shown in Fig. 318.
13^
Puzzles Old and New.
No. XXVlii.The Yankee Square. Solution.
<X/
e^
c \
<L
\
r
o\
£
CO
e
Fig. 319.
The segments are arranged as shown in Fig. 319. .
No. XXix.—Another Cross Puzzle. Solution.
Fig. 320.
Arrange the segments as shown in Fig. 320.
No. XXX.— The Carpenter's Puzzle. No. 1.
Solution.
Fig. 321.
The piece of wood is cut as indicated in Fig. 321, and the
Key to "Dissected" or Combination Puzzles, 137
pieces rearranged as shown in Fig. 322.
Fig. 322,
No, xxxr— Carpenter's Puzzle. No. 2. Solution.
He cuts the board as shown in Fig. 323 — viz. — from a
to b (halfway across) ; from c to d, and then along the middle
1
Fig. 323.
Fig. 324.
from b to d. He then reunites the two pieces as shown in
Fig. 324.
No. XXXII.— The CabinetMaker's Puzzle.
Solution.
Finding the centre of the circle,* he describes a second
circle exactly half the diameter of the first, and then divides
the whole into eight parts, a, a, a, a, and b, b, b, b, by means
of two lines drawn at right angles to each other (see
325). He then cuts the
veneer
through
Fig.
the lines thus
* To find the centre of a given circle, draw a straight line from any
one point to any other point of the circumference. Bisect this line,
and draw another line at right angles to it through the point of section,
terminated, at each end by the circumference. Bisect this lastmen
tioned line, and the point of section will be the centre of the circle.
133
Puzzles Old and New.
described, and rearranges the pieces as shown in Fig. 326.
The space in the centre forms the handhole.
/ ^
CL \
h
b
b
b
\ cu
Cb J
Fig. 325.
Fig. 326.
No. XXXIII. The Bonbon Nuf> Puzzle. Solution.
The six pieces of which the puzzle is composed are. all
shaped like a in Fig. 327, with one exception. This has a
triangular notch, cut out of it, as b in the same figure.
The first step is to examine the puzzle carefully, in order
to discover the notched piece. This, like the rest, is held
between the projecting beads of two of the other pieces, but
Fig. 327.
by reason of the notch may, by a little gentle pressure, bo
forced outwards from between them. This frees the two
pieces lying at right angles to this, and these being re
moved, the puzzle falls to pieces.
To reconstruct the nut, you have only to replace the
pieces in reverse order, the " key " or notched piece last ;
but considerable dexterity is needed to do this, the pieces
having an aggravating knack of suddenly collapsing and
falling into hopeless confusion, just as you have all but
" got it right ; " indeed, ifc seems as if four hands at least
were necessary to hold the pieces in position. With per
severance, however, success should be only a question of
time, and the neophyte must persevere accordingly.
Key to "Dissected" or Combination Ptizzles. 139
No. XXXIV —The Rattle Puzzle. Solution.
Any one who lias solved the Bonbon Nut Puzzle (last
described) will have little difficulty with the Rattle, the
principle being identical. One of the end pieces has a notch
in it, and this being removed, the remaining pieces come
apart almost spontaneously.
No. XXXV.— The Cross Keys (op Threepiece)
Puzzle. Solution.
The three pieces of which the puzzle is composed are
shaped as a, b, and c (Fig. 328) respectively.
To put them together, take a upright between the fore
finger and thumb of the left hand. Through the slot push b,
with the crosscut uppermost, till the farther edge of the'
cu
r\
C/
central slot comes all but flush with the outer face of a.
Then take c, with the short arm of the cross towards you,
and lower it gently down over the top of a, the uncut centre
portion (next the short arm of the cross) passing through
the crosscut in b. You have now only to push b onward
through a till the transverse cut is hidden, and the cross is
complete.
To separate the parts, reverse the process.
No. XXXVI.— The Nut (or Sixpiece) Puzzle.
Solution.
Of the several pieces of which the puzzle is composed, one,
known as the "key," is square from end to end, as a in Fig.
140
Puzzles Old and New.
329. This key is, in all puzzles of this class, the first piece
to be removed, and the last to be replaced.
The first step is to discover the key. This is done by
pressing on the ends of the various pieces, the key being the
only one that yields to the pressure. Having ascertained
which it is, take the puzzle in the left hand, holding it in
such a position that the key shall be horizontal, lying from
right to left, and uppermost of its own pair. Push this out,
and you will then be enabled to lift out one of the pair
pointing towards you. This is shaped like b in the figure.
Push the two uprights a little way from right to left (or
u
f
Fig. 329.
left to right, according as you may be holding the puzzle) .
This will release the second piece, c, of the pair pointing
towards you. The two uprights, d and e, may now be re
moved, and / is left alone.
With a view to the subsequent putting together of the
puzzle, note carefully the shape of the last three pieces.
The upright pair, d, e, have each a projecting piece, or tongue,
in the centre, one going half across, the other only halfway
across. The two faces shown in our illustration are to be
brought together. The third piece, /, has simply an oblong
block cut out of its centre, halfway through, being next in
simplicity to the key piece, a.
To reconstruct the puzzle, take this piece, /, with the cut
part uppermost, and to it fit the pair d, e. From this
point hold the puzzle by the lower ends of d, e. Next insert
c across /, and push d and e to right or left (as the case may
be), so as to lock it in position. Now insert 6, with the smaller
cut uppermost. Finally, insert a, and the deed is done.
Key to "Dissected" or Combination Puzzles. 141
No. XXXVII.— The Fairy Teatable. Solution.
This, notwithstanding the large number of pieces of which
it is composed, is by no means a difficult puzzle.
The first step, in taking it to pieces, is to draw out the
two bolts a, a, which secures the top to the pillar. The next
is to push or pull out the little piece b, when the three other
pieces combined with it will drop out of their own accord.
The pillar may then be separated into four different portions,
thereby releasing the feet, c, c, c, c, and the two little crosses
(each in two portions) d, d, and e, e.
To reconstruct the table, exactly reverse the process.
No. XXXVIII.— The Mystery. Solution.
This is one of the best puzzles of its kind, the clue being
concealed with great ingenuity. When, however, the first
step is discovered, the remaining steps of the process are by
no means difficult.
Fig. 330.
The secret lies in the fact that, as in the case of the Six
piece Puzzle (p. 139), one of the cross*pieces is simply pushed
through from side to side, forming a " key." This found,
hold the puzzle (as in the case of the Sixpiece) in the left
hand, with the key, a, horizontal from right to left, and push
it out. This done (see Fig. 330), you are enabled to remove
the two horizontal semicircles, 6, c. Now lift out the two
[42
Puzzles Old and New,
pieces, d, e, which lie pointing towards you, and remove them
(laying them side by side for greater convenience in recon
struction). Their removal in turn releases the two vertical
semicircles, /, g. Withdraw the remaining horizontal piece,
h, and you have left only the two uprights, ij, supporting
the two remaining semicircles, h, I.
To reconstruct the " Mystery," insert the pieces in the re
verse order, remembering to begin with the only two straight
pieces that are exactly alike (i,j), holding them upright, the
one behind the other. Insert a pair of the semicircles
(these are all alike) at top and bottom, then the horizontal
piece, h, and so on.*
No. XXXIX.— The Diabolical Cube. Solution.
Stand the piece c (the one which looks like a flight of
Fig. 331.
* It will be found a great assistance, in attempting to solve one of the
more difficult puzzles of the " Cluster " class for the first time, before
taking it to pieces, to make pencil marks here and there to show which
Key to "Dissected" or Combination Ptizzles. 143
steps) tip on end, and beside it the piece a, with its project
ing portion uppermost, but farthest away from the highest
step of c. Against the nearer side of c place the square block
e, and stand the small block / on end beside it. The state
of things will now be as shown in Fig. 331. Fix d in beside
a, with one of its projections pointing downward and the
other resting on /. Yon will find that you have now only
room left for the remaining piece, b, whose cutout central
space just fits the projecting top of c, Place this in position,
and the cube is complete.
No. XL.— The Chinese Zigzag 1 . Solution.
To reconstruct the block, yon must follow as closely as
possible, but in reverse order, the process by which it was
Fig. 332.
taken to pieces. First put together one of the external
layers, and, having completed this, lay it, with the flat 'side
undermost, to form the bottom of the block. Then pnt to
gether the layer next in order, and slide it into position,
and in like manner with the two remaining layers.
pair of pieces, and which faces of such pair, come together ; in one case,
say, a single stroke across the line of junction ; in another, tivo strokes
side by side; in another, three ; and so on.
144
Puzzles Old and New.
A novice usually endeavours to reconstruct the block
haphazard, so to speak, instead of layer by layer, under which
conditions success is impossible.
Fig. 332 shows the four layers in readiness for the final
reconstruction, b sliding (from left to right, or vice versa)
over a, and d c over b in like manner.
No. XLL— The Man of Many Parts. Solution.
The secret lies in so arranging the cards that one half of
each shall lap over half of the one next to it, thereby cover
ing up its more eccentric features, and leaving visible only
such half of the card as contributes to the desired result,
which is as shown in Fig. 333.
CHAPTER IV.
ARITHMETICAL PUZZLES.
No. I.— The " Fortyfive " Puzzle.
The number 45 has some curious properties. Among
others, it may be divided into four parts, in such manner
that if you add two to the first, subtract two from the second,
multiply the third by two, and divide the fourth by two, the
result will in each case be equal.
What are they ?
No. II.— A Singular Subtraction.
Required, to subtract 45 from 45 in such manner that there
shall be 45 left.
No. III.— A Mysterious Multiplicand.
Required, to find a number which, multiplied by 3, 6, 9,
12, 15, 18, 21, 24, or 27, shall in each case give as product
the same digit, three times repeated.
No. IV.— Counting" the Pigs.
A youngster asked a farmer how many pigs he had.
" You shall reckon for yourself," said the farmer. " If I had
as many more, and half as many more, and seven to boot, I
should have 32."
How many had he ?
U3 L
146 Puzzles Old and New.
No. V.— Another "Pig*" Problem.
A farmer, being asked the same question, replied, " If I
had as many more, and half as many more, and two pigs
and a half, I should have just a score."
How many had he ?
No. VI.— A Little Miscalculation.
A marketwoman bought 120 apples at four a penny, and
the same number of another sort at six a penny ; but finding
that they were beginning to spoil, determined to sell them off
at costprice. To save trouble, she mixed them together
and sold them at ten for twopence, expecting to just get her
money back again. But when all were sold, she found, to her
surprise, that she had lost twopence over the transaction.
How did this happen ?
No. VII.— A Simple Magic Square.
Required, to arrange the numbers 1 to 9 inclusive in the
form of a square, in such manner that the total of each
line, whether horizontal, vertical, or diagonal, shall be the
same — viz., 15.
No. VIII.— The "Thirtyfour" Puzzle.
Required, to arrange the numbers 1 to 16 inclusive in the
form of a square, in such manner that the total of each
line, horizontal, vertical, or diagonal, shall be the same — viz.,
34.
No. IX.— The "Sixtyfive" Puzzle.
Required, to arrange the numbers 1 to 25 inclusive in the
form of a square, under the same conditions, the total being
65 each way.
No. X.— The "Twentysix" Puzzle,
This is a magic square with a difference, the four corner
places being omitted. The problem is to arrange the
numbers 1 to 12 inclusive in the form of a cross, as shown
in Fig. 334, so as to make 26 in seven different ways — viz.,
Arithmetical Puzzles.
H7
the two horizontal and the two vertical rows, the group
of squares marked aaaa, the group marked bbbb, and
b
b
a
c
c
a
a
c
c
a
b
b
Fig. 331.
the group marked cccc, each making the abovementioned
total.
No. XI.— An Unmanageable Legacy.
An old farmer left a will w^hereby he bequeathed his
horses to his three sons, John, James, and William, in the
following proportions : John, the eldest, was to have one half,
James to have onethird, and William oneninth. When he
died, however, it was found that the number of horses in
his stable was seventeen, a number which is divisible neither
by two, by three, or by nine. In their perplexity the three
brothers consulted a clever lawyer, who hit on a scheme
whereby the intentions of the testator were carried out to
the satisfaction of all parties.
How was it managed ?
No. XII.— Many Figures, but a Small Result.
Required, of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, to com
pose two fractions, whose sum shall be equal to unity. Each
number to be used once, and once only.
No. XIII.— Can You Name It?
Required, to find a number which is just so much short of
50 as its quadruple is above 50.
148 Puzzles Old and New.
No. XIV.— Squares, Product, and Difference.
Required, to find two numbers the sum of whose squares
is greater by 181 than their product, and whose product is
greater by 161 than their difference.
No. XV.— A Peculiar Number.
Required, to find a number of six digits of such a nature
that if you transfer the two lefthand digits (28) to the oppo
site end, the new number thus formed is exactly double the
original number.
No. XVI.— A Novel Century.
Required, by multiplication and addition of the numbers 1
to 9 inclusive, to make 100, each number being used once, and
once only.
No. XVII.— Another Century,
Required, by addition only of the numbers 1 to 9 inclusive
to make 100, each number being used once, and once only.
No. XVIII.— Another Way to make a Hundred.
Required, with six nines to express the number 100.
No. XIX.— The Lucky Number.
Many persons have what they consider a " lucky " number.
Show such a person the row of figiires subjoined —
1, 2, 3, 4, 5, "6, 7, 9,
(consisting of the numerals from 1 to 9 inclusive, with the 8
only omitted), and inquire what is his lucky or favourite
number. He names any number he pleases from 1 to 9, say
7. You reply that, as he is fond of sevens, he shall have
plenty of them, and accordingly proceed to multiply the series
above given by such a number that the resulting product con
sists of sevens only.
Required, to find (for each number that may be selected)
the multiplier which will produce the above result.
Arithmetical Puzzles. 149
No. XX.— The Two Ages.
Father and son are aged 71 and 34 respectively. At what
age was the father three times the age of his son ; and at
what age will the latter have reached half his father's age ?
No. XXL— The Graces and the Muses.
The three Graces, each bearing a like number of roses,
one day met the nine Muses. Each Grace gave to each
Muse the eighteenth part of her store, when it was found
that each Muse had twelve roses less than each of the three
Graces.
What number of roses had each Grace originally?
No. XXII.— The Graces and the Muses Again.
Sometimes the puzzle is stated in another form. The
three Graces, laden with roses as before, meet the nine
Muses, and each Grace gives to each Muse such a pro
portion of her store that when the division is complete,
each Grace and each Muse has an equal share.
How many roses had each Grace at first ?
No. XXIII. Just One Over.
A man, being asked how many sovereigns he had in his
pocket, replied, " If I divide them by 2, by 3, by 4, by 5, or
by 6, I shall always have one over."
What number had he ?
No. XXIV.— Scarcely Explicit.
Another person, being asked a similar question, replied,
" If I had half as much more, twothirds as much more,
threefourths as much more, fourfifths as much more, five
sixths as much more, and nine sovereigns to boot, I should
have exactly £100."
How much had he ?
No. XXV.— Making" Things Even.
Two children were discussing their pocketmoney. " If
you were to give me a penny," said Johnny, " I should have
150 Puzzles Old and New.
twice as much as you." "That would not be a fair division,*'
aid Tommy; " you had better give me a penny, and then we
ball be just alike."
How much money had each ?
. No. XXVI.— A Rejected Proposal.
A little later Johnny and Tommy met again. " I have
now just twice as much as you have," said Johnny; '' but if
you were to give me a penny I should have three times as
much." " No, thank you," said Tommy; " but give me two
pence, and we shall be equal."
How much had each ?
No. XXVII.— The Marketwomap/and hergtock.
A woman selling apples met three boys. TKe first bought
half her stock, and gave her back 10 ; the second bought a
third of what she had then remaining, and gave her back 2 ;
and the third bought half of her then remaining store, and
gave her back 1 ; after which she found that she had 12
apples left.
How many had she at first ?
An
No. XXVIII.— The Captiw
An elderly queen, her daughter, and little son, weighing
Q/' 195 pounds, 105 pounds, and 90 "pounds respectively, were
kept prisoners at the top of a high tower. The only com
munication with the ground below was a cord passing over
a pulley, with a basket at each end, and so arranged that
when one basket rested on the ground the other was oppo
site the window. Naturally, if the one were more heavily
loaded than the other, the heavier would descend ; but if the
excess on either side was more than 15 pounds, the descent
became so rapid as to be dangerous, and from the position
of the rope the captives could not check it with their hands.
The only thing available to help them in the tower was a
cannonball, weighing 75 pounds. They, notwithstanding,
contrived to escape.
How did they manage it ? K^,
Arithmetical Puzzles. 151
No. XXIX.— Father and Son.
A father aged 45 has a son of 12.
How soon will the father be only three times the age of
the son ?
No. XXX.— A Complicated Transaction.
William gives Thomas as many shillings as Thomas has.
Thomas then gives William as many shillings as William
lias left. This done, William has 36 shillings, and Thomas
42 shillings.
How much had each at first ?
No. XXXI.— A Long* Family.
A farmer and his wife have fifteen children, born at
regular intervals, there being a difference in each case of a
year and a half. The eldest is eight times the age of the
youngest.
How old must the latter be ?
No. XXXII.— A Curious Number.
A certain number is divisible into four parts; in such
manner that the first is 500 times, the second 400 times, and
the third 40 times as much as the last ami smallest part.
What is the number, and what are tha several parts ?
No. xxxill.— The Shepherd and his Sheep,
A shepherd was asked how many sheep he had in hi;
flock. He replied that he could not say, but he knew if he
counted them by twos, by threes, by fours, by fives, or by
sixes, there was always one over ; but if he counted them by
sevens, there was none over.
What is the smallest number which will answer the above
conditions ?
No. XXXIV.— A Difficult Problem.
What is the smallest number which, divided by 2, will
give a remainder of 1 ; divided by 3, a remainder of 2 ;
divided by 4, a remainder of 3 ; divided by 5, a remainder of
>
152 Puzzles Old and New.
4 ; divided by 6, a remainder of 5 : divided by 7, a remainder
of 6 ; divided by 8, a remainder of 7 ; divided by 9, a remain
der of 8 j and divided by 10, a remainder of 9.
No. XXXV.— Well Laid Out.
A lad went into a shop to buy drawing materials. He
found that pencils cost twopence each, sheets of paper three
pence, and drawingpins a halfpenny, while indiarubber
was fourpence. He bought a supply of each, spending two
shillings, and found that he had exactly twentyone articles.
What were they ?
No. XXXVI.— The Two Travellers.
A and B are travelling the same road, A going four miles
an hour, B five miles an hour. But A has two and a half
hours' start. 
In what length of time will B overtake A, and how far
from the startingpoint ?
No. XXXVII.— Measuring* the Garden.
A garden, oblong in shape, is thres times as long as it is
wide. If it were a yard more each way, its area would be
increased by 149 square yards.
What are its dimensions ?
No. XXXVIIL— When Will They g*et It?
Seven guests at a restaurant came, the first every day, the
second every other day, the third every third day, and so on
to the seventh, who came once a week only. The host, in a
liberal mood, declared that on the first day all came to
gether he would treat them to a dinner gratis.
How soon, according to the above order of rotation, would
they be in a position to claim his promise ?
No. XXXIX.— Passing* the Gate,
It was the rule in a certain continental town that any one
passing through either of the four city gates, whether going
out or coming in, should pay a penny. A stranger arrived
Arithmetical Puzzles. 153
one day at the town, paid his penny and passed through the
first gate, He spent in the town one half of the money he
had left, and then went out again by the same gate, again
paying a penny. The next day he did the like, entering and
passing out by the second gate, and meanwhile spending
half his available cash in the town. On the following two
days he did the same, entering and leaving by the third and
fourth gates respectively. When he left the town for the
fourth time he had only one penny left,
How much had he at first ?
No. XL.— A Novel Magic Square.
Required, to arrange the numbers 1 to 81 in the form of a
magic square, in such manner that after removing the outer
most rows you still have a magic square, and so on, removing
row by row with the same result until only the number
occupying the central square remains, which number shall
be the greatest common divisor of the sums of the several
squares,
No. XLL— Another Magic Square.
Required, to form with the series of numbers 0, 1, 2, etc.,
to 63 inclusive, a magic square of 64 places, in such manner
that the total of each horizontal or vertical line shall be 252;
if the principal square be broken up into four smaller ones
of sixteen places each, the total of each horizontal or verti
cal line shall be 126 ; and if each of these smaller squares be
again broken up into four squares of four places each, the
total of the numbers in each such four places shall again be
126.
No. XLII.The Set of Weights.
With what five weights can a man weigh any quantity
(proceeding by steps of half a pound) from half a pound up
to 60 lbs. ?
No. XLIIL— What Did He Lose?
A man goes into a shop and buys a hat, price one guinea.
He offers in payment for it a £5 note. The hatter gets
the note cashed by a neighbour, the purchaser pocketing his
change, £3 19s., and walking off with the hat, No sooner
154 Puzzles Old and New,
has he left, however, than the neighbour who changed the
note comes in with the news that it is a counterfeit, and the
hatter has to refund the value.
How much is the hatter out of pocket by the trans
action ?
No. XLIV.— A Difficult Division.
A wine merchant has in his cellar 21 casks. * Seven are
full of wine ; seven halffull, and seven empty.
How can he divide them (without transferring any por
tion of the liquid from cask to cask) among his three sons
— Dick, Tom, and Harry — so that each shall have not only
an equal quantity of wine, but an equal number of casks ?
No. XLV.The Hundred Bottles of Wine.
An innkeeper sold in eight days 100 bottles of wine,
each day overpassing by three bottles the quantity sold on
the previous day.
How many did he sell on the first, and on each of the
succeeding days ?
No. XLVI.The Last of Her Stock.
An old market woman, finding that she had but a few
apples left, divided them among her three grandchildren,
as follows : to Willie she gave half her stock and one
apple over ; to Tommy half what she had then left and one
apple over ; and to Jennie half what she had still left and
one apple over. This done, she had none left.
How many apples did she divide ?
No. XLVIL— The Walking Match.
persons, A, B, G, D, start from the same point to
walk round a circular piece of ground, whose circumference
is one mile. A walks five miles an hour, B four miles, G
three miles, and D two miles an hour.
How long will it be before all four again meet at the
startingpoint ?
Arithmetical Puzzles. 157
collections, that of Bachet de Mezeriac, Eecueil de Problemes
plaisants et delect ables qui se font par les nombres, bearing
date as far back as 1613. Though not exactly arithmetical,
in the ordinary sense, they depend npon arithmetical prin
ciples, and are therefore inserted in this chapter.]
No. LVI— The Three Travellers.
Three travellers, accompanied by their servants, arrive at
the bank of a river and desire to cross. The only means of
transit is a boat which carries two persons. The travellers
have reason to believe that the servants have entered into a
conspiracy to rob and murder them, should they be able to
get the upper hand. It is therefore essential that a single
master should not be left alone with two of the servants, or
two masters with all three of the servants.
How can the transit be arranged so as to avoid either of
the above conditions ?
No. LYII.The Wolf, the Goat, and the Cabbages.
A boatman has to ferry across a stream a wolf, a goat,
and a basket of cabbages. His boat is so small that only
one of the three, besides himself, can be contained in it.
How is he to manage, so that the wolf shall have no oppor
tunity of killing the goat, or the goat of eating up the
cabbages ?
L o
No. LVIII — The Three Jealous Husbands.
Three jealous husbands, travelling with their wives,
find it necessary to cross a stream in a boat which only
holds two persons. Each of the husbands has a great
objection to his wife crossing with either of the other male
members of the party unless he himself is also present.
How is the passage to be arranged ?
No. LIX.— The Captain and His Company.
The captain of a company of soldiers, in the course of his
day's march, comes to a river which must be crossed. The
1 58 Puzzles Old and New.
only means of transit is a boat wherein two children are
paddling about, and which is so small that it will only hold
the two children or one grown person.
How is the transit to be effected ?
No. LX.— The Treasure Trove.
An Irishman and a Scotchman, digging together in a field,
came upon a number of gold coins. When they were about
to divide them, the Scotchman, who was of an avaricious
turn, conceived a plan to outwit the Irishman and secure
the whole for himself. He therefore proposed to the Irish
man that if, without asking any question, he could name
the exact number of coins he should take the whole ; if he
failed, the other should take all. The Irishman readily
agreed, and counted the money, taking special care that
the Scotchman should not see how much it was. "Kow
add 666 to it," said the Scotchman. " Done," replied the
Irishman. "Now, ye'll maybe subtract the whole amount
from 999." "Done again," replied the Irishman ; "but the
divil a bit are ye nearer ! " " Bide a wee," said the Scotch
man. "Now jist pit down 333, and tak' awa the last
figures from it, and ye'll no be far off the tottle of the bit
money." "Mother o' Moses!" exclaimed the Irishman,
" somebody must have tould ye;" and the Scotchman Avalked
off with the treasure trove accordingly.
How did the Scotchman get at the right total ?
No. LXI.— The Row of Counters.
This is in principle very similar to the. last puzzle. In
point of fact it is the same thing, though in a different
form, and we therefore insert it next in order.
A spectator is invited to place upon the table two rows
of counters, unequal in number. The actual numbers are
immaterial, but may range, say, from ten to twenty. The
person who performs the trick is, meanwhile, blindfolded,
so that he can have no knowledge what the numbers are.
The person who has laid out the counters is then requested —
1. To subtract the smaller number from the larger, and
state the difference. (We will suppose that he states such
difference to be 3.)
Arithmetical Puzzles. 159
2. To remove a certain number (say 5) from the smaller
row.
3. To subtract the number remaining in the smaller row
from the larger row. The number subtracted is to be re
moved altogether, as also what now remain of the smaller
row.
4. The operator, without asking any further question, at
once names the number left in the larger row.
How does he ascertain it ?
No. LXIL— A Loan and a Present.
This is another puzzle on the same principle. It is usually
presented in the shape of a conjuring trick. The operator
requests some one to think of a given number of shillings,
large or small, as he pleases. He is then in imagination to
borrow the same amount from some member of the company,
and add it to the original number. " Now please suppose,"
says the operator, " that I make you a present of fourteen
shillings, and add that also. Now give half the total
amount to the poor ; then return the borrowed money, and tell
the company how much you have remaining. I know
already what it is ; in fact, I hold in my hand the precise
amount." " Seven," is the reply. The operator opens his
hand and shows that it contains exactly seven shillings.
How is the amount ascertained ?
No. LXIIL— Eleven Guests in Ten Beds.
An innkeeper had a sadden influx of guests, eleven
arriving in one party, and demanding beds. The host had
only ten beds at his disposal, but he, notwithstanding,
managed to accommodate them as follows : he put two in
the first bed, with the understanding that the second should
have a bed to himself after a brief interval ; he then put
the third in the second bed, the fourth in the third bed,
and so on, the tenth being accommodated in the ninth bed.
He had thus one bed still left, which the eleventh man, now
sleeping double in the first bed, was invited to occupy.
It is clear that there must be a fallacy somewhere, but
where does it lie ?
160 Puzzles Old and New.
No. LXIV.— A Difficult Division.
A and B have purchased an eightgallon cask of wine,
and desire to divide it equally; but they have only two
measures wherewith to do so — one a fivegallon and the
other a threegallon.
How are they to manage ?
No. LXV.— The Three MarketWomen.
Three peasant women went to market to sell apples. The
first had 33, the second 29, and the third 27 only. Each of
them gave the same number of apples for a penny, and yet,
when they got home, they found that each had received an
equal amount of money.
How could such a result come to pass ?
No. LX VI — The Farmer and His Three Daughters,
This is a puzzle of the same kind, differing only in the
figures.
A farmer sent his three daughters to the market to sell
apples. The elder had 50, the second daughter 30, and the
younger 10. The farmer jokingly told them all to sell at
the same price, and bring home the same amount of money,
and, to his surprise, they actually did so.
How did they manage it ?
No. LXVIL— How Many for a Penny?
A boy purchased a pennyworth of apples. He gave to a
playmate onethird of his store, and onethird of an apple
over, after which he had exactly one apple left.
How many did he get for his penny ?
No. LXVIIL— The Magic Cards.
These are usually presented as a conjuring trick, but they
also form a very effective puzzle, for it is clear that the
secret must lie in the cards themselves, and, given sufficient
acuteness, must be discoverable.
Prepare seven cards with numbers on them as follows :—
A rith metical Puzzles.
161
I.
1 33 65 97
3 35 07 00
5 37 69 101
7 39 71 103
9 41 73 105
11 43 75 107
13 45 77 109
15 47 79 111
17 49 81 113
/l9 51 83 115
r 21 53 85 117
23 55 87 119
25 57 89 121
27 59 91 123
29 61 93 125
31 63 95 127
IT.
2
34 66 98
3
35 07 00
6
38 70 102
7
30 71 103
i 10 42 74 106
! 11
43 75 107
1 14 46 1^ 110
15
47 79 111
18
50 82 114
r 22
51 83 115
54 86 118
23
55 87 119
26
58 90 122
27
59 91 123
30 62 94 126
31 63 95 127
h~
III.
4 36 63 100
5 37 69 101 I
6 38 70 102
7 39 71 103
12 44 76 103
13 45 77 100
14 46 73 110 j
15 47 79 111
20 52 84 116
21 53 85 117
22 54 86 118
23 55 87 119
28 60 92 124
29 61 93 125
30 62 94 126
31 63 95 127
IV.
8 40 72 104
41 73 105
10 42 74 106
11 43 75 107
12 44 70 103
13 45 77 109
1«H6 78 110
15 47 79 111
24 56 38 120
25 57 89 121
26 58 90 122
27 59 91 123
23 60 92 124
29 61 93 125
30 02 94 126
31 63 95 127
V.
16 43 80 112
17 49 81 113
18
50 82 114
,19
/20
51 83 115
52 84 116
21
53 85 117
22
54 86 118
23
55 87 119
24
56 88 120
25
57 89 121
20 58 90 122
27 59 91 123
28 00 92 124
29 61 93 125
30 62 94 120
31
63 95 127
VI.
32
48 96*112
33
49 97 113
34 50 98 114
35
51 99 115
36
52 100 116
37
53 101 117
38 54 102 118
39
55 103 119
40 56 104 120
41
57 105 121
42
58 106 122
43
59 107 123
44 60 108 124
45 61 109 125
46 62 110 126
47
63 111 127
i
VII.
04 80 90 112
05 81 97 113
66 82 98 114
67 83 99 115
68 84 100 116
69 85 101 117
70 86 102 118
71 87 103 119
72 88 104 12U
73 89 105 121
74 90 100 122
75 91 107 123
76 92 108 124
77 93 109 125
78 94 110 120
79 95 111 127 
A person is requested to think of any number, from 1 to
127 inclusive, and to state on which one or more of the seven
cards it is to be found, Any one knowing the secret can
instantly name the chosen number.
How is the number ascertained ?
No. LXIX.— The " Fifteen" or "Boss" Puzzle.
This, like a good many of the best puzzles, hails from
America, where, some years ago, it had an extraordinary
yogue, which a little later spread to this country, the
H
l62
Puzzles Old and New.
British, public growing nearly as excited over the mystic
"Fifteen " as they did at a later date over the less innocent
" Missing Word " competitions.
The "Fifteen" Puzzle consists of a little flat wooden or
cardboard box, wherein are arranged, in four rows, fifteen
little cubical blocks, each bearing a number, from 1 to
15 inclusive. The box being square, would naturally accom
modate 16 such cubes,* and there is therefore one space
always vacant, and by means of such vacant space, the
cubes may be shifted about in the box so as to assume
different relative positions.
1
2
3
4
5
6
7
8
9
10
11
12
13
15
14
Fig. 335,
The ordinary " Fifteen " Puzzle is, having placed the cubes
in the box haphazard, so to move them about (without lifting',
but merely pushing one after another into the space for the
time being vacant) as to bring them into regular order from
1 to 15, leaving the vacant space at the righthand bottom
corner.
In what is known as the " Boss " or Master Puzzle all
the blocks are in the first instance placed in regular order,
with the exception of those numbered 14 and 15, which are
reversed, as in Fig. 335.
* As a matter of fact, 16 cubes are usually supplied, tie set thereby
being made available for the " Thirty four " Puzzle (referred to at p, 146).
When the " Fifteen" Puzzle is attempted, the cube bearing the number
16 is removed from the bos.
Arithmetical Puzzles.
16
No. LXX.— The Pegaway Puzzle.
The "Pegaway" Puzzle (Perry & Co.) is a variation or
modification of the "Fifteen " Puzzle, the difference being that
there are nine instead of sixteen cells, and that eight instead
1
2
o
O
4
5
G
7
8
Fig. 336.
of fifteen numbered blocks or " pegs " are used. These
being arranged liaphazard in the cells, the problem is to
bring them into consecutive order, as shown in Fig. 336.
No. LXXL— The OverPolite Guests.
Seven gentlemen met to dine at a restaurant, when a
question arose as to precedence, no one desiring to take
what were regarded as the more honourable seats. To settle
the matter, one of them proposed that they should dine to
gether every day until they had respectively occupied all
possible positions at the table ; and the suggestion was
accepted.
How often must they dine together to answer the above
conditions ? *
* This problem is sometimes propounded in another shape, as follows :
A party of seven students, with more wit than money, agreed with a
restaurantkeeper to pay him £10 per head so soon as they should have
occupied all possible positions at the table, he undertaking to entertain
them daily in the meantime with a dinner costing half a crown per head.
Query, how much the host made or lost by the transaction?
164
Puzzles Old and New.
No. LXXIL— The " Royal Aquarium" Thirteen
Puzzle.*
This is an adaptation of the "Magic Square" idea, but
modified in a very ingenious manner, the ordinary processes
for forming a magic square being here quite inapplicable.
The puzzle consists of nine cards, not quite 1J inch each
way, each bearing four numbers, radiating from the centre,
after the manner shown in Fig. 337.
CT
m
CO
V
to
10
*».
^
CO
rf*
«*
CTT
Fig. 337.
The figures shown in heavy type in the diagram are in
the original printed in red.
The experimenter is required to arrange the nine cards in
a square, the red numbers forming perpendicular lines, and
the black numbers horizontal lines, the three figures in
each line, whether horizontal or perpendicular, making,
when added together, 13.
* Procurable at the Royal Aquarium, Westminster,
post, with solution, 3d,
Price Id., or by
Arithmetical Puzzles. 165
No. LXXIIL— An Easy Creditor.
A gentleman being in temporary need of money, a friend
lent him £60, telling him to repay it in such sums as
might suit his convenience. Shortly afterwards he made a
payment on account. His second payment was half as much
as the first; his third threequarters as much; his fourth
onequarter as much, and his fifth twofifths as much. It
was then found, on striking a balance, that he still owed
£2.
What was the amount of the first payment ?
No. LXXIV.— The Three Arabs.
Two Bedouin Arabs halted in the desert to eat their __
midday meal.. Their store consisted of eight small loaves, •/
of which five belonged to the first, and three to the second.
Just as they sat down, a third Arab overtook them, and
asked to be permitted to share their meal, to which they
agreed. Each ate an equal portion of the eight loaves, and
the third Arab, at the close of the meal, handed the others
eight pieces of money in payment. A dispute arose as to
the division of the money, the first Arab maintaining that
as he had had five loaves and the other three only, the
money should be divided in the same proportion. The other
maintained that as all had eaten equally, each should take
half the money between them. Finally they agreed to refer
the matter to the third Arab, who declared that both were
in the wrong, and pointed out the proper division.
What was it ?
No. LXXV.— An Eccentric Testator.
An eccentric old gentleman left a will, whereby he be
queathed to his eldest son £110, and oneninth of what
remained ; to his second son £220, and oneninth of what
then remained; to his third son £330, and oneninth of the
remainder ; and so on, each junior in turn taking £110 more
by way of original gift, and oneninth of the portion still
remaining.
The legatees at first complained of so unequal a disposi
tion, but on ascertaining the value of the estate and proceed
ing to a division, they found, to their surprise, that the
1 66 Puzzles Old and New.
division exactly exhausted the estate, and that the share of
each was of exactly the same value.
How many legatees were there, and what was the total
value of the estate ?
No. LXX VI.— Another Eccentric Testator.
A testator, having five sons, left his property as follows :
To the eldest one sixth of the whole, and £240 in addition.
To the second onefifth of the residue, and £288. To the
third onefourth of the then residue, and £360. To the
fourth onethird of the then residue and £480. And to the
fifth onehalf of what still remained, and £720. There was
then nothing left.
What was the amount of the property, and what was the
share of each son ?
No. LXXVIL— An Aggravating Uncle.
An uncle with a turn for figures presented his youthful
nephew with a box of soldiers, but made it a condition that
he should not play with them till he could discover, on
arithmetical principles, how many the box contained. He
was told that if he placed them three in a row, there would
be one over ; if he placed them four in a row, there would
be two over ; if five in a row, three over ; if six in a row, he
would have four over. The total number was under 100.
How many soldiers did the box contain ?
No. LXXVIIL— Apples and Oranges.
A paterfamilias brought home a quantity of apples and
oranges, the same number of each, and distributed them
among his children. After each child had received 12
apples there were 48 over, and after each child had received
15 oranges there were 15 over.
How many were there of each kind of fruit, and among
how many children were they divided ?
No. LXXIX.— The Two Squares.
A certain number of counters, if arranged in rows so as to
form a square, leave a remainder of 146 counters unem
A rithmetical Puzzles. 1 6 7
ployed. To enlarge the square bjran additional row each
way 31 more counters would be required.
What is the number of counters ?
No. LXXX.— A Curious Division.
Required, to divide 7890 counters into three heaps, in such
proportions that if the first heap be divided by three, the
second by six, and the third by nine, the quotient shall in
each case be the same number.
No. LXXXL— A Curious Multiplication.
Required, to divide the same number (7890) of counters
into four heaps, in such proportions that if the first be multi
plied by three, the second by four, the third by six, and the
fourth by twelve, the product shall in each case be the same.'
No. LXXXIL— The Two Schoolmasters.
Two country schoolmasters were discussing their respec
tive schools. The first said, " Onesixth of my pupils are
away ill, eleven are haymaking, seven are gone to the fair,
and I have thirtyseven here at work." The other replied,
"You have a pretty large school, but mine is larger than
yours, for I have seventytwo pupils."
What was the difference of their numbers ?
No. LXXXIIL— Nothing" Left.
There is a certain number from which, if you subtract ten,
multiply the remainder by three, find the square root of the
product, and from sucli square root subtract eighteen,
nothing is left.
What is the. number ?
No. LXXXIV.— The Three Generations.
An old man was asked how old he was. He replied,
" The united ages of my son and myself are 109 years ; those
of my son and my grandson are 56 years j and my grandson
and myself together number 85 years."
How old was each ?
1 68 Puzzles Old and New.
No. LXXXV.— The Two Brothers.
" How old is your brother ? " a man was asked. " Two
thirds of his age," he replied, "is just fivetwelfths of mine,
and I am nine years older than he." "tfq^ 4^5
What was the age of each ? '
No. LXXXVL— The Two Sons,
An elderly mathematician was asked what were the
nges of his two sons. He replied, " The one is five and a
quarter years older than the other, and six times the age of
the elder, added to five times the age of the younger, would
be 301."
What was the age of each?
No. LXXXVIL— The Two Nephews.
The same man was asked the ages of his two nephews.
"The elder," he replied, "is just three times as old as the
younger, and if you add together the squares of their ages,
the total will be 360."
What was the age of each ?
No. LXXXVIIL— The Reversed Number.
There is a number consisting of two digits; the number
itself is equal to five times the sum of its digits, and
if nine be added to the number, the position of its digits
is reversed.
What is the number ? *
No. LXXXIX — Another Reversed Number.
There is another number of two digits, which itself is
equal to seven times the sum of its digits. If 18 be sub
tracted from it, the position of its digits is reversed.
What is the number ?
No. XC— The Shepherd and his Sheep.
A shepherd was asked how many sheep he had. He re
plied, " I have 100, in five sheepfolds. In the first and
second there are altogether 52 sheep ; in the second and
third, 43 ; in the third and fourth, 34 ; and in the fourth
and fifth, 30."
How many sheep had he in each fold ?
Arithmetical Puzzles. .169
No. XCL— The Shepherdess and her Sheep.
A shepherdess had the care of a number of sheep, in four
different folds. In the second were twice as many as
the first, in the third twice as many as the second, and in
the fourth twice as many as the third. The total number
was 105.
How many sheep were there in each fold ?
No. XCIL— A Weighty Matter.
With how many weights, and of what denominations
respectively, can you weigh any number of pounds from 1 to
127 inclusive?
No. XCI1L— The Three Topers.
Three topers were discussing how long it would take each
of them to drink forty quarts of lager beer. Peter under
took to do it between four in the morning and twelve at
night; Paul between ten in the morning and twelve at
night ; and Roger between two in the afternoon and twelve
at night.
Assuming that they could do as they boasted, how long
would it take them to get through the same quantity, all
drinking simultaneously ?
No. XCIV.— The False Scales.
A cheese put into one of the scales of a false balance was
found to weigh 16 lbs. When placed in the opposite scale it
weighed 9 lbs. only.
What was its actual weioht ?
No. XCV.— An Arithmetical Policeman.
A belated reveller, hearing the clock strike, but being too
obfuscated to be quite sure as to the number of strokes, pro
ceeded to "ask a policeman" w T hat time it was. The
policeman replied, " Take half, a third, and a fourth of the
hour that has just struck, and the total will be one larger."
What was the hour ?
i 70 Puzzles Old and New.
No. XCVL— The Flock of Geese.
Two friends, passing a woman with a flock of geese, made
a wager as to who should guess nearest at their number,
without actually counting, one maintaining that there were
not more than thirty, the other that there were over
forty of them. On asking the marketwoman which was
right, she replied, " If I had as many more, and onehalf as
many more, and onefourth as many more, I should have
one short of a. hundred. Now puzzle it out for yourselves."
What was the number of the flock ?..
No. XCVIL— The Divided Cord.
A piece of cord. is thirtysix inches in length. Required,
so to divide it into two parts that one of them shall be
exactly fourfifths the length of the other.
No. XC VIII.— The Divided Number.
Divide the number 46 into two parts in such manner that
if the one be divided by 7 and the other by 3, the sum of the
quotients shall be 10.
No. XCIX.— The Two Numbers.
There are two numbers, such 'that twice the first plus the
second = 17, and twice the second plus the first = 19.
Find the numbers.
No. C— The Horse and Trap.
A man purchased a horse and trap. Five times the price
of the horse was just equal to twelve times the price of the
trap, and the two together cost £85.
What was the price of each ?
No. CI.— The Two Workmen.
A, working 7 hours a day, can do a piece of work in 10
days, and. B, working 8 hours a day, can do it in 7 days.
Supposing both employed together, how many hours a day
must they w T ork in order to complete it in 5 days ? .
Arithmetical Puzzles. 171
No. CIL— Another Divided Number.
Required, to divide the number 237 into three parts, in
such manner that 3 times the first shall be equal to 5 times
the second and 8 times the third.
No. CIIL— The Three Reapers.
A, B, and (7, working together, can reap a certain field in
5 days. B, working alone, would take twice as long as A
and G together; and G, working alone, would take three
times as long as A and B together.
How long would each take to do the work separately ?
No. ClV.The Bag of Marbles.
Three boys have a bag of marbles given to them, and it is
agreed that they shall be divided in proportion to their aoes,
which together amount to 17J years. The bag contains 770
marbles, and as often as Tom takes 4 Jack takes 3, and as
often as Tom takes 6 Dick takes 7.
How many marbles will each get, and what are their re
spective ages ?
No. CV — The Expunged Numerals. A.
Given, the following sum in addition : —
ill
117
909
Required, to strike out six of these numbers, so that the
total of the remaining numbers shall be 20 only.
No. CVT— The Expunged Numerals.
Given, the sum following :—
] J i
3: ::>
999
Required, to strike out nine of the above figure?, so that
the total of the remaining figures shall be 1111.
172 Puzzles Old and New.
No. CVIL— A Tradesman in a Difficulty.
A man went into a shop in New York and purchased
goods to the amount of 34 cents. When he came to pay, he
found that he had only a dollar, a threecent piece, and a two
cent piece. The tradesman had only a half and a quarter
dollar. A third man, who chanced to be in the shop, was
asked if he could assist, but he proved to have only two
dimes, a fivecent piece, a twocent piece, and a onecent
piece.* With this assistance, however, the shopkeeper
managed to give change.
How did he do it ?
No. CV HI.— Profit and Loss.
A tradesman sells a parcel of soiled goods at a loss for
£2 16s. The market price was £3 5s., at which price he
would have made three times as much by them as he actu
ally lost.
What did they cost him originally ?
No. CIX.— A Curious Fraction.
What like fractions of a pound, of a shilling, and of a
penny will, when added together, make exactly a pound?
No. CX.— The Menagerie*
The proprietor of a menagerie was asked how many birds
and how many beasts it included. He replied, "Well, the
lot have 36 heads and 100 feet."
How many of each were there ?
* For the assistance of readers who may not be familiar with the
American coinage, we may mention that the cent is equivalent to our
halfpenny. Of these 100 go to the dollar, which is therefore worth
4s. 2d., while the half and quarter are equivalent to 2s. Id. and Is* 0%d,
(50 cents and 25 cents) respectively. The dime is the tenth part of a
dollar, and is therefore worth 10 cents, or 5d.
Arithmetical Puzzles. ' 173
No. CXI.— The Marketwoman and her Eggs.
A marketwoman, selling eggs, sold to her first cus
tomer the half of her stock and half an egg over. To
her second customer she sold onehalf of the remainder and
half an egg over. To a third customer she sold half her yet
remaining stock and half an egg over, when she found that
she had none left.
How many eggs had she originally ?
No. CXIL— The Cook and his Assistants.
This is a problem of the same kind, differing only in the
figures. A cook distributes eggs to his three assistants, to
the first onehalf and half an egg over, to the second one,
half of the remainder and half an egg over, and to the third
onehalf of what still remains and half an egg over. He has
still four eggs left.
How many had he at first ?
KEY TO CHAPTER IV.
ARITHMETICAL PUZZLES.
The solution of problems of this class is greatly facilitated
by an elementary knowledge of the Properties of Numbers.
We append, therefore, a statement of some of such properties,
not limited to the particular examples we have given (the
majority of which depend upon much simpler considera
tions), but of general utility in relation to such problems.
We offer in each case a brief demonstration of the fact
stated, though for the purpose of the present work it will be
quite sufficient if the reader is content to accept the proposi
tions laid down w r ithout demanding mathematical proof of
their ac curacy „
Elementary Properties op Numbers.
Proposition 1. — The sum or difference of two even numbers
is alivays an even number.
Proof.— This becomes selfevident if the pro
position be expressed in algebraic form. Thus —
Let 2x = the larger number,
and 2y — the smaller number.
Then their sum will be 2x + 2y,
and their difference will be 2x — 2y ;
each of which is obviously divisible by 2, with
out fraction or remainder — i.e., an even number.
Prop. 2. — The sum or difference of two odd numbers is al
ways an even number.
Proof. — Let 2x + 1 = the larger number,
and 2y + 1 = the smaller number.
Then their sum will be 2x + 2y + 2 ;
* and their difference 2x — 2y ;
each of which is obviously divisible by 2 : i.e.,
an even number.
i7i
Key to Arithmetical P tizzies. 175
Prop. 3. — The sum of an even and an odd number is alivays
an odd number.
Proof. — Let 2x = the even number,
and 2y + 1 = the odd number.
Then their sum will be 2x + 2y + 1, which is
obviously not divisible by 2 without a fraction or
remainder — i.e., is an odd number.
Prop. 4. — The difference of an even and an odd number is
always an odd number.
Proof. — Here there are two possible cases to
be considered, as the odd or the even number
may be the larger. In the first case —
Let 2x + 1 = the larger number,
and 2y =the smaller number.
Here their difference will be represented by
2x + l—2y, which is obviously an odd number.
In the second case —
Let 2x =the larger number,
and 2y + l= the smaller number.
Then their difference will be represented by
2x — 2y — 1, which is obviously an odd number.
Prop. 5. — If the sum of two numbers be an even number, then
their difference is also an even number.
Proof. — This follows from preceding propo
sitions. The sum of the two numbers being, ex
hypothesi, even, the numbers themselves must
either (by Props. 1, 2, 3) be both even or both
odd; and if so, their difference (by Props. 1, 2)
must be even.
Prop. 6. — If the sum of two numbers be an odd number, then
their difference is also an odd number.
Proof. — This follows from Props. 3 and 4. The
sum of the two numbers being odd, the numbers
themselves must (by Prop. 3) be one odd, and
one even, in which case their difference (by Prop.
4) will be odd.
Prop. 7. — For a number to be divisible by 2, its last digit
must be even. »
Proof.— (Selfevident. All odd numbers divided
by 2 leave a remainder of 1.)
176 Puzzles Old and New.
Prop. 8. — If a number to be divisible by 3, the sum of its
digits iv ill also be divisible by 3.
Proof. — Let a represent the units, b the tens,
c the hundreds, d the thousands, and so on. Then
the whole number may be expressed in the form
of a sum in addition, as follows :
a
60
cOO
d 000
d c b a
Now6 =6 10 =6(9 + 1) = 6 9 +6
and c 00 =c 100 = c (99 + I)=c99 +c
and d 000 = cl 1000 = d (999 + 1) = d 999 + cZ.
The whole number is therefore equal to
a
6 9 + 6
c99 + c
d 999 + d
And as 69 + c99 + d999 is obviously divisible by
3, it follows that, to make the whole number ex
actly divisible by 3, a + b + c + d (the sum of its
digits), must be divisible by 3 in like manner.
Prop. 9. — Any number, less the sum of its digits, is divisible
by 3.
Proof. — This is incidentally proved in the
course of the demonstration of Prop. 8, where we
have seen that any number may be thrown into the
form 19 + c99 + d 999 + a + b + c + d. Deducting
from this a + b + c + d, representing the sum of
the digits of the original number, it is clear that
the remainder is a multiple of, and therefore
divisible by, 3.
Prop. 10. — Any number, to be exactly divisible by 4, must have
the number formed by its last two digits divisible by 4.
Proof. — Any number of more than two digits
consists of so many hundreds as may be expressed
by the preceding digits, plus the number ex
pressed by the last two digits. (Thus 532 is equi
valent to 500 + 32 ; 1429 to 1400 + 29, and so
Key to Arithmetical Puzzles. 177
on). Each hundred, being a 'multiple of 4, is
necessarily divisible by that number. If, there
fore, the remaining digits (the last two) are
divisible by 4, the whole number must be so
divisible.
Prop. 11. — For a number to be divisible by 5, it must have
either 5 or for its units digit.
Proof. — When any given number is divided by
5, the only possible remainder is 0, 1, 2, 3, or 4.
Whichever of such remainders be left over from
the division up to the tens place, it is clear that
there must be a or a 5 in the units place to
enable the division to be completed without re
mainder.
Prop. 12. — For a number to be divisible by 6, it must be an
even number, and the sum of its digits must be divisible by 3.
Proof. — As 6 = 2x3, the divisibility of a given
number by 6 depends upon its divisibility by 2
and 3; for the tests of which divisibility see
Props. 7 and 8.
Prop. 13. — (There is no general criterion as to divisibility
by 7.*)
Prop. 14. — For a number to be evenly divisible by 8, its last
three digits must be divisible by 8.
Proof. — The proof here follows the same line
of argument as that of Prop. 10. Any number
of more than three digits consists of a given
number of thousands, plus the number expressed
by the last three digits. Each thousand, being
a multiple of 8, is necessarily divisible by 8, and
if therefore the number represented by the re
maining three digits is divisible by 8, the whole
number will be so divisible.
* It should, however, be noted that every cube number is either ex
actly divisible by 7, or may be made so by adding or subtracting 1 to or
from it.
N
178 Puzzles Old and New,
Prop. 15. — For a number to be evenly divisible by 9, the sum
of its digits must be evenly divisible by 9.
Proof. — See the proof of Prop. 8, where ifc is
shown that any given number can be expressed
in the form d 999 + c 99 + b 9 + a + b + c + d. The
portion d 999 + c 99 + b 9 being obviously divisible
by 9, it follows that to make the whole sum divi
sible by 9, the remaining portion of the number,
a+b+c+d (representing the sum of its digits),
must also be divisible by 9.
Prop. 16. — For a number to be divisible by 10, its last digit
must be 0,
Proof. — This proposition is practically self
evident. Any given number represents so many
tens plus the number of units expressed by the
last digit. If such last digit be 1, 2, 3, 4, 5, 6, 7,
8, or 9 (i.e., any number save 0), it will represent
a remainder of that amount, and the whole num
ber cannot be divided evenly by 10.
Prop. 17. — The product of any two consecutive numbers (as
4 mid 5, 14 and 15, or 26 and 27) is always divisible by 2.
Proof. — Of any two consecutive numbers one
must be an even number, and therefore a mul
tiple of 2. By whatever other number it may
be multiplied, the product will therefore also be
divisible by 2.
Prop. 18. — The product of any three consecutive numbers is
alivays divisible by 6.
Proof. — This is demonstrable in like manner,
for of three consecutive numbers one must be a
multiple of 3, and one of 2 * ; and if a number is
divisible by 3 and by 2, it is also divisible by 6.
Prop. 19. — The product of any four consecutive numbers is
alivays divisible by 24; of any five, by 120, and so on (the new
divisor in each case being obtained by multiplying the last
divisor by the number of consecutive figures) .
Proof. — The proof is a mere extension of that
of Prop. 18.
* Both characters may be united in the intermediate number, as in
the case of 11, 12, 13 ; 23, 24, 25.
Key to Arithmetical Puzzles. 179
Prop. 20. — The difference "between the squares of two numbers
is equal to the product of their sum and difference.
e.g., 5 2 3 2 =259 = 16
(5f3)x(53)=8x2 = 16.
Proof. — This is but a restatement of the "well
known algebraical formula —
But it may be made equally clear without re
course to algebra, thus : —
Take any two numbers — say 5 and 3.
(5 — 3) multiplied by (5 + 3) means (5 — 3)
multiplied by 5 and (added on to this) (5 — 3)
multiplied by 3.
rTow (5 — 3) multiplied by 5 gives 5 3 — (5 x 3)
and (53) „ „ 3 „ (5x3)3 3 . .
Adding the two results together,
(53) multiplied by (5 + 3)=5 3 3 3 .
And this holds with any numbers, so that gene
rally n 2 — m 3 = (ii + m) x (ii — m) .
Prop. 21. — The difference between the squares of any two
numbers is always divisible by the difference of the tivo numbers.
It is also divisible by the sum of the two numbers.
Proof. — This is a necessary consequence of the
preceding proposition.
Prop. 22. — The difference of the squares of any tivo odd num
bers is always divisible by 8.
Proof. — Suppose (2x + 1) and (2y + l) to be
the two odd numbers.
Then (by Prop. 20) the difference of their
squares equals the product of their sum and
difference: i.e. (2x + 1) 2  (2y + 1) 2 = {(2oj + 1) +
(2y + l)}x{(2a + l)(2y + l)}
= {2x + 2y + 2}{2x2y]
— 4i{xty + l}{x — y}
Now if x and y are both odd or both even, then
(by Prop. 1) (x—y) is even, in which case
4(x + y + l) (x — y) is divisible by 4 and further
by 2 ; that is to say, by 8.
But if x and y are unlike, — that is to say one
odd and the other even, — then x + y is odd, and
i8o
Puzzles Old and New.
therefore (x + y + 1) is even; in which case
4(x + y + l) (x—y) is again divisible by 4 and
further by 2 ; that is to say, by 8.
Wherefore (2x + 1) 3  (2y + l) 2 is divisible by 8.
Prop. 23. — The difference betiveen a number and its cube is
the product of three consecutive numbers, and is consequently
divisible by 6.
Suppose n z and n to be the cube and the
number; then difference between them is n 3 — n.
This evidently = w(w 2 — 1).
But looking upon 1 as a square number (the
square of unity) we have (?i 2 — 1) = (n — 1)
x (n + 1), as in Prop. 20.
So that n 3 — n = n(n — l)(n+l).
But (n — 1), n, and (n + 1) are obviously three
consecutive numbers, and therefore (by Prop. 18)
divisible by 6.
Prop. 24. — Any prime number which, divided by 4, leaves a
remainder 1 is the sum of two square numbers.
The mathematical proof of the universal truth of this pro
position is very complicated, and would be beyond the scope
of a work like the present. We subjoin a list of all such
numbers below 4.00, and from the evident truth of the
theorem in this large number of cases the reader may be
content to infer its general accuracy.
List
D
13
17
29
37
41
53
6L
73
89
97
101
109
OF ALL PRIME
BY
■NUMBERS BELOW 400 WHICH, BEING DIVIDED
A REMAINDER OE 1.
4 +
9 +
16 +
25 +
36 +
25 + 16 =
49+ 4 
36 + 25 =
64+ 9 =
64 + 25 
81 + 16 =
100+ 1 =
100+ 9 =
, EEAVE
2 2 + l 2
3 2 + 2 2
4 3 +l 2
5 2 + 2 2
6 2 + l 3
5 3 + 4 2
7 2 + 2 3
6 2 + 5 2
8 3 + 3 2
8 2 + 5 2
9 2 + 4 2
10 2 + 1 2
10 3 + 3 3
113
137
149
157
173
181
193
197
229
233
241
257
269
= 64 +
= 121 +
= 100 +
= 121 +
 169 +
 100 +
= 144 +
 196 +
= 225 +
= 169 +
 225 +
= 256 +
49
16
49
36
4
81
49
1
4
64
16
1
= 169 + 100 =
= 8 2 + 7 3
= 112+ 42
= 10 2 + 7 2
= ll 2 + 6 2
= 13 2 + 2 2
= 10 2 + 9 2
 12 2 + 7 2
 14 2 + l 2
= 15 2 + 2 2
= 13 2 + 8 3
= 15*+ 4 2
==' 16 2 + I 2
13 2 + 10 3
Key to Arithmetical Puzzles.
181
277 = 196+ 81 =
= 14 2 + 9 2
349
= 324+ 25 =
18 2 + 5 2
281 = 256+ 25 =
= 16*+ 52
353
= 289+ 64 =
17 2 + 8 2
293 =289+ 4 =
= 172 + 2 2
373
= 324+ 49 =
182+ 72
313 = 169 + 144 =
= 13 2 + 12 3
3S9
= 289 + 100 =
172 + 102
317 = 196 + 121 =
= 14 2 + 112
397
 361+ 36 =
192+ 6 2
337 = 256+ 81 =
= 102+ 9 2
With this brief introduction we proceed to give the solu
tions of the various puzzles propounded in the preceding
pages.
It is generally very much more easy to solve puzzles of this
class by the aid of algebra, and in some instances we have
exhibited the solution in this form. But this is scarcely re
garded as a legitimate method of solving an arithmetical
puzzle, and where the reader can solve such a puzzle by step
bystep argument, without the use of algebraic symbols, he
may consider that he has attained a considerably higher
measure of success. Even if algebra be used in the first
place, the problem should, if possible, be worked out arith
metically afterwards.
No. I.— The Fortyfive Puzzle. Solution.
The first of the required numbers is 8.
8 + 2 = 10
The second is 12.
122 = 10
The third is 5.
5x2 = 10
The fourth is 20.
20 + 2 = 10
8 + 12 + 5 + 20 = 45.
No. II.— A Singular Subtraction. Solution.
This is somewhat of a quibble. The number 45 is the
sum of the digits 1, 2, 3, 4, 5, 6, 7, 8,9. The puzzle is
solved^ by arranging these in reverse order, and subtracting
the original series from them, when the remainder will be
1 82 Puzzles Old and New.
found to consist of the same digits in a different order,
and therefore making the same total — viz., 45.
987654321  45
123456789 == 45
864197532 ^ 45
No. III.— A Mysterious Multiplicand. Solution.
The number 37 will be found to answer the conditions of
the problem. Multiplied by. 3, it is 111; by 6, 222; by 9,
333; by 12, 444; by 15, 555; by 18, 6Q6 ; by 21, 777; by
24, 888 ; and by 27, 999.
No. IV.— Counting* the Pigs. Solution.
Answer. — He had ten.
Proof: 10 + 10 + 5 + 7 = 32.
No. V.— Another Pig Problem. Solution,
He had seven.
Proof: 7 + 7 + 3 + 2 = 20.
No. VI.— A Little Miscalculation. Solution.
By the time she has sold 20 lots at 10 for twopence (four
of the one kind and six of the other), the 120 cheaper apples
are exhausted, while she has only sold 80 of the dearer. So
far, she has neither gained nor lost, but she has still on hand
40 of the dearer apples, worth, at cost price, tenpence. By
selling these at 10 for twopence she only gets eightpence,
leaving a deficit of twopence.
No. VII.— A Simple Magic Square. Solution.
The arrangement depicted in Fig. 338 answers the con
ditions of the problem, the total being 15 in each direction.
Key to Arithmetical Puzzles.
1 8
No. VIII.— The Thirtyfour Puzzle. Solution.
It is computed that no less than 3456 arrangements of
the sixteen figures will answer the required conditions.
Whether this total be exact we will not undertake to decide,
2
9
4
7
5
3
6
1
8
..
Fig. 338.
bat it is certain that such arrangements may be numbered
by some hundreds. The fact that, with so many possible
solutions, they were not oftener hit upon by the hundreds of
thousands who, during its temporary " boom," racked their
brains over the " Thirtyfour " Puzzle arises from the fact
that, numerous as they are, they are but as a drop in the ocean
compared with the number of combinations (amounting to
over twenty billions) of which sixteen given articles are
capable.
We append, hj way of specimens, a few of the possible
solutions (Figs. 339344).
1
15
14
4
12
6
7
9
8
10
11
5
13
3
2
16
1
14
15
4
8
11
10
5
12
7
6
9
13
2
3
16
i
Fig. 339.
Fig. 340.
1 84
Ptizzles Old and New.
1
12
6
15
13
8
10 3
IG
5
11
2
4
9
7
14
1
8
15
10
14
11
4
5
12
13
G
3
' 7
2
9
IG
i
Fig. 341.
Fig. 312.
1
7
IG
10
12
14
5
3
6
4
11
13
15
9
2
8
1
11
6
16
8
14
3
9
15
5
12
2
10
j i
4 13 1 7
1 i
i i
Fig. 343.
Fig. 344.
In eacli of these cases the number 34 can be counted in
ten different directions — viz., four horizontal, four perpen
dicular, and two diagonal. The simplest method of con
structing such squares is as follows :— First arrange the
figures (which may be any series in arithmetical progression)
in the form of a square in their natural order, as, for example,
the series shown in Fig. 345. Regard this as tivo squares,
one within the other, as indicated by the heavy black lines.
Now rearrange the series, reversing the diagonals of each
such square, when the figures will be as in Fig. 346, counting
46 in each of the directions abovementioned.
By marshalling, in the first instance, the figures in vertical
instead of horizontal rows, the resulting arrangement will be
slightly different}, but yet producing the same totals (see
Figs. 347 and 348).
Key to Arithmetical Puzzles. 185
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Fig.
345.
4
5
8
12
1G
9
13
17
6
10
14
as
7
11
15
19
19
5 6
16
8
14
13
11
12
10
9
15
7
17
18
4
Fig. 346.
19
8
12
7
5
14
10
17
6
13
9
13
1G
11
15
4
Fig. 347.
Fig. 348.
Magic squares may, however, be so arranged as to produce
the same total in a far greater number of ways. The
1 /
JS
10
\8
32
6
/3 "
1
a
*4
2
;K
5
/h
Fig. 349c
American "34" Puzzle required that the total in question
(34) should be produced in twenty different directions.
1 86 Puzzles Old and Nezv.
The problem in such a shape is beyond the scope of any but
skilled mathematicians ; but our readers may be interested
to know that such a result has not only been achieved, but
even surpassed, as in the arrangement given in Fig. 349,
where 34 is obtained in no less than 24 different ways.
34 may here be counted —
Horizontally . . . . .4 ways.
Perpendicularly . . . . 4 ,,
Diagonally . . . . . 2 ,,
The four corner squares (1, 8, 14, 11) 1 „
The four centre squares (6, 3, 9, 16) 1 ,,
The four corner groups of 4 squares,
each 1, 15, 12, 6 ; 10, 8, 3, 13 ; 7,
9, 14,4; and 16, 2, 5, 11 . .4 ,,
The four groups of 4 forming the centre
of each side — 15, 10, 6, 3 ; 12,6, 7, 9 ;
3, 13, 16, 2 ; and 9, 16, 4, 5 . .4 „
Total 20 ways.
These are the twenty totals required by the conditions of
the American puzzle. But there are yet four other ways in
which 34 can be made — viz., by means of adding together
the groups on opposite sides of the "octagon " — 15, 10, 4, 5 ;
10, 13, 7, 4; 13, 2, 12, 7; and 12, 15, 5, 2, making in all
twentyfour.
Any reader caring to follow the subject further will find
much interesting matter in relation to it in a little penny
pamphlet published by Heywood of Manchester, entitled
The Curiosities of the Thirtyfour Puzzle, and giving no less
than 150 solutions of the problem, with instructions for
obtaining the whole number of which it is capable.
No. IX.— The SixtyFive Puzzle. Solution.
The arrangement shown in Fig. 350 answers the condi
tions of the problem, the total being 65 in. each direction.
We have given in the last answer the method of forming
the simpler magic squares where they consist of an even
number of cells. Where the square consists of an odd
number of cells, another method is employed. We will take,
Key to Arithmetical Puzzles.
187
in the first place, the simplest possible example : a square
of nine cells, and an arithmetical series * rising by 1 only,
say the numbers 3 to 11 inclusive.
3
20
7
24
11
16
8
25
12
4
9
21
13
5
17
22
14
1
18 10
15
2
19 6
23
Fig. 350.
Having drawn a square of the required number of cells,
add an additional cell on either side, contiguous to the
centre square on that side. It will be seen that this gives
three parallel diagonals, of three squares each, in four
different directions. Arrange your series of numbers in
3
11
4
6
10
8
7
9
9
7
5
8
10
6
4
11
3
Fig, 351.
Fig. 352.
* It was formerly believed that only series of figures in arithmetical
progression were capable of being formed into magic squares ; but this
belief has loDg since been shown to be unfounded.
i88
P itzzles Old and New.
4
9
11
G
7
5
8
3
10
10
3 8
11
Fig. 353.
Fig, 354.
1
2
6
o
7
11
4
8
12
16
5
9
13
17
21
10
li
18
22
15
19
23

20
24
25
Fig. 355.
consecutive order in either set of parallels, say as in Fig.
351 or as in Fisr. 352. Then transfer the number in each
Key to Arithmetical Puzzles.
189
external cell to the vacant cell on the opposite side of the
square. The respective results will be as shown in Figs. 353
and 354 respectively, either of which will answer the re
quired conditions, giving a total of 21 each way.
With a larger number of cells the process is a little more
elaborate, but the principle is the same. Thus, suppose it
be required to form a magic square of 25 cells with the
numbers 1 to 25 inclusive. We now add extra cells so as to
3
20
7
21
11
16
8
25
12
■i
9
21
13
5
17
/
22
11
1
18
10
15
2
19
G
23
Fig. 356.
form five parallel diagonals, as in Fig. 355. Proceeding as
before, we arrange the series as shown in the same figure,
and then transposing each o£ the outlying numbers to the
vacant square on the opposite side, we have the result shown
in Fig. 356, giving a total of 65 each way.
It is to be observed that we have in each case moved the
outlying number^ye places onwards, i.e. the exact length of
one side of the square. Bearing this fact in mind, it becomes
tgo
Puzzles Old and New.
49
48
42
47
41
35
46
40
34
28
45
39
33
27
21
44
38
32
.26
20
13
14
43
I 37
31
25
19
7
I 36
30
23
24
18
1G
12
6
!
i 29
17
11
5
22
16
4

■I
15
9
3
!
8
2
1
Fig. 357.
46
15
40
8
9
33
31
3
28
45
20
21
39
2
27
38
14
32
17
1
25
49
26
43
44
13
31
19
37
30
6
24
48
18
36
12
5
22
23
7
42
11
29 ,
47
16
41
10
35
4
Fig. 358.
equally easy to deal with squares of a still larger number
of cells (provided always that they consist of an odd
Key to Arithmetical Puzzles.
191
number). Nos. 357 and 358 illustrate the same process as
applied to the numbers 1 to 49, arranged to form a magic
square of 49 cells, with a total in each, direction of 175.
(We have iu this case started from the bottom, and worked
upwards from left to right.) The square in this case having
seven cells on a side, each of the outlying figures is moved
forward seven places, with the result shown in Fig. 358, #
No. X.— The TwentySix Puzzle, Solution
We append, in Figs, 359, 360, two solutions, or varia
1
4
11 6
7
2
8 10
3
5
9
I
12
12

9 .
I
2
7
6
11
5
1
10
8
4
1
Fig. 359.
Fig. 360.
tions of the same solution. It is not impossible that there
are many others which would equally answer the conditions
of the problem.
No. XI.— An Unmanageable Legacy, Solution.
The lawyer had a horse of his own, which he drove into
the stable with the rest. " Now," he said to John, " take
your half." John took nine horses accordingly. James and
William were then invited to take their shares, which they
did, receiving six and two horses respectively. This division
exactly disposed of the seventeen horses of the testator ; and
the lawyer, pocketing his fee, drove his own steed home
again.
N.B. — The above solution rests on the fact that the sum of
* Further information on the subject of Magic Squares will be found
in Hut toil's Mathematical Recreation?, and in the Encyclopedic des Jeux
of M. de Monlidars, p. 410.
192 Puzzles Old and New.
the three fractions named, *, ■§, and ,}, when reduced to a com
mon denominator, will be found not to amount to unity, but
only to \\. The addition of another horse ( = tV) bringing
the total number up to eighteen, renders it divisible by such
common denominator, and enables each to get his proper
share, the lawyer then resuming his own ^ which he had
lent for the purpose of the division.
In the administration of the Mahomedan Law of In
heritance, which involves numerous and complicated frac
tious, this expedient is frequently employed.
No. XII —Many Figures, but a Small Result.
Solution.
3.A 4 J_±S.
7 ' 2 6*
(Reducing each, fraction to its lowest denominator, it will
be found to be equal to §, and } + ± = 1.)
No, XIII.— Can You Name it? Solution.
Answer, 20.
50  20  30. 80  50 = 30.
No. XIV.— Squares, Product, and Difference.
Solution.
Answer, 11 and 15.
Their product is 165, and their difference 4. The former
exceeds the latter by 161. The sum of their squares is
346, and 346  165 = 18L
No. XV.— A Peculiar Number. Solution.
The required number is 285714, which, multiplied by 2,
becomes 571428. \
From the conditions of the question, it is clear that the
required number must begin with 28, and end with 14 (one
half of 28). As it has six digits, it must therefore take the
form 28 . . 14, the two intermediate digits being unknown.
It is equally clear that the double number must end with
1428, the last two figures being affixed to the final 14 of the
original number,, Again, it must begin with either 56 or 57,
for any higher number divided by 2 would have a larger
quotient than 28, and therefore is by the terms of the
problem inadmissible. This gives us as the double number
either 561428 or 571428,
Key to Arithmetical Puzzles. 19,
Dividing each by 2, we have 280714 and 285714, the
second of which is found to answer the required conditions.*
No. XVI.— A Novel Century. Solution.
9x8 + 7 + 6 + 5 + 4+3 + 2+1= 100.
No. XVII.— Another Century. Solution.
There are several ways of fulfilling the conditions of the
puzzle. The first takes the form of a sum in addition :
15
36
47
98
2
100
Ofcher solutions are as follow :
If + 98H + = 100. gOfi + l?t = 100.
70 + 24 T 9 s + H = 100. 87 + 9f + 3J$ = 100.
No. xviii.— Another Way to Make 100. Solution
qq 99
No. XIX.— The Lucky Number. Solution.
Multiply the selected number hyni?ie, and use the product
as the multiplier for the larger number. It will be found
that the results will be respectively as under : — ■
12345679 x 9 = 111 111 111
x 18  222 222 222
„ x 27 = 333 333 333
„ x 36 = 444 444 444
., x 45 = 555 555 555
x 54 = 666 666 666
x 63 = 777 777 777
x 72 = 888 888 888
x 81 = 999 999 999
It will be observed that the result is in each case the
" lucky " number, nine times repeated.
* See also Puzzle No. XLV,
O
j 94 Puzzles Old and New.
No. XX.— The Two Ages. Solution.
The father was three times the age of his son 15 years
earlier, being then fiftyfive and a half, while his son was
eighteen and a half. The son will have reached half his
father's age in three years' time, being then thirtyseven,
while his father will be seventyfour.
No. XXI.— The Graces and the Muses. Solution.
The number each Grace had originally was 36, for —
She gave to each Muse yg of her store : i.e., to the nine
T 9 g = , leaving \ still in her own possession ; and each Muse
receiving ~ w from each Grace, had in all T 3 g = ~.
But the  left to each Grace exceeded the ^ held by each
Muse by 12, and if •§ — £ (= •§■) be 12, the whole number
must have been 36.
No. XXII.— The Graces and the Muses again.
Solution.
Twelve, or any multiple of twelve, will answer the
conditions of the puzzle. Assuming the number to be
twelve, each Grace gives one rose to each Muse. Each Muse
will thus have three (one received from each Grace), while
the Graces have each three roses left.
No. XXIII.— Just One Over. Solution.
61, being the least common multiple of 2, 3, 4, 5, and 6
(60) + 1.
No. XXIV.— Scarcely Explicit. Solution.
He had £20, as may be seen by the following demonstra
tion : —
Let x be the number of pounds in his pocket. Then, ac
cording to the conditions of the puzzle,
x + \x +%a? + f us + .■*■«•'+ fss + 9 =100
Reducing the fractions to a common denominator :
X
%,e.
Key to Arithmetical Puzzles. 195
(_3O_ + _a±45j 1 i8j 1 60 : J ] ~ 1(K) _ 9
„ , 213.i! / , 71 \
* + ~60 .(=' + 20*) = 91
20 a? +
71 aj
= 91
20
91 x
20
= 91
ra
20
= 1
a?
= 20
No. XXV.— Making Things Even. Solution.
Johnny had sevenpence, and Tommy fivepence.
No. xxvi.— A Rejected Proposal. Solution.
Johnny had eightpence, and Tommy fourpence.
No. XXVII.— The Market Woman and her Stock.
Solution.
Her original stock was 40 apples. Her first customer,
buying half her stock, and giving back 10, left her with
30 ; the second, buying onethird of 30, and giving her back
2, left her with 22 ; and the third, buying half of these,
and giving her back 1, left her with 12".
To solve the problem, however, it is necessary (unless
algebra be used) to work the process backwards. Take
away 1 (given back by the last boy) from her ultimate re
mainder, 12, thus leaving 11. It is clear that as he purchased
half her stock, she must before he did so have had 22 apples.
Of these, 2 had been given back to her by the second boy, so
that prior to his so doing she must have had 20. As he
bought onethird of her stock, the number previous to his
purchase must have been 30. Of these 30, 10 were given
back to her by the first boy, prior to which she must have
had 20 ; and as this 20 represents half her original stock (for
the first boy bought the other half), she must at the outset
have had 40.
196 Puzzles Old and New,
No. XXVIII.— The Captives in the Tower. Solution.
The boy descended first, using the cannonball as a counter
poise. The queen and her daughter then took the cannon
ball out of the upper basket, and the daughter descended,
the boy acting as counterpoise. The cannonball was then
allowed to run down alone. When it reached the ground,
the daughter got into the basket along with the cannonball,
and their joint weight acted as counterpoise while the
queen descended. The princess got out, and the cannonball
was sent down alone. The boy then went down, the cannon
ball ascending. The daughter removed the cannonball and
went down alone, her brother ascending. The latter then
put the cannonball in the opposite basket, and lowered him
self to the ground.
No. XXIX.— Father and Son. Solution.
In four years and a half, when the son will be sixteen and
a half, the father fortynine and a half. When the son
reaches 16, the father will be 49 — i.e., still a little more than
three times the son's age. But when the son reaches 17, his
father will be 50, which is not quite three times 17. It is,
therefore, clear that the required age is between those two
points, and a little reflection will show that only the ages
stated exactly answer the conditions of the problem.
No. XXX.— A Complicated Transaction. Solution.
William had 48 shillings, and Thomas 30.
The arithmetical solution of this question is somewhat in
tricate, but by the algebraic method it is simple enough*
Thus : Let w = William's number,
and t = Thomas' number.
Then the state of William's finances at the close of the trans
action will be represented by w — t + (w — t), and by the
terms of the question this = 36. We also have it stated in the
question that the joint finances w + t = 36 + 42 = 78, so
that w = 78  t.
Reducing the first equation to simpler form, we have
2w  2t = 36
or w — t = 18
and w = 18 + t
Key to Arithmetical Puzzles. 197
Comparing the values of w thus ascertained, we have
18 + * = 78  t
2t = 78  18 = 60
.. * = 30
and w being = 18 + * = 18 + 30 = 48.
No. XXXI.— A Long Family. Solution.
Their respective ages are as follows : the youngest 3, and
the eldest 24.
Here again the assistance of algebra is needed for a ready
solution.
Let x = the age of the youngest (fifteenth). Then 8x ==
the age of the eldest.
And by the terms of the question —
8x = (x + \\ + 1\ + 1 + H + H + 1 + 14 + 11
+ 14 + 1J + 11 + 1 + lj + H = » + 21.
i.e. 7x = 21.
.*. x = 3 : the age of the youngest,
and 8oj = 24 : the age of the eldest.
No. xxxii.— A Curious Number. Solution.
This problem seems at first sight somewhat formidable,
but it is in reality very easy.
It is susceptible of various answers, equally correct,
according to the value assigned to the smallest part, or
unit of measurement. Thus, if such smallest part be 1, the
number will be
1 + 40 + 400 + 500 = 941
If such unit be 2, the number will be
2 + 80 + 800 + 1000 = 1882
and so on, ad infinitum.
No. xxxiii.— The Shepherd and his Sheep.
To ascertain the number of the flock, find in the first
place the least common multiple of 2, 3, 4, 5, and 6 — i.e., 60.
Then take the lowest multiple of this, which, with 1 added,
will be divisible by 7. This will be found to be 301, which
is the required answer.*
* See also answer to No. L.
198 Ptizzles Old and New.
No. XXXIV.— A Difficult Problem. Solution.
The solution of this problem depends upon certain of the
Properties of Numbers referred to at the commencement of
this section (see pp. 174181), where we find it stated that
the condition that a number shall be evenly divisible by —
(I.) 2, is that its last digit shall be an even number.
(Prop. 7.)
(II.) 3, is that the sum of its digits shall be divisible by
3. (Prop. 8.)
(III.) 4, is that its last two digits shall be divisible by 4.
(Prop. 10.)
(IV.) 5, is that it shall end with 5 or 0. (Prop. 11.)
(V.) 6, is that the sum of its digits shall be divisible by
3, and its last digit be an even number. (Prop.
12 )
(VI.) 8, is that its last three digits shall be divisible by 8.
(Prop. 14.)
(VII.) 9, is that the sum of its digits shall be divisible by 9.
(Prop. 15.)
(VIII.) 10, is that its last digit shall be 0.* (Prop. 16.)
To find the required number, let a be the digit occupying
the " units " place, b the digit occupying the " tens " place,
c that occupying the "hundreds" place, d the "thousands,"
and so on, as far as may be necessary.
Now,, bearing in mind the conditions of the problem, we
are enabled at once to fix the value of a. As the required
number when divided by 2 is to leave a remainder of 1, we
know (by I.) that a must be an even number + 1 — i.e., an
odd number. As the required number is to be divisible by 5,
with a remainder of 4, a must (by IV.) be 5 + 4 or + 4 — i.e.,
9 or 4. But as it is to be divisible by 10, with remainder
9, it can only (by VIII.) be + 9 = 9.
From III. we gather that the required number being
divisible by 4, with remainder 3, b a must be a multiple
* Some of these conditions might in the actual working out of the prob
lem be disregarded — e.g., if a number be divisible by 8, with remainder
7, it will, as a matter of course, be divisible by 4, with remainder 3, or
by 2, with remainder 1. If divisible by 6, with remainder 5, it will
necessarily be divisible by 3, with remainder 2 ; and if divisible by 10,
with remainder 9, it will in like manner be divisible by 5, with re
mainder 4.
Key to Arithmetical Puzzles. 199
of 4, + 3. As a = 9, a — 3 must = 6, and such multiple of 4
must therefore be one ending with 6. Now of such multiples
(consisting of two digits only) there are but five — viz., 16, 36,
56, 76, and 96. b a must therefore be one or other of these
+ 3: i.e., either 19, 39, 59, 79, or 99. But neither of these
standing alone answers the remaining conditions of the
problem. The required number must therefore be one of
more than two digits, terminating with one or other of the
double numbers abovementioned.
As the required number is divisible b y 9, with remainder
8, we know (by VII.) that the sum of its digits, less 8,
must be exactly divisible by 9. Deducting 8 from each of
the five possible values of b a, we have as remainders 11, 31,
51, 71, and 91 ; the sum of whose digits is 2, 4, 6, 8, and 10
respectively. If, therefore, the required number be one
of three figures, c must be either 9— 2 = 7;9 — 4 = 5; 9
6 = 3; 98 = 1; or 18  10 = 8 ; and c b a must
be either 719, 539, 359, 179, or 899.
Neither of these, however, is divisible by 7, with remainder
6 ; and it is therefore clear that no number of three digits
only will answer the required conditions.
Proceeding to the consideration of numbers of four
digits, we naturally begin with such as have 1 as the
value of d. Now (by VII.) we know that, to answer the
required conditions, (1 + c + b + a) must be a multiple of 9,
with 8 over ; or, substituting the known value of a, we have
1 + c + b + 9 = (9 or 18)* + 8
= 17 or 26
.*. c +'. b= 7 or 16
And as we already know that b is either 1, 3, 5, 7, or 9, c
must be either 6, 4, 2, 0, or 7. The only possible solutions
commencing with 1 are therefore 1619, 1439, 1259, 1079, or
1799. Of these, however, none save 1259 is divisible by 7,
with 6 over, and this number is not divisible by 8, with 7
over.
We proceed to try 2 as the value of d. Then, as before :
2 + c + b + a = (9 or 18) + 8
2 + c + b + 9 = 17 or 26
c + b =■ 6 or 15
* 18 is the highest possible multiple of 9 that will suit the equation.
For, even supposing the unknown values of c and b to be each 9, 1 + 9
+ 9 + 9 would obviously not = 27 (the next highest multiple) + 8.
200 Ptizzles Old and New.
in which case, b being known to be either 1, 3, 5, 7, or 9,
c must be either 5, 3, 1, 8, or 6, and the only possible solu
tions commencing with 2 are —
2519, 2339, 2159, 2879, and 2699.
Of these, 2519 is the only one which answers all the con
ditions of the problem, and 2519 is therefore the required
number.
No. XXXV. Well Laid Out. Solution.
Either of the solutions following will answer the conditions
of the problem : —
s. d.
2 pieces indiarubber @ M. = 8
2 sheets of paper... @ 3d = 6
1 pencil @ 2d. = 2
16 drawingpins ... @ \d. = 8
21 articles. Cost 2.
s. d.
or, 5 pencils @ 2d. = 10
1 sheet of paper... @ 3d. = 3
1 piece indiarubber @ 4<d. — 4
14 drawingpins ... @ \d. = 7
21 articles. Cost 2.
To arrive at the result we, in the first place, consider
what one article of each kind would together amount to.
The result is as under : —
1 piece indiarubber @ Aid.
1 sheet of paper ... @ 3d.
1 pencil ... ... @ 2d.
1 drawingpin ... @\d.
i M
We have thus only four articles (less than onefifth of the
required number), for 9^d., which is more than onethird of
the total sum to be expended. It is therefore clear that to
make up the total of 21 articles a considerable proportion,
numerically, must be pins ; and it is farther clear that, to
Key to Arithmetical Puzzles, 201
avoid an odd halfpenny in the total, the number of pins must
of necessity be even.
We proceed to try one of each of the three more costly
articles (valued. + 3d. + 2d: = 9d.), with such a number of
pins as will bring the total number nearer to the required 21.
18 would bring it to exactly that number, but the cost
would be only 9d. + 9d. = Is. 6d. ; so that the quantity of the
cheaper items (the pins) is obviously now in excess. With
16 pins we have 3 + 16 = 19 articles, value Is. hd.
This is 2 articles and 7 pence short of the limit. We have
therefore to consider what 2 additional articles would to
gether represent 7d. in value, and we find that a second
piece of rubber, value 4sd. t and a second sheet of paper, value
3d., just answer the required conditions. This leads us to
the first of the solutions above given.
But we have further to consider whether this is the only
solution — and we accordingly proceed to try 14 drawingpins
(value Id), with one of each of the other articles. This gives
us a total of 17 articles, value Is. 4<d. This is 8d. short in
value, and 4 short in number of the required result. The
only 4 articles that would make up the required value are
4 additional pencils. Adding these, and their value, we have
the second solution.
We proceed to try whether the required conditions can be
answered with twelve pins, and one of each of the more
costly articles. This gives us 15 articles, value Is. 3d., being
6 short in number, and 9d. short in value. Now, 6 pencils
(the cheapest of the more costly articles) would cost Is., being
3d. in excess of our limit in price ; and any further diminution
of the number of drawingpins would bring us still more in
excess of such limit. It follows that the abovementioned
are the only possible solutions.
No. XXXVI.— The Two Travellers. Solution.
A, in his 2 hours' start, has travelled 10 miles. As B
gains on him at the rate of a mile an hour, it will take him
ten hours to recover this distance, by which time A will
have been travelling 12 hours, and will be 50 miles from
the point whence he started.
202 Puzzles Old and New.
No. XXXVII.— Measuring* the Garden. Solution.
The garden is 111 yards long by 37 wide, and contains,
therefore, 4107 square yards. If it were a yard more each
way (112 yards by 38) its area would be 4256 square yards.
4256  4107 = 149.
The problem may be solved algebraically as follows : —
Let x = breadth in yards
Then 3x — length ,,
And Sx 2 = area.
And (aj + 1) (3x f 1) = (3re 3 + 4a? + 1) = area if length and
breadth increased by one yard each way.
By the terms of the question, such lastmentioned area —
O
OX"
+
4<x
+ 1
=
3x*
+ 149
4x
+ 1
=
149
i.e.,
4<x
=
148
.*. X
=
37
the requii
:ed breadth.
and 2>x
=
141
j?
length.
No. XXXVIII.— When Will They Get It ? Solution.
In 420 days ; 420 being the least common multiple of 1, 2,
3, 4, 5, 6, and 7.
No. XXXIX.— Passings the Gate. Solution.
He had at the outset five shillings and one penny.
On the first day he pays one penny at the gate, spends
half a crown, and pays a penny on going out, leaving him
with two and fivepence.
The second day he pays a penny on entering, spends four
teenpence, and after paying a penny, on going out is left
with thirteenpence.
The third day he brings in one shilling, spends sixpence,
and is left, on going out, with fivepence.
The fourth day he brings in fourpence, spends twopence,
and after paying the toll to go out, is left with one penny
only.
To solve the problem, the calculation must be worked
backwards. Thus, on the fourth day he pays a penny on
coming out, and has still one left, together making twopence.
He had spent half his available money in the town. The
Key to Arithmetical Puzzles.
203
total must therefore on that clay have been fourpence,
exclusive of the penny he paid to come in. This gives us
fivepence as the amount with which he came out the previous
evening. The penny he paid to get out brings this amount
to sixpence, and as he had first spent a like amount, he must
previously have had a shilling, exclusive of the penny to
come in.
By continuing the same process, it is easy to arrive at his
original capital.
No XL.— 'A Novel Magic Square. Solution.
(See Fig. 361.) This is an interesting example of what is
called the "bordered" magic square, the repeated removal
of the border not affectiug its magic quality. The sums of
the various rows are 369, 287, 205, and 123, of which the
central number, 41, is the greatest common divisor. As to the
5
80
59
73
61
3
63
12
13
1
20
55
30
57
29
71
26 J 81
4
14 1 31
50
29
60
35 I 68
78
76
7
58 j 46
38
45
40
36  24
6
«f
33
43
41
39
49
17
75
74
67
64
48
42
37
44
34
18
8
10
47
32
53
22
51
72
15
66
56
27
52
25
54
11
62
16
77
69
2
23
9
21
79
19
70
A* Fig. 361.
mode of constructing, such squares, see Iluttotis Mathematical
Recreations, and other works of the same kind.
We append, for the purpose of comparison (Fig. 362), the
ordinary magic square of the numbers 1 to 81, constructed
according to the rule given on pp. 187189. It will be seen
that this, though perfect as an ordinary magic square,
wholly fails to meet the special conditions of the problem.
The total of each row is in this case 369.
204
Puzzles Old and New.
5
46
54
14
13
63
62
21
70
30
80
29
78
37
6
47
22
71
79
39
8
38
7
15
56
25
66
55
23
72
31
24
65
64
33
32
81
40
48
17
5S
27
68
28
16
57
26
67
36
77
73
41
9
49
18
34
74
42
1
50
35
76
45
75
44
4
43
2
5f
10
59
3
52
11
60
19
69
53
12
61
20
Fig. 362.
jNo. XLL— Another Magic Square. Solution,
45
17
58
6
55
11
32
28
10
54
29
33
16
44
7
59
23
43
60
13
49
26
38
48
12
39
27
42
22
61
1
62
2
41
21
36
3
24
51
15
25
37
14
50
63
20
40
4
56
19
47
30
£4
9
53
35
31
52
8
57
!
5
46
18
Fig. 363.
Key to Arithmetical Puzzles.
205
The solution is as shown in Fig. 363. This is a very
pretty example of what is known to mathematicians as the
Tesselated Magic Square.*
No. XLIL— The Set of Weights. Solution.
The five weights are as under : —
Jib., ljlb., 4lbs., 13 J lbs., and40Jlbs.
To weigh the intermediate quantities from 1 lb. upwards,
the weights are distributed between the two scales as
follows : —
To weigh 1 lb.
2 lbs.
2i
3
5
^ i
6
• o
In "Weight Scale.
1 lb. weight
8J
9
In Goods Scale.
The goods to be weighed
plus \ lb. weight.
1 J and \ lb. weight.' Goods only.
Goods + \\ and \ lbs.
Goods + li lb.
Goods only.
Goods +Jlb.
Goods only.
2
^2
and \\
4i lb w r eight.
^2 •>•>
41 and \ lb
_ lb.
4f and If lb.
4§, l,and
13lb.
I3ib.
13 \ and
13 It
13 lb.
*lb.
57
Goods, 4, 1 J, and f lbs.
„ 4t\ and If lbs.
,, 4J and If lbs.
, y 4^ and J lbs.
„ 4 lbs.
And so on ; to w r eigh 60 lbs., all of the weights, with the
exception of the \ lb., being placed in the weight scale.
IST.B. — If preferred, the weighing of any given quantity up
to 60 lbs. with the weights named may be propounded as an
independent puzzle.
No. XLIIL— What Did He Lose? Solution.
The reply of most people is, almost invariably, that the
hatter lost £3 19s. 0d. and the value of the hat, but a little
consideration will show that this is incorrect. His actual
* We are indebted for this particular Magic Square to a recent issue
(May 13th, 1893) of the Queen, where it appears in connection with
a more than ordinarily complex example of the Knight's Tour Puzzle.
2o6 Puzzles Old and New,
loss was £3 19s. Oil., less his trade profit on the hat ; the nett
value of the hat, plus such trade profit, being balanced
by the difference, 21s., which he retained out of the proceeds
of the note.
No. XLIV.— A Difficult Division. Solution.
According to the conditions of the problem, each son's
share will be seven casks (irrespective of contents), and of
wine, 3 1 casks.
The division can be made in either of two ways, as under : —
Dick and Tom each take 2 full, 2 empty, and 3 halffull
casks; and
Harry, 3 full, 3 empty, and 1 halffull ;
or
Dick and Tom each, take 3 full, 3 empty, and 1 halffull
cask j and Harry 1 full, 1 empty, and 5 half filled casks.
No. XLY.— The Hundred Bottles of Wine. Solution.
He sold on the first day 2 bottles only ; on the second, 5 ;
on the third, 8; on the fourth, 11 ; on the fifth, 14; on the
sixth, 17 ; on the seventh, 20 ; and on the eighth day, 23.
2 + 5 + 8 + 11 + 14+17 + 20 + 23 = 100
To ascertain the first day's "sale, or first term of the series,
take the ordinary formula for ascertaining the sum of an
arithmetical progression.
8 =_n (2a + n~^Tb)
2
Now 8 (the sum of the series), n (the number of terms), and h
(the daily rate of increase), are known, being 100, 8, and 3
respectively. Substituting these known values in the
formula, that of the first term, a, is readily ascertained.
Thus :—
100
.00
:=
f (2d + 21)
4 (2a + 21)
—
8a + 84
84
=
8a
16
=
8a
a
=
2
Key to Arithmetical Puzzles. 209
Applying this rule to the question under consideration,
we find that the L.C.M. of 2, 3, 4, 5 and 6 is 60 ; but 60 +
1 (= 61) is not evenly divisible by 11. We proceed to try
60 x 2 + 1 (= 121), which is found to be so divisible, and
is therefore the number required.
No. LI.— The Three Legacies. Solution.
As the amount of each share is to correspond with length
of service, it is plain that the housemaid will receive one
share, the parlourmaid three, and the cook six — in all, ten
shares. The value of a single share is therefore onetenth
of £70, or £7, which is the portion of the housemaid, the
parlourmaid receiving £21, and the cook £42.
No. LIL— Another Mysterious Multiplicand, Solu
tion.
The conditions of the problem are answered by the'
number 73, which, being multiplied by 3, 6, and 9, the pro
ducts are 219, 438, and 657 respectively. The correctness
of the solution is obvious. The rationale of the fact and
the process whereby the required number is ascertained, we
must confess ourselves unable to explain.
No. LIU.— How to Divide Twelve among Thirteen
O)
cv
X
<V
a
1 Fig. 361.
2io P tizzies Old and New.
It will be found that, counting as described in the prob
lem, the person standing eleventh from the point at which
you begin will be excluded,, The distributor will therefore
begin ten places farther back, or (which is the same thing)
three places farther forward, in the circle. Thus, if x (see
Fig. 364) be the person to be excluded, the distributor will
begin to count at the point a. The numbers placed
against the various places show the order in which the
gifts will be distributed and the men drop out of the circle.
No. LIV.— Tenth Man Out. Solution.
The arrangement was as shown in Fi°\ 365, the white
re
'©..
o
©
Q
(3
®Q*U&®
Fig. 365.
spots representing the white men, and the black spots the
black men. The counting begins with the man marked a.
No. LV.— Ninth Man Out. Solution,
In this case the arrangement will be as shown in Fig,
366,
The proper order may be readily remembered by the aid
of the Latin line —
. PojJuleam Virgam Mater Begina Ferelat,
signifying, " The Mother Queen carried a poplar switch."
Key to Arithmetical Puzzles. 2 1 1
The interpretation lias, however, nothing to do with the
matter. The significance of the formula lies in the vowels
m^ ^0
© Q
©
0.
.0
Fig. 366.
which occur in it, and which are taken to mean — a, 1 ; e,
2 ; i, 3 ; o, 4 ; and u, 5, respectively. The line, therefore, to
the initiated, reads as under : —
Populeam Virgam Mater Begina Ferebat ;
4 5 21 3 t 1 12 231221
which will be found" to correspond with the grouping of
whites and blacks around the circle. Thus, the o in Po
indicates that the first group is to consist of 4 whites, the
u in pto that the next group is to consist of 5 blacks, and
so on, black and white alternately throughout.
No. LVL— The Three Travellers. Solution.
The plan adopted is as follows : —
1. Two of the servants are sent over.
2. One of the servants brings back the boat, and takes
over the third servant.
3. One of the servants brings the boat back, lands, and
two of the masters go over.
4. One of the masters and one of the servants return.
The servant lands, and the third master crosses with the
second.
The position *of matters is now as follows : The three
2io P tizzies Old and New.
It will be found that, counting as described in the prob
lem, the j>erson standing eleventh from the point at which
you begin will be excluded*, The distributor will therefore
begin ten places farther back, or (which is the same thing)
three places farther forward, in the circle. Thus, if x (see
Fig. 364) be the person to be excluded, the distributor will
begin to count at the point a. The numbers placed
against the various places show the order in which the
gifts will be distributed and the men drop out of the circle.
No. LI V.— Tenth Man Out. Solution.
The arrangement was as shown in Fig. 365, the white
•G)
Fig. 365.
v0'
spots representing the white men, and the black spots  the
black men. The counting begins with the man marked a.
No. LV.— Ninth Man Out. Solution,
In this case the arrangement will be as shown in Fig,
366,
The proper order may be readily remembered by the aid
of the Latin line —
. Pojpuleam Virgam Mater Eegina Ferebat,
signifying, " The Mother Queen carried a poplar switch."
Key to Arithmetical Puzzles. 211
The interpretation lias, however, nothing to do with the
matter. The significance of the formula lies in the vowels
'Ft
w
s>
©
Q
Fig. 366.
which occur in it, and which are taken to mean — a, 1 ; e,
2 ; i, 3 ; o, 4 ; and u, 5, respectively. The line, therefore, to
the initiated, reads as under : —
Populeam Virgam Mater Regina Ferebat ;
4 5 21 3 t 1 12 231221
which will be found* to correspond with the grouping of
whites and blacks around the circle. Thus, the o in Po
indicates that the first group is to consist of 4 whites, the
u in pu that the next group is to consist of 5 blacks, and
so on, black and white alternately throughout.
No. LVL— The Three Travellers. Solution.
The plan adopted is as follows : —
1. Two of the servants are sent over.
2. One of the servants brings back the boat, and takes
over the third servant.
3. One of the servants brings the boat back, lands, and
two of the masters go over.
4. One of the masters and one of the servants return.
The servant lands, and the third master crosses with the
second.
The position x>f matters is now as follows : The three
2 1 2 Puzzles Old and New.
masters are on the farther side, and one of the servants,
who is sent back with the boat, and fetches, one at a time,
the other two servants.*
No. LV1L— The Wolf, the Goat, and the Cabbages.
Solution, ,
This is a very simple problem. It is solved as nnder : —
1. He first takes across the goat, and leaves him on the
opposite side.
2. He returns and fetches the wolf, leaves him on the
opposite side, and takes back the goat with him.
3. He leaves the goat at the startingpoint, and takes over
the basket of cabbages.
4. He leaves the cabbages with the wolf, and returning,
fetches the goat.
Or,
1. He takes over the goat.
2. He returns and fetches the cabbages.
3. He takes back the goat, leaves him at the starting
point, and fetches the wolf.
4. He leaves the wolf on the opposite side with the basket
of cabbages, and goes back to fetch the goat.
No. LYIIL— The Three Jealous Husbands. Solution.
For the sake of clearness, we will designate the three
husbands A, B, and 0, and their wives, a, b, and c, respec
tively. The passage may then be made to the satisfaction
of the husbands in the following order : —
1. a and b cross over, and b brings back the boat.
2. b and c cross over, c returning alone.
* It will be found a great advantage in this class of puzzles to have
some material representative of each character. This requirement is
met by Messrs. Jaques & Son in the present case by a little mechanical
arrangement, under the name of " The Boat Puzzle. " A little boat is
mounted on a grooved board, representing the stream, while the masters
and servants are represented by six movable figures, three white and
three black.
For lack of such an appliance, counters or cardboard discs, with
the names of the various persons or objects written upon them, will Tbo
found useful.
Key to Arithmetical Puzzles. 213
3. c lands, and remains with her husband, while A and B
cross over. A lands, B and b return to the startingpoint.
4. B and G cross over, leaving b and c at the starting
point.
5. a takes back the boat, and b crosses with her.
6. a lands, and b goes back for c.
Arithmeticians have racked their brains to devise a means
of transit for four husbands and four wives under like con
ditions, but, with a boat holding two persons only, the
problem is insoluble. If we suppose, however, that the
boat contains three persons, it may be solved as follows : —
(Distinguishing the four husbands as A, B, G, and D, and
the four wives as a, b, c, and d, respectively.)
1. a, b, aud c cross over; c brings back the boat.
2. c and d cross over, and d brings back the boat.
3. A, B, and G cross over ; G and c bring back the boat.
4. G, D, and c cross over.
5. c takes back the boat and fetches d.
No. LTX.— The Captain and his Company. Solution.
The captain orders the two children to pass to the farther
side. One of them then brings back the boat, lands, and a
soldier crosses alone to the farther side. The second child
then brings back the boat.
The state of things (save that one man has crossed) is
now just as at first, the boat and the two children being on
the hither side of the stream. The process is repeated until
the whole of the company have passed over.
No. LX.— The Treasure Trove. Solution.
The secret lies in the fact that the process indicated by
the Scotchman simply brings back the number with which it
started ; i.e., the number of coins in the " find," as will be seen
by the following demonstration. Let us suppose x to be the
unknown number. Then x + 666 will represent the result
of the first step ; 999  (x + 666) or 333  x that of the
second step ; and 333 — (333 — x) = x that of the third step.
The real value of x is, throughout, unknown to the
Scotchman ; but the Irishman, finding that the suggested
process has brought out the required amount, rashly assumes
that his antagonist must be acquainted with it.
214 Puzzles Old and New.
No. LXL— The Row of Counters. Solution.
The required number will be the original difference plus
the number removed, at the second stage of the puzzle, from
the smaller row ; in this case, therefore, 3 + 5 ( = 8). This re
sult will be correct whatever were the actual numbers
of the two rows originally. This may be illustrated alge
braically as follows : —
Let x = number in larger row,
y = „ ,, smaller row,
and B = ultimate remainder of larger row.
As the difference of the two rows is stated to be 3, it
follows that x = y + 3.
And the smaller row, after deduction of the special
number indicated, which we will call z, will be y — z, and B,
the ultimate remainder of the larger row, after deduction of
y — z, will be x — (y — z), or x + z— y.
Substituting for x the value above found, we have
B = (y + 3) + Z  y
= 3 +
And z being a known number (in this case 5) ; B = 8 ; or, in
other words, the original diff erence plus the number removed
from the smaller row prior to the second deduction.
An example may render the process clearer. Thus, suppose
that the first row consists of eighteen, and the second of
thirteen counters. The difference is in such case 5. Sub
tract what number you please, say 7, from the smaller row.
This leaves 6. 6 being subtracted from 18, the number of
the larger row, the remainder will be 12. (5 + 7).
No. LXIL— A Loan and a Present. Solution.
The remainder in this case will be one half of the amount
added by way of "present." This is very easily demon
strated.
Let x equal the number thought of ; then 2x + 14 will
equal that number plus the imaginary loan and present.
Half that amount being professedly given to the poor, will
leave a remainder of x + 7, and on the repayment of the im
aginary loan the figures will stand as x ■ + 7 — x (= 7), the
value of x having no bearing whatever on the result.
Key to Arithmetical Pttzzles. 215
No. LXIIL— Eleven Guests in Ten Beds. Solution.
The fallacy lies in the fact that the real eleventh man
remains unprovided with a restingplace. The tenth man
having taken possession of the ninth bed, the eleventh man
should in due course occupy the tenth bed, but he does not
do so. The man who is called from sleeping double in the
first bed to occupy this is not the eleventh, but the second
man, and the real eleventh man goes bedless.
No. LXIV.— A Difficult Division. Solution.
8 gal. 5 gal. 3 gal.
1. They begin by filling the fivegal
lon from the eightgallon mea
sure.
The contents of the three vessels
are then — 3 ,, 5 ,, ,,
2. They fill the three gallons from
the five gallons, making — 3 „ 2 „ 3 ,,
3. They empty the three gallons into
the eightgallon measure 6 „ 2 „ „
4. They empty the two gallons from
the fivegallon into the three
gallon measure 6 s , „ 2 ,,
5. They fill the fivegallon from the
eightgallon measure 1 „ 5 ,, 2 „
6. They fill up the threegallon
from the fivegallon measure 1 ,, 4 „ 3 ,,
7. They empty the three gallons into
the eightgallon measure 4 ,, 4 „ „
Making the required equal division.
No. LXV.— The Three Market Worn en. Solution.
They began by selling at the rate of three apples for a
penny. The first sold ten pennyworth, the second eight
pennyworth, and the third seven pennyworth. The first had
then left three apples, the second five, and the third six.
These they sold at one penny each, receiving, therefore, in
the whole —
The first, lOd. + 3d. = lSd.
The second, 8d. + hd. = 13c7.
The third, Id. + 6d. = 13d.
216 Puzzles Old and New.
No. LXVI^— The Farmer and his Three Daughters.
Solution.
They began by selling at the rate of seven a penny, the
first selling seven pennyworth, the second four, and the
youngest one pennyworth. But they had saved the choicest
of the fruit, the first having one apple left, the second two,
and the youngest three. Meeting a liberal customer, they
sold these at threepence each, and the respective amounts
received by them were therefore as under —
The first, 7<l + 3d, = lOd.
The second, 4a 7 . + Qd. = lOd.
The third, If?. + 9d. = lOd.
No. LXVIL— How Many for a Penny. Solution.
He bought two apples for his penny.
To arrive at the result, let x = the whole number of
x 1
apples. He gives away ^ + o» and nas stl11 one a PP le left 
Then—
or x + 1 + 3 ( = x + 4) = 3%
.*. &b. = 4
x = 2
No. LXVIIL— The Magic Cards. Solution.
The seven cards are drawn up on a mathematical prin
ciple, in such manner that the first numbers of those in
ivhieli a given number appears, when added together, indi
cate that number.
Suppose, for instance, that the chosen number is 63, This
appears in cards I., II., Ill, IV., V., and VI. The key
numbers of these are 1, 2, 4, 8, 16, and 32 ; and 1 + 2 + 4 + 8
+ 16 + 32 = 63.
If the number 7 were selected, this appears only in cards
I, II, and III, whose key numbers are 1, 2, and 4 = 7.
The principle of construction seems at first sight
rather mysterious, but it is simple enough when explained.
The reader will note, in the first place, that the first or
" key " numbers of each card form a geometrical progression,
Key to Arithmetical Puzzles. 217
being 1, 2, 4, 8, 16, 32, 64. The total of these is 127, which
is accordingly the highest number included.*
It is further to be noted that by appropriate combinations
of the above figures any total, from 1 up to 127, can be
produced.
The first card consists of the alternate numbers from 1 to
127 inclusive. The second, commencing with 2 (the second
term of the geometrical series), consists of alternate groups
of two consecutive figures — 2, 3; 6, 7; 10, 11, and soon.
The third, beginning with 4, the third term of the series,
consists of alternate groups of four figures — 4, 5, 6, 7 ; 12,
13, 14, 15; 20, 21, 22, 23 ; and so on. The fourth, commenc
ing with 8, consists in like manner of alternate groups of
eight figures. The fifth, commencing with 16, of alternate
groups of sixteen figures. The sixth, commencing with 32,
of alternate groups of thirtytwo figures ; and the last, com
mencing with 64, of a single group, being those from 64 to
127 inclusive.
It will be found that any given number of cards arranged
on this principle will produce the desired result, limited
by the extent of the geometrical series constituting the
first numbers.
No. LXIX.— The "Fifteen" or "Boss" Puzzle.
Solution.
Notwithstanding the enormous amount of energy that has
been expended over the " Fifteen " Puzzle, no absolute rule
for its solution has yet been discovered, and it appears to
be now generally agreed by mathematicians that out of
the vast number of haphazard positions in which the fifteen
cubes may at the outset be placed, f about half admit of tlio
blocks being so moved as to finally assume their proper
order.
To test whether a given arrangement admits of such a
possibility, the following rule has been suggested. Reckon
how many transpositions of given pairs are necessary to
* If there had been six cards only, the series would Lave terminated
•with 32, and the highest number would have been 63. If eight cards
were used, the final term of the series would have been 128, and its total
255, which would accordingly have been the maximum number.
f The possible number of such positions is only 1,307,674, 363,000.
2l8
Puzzles Old and New.
bring the blocks into the required order. If the total
number of such transpositions be even, the desired rearrange
ment is possible. If the number be odd, such rearrange
ment is impossible.
Applying this rule to the "Boss " Puzzle (see Fig. 335), it
will be seen that only one transposition (that of blocks 14
and 15) is here needed, and one being an odd number, the
problem is insoluble ; as, in fact, the " Fifteen " Puzzle in
this form has been found by all who have hitherto tried it.
No. LXX.— The Pegaway Puzzle. Solution.
The possibility of success in solving this puzzle appears
to be governed by precisely the same rule as the " Fifteen "
Puzzle — viz., that if the number of needful transpositions is
[
3
2
1
1
2
3
4
5
6
4
5
6
8
7
8
7
Fig. 367.
Fig. 368.
even, the puzzle can be solved ; if odd, it is insoluble.
Thus, if the "pegs " be arranged at starting as in Fig. 367.
the aspirant will probably succeed ; if as shown in Fig. 368,
he will fail.
No. LXXL— The OverPolite Guests. Solution.
To obtain the answer, all that is needed is to find the
number of permutations of seven objects — viz., 5040. It
would take, therefore, 5040 days, or nearly fourteen years,
to exhaust the possible positions.
In the alternative form of the problem, the host would
supply 5.040 dinners at half a crown (value £630) for a
payment of £70.
Key to Arithmetical Puzzles. 221
No. lxxvl— Another Eccentric Testator.
I Solution.
The estate amounted to the sum of £7200, and each son
took an equal share — i.e., £1440.
For convenience of reference, we will call what remained
after the eldest son had taken his share the first residue ;
what remained after the second had taken his share, the
second residue ; what remained after the third had taken
his share, the third residue ; and what remained after the
fourth had taken his share, the fourth residue.
JNow, according to the conditions of the problem, the
fifth son took one half of the fourth residue, plus £720. As
the fourth residue was thereby exhausted, with no remainder,
it is clear that the £720 constituted the remaining half of
such residue, and that the fourth residue therefore consisted
of twice £720, or £1440.
The fourth son took  of the third residue, plus £480, and
what remained after these two deductions constituted the
fourth residue, whose value we have found to be £1440. In
other words, f of the third residue, less £480 — £1440, and §
of the third residue = £1440 + £480, or £1920. Twothirds
of the third residue being £1920, the whole, or §, of such
residue, must be £2880.
As the third son took \ of the second residue, plus £360,
and there then remained £28S0, f of the second residue must
in like manner = £2880 + £360, or £3240, and the whole
of such second residue must be £3240 + £1080 = £4320.
As the second son took i of the first residue, plus £288,
and there was then left £4320, we deduce, by a like process
of reasoning, that the amount of the first residue was £5760.
As the eldest son took \ of the whole property, plus £240,
and there was then left £5760, it is clear, in like manner, that
f of the whole inheritance was £5760 + 240, or £6000 ; and if
£6000 represent §, the whole will necessarily be £7,200.
Testing the correctness of our demonstration by the
conditions of the puzzle, we find that —
7200
The eldest son takes g + 240 = 1200 + 240 = £1440.
First residue 7200  1440 = 5760.
The second son takes £■ + 288 = 1152 + 288 = 1440.
o
222 Puzzles Old and New.
Second residue 5760  1440  4320.
4320
The third son takes j* + 360 = 10S0 + 360 = 1440.
Third residue 4320  1440 = 2880.
2880
The fourth son takes g + 480 = 960 + 480 = 1440.
Fourth residue 2880  1440 = 1440.
1440
The fifth son takes pr + 720 = 720 + 720 = 1400.
Total £7200.
No. LXXV1L— An Aggravating* Unele. Solution.
The number of soldiers was 58.
On examination of the conditions or the puzzle, it will be
found that in each case, whether divided by 3, 4, 5, or
6, there are always two short of an even division. All
that is needed, therefore, is to find the least common mul
tiple of 3, 4, 5, and 6, and deduct 2 from it. The L.C.M.
of 3, 4, 5, and 6 is 60, and 60 — 2 = 58, the required number.
No. LXX VIII.— The Apples and Oranges. Solution.
As each child had 3 more oranges than apples, and this
caused a difference of 33 (48 — 15) in the number left over, it
follows that the number of children must have been 11. As
each child received 12 apples, and there were 48 over, the
total number of apples must have been (11 x 12) +48 =
132 + 48 = 180. As each child received 15 oranges, and
there were 15 over, the total number of oranges must have
been (11 x 15) + 15 = 165 + 15 = 180.
No, LXXIX.— The Two Squares. Solution.
As the 146 counters remaining from the first arrangement
were insufficient by 31 to make the two additional rows
desired, the actual number needed for such two rows must
have been 146 + 31 = 177. Of these, one (the counter at
Key to Arithmetical Puzzles. 223
the corner) would be common to both rows, and the number
in one row would therefore be —^~ + 1 = 89.
A square of 89 counters each way would be 89 x 89 = 7921.
But according to the terms of the problem the actual
number of counters was 31 short of this — i.e., 7890.
The correctness of this conclusion may be tested by cal
culating the number in the original square, which was one
counter less (viz., 88) each way. The square of 88 is 7744
But in this case there were 146 counters left over. 7744 +
146 = 7890, the same number arrived at by the first process.
No. LXXX.— A Curious Division. Solution.
To ascertain the common quotient, add together the
three divisors, 3, 6, and 9, and divide 7890 by their total,
18. The quotient is 438^. Multiply the number thus ob
tained by 3, 6, and 9 respectively, and the three products
will give the numbers contained in the three heaps.
Thus— 438 x 3 = 1315 
438^ x 6 = 2630
438^ x 9 = 3945
7890
No. LXXXL— A Curious Multiplication. Solution.
This is a very much easier problem, as the proportions of
the four heaps are almost selfevident. Taking the fourth,
or smallest heap, as the unit of measurement, and calling it
1 accordingly, the third, which is of obviously double the
size (as requiring to be multiplied by 6 only, instead of 12,
to raise it to the same amount), will = 2. The second heap
will in like manner = 3, and the first = 4, and their collective
values will bel +2 + 3 + 4 = 10. Divide the whole number
by this amount ; the quotient is 789. This gives us the
value of our unit of measurement, and from it we may de
duce the value of all four heaps, thus : —
The fourth heap contains ... ... 789 counters.
The third „ „ 789 x 2 = 1578 „
The second „ „ 789 x 3 = 2367 „
The first „ „ 789 x 4 = 3156 „
Total 7890
224 Puzzles Old and New.
No. LXXXIL— The Two Schoolmasters. Solution.
The difference was 6, the smaller school having 66 pupils
only.
For, inasmuch as onesixth of the pupils were away ill, the
remainder, viz. —
11 haymaking,
7 at the fair,
and 37 at school,
together making 55 — must have been fivesixths of the whole
number, and 55 — 5 (= 11), onesixth. The whole number
of the smaller school was therefore 11 x 6 = 66.
No. LXXXIIL— Nothing 1 Left. Solution.
The required number is 118.
To obtain it, work the process indicated in reverse order,
as follows : —
+ 18 = 18
18 2  324
324 + 3 = 108
108 + 10 = 118
No. LXXXIV.— The Three Generations. Solution.
The old man was 69, his son 40, and his grandson 16.
As the old man and his son were together 109 years old,
and the old man and his grandson only 85 years old, it fol
lows that the age of the son was 109 — 85 ( = 24) years
greater than that of the grandson. As the son and grand
son were together 56 years old, and the former was 24 years
older than the latter, it follows that the grandson's age was
5624 32 ia
— 2 = 2" = yearS '
The son's age was therefore 16 + 24 = 40 years.
As the united ages of the old man and his son were to
gether 109, the age of the former must have been 109 — 40 =
69 years.
No. LXXXV.— The Two Brothers. Solution.
As f of the age of the younger is T V that of the elder,
the actual age of the younger must be • , o a g= l2 = 24 == 8
Key to Arithmetical Puzzles. 225
3
that of the elder, and the difference between them ~ that of
the elder. By the terms of the question, we know that this
3 1
= 9 years, in which case^ must = 3 years, and the whole
8 o
age of the elder 3 x8(= 24) years. The age of the younger
will consequently be 24 — 9 = 15 yeai^s.
No. lxxxvi.— The Two Sons. Solution.
The younger son is 24 ; the elder, 29 years old.
The solution is most easily got at by means of a simple
equation, thus : —
Let y = age of younger.
Then^ + 5 J = age of elder,
By the terms of the question —
5 y + 6 (y + 5J) = 301
5 y + 6 y + 31 j = 301
11 y = 301  31i = 269i
V = 24i
The younger son is therefore 24 years old, and the elder
24i + 5J = 29f .
No. lxxxvtl— The Two Nephews. Solution.
The one was 18, the other 6 years old.
It is clear that as the larger number is three times
the smaller, it must be a multiple of 3. It cannot be a
larger number than 18, because the square of the next
multiple of 3 — viz., 21 — would alone exceed 360. We
proceed to try 18 accordingly as the larger number, and
18
g ( = 6) as the smaller number, when we find that 18 2 +6 2
= 324 + 36 = 360, the total required. 18 and 6 years are
therefore the respective ages.
226 Puzzles Old and Nezv.
No. LXXXVIII.— The Reversed Number. Solution.
The required number is 45. Ifc may be got at either alge
braically or arithmetically. In the first case —
Let x — the first, or lefthand, digit,
and y — the second, or righthand, digit.
Then (as x represents tens) 10 x + y will be the total
number, and 10 y + x the number produced by reversing
its digits, while x + y will be the sum of such digits. Now,
by the terms of the question,
10 x + y — 5 (a? + y)
= 5 x + h y
.". 5 x = 4 y
And x =
4?/
5
Further, by the terms of the question —
10 x + y + 9 = 10 y + x.
Substituting the value above obtained for x —
we have 4 ^' + y + 9 = lOy + ^
or, 40 y f 5 y + 45 = 50 y + 4 y
45 y + 45 — 54 y
9 y =z 45
y = 5
4 1,\ A
Then se ( = ~r ) = 4
and 10 x + y (the whole number) = 45.
The arithmetical process is, however, in this case the
simpler, thus —
As the required number is equal to five times the sum of
its digits (i.e., a multiple of 5), its final digit must be 5 or
0*. But such final digit is also the first digit of the reversed
number. would not answer this condition, and the
final di<nt must therefore be 5. Nine added to any number
ending in 5 mates the final digit 4. This gives us the
reversed number as 54, and the actual number consequently
as 45, which is found to fully answer the required conditions.
* See Properties of Numbers, p. 177.
Key to Arithmetical Puzzles. 227
No. LXXXIX— Another Reversed Number.
Solution,
The required number is 42.
Here again either the algebraical method or the arith
metical method may be used. Let x = the first digit, and
y the second digit. Then 10 x + y = the required number,
and 10 y + x = the reversed number.
By the terms of the problem —
(1) 10 x + y = 7 (x + y)
(2) 10 x + y  18 = 10?/ + a
10 x + y = 1x + 7y
Sx = 6y
x = 2y
Substituting this value in equation (2), we have
20*/ + y  18 = lOy + 2y
2ly  18 = 12y
% = 18
y = 2
and as x — 2y the number must be 42.
The arithmetical or logical process is as follows : —
By the terms of the question, the required number is a
multiple of 7. As it has only two digits, and its remainder
less 18 is also a number of two digits, this limits the selec
tion to 35, 42, 49, 56, G3, 72, 77, 84, 91, or 98. Among
these the only number which answers the remaining con
ditions is 42, which is accordingly the answer.
No. XC— The Shepherd and his Sheep, Solution,
There were
in
the first fold
27
55 55
5) 55
,, second
„ third
,, fourth
25
18
16
» >>
„ fifth
14
Total 100
228 Puzzles Old and New.
For as there were —
In the first and second folds 52
„ second and third 43
„ third and fourth 34
fourth and fifth 30
The number in the first and fifth, with
double the respective numbers in the
second, third and fourth, will be 159
As the actual number is 100, it follows that the number
in the second, third and fourth folds == 159 — 100 = 59.
. And as there were in the second and third 43, the num
ber in the fourth fold must have been 59 — 43 = 16.
As there were in the fourth and fifth 30, the number in
the fifth fold must have been 30 — 16 = 14.
As in the third and fourth there were 34, the number in
the third fold must have been 34 — 16 = 18.
In like manner, the number in the second fold was 43 — 
18 = 25.
And the number in the first fold was 52 — 25 = 27.
27 + 25 + 18 + 16 + 14 = 100,
No. XCL— The Shepherdess and her Sheep.
Solution.
The numbers are 7, 14, 28, and 56 respectively.
By the terms of the question, it appears that the num
bers in the four folds are in geometrical progression, with a
common ratio of 2. Taking experimentally the smallest
possible such progression, we find it to be 1 + 2 + 4 = 8 == 15.
But the total of the actual progression is stated to be 105.
Dividing this by 15, we have as quotient 7, which we use as
a common multiplier to bring the series up to the required
total. Thus:—
7
X
1 = 7
7
X
2 = 14
7
X
4 = 28
7
X
I
8 = 56
tal
105
Key to Arithmetical Puzzles. 229
No. XC1I.A Weighty Matter. Solution.
Seven weights are required, of 1, 2, 4, 8, 16, 32, and 64
lbs. respectively, together making 127 lbs. It will be found
that, by using one, two, or more of these, any weight from
1 to 127 lbs. can be weighed.*
No. XC1IL— The Three Topers. Solution.
The quantity would be consumed in 4f hours.
For — as Peter can drink 40 quarts in 20 hours, he would
drink in 1 hour f§ = 2 quarts.
As Paul can drink 40 quarts in 14 hours, he would drink
in 1 hour yf = 2f quarts.
As Roger can drink 40 quarts in 10 hours: he would
drink in 1 hour x§ = 4 quarts.
And they would together drink in 1 hour 2 + 2f + 4 ( =
8f) quarts, and 40 quarts in (40 ~ 8f) = 4^f hours.
No. xciv.— The False Scales. Solution.
Answer, 12 lbs.
Problems of this class are solved by ascertaining the
square root of the product of the two weights.
In this case 9 x 16 = 144, and a/ 144 = 12, the required
answer.
No. XOV.— An Arithmetical Policeman. Solution.
The hour struck was twelve.
The fractions mentioned, , , and \, if regarded as
fractions of a single hour, will be found, when added
together, to amount to ^f of an hour. To raise the surplus
•j2 to the value of a complete hour, it must be multiplied
by 12, and 12 is therefore the hour that was struck.
* The numbers, it will be observed, are in geometrical progression.
For another illustration of the fact here stated, see The Magic Cards,
p. 161,
230 Puzzles Old and New.
No. XCVLThe Flock of Geese. Solution.
The number o£ the flock was 36.
For, taking the lowest number (4), which is divisible by
2 and by 4 (as, from the conditions of the problem, it is
clear that the required number must be), and going through
the process suggested with such number, we have the
following result : —
4 + 4 (as many more) + 2 (half as many more) + 1 (one
four fch as many more) = 11.
Dividing 99 (the total to be obtained after going through
the same process with the actual number in the flock) by
the number thus obtained, we find the quotient to be 9.
4, therefore, multiplied by 9 ( = 36) should be the required
number. Putting it to the test, we find that 36 + 36 +
18 + 9 = 99, exactly answering the conditions.
No. XCVIL— The Divided Cord. Solution.
The one portion is 16, and the other 20 inches.
As the one segment is to be f of the other, their re
spective proportions will be 4 to 5, or in all 9 parts.
One ninth part of 36 inches is 4 inches. Taking this as
the unit of measurement, we find that —
the longer segment is 5" x 4" = 20 inches,
and the shorter „ „ 4"x4" = 16 ,,
Together, 36 „
No. XCVIIL— The Divided Number. Solution.
The parts are 28 and 18 respectively.
It is clear from the conditions of the problem that the
first part is a multiple of 7, and the second a multiple of 3.
Now there are in the number 46 six multiples of 7 — viz.,
7, 14, 21, 28, 35, and 42. In like manner there are in the
same number fifteen multiples of 3 — viz., 3, 6, 9, 12, 15,
18, 21, 24, 27, 30, 33, 36, 39, 42, and 45. Of these, the only
pairs which together make 46 are 7 and 39, and 28 and 18.
The first pair clearly does not answer the conditions of the
Key to Arithmetical Puzzles. 231
question, for \ + ^ = 1 + 13 = 14. We proceed to try the
second pair, and find that V s  + ¥" = 4 + 6 = 10. 28 and 18
are therefore the required numbers.
By the aid of algebra the problem may be solved much
more neatly. Thus —
Let x = the portion to be divided by 7, and y the portion
to be divided by 3.
Then, by the terms of the question—
x r y = 46
i.e. x == 46 — y.
Further, % + f = 10
or, substituting the value above found for as—
46 — ?/ ii
3 (46  y) + 7y = 210
138  By + 7y = 210
4//  210  138  72
2/ = 18
And as x — 46 — y
x = 46  18
■ 28.
No, XCIX.— The Two Numbers. Solution.
The required numbers are 5 and 7. For if twice the
first + the second = 17, and twice the second + the first =
19, then the above added together — i.e., three times the first
+ three times the second must be 17 + 19 — 36, and the
sum of the numbers themselves must be V 6  = 12.
And since twice the first + the second is an odd number,
the second is also an odd number, and the first, being an
even number (12) less an odd number, must also be an odd
number.*
* Because twice any whole number is always an even number ; and
an even number plus an even number is always an even number. See
p. 174.
232 Puzzles Old and New.
Now the only pairs of odd numbers which together make
12 are 1 and 11, 3 and 9, and 5 and 7. Of these, we find by
experiment that 5 and 7 are the only two that answer the
conditions ; 5 x 2 + 7 = 17, and 7x2 + 5 = 19.
Here again the problem is much more readily solved by
algebra. Thus —
Let x = first number,
And y == second number.
We have then the equations
2x + y = 17
3y + a = 19
From the latter equation we deduce that
x = 19  2y.
Substituting this value in the first equation, we get
2(19  2y) + y = 17
38  % + y = 17
or Sy = 38  17 = 21
2/ = 7.
Substituting this value in the second equation —
%j + x = 19
we have 14 + x = 19
x = 19  n
=5
No. C— The Horse and Trap. Solution.
The horse cost £60, and the trap £25.
As five times the value of the horse = twelve times the
value of the trap, it is clear that the latter was worth ^ as
much as the former. The value of the two together may,
therefore, be expressed as 1 + T \ (or if), and this by the
conditions of the problem =£85. If ■££ of the value of the
horse be £85, jV of such value will be T y of that amount —
i.e., £5. The total value of the horse is therefore £5 x 12 =
£60. The value of the trap, being T \ of £60, is £25.
i\.ey to sirimnieticaL ruzztes. 233
No. CI.— The Two Workmen. Solution.
They must work 6f hours per day.
For A, doing the whole in 70 hours, does T V * n 1 tour.
And B, doing the whole in 56 hours, does ~ in 1 hour.
Or, together, in 1 hour /<, + T \ = ¥ fp + ¥ f ^ = ^f q.
They would, therefore, do the whole in ^{p = 31^ hours.
And to complete it in 5 days, they must work — £ = 6
hours per day.
No. Cll.—The Divided Number. Solution.
The required numbers are 120, 72, and 45.
For since 3 times the first = 5 times the second, the second
must be f of the first.
And since 3 times the first=8 times the third, the third
must be f of the first.
The three parts are therefore as 1, f, and f, or as 40, 24,
and 15.
Now, 40 + 24 + 15 = 79, and 237 + 79 = 3.
Using this as a common multiplier, we have
120 = first part.
72 = second part.
45 = third part.
237
40
X
3
:=
24
X
3
=
15
X
3
z=
No. CIIL— The Three Reapers. Solution.
As A, B, and C, working simultaneously, reap the field
in 5 days, it is obvious that they together reap in one day \ of
it. Now A's and C's day's work are together equal to twice
B's. Therefore B's plus twice B's (or three times B's) also
= \; and B's day's work = one third of \ = T x ? .
Further, as A's and B's day's work together = three times
C's, then C's plus three times C's ( = four times C's) also =
I, and C's day's work = \ of ^ = £$.
Again, as all three together in one day do \, A must in
one day doidV+zV)
 12 (4 + 3) _. 127 h_ 1
60 "60 ~ 60 12'
234 Puzzles Old and Neiv.
A would therefore take 12 days, B 15 days, and G 20 days
to do the work singly.
No. ClV.The Bag" of Marbles. Solution.
If when Tom takes 4, Jack takes 3, and if when Tom takes
6, Dick takes 7, then when Tom takes 12 (the least common
multiple of 4 and 6), Jack and Dick will take 9 and 14
respectively, and they will together have taken 12 + 9 + 14
= 35 marbles. Now 35 is contained in 770 22 times, and
therefore : —
Tom's share will be 12 x 22 = 264
Jack's „ 9 x 22 = 198
Dick's „ 14 x 22 = 308
770
Again, their respective ages will be as 12, 9, and 14 ; but
12, 9, and 14 together make 35, and the total of their ages is
only 17 years, or onehalf of 35. Their ages are therefore
onehalf the above figures, or 6, 4, and 7 respectively.
No. CV.— The Expunged Numerals. A. Solution.
The puzzle is solved by striking out the first figure of the
top row, the whole of the second row, and the two first figures
of the last row. The sum will then stand as under : —
. 11
J." 9
20
No. CVL— The Expunged Numerals. B. Solution.
The dots indicate the figures to be expunged.
.11
33.
77
1111
Key to Arithmetical Pttzzles. 235
No. CVIL— A Tradesman in a Difficulty. Solution.
1. The purchaser gives his onedollar and his twocent
piece to the tradesman, and his threecent piece to the
stranger.
2. The tradesman gives his halfdollar to the purchaser,
and his quarter dollar to the stranger.
3. The stranger gives his two dimes and his onecent piece
to the purchaser, and his five cent piece and his two cent
piece to the tradesman, when each has his right amount.
The correctness of the arrangement is not at once obvious,
but it is easily proved. For the sake of brevity, we will call
the purchaser P, the tradesman T, and the stranger 8.
P has at the outset 1 dollar, 3 cents, and 2 cents, together
— 105 cents. He should, therefore, have left, after paying for
his goods, 105 — 34 = 71 cents.
T has at the outset a halfdollar and a quarter dollar, to
gether =75 cents. After receiving payment for his goods he
should, therefore, have 75 + 34 = 109 cents.
8 has 2 dimes, 5 cents, 2 cents, and 1 cent, in all 28 cents.
He has neither to gain nor lose by the transaction, so should
be left at the close with the same amount. Let us now see
how these figures correspond with the result of the transac
tion.
P parts w T ith 1 dollar and 2 cents to T, and 3 cents to 8,
in all 105 cents. He has therefore no cash of his own left ;
but he receives from T 50 cents and from S 21 cents. He
has, therefore, at the close of the transaction 50 + 21 = 71
cents. T hands to P half a dollar, and to 8 a quarter dollar ;
but, on the other hand, he receives from P 1 dollar and 2 cents,
and from S 7 cents — in all, 109 cents.
8 has given 2 dimes and 1 cent to P, and 7 cents to the
tradesman. This ( = 28 cents in all) clears him out. On the
other hand, he receives from 8 a quarter dollar, and from P
3 cents, which also together amount tc^28 cents.
Each has therefore received the precise amount to which
he was entitled.
No. CVIII.— Profit and Loss. Solution,
The original cost was £2 18s. 3d.
The difference between the marked price (65 shillings) and
236 Puzzles Old and New.
the price obtained (56 shillings) is 9 shillings. This sum
represents the amount of the actual loss, plus three times the
amount of such loss (which latter item would have been the
profit) — in all, four times the amount of the loss. The loss
was, therefore, £ shillings, or 2s. 3d. Add this amount to
the price obtained, £2 16s., and we have £2 18s. 3d., the
cost price.
No. Cix — A Curious Fraction. Solution.
The fraction in question is f ~§ .
To obtain the result, take the three denominations named,
£1, Is., and Id : reduce all to pence, and add them together.
The total (240 + 12 + 1 = 253) will give us the denominator of
the required fraction, while the value of £1 in pence (240)
will be the numerator. The process is perhaps clearer in
algebra. Thus : —
Let  be the required fraction.
y
rp q* fit
Then pounds +  shillings +  pence =3 1 pound.
Or, reducing all to the " pence " denomination—
240a? + 12a; +x __ ^aq
y
253a? " = 240
y
X
2 4
Examining the correctness of the above result, we find that
m of £i = is*, nitp.
Iff Of Is. = ilTHJ.
2I3 * 1°^ = a~5"3 ( *'
£10
No. ex.— The Menagerie. Solution.
There being 36 heads (i.e., 36 creatures in all), if all had
been birds they would have had 72 feet. If all had been
beasts, they would have had 144 feet. It is clear, therefore,
Key to Arithmetical Puzzles. 237
that there were some of each. Suppose the numbers equal,
the feet would then count as under :—
18 birds : 36 feet.
18 beasts : 72 feet.
36 108 feefc (being an excess of 8 over the
stated numbei\)
Each bird added to the " bird " half (involving at the
same time the deduction therefrom of one beast) produces
a diminution of 2 in the number of feefc. As the equal
division gives an excess of 8 feet, we must therefore deduct
4 beasts and add 4 birds.
This gives us 18 + 4 = 22 birds, having 44 feet.
184 = 14 beasts „ 56 „
36 100
No. CXI.— The Market Woman and her Eggs.
Solution.
Her original stock was seven.
To discover it, it is sufficient to note that she gave her
last customer half her remaining stock, plus half an egg. As
this left her with none, the half egg must have been equal
to the half of her then stock, which must, therefore, have been
1 egg only. She gave the second customer half her then
stock, phis half an egg ; and as this left her with one egg only,
it is obvious that the half in question must have been 1 egg.
She had, therefore, prior to this second transaction 3 eggs
left. At the first sale she gave half her original stock, plus
half an egg, and as this left her with 3 eggs, it follows that
her original stock must have been 7.
No CXII— The Cook and his Assistants. Solution.
His original stock was 39. He gave half plus half an egg
(i.e., 20) to the first assistant. . This left him with 19. He
gave half of these plus half an egg (9 + \ — 10) to the second
assistant, and had then 9 left. He gave half of these plus
half an egg (4 + ■§■ = 5) to the third assistant, and had 4
left.
The method of obtaining the solution is the same as in the
last case.
238 Puzzles Old and New.
The reader may be interested to know how problems of
this class are constructed. As the division is by moieties, 2
is taken as the basis number, and this is raised to the power
corresponding to the number of divisions. Thus in the two
cases supposed, of three divisions, we take 2 3 = 8. Multiply
this by any number you please ; subtract 1 from the result.
The ultimate remainder, when the original number has been
three times diminished as described, will be less by 1 than the
number you multiplied by. Thus if we take as the starting
point, —
7, the ultimate remainder will be nil.
55 55 55 ■*■•
J5 55 55 &•
55 55 5) "•
4
11 11 1» ^2.
Why the process indicated should have this peculiar result
is a further puzzle, which we will leave to the ingenuity of
our mathematical readers.
(8 x 1)
 1 = 7, t
(8 x 2)
 1 = 15,
(8x3)
123,
(8 x 4)
 1 = 31,
(8 x 5)
 1 = 39,
and so on.
CHAPTER V.
WORD AND LETTER PUZZLES.
No. I.— A Puzzling* Inscription.
The following queer inscription is said to be found in the
chancel of a small church in Wales, just over the Ten Com
mandments. The addition of a single letter, repeated at
various intervals, renders it not only intelligible, but appro
priate to the situation :
PR S V R Y P R F C T MN
VRKPTHSPRCPTSTN.
What is the missing letter ?
No. II.— An Easy One.
E D R N W.
Make one word of the above letters.
No. III.— Pied Proverbs.
Each of the following series of letters, duly arranged, will
be found to form a popular proverb.
A. aeegghillmnnnoooorrsssstt.
B. aaceeeffhhiiiiimnoooprrssttttt,
C aaaddeefiiimmnnnoorttw.
D, aabbddeehhhhhiiiinnnoorrssttttuww.
239
240 Puzzles Old and New.
E. aadegghiillllnoorssttttt.
F. abdeefiinnnoopprrrsssttuw.
G. aabdeeeeefffhiiikmnnrrsst.
H. aadeeehllllllnssfctww.
No. IV.— Scattered Sentiment,
Daruno em lislal vcrlio,
Ni dasesns ro lege,
Lilt silfe rdaems eb vero,
Twees riemem's fo ethe.
The above, duly rearranged, v\ r ill be found to form a
couplet suitable for a valentine.
No. v.— DroppedLetter Proverbs.
Supply the missing letters, and each of the series following
will be found to represent a popular proverb.*
a. Atthntmsvsnn.
b. Fithateewnaray.
c. Srkwiehiosht.
cl, H1gsbswoagslt.
e. Brsfftrfctgtr.
/. Hwogsbrwggssrwg.
g. C1rndfospkhtth.
li . Wetewnsnhwtst.
* Each dash represents either a dropped letter or ihe space between
two words. In some of the later examples one dash stands for two
dropped letters.
Word and Letter Puzzles. 241
i. Srrknnsnik1nfns.
h. Hnsystbsp1c.
I. Apndysgtyr.
m. Tkcrfhpnntepnswltkcrftesls.
No. VI.— DroppedLetter Nursery Rhymes.
The following, the missing letters being duly supplied,
will be found to represent familar quotations of the juvenile
order : —
(1.) Hwohhltlbsbe
Ipoeahhnnhu ;
Hgtesoealhdy
Eoeeypnnfoe.
(2.) Jcadiletphh1
Tfthpiowtr :
Jcf1dwadrkhsrw
Adi1aeubigfe.
(3.) Hyideidehctnteid1
Teojmeoeteon
Teiteoluhdoescfnsot
Adhdsrnwyihhson.
No. VII —Transformations.
This is a form of wordpuzzle that deserves to be better
known, as it may be made productive of considerable amuse
ment. It consists in taking a word of a given number of
letters, and trying in how many "moves " or transpositions,
altering only one letter each time, you can transform it into
some other prearranged word of the like number of letters,
but of different or opposite meaning ; as Light into Heavy,
Rose into Lily, Hard into Easy, or the like. Each step of the
process must be a known word. We will take the lastnamed
pair as an example. Five " moves " will in this case suffice,
as under : —
R
242 Pussies Old and New.
Hard— (1) card, (2) cart, (3) cast, (1), east, (5) Easy.
This, however, is a more than usually favourable specimeu,
one of the letters, a, being common to both words, and,
therefore, requiring no change. A considerably larger
number of moves will usually be found necessary.*
The reader is invited to transform : —
Hand into Foot — in six moves.
Sin into Woe — in three moves.
Hate into Love — : in three moves.
Black into White— in eight moves.
Wood into Coal— in three moves.
Blue into Pink — in four moves.
Cat into Dog — in three moves.
More into Less — in four moves.
Rose into Lily — in five moves.
Shoe into Boot — in three moves.f
No. VIII.— Beheaded Words.
1. Behead a tree, and leave the roof of a vault.
2. Behead "on high," and leave the topmost story.
3. Behead "thrown violently," and leave an organ of the
body.
4. Behead a preposition, and leave a contest.
5. Behead your own property, and leave ours.
6. Behead to delete, and leave to destroy.
7. Behead a reproach, and leave a relative.
8. Behead to annoy, and leave comfort.
9. Behead an occurrence, and leave an airhole.
The deleted initials, taken in the above order, will give the
name of an American genera], after whom a wellknown
street in Paris is named.
* Unless one or more letters are common to both words, the number of
moves cannot possibly be less than the number of letters in each word.
+ Where several persons take part, this may be made a very amusing
game. Certain pairs of words having been agreed upon, each takes the
list, and tries in how few moves he can effect the required transforma
tions, the player with the smallest total vanning the game.
Word and Letter Puzzles. 243
No. IX.— Anagrams.
An anagram is defined by Ogilvie as "the transposition
of the letters of a name, by which a new word is formed."
This definition hardly goes far enough, inasmuch as it ig
nores the far more interesting class of anagrams, in which
the letters of a whole sentence are rearranged so as to assume
a different sense. To be worthy of serious consideration,
however, the anagram must have a further quality — viz., that
the new rendering must have some sort of relation to the
original. In some cases a new rendering of this kind has hap
pened to be singularly appropriate ; so much so, indeed, that
in less enlightened times people have claimed for anagrams
a sort of inspiration, or magical significance. There is a
historic instance in the case of James I., of England, whose
name, James Stuart, was transposed by his courtiers, to his
great delight, into A just master, and who was more than half
persuaded of his descent from the mythical King Arthur,
on the ground that his full name, Charles James Stuart, was
capable of transposition into Claims Arthur's Seat. Another
wellknown instance is that of Lady Eleanor Davies,
wife of the poet, Sir John Davies (temp., Charles I.),
who claimed to be a prophetess, on the somewhat unsub
stantial ground that the letters of her name, Dame Eleanor
Davies, duly shuffled, form the sentence Reveal, Daniel.
Her pretensions ultimately caused her to be arraigned
before the Court of High Commission, when the Dean
of Arches pointed out that the same letters might also
read Never so mad a ladie. The public preferred the latter
rendering, and no more was heard of the soidisant prophetess.
A recent prize competition among readers of TitBits, for
the best anagrammatic rendering of the title of either of
the articles in a given number, produced the following,
" Dangers of amateur physicking," " The sick men pay
for drugs again." This is a model of what an anagram
should be, and we can well understand a credulous person
believing that there must be something more than mere
chance in so pregnant a warning.
The following are instances of specially meritorious ana
grams, in various languages : —
Ie cliarme tout, an anagrammatic rendering, attributed to
244 Puzzles Old and New.
Henri IV., of the name of Marie Touchet, the mistress of
Charles IX.
Honor est a Nilo — Horatio Nelson.
Quid est Veritas ? (the Latin rendering of Pilate's ques
tion, " What is truth ? "). Est vir qui adest — " It is the man
who is before you."
Flit on, cheering angel. — Florence Nightingale.
The Tichborne trial gave rise to a somewhat complicated
anagram. Sir Roger Charles Doughty Tichborne, Baronet —
You horrid butcher, Orton, biggest rascal here.
We append a brief selection for our readers to exercise
their ingenuity upon.
a. Bare mad frolic. Transposed, represents — a political
cry.
b. Got a scant religion : — the name of a prominent division
of Nonconformists.
c. Best in prayer : — ditto, ditto.
d. Lady mine : — what every unmarried lady should be.
e. City life : — happiness.
/. Tournament : — a description of tilting.
g. Melodrama : — what melodrama ought to be.
h. Misanthrope : — what he deserves.
i. Old England :— the same country poetically described.
j. Telegraphs : — what they are to commerce.
h. Lawyers : — a satirical description of themselves.
1. Astronomers : — ditto.
m. Astronomers : — their occupation gone.
No. X.— Word Squares.
The problem in this case is to arrange a series of words,
having the like number of letters each, one above the other,
in such manner that they shall read alike, whether in a hori
zontal or vertical direction. The following are examples of
sixteen letter squares.
Take the equivalent (a word in four letters) of—
Word and Letter Puzzles.
245
1. A narrow road.
2. A plane surface.
3. A preposition signifying
propinquity.
4. Two parts of the body.
1.
2.
3.
4.
b.
Not any.
Across.
Not far.
Does wronq*.
C.
1. Halting.
2. Dry.
3. A possessive pronoun.
4. Paradise.
The next two examples
letters : —
g '
1. To squander.
2. A stage player.
3. A mineral concretion.
4. A pickmeup.
5. Upright.
d.
1. A burden.
2. A river.
3. Begs.
4. A piece of writingfurni
ture.
e.
1. A puppet.
2. A river.
3. A wild beast.
4. Solitary.
./•
1. A noted city.
2. A stone, reputed unlucky.
3. To knock about.
4. A girl's name.
are of words of five and six
h.
1. A shepherd.
2. Dress.
3. Thickheaded.
4. Walking on the toes.
5. A bird.
6. To ransom.
No. XI.— Word Diamonds.
Sometimes, instead of a square, the letters are required to
be arranged in the form of a diamond, say, one letter in the top
line, three in the second, five in the third, three in the fourth,
and one in the last line, subject to the same condition — that,
horizontally or vertically, they shall read alike. The follow
ing are examples : —
a.
1. A single letter.
2. The juice of the olive.
3. Fir trees.
4. A meadow.
5. A single letter.
6.
1. A single letter.
2. A garden tool.
3. Substantives.
4. An extremity.
5. A single letter.
246
Puzzles Old and New.
1.
2.
3.
4.
5.
c.
A single letter.
The cry of a sheep,
A sweetmeat.
A girl's name.
A single letter.
1.
2.
3.
4.
0.
A single letter.
Smoked pigmeat.
A male Christian name.
The title of a married lady.
A single letter.
1.
2.
3.
4.
5.
d.
A vowel.
An animal.
Eve's bane.
A tree.
A vowel.
€.
1.
2.
3.
4.
5.
9'
A consonant.
To place.
A fruit.
An adverb denoting ex
cess.
A consonant.
h.
I.
2.
3.
4,
5.
A consonant.
The ocean.
Arab dwellings.
Consumed.
A consonant.
1.
2.
3.
4,
5.
A consonant.
A feature.
A boundary.
A deep hole.
A consonant.
The next pair are a trifle more elaborate, the longest word
having seven, instead of five, letters.
1. A consonant.
2. A point.
3. The Papal crown.
4. A precious stone.
5. Haughty.
6. A conjunction.
7. A consonant.
In the next pair the longest
h.
1 . A consonant.
2. A precious stone.
3. Danger.
4. A military officer.
5. A performer of nocturnal
music.
6. A marvel.
7. A long spoon.
8. Sheltered from the wind.
9. A consonant.
3
1. A consonant.
2. Phoebus.
3. Tasty.
4. Jove.
5. Saltpetre.
6. A river.
7. A consonant.
word is of nine letters.
Z.
1. A consonant.
2. Instead of.
3. Apple centres.
4. Pincers.
5. Chinaware.
6. To choose again.
7. Vacancy.
8. To occupy a seat.
9. A consonant.
Word and Letter Puzzles. 247
No. XII.— A Cross of Diamonds.
This is an extremely ingenious puzzle. Required, to form
a cross consisting of four diamonds of five words each, united
in the centre by five additional letters, forming a smaller
diamond, after the fashion shown in Fig. 370 : —
*
^ w ^p
•Xf . top w *3p T&
4fc> «tf» Jg.
"3P w ■ "IP
^ # *
^P "J»' W 7P *2P ^ W
■agt 4&^k4& " je* '' ji. j,' y* j> oi ^i
TlP ^P W ^P W^7 TfP ^JP Jp TP ^/P Jp «jp
*.""* \^ J£, Ji. Ji. A»»
TP "7P *JP ^p ^p «3P *fr
rfp w
* # *
•ap •«* tp w w
# # #
Fig. 370.
The diamond at top is to be made up as follows: 1, a single
letter; 2, the queen of the fairies; 3, a title of courtesy
applied to ladies ; 4, wicked ; 5, a single letter.
The righthand diamond as follows : 1, a single letter; 2,
past tense of a verb meaning to possess ; 3, a colourless
fluid; 4, the abode of a wild animal ; 5, a single letter.
The lefthand diamond : 1, a single letter ; 2, a fruit ; 3, a
flower; 4, a metal ; 5, a single letter.
The bottom diamond: 1, a single letter ; 2, to strike; 3,
close ; 4, an article ; 5, a single letter.
The central diamond, read in conjunction with the bottom
letter of the top diamond, the top letter of the bottom
diamond, the lefthand letter* of the righthand diamond,
and the righthand letter of the lefthand diamond, will
form as follows: — From centre to top, a male sheep; from
centre to bottom, a. small animal ; from centre to right,
crude ; from centre to left, a quick blow ; from top to centre,
to deface ; from bottom to centre, a resinous substance ; from
right to centre, open hostility ; from left to centre, equal
value.
248
Puzzles Old end New.
Each diamond (other than that in the centre) mnst be
perfect in itself, forming the same words both horizontally
and vertically.
No. XIII.— Knight's Tour Letter Puzzles.
The principle of the " Knight's Tour " on the chessboard
has been made the foundation of another and different class
of puzzles. The squares of the chessboard are here occu
pied by letters or words (one in each square), which, if read
B
L
T
E
Y'
L
E
Y
H
L
T
B
T
A
T
A
A
A
H
T
I
E
f
E
I
N
E
D
H
W
Y
E
S
Y
E
T
E
s.
D
B
W
Y
N
E
s
I?—
E
II
/
A
A
A
W
I
D
E
A. Fig. 371.
in due sequence, according to Knight's Tour rules, form a
proverb, a verse of poetry, or a wellknown quotation.
A (Fig. 371) is an example of this class. The letters, read
aright, will be found to form a popular proverb.
Word and Letter Puzzles.
249
B (Fig. 372), furnishes another example, and C (Fig. 373),
a third.
i
B
B
I
N ! .
F
D
E
I
A
S
F
D !
N
s
F
A
1
1
1
B. Fig. 372,
E
E
T
L
H
E
]A
s
!E
A
S
D
i E
1 —
s
p
s
C. Fig. 373.
250
Puzzles Old and New,
No. XIV.— Knight's Tour Word Puzzle.
The words of the following puzzle (Fig. 374), duly read
sym
pathy
man
the
link
in
granted
body
it
is
the
the
in
be
neath
liveth
fierce
soon
as
to
secret
alone
not
silver
and
fly;
in
silken,
it
given
the
soul
mind,
wishes
desire,
has
heaven
it
to
the
fan
tasy's
can
to
die;
tie
gift
not
desire
heart,
with
whose
it
God
heart
doth
hot
love's
mind
bind.
which
not
is
which
and,
dead
fire,
true
Fig. 374.
according to Knight's Tour rules, will be found to form a
wellknown passage from the "Lay of the Last Minstrel."
Word and Letter P tizzies.
2^1
No. XV.— Hidden Proverbs.
REKOf NEDTHANW
SYOURCAKEANDA
STETOBEFEARHR
E AR KSSPOILEAE
LEOOHERSNTDVO
OTMOTLINOHTEU
kpSbiliAGMEHIR
SNIYGORSOBATS
ENGKENOTSRNPA
IAOAMOOTSOAEW
RODEVILAHTDAS
OUOY^NO ILDAECA
TCIVREHHTAHEZ
The above apparent jumble contains five wellknown pro
verbs, arranged in a systematic order. When the clue is
once discovered, the proverbs can be read without difficulty.
Puzzle : To find it.
2^2
Puzzles Old and New.
No. XVI.— The Five Arab Maxims.
The subjoined table (Fig. 375), to any one who can read
never
all
A*.
forhe who
all
often
more than
tell
you may
know
tells
he knows
tells
he knows
attempt
you can do
attempts
he can do
attempts
he can do
believe
you hear
believes
he hears
believes
he hears
spend
you can
afford
spends
he can
afford
spends
he can
afford
decide
upon
you may
see
decides
upon
he sees
decides
upon
he sees
Fio. 375.
it rightly, contains five maxims, said to be held in great
esteem by the Bedouin Arabs.
KEY TO CHAPTER V.
WORD AND LETTER PUZZLES.
No. I.— A Puzzling* Inscription. Solution.
The letter e, which, inserted at the proper intervals, makes
the inscription read as under : —
PERSEVERE YE PERFECT MEN,
EVER KEEP THESE PRECEPTS TEN,
No. II— An Easy One.
This is a problem of the " quibble " order. The seven
letters duly arranged form one word.
No. III.— Pied Proverbs. Solutions.
A. Rolling stones gather no moss.
B. Procrastination is the thief of time.
C. Time and tide wait for no man.
D. A bird in the hand is worth two in the bush.
E. All is not gold that glitters.
F. Fine words butter no parsnips.
G. Fine feathers make fine birds.
H. All's well that ends well.
No. IV.— Scattered Sentiment. Solution.
The lines should read as follows : —
Around me shall hover,
In sadness or glee,
Till life's dreams be over,
Sweet mem'ries of thee.
253
254 Puzzles Old and New,
No. v.— DroppedLetter Proverbs.
a. A stitch in time saves nine.
b. Faint heart never won fair lady.
c. Strike while the iron's hot.
d. He laughs best who laughs last.
e. Birds of a feather flock together.
/. He who goes a borrowing goes a sorrowing.
g. Children and fools speak the truth.
h. When the wine is in, the wit is out.
i. Short reckonings make long friends.
h. Honesty is the best policy.
I. A pin a day is a groat a year.
?52. Take care of the pence, and the pounds will take care of
themselves.
No. VI.— DroppedLetter Nursery Rhymes.
(1) How doth the little busy bee
Improve each shining hour ;
He gathers honey all the day
From every opening flower.
(2) Jack and Jill went up the hill
To fetch a pail of water ;
Jack fell down and broke his crown,
And Jill came tumbling after.
(3) Hey diddle diddle, the cat and the fiddle,
The cow jumped over the moon,
The little dog laughed to see such fine sport,
And the dish ran away with the spoon.
No. VII.— Transformations. Solutions.
Hand; hard; lard;* lord; ford; fort; Foot.
Sin ; son ; won ; Woe.
Hate ; have ; lave ; Love.
Black; slack; stack; stalk;* stale; shale; whale; while:
White.
* These are examples of a necessity, which frequently arises, of inttr
Key to Word and Letter Puzzles. 255
"Wood ; wool ; cool ; Coal,
v 1 Blae_;Jhile ; pile ; pine ; Pink.
"Cat; cot; cog; Dog.
More ; lore ; lose ; loss ; Less.
Rose ; lose ; lost ; list ; lilt ; Lily.
Shoe ; shot ; soot ; Boot.
No. viii— Beheaded Words.
1. Larch.
2. Aloft.
3. Flung.
4. About.
5. Yours.
6. Erase.
7. Taunt.
8. Tease.
9. Event.
The initials, as will be seen, give the word Lafayette.
No. IX.— Anagrams.
a. Radical reform.
b. Congregation alist.
c. Presbyterian.
d. Maidenly.
e. Felicity.
/. To run at men.
g. Made moral.
h. Spare him not.
i. Golden Land.
j. Great helps.
Jc. Sly ware.
I. Moonstarers.
ra. No more stars.
posing a inove which does not directly aid the transformation, but indi
rectly as a link with some more desirable word.
In the first example, the word " food " might (in place of " fort") form
the intermediate step between " ford" and "foot."
256 Puzzles Old and New.
No. X.— Word Squares.
a. L A N E J. NONE
AREA OVER
NEAR NEAR
EARS ERRS
c. L A M E d. L A D
A R ID OUSB
MINE ASKS
EDEN DESK
e. DOLL /.ROME
OHIO OPAL
LION MAUL
LONE ELLA
?1 WASTE L PASTOR
ACTOR ATTIRE
STONE STUPID
TONIC TIPTOE
ERECT ORIOLE
REDEEM
No. XI.— Word Diamonds, Solutions.
In solving puzzles of this class, endeavour, if the indica
tions be sufficient, to guess the centre or key word.
Arrange this word in the form of a cross, thus —
P
I
PINES.
E
S
If you have guessed rightly so far, the discovery of the
remaining words is a comparatively easy matter.
If the indication given as to the key word is too vague to
guide you, endeavour to discover one or both of the next
Key to Word and Letter Puzzles. 257
Sff woTd^' Whi0h WiU frequentI y ^ a cUie *» «■» P™
c. p
S
* C
Y
S
1ST
i, D
D
&. N
J L HOE
PINES NOUNS
L E A END
S
d. A
BAA A p E
CANDY APPLE
ADA ELM
E
T /• , H
SEA HAM
TE "S HARRY
A T E MRS
Y
M h. L
SET LI p
MELON LIMIT
TOO PIT
T
J
TIP SUN
TIARA SAPID
DIAMOND JUPITER
PROUD NITRE
A N D DEE
R
258 Puzzles Old and New.
A.. S I. P
G E M FOR
PERIL CORES
GENERAL FORCEPS
SERENADER PORCELAIN"
MIRACLE REELECT
LADLE. SPACE
LEE SIT
R N
No. XII.— A Cross of Diamonds. Solution.
M

T
M A B
MADAM
BAD
M
W
1ST
U T
A
H
A
P
T
U L I
P A R A W A T
E R
T
I N
P
A
T
HIT
TIGHT
THE
T
D
E
R
N
Key to Word and Letter P tizzies. 259
No. Xlll.Knight's Tour Letter Puzzles.
Solution.
A. Fig. 376 indicates the order in which the letters are
to be taken, when they will be found to read as follows : —
" Early to bed and early to rise
Is the way to be healthy, and wealthy, and wise."
*>
40
19
36
5
50
21
31
18
37
4
51
20
35
6
49
41
2
39
54
52
33
22
33
17
1
53
55
48
7
13
42
57
53
60
23
32
16
27
14
59
56
8
47
43
12
29
26
45
10
31
24
23
L
15
44
11
30
25
46
9
Fig. 376.
260
Puzzles Old and New.
In B the order of the letters is as shown in Fig. 377, the
hidden proverb being, " Safe bind, safe find."
5
7
6
8
4
13
12
10
16
14
1
9
2
11
15

i
Fig. 377.
Key to Word and Letter Puzzles. 261
In G the order is as shown in Fig. 378, and the proverb,
" More haste, less speed."
9
4
8
10
5
17
13
7
11
6
18
16
2
14
12
15
1
Fio. 378.
262
Puzzles Old and Nezu.
No. XIV.— The Knight's Tour Word Puzzle.
Solution.
43
40
10
47
41
43
61
23
13
52
17
23
12
45
21
27
59
30
25
22
26
23
53
44
9
42
11
60
48
39
8
62
57
31
56
21
7
14
35
4
51
16
3
18
63
38
49
34
53
2
33
32
15
6
36
19
55
20
C4
50
37
5
54
1
Fig. 379.
The key to this puzzle will be found in Fig. 379. Read
according to the order here indicated, it will be found that
the words make the following stanza : —
" True love's the gift which God has given
To man alone beneath the heaven :
It is not fantasy's hot fire,
Whose wishes, soon as granted, fly ;
It liveth not in fierce desire,
With dead desire it doth not die :
It is the secret sympathy,
The silver link, the silken tie,
Which heart to heart, and mind to mind,
In body and in soul can bind."
Lay of the Lad Minstrel. Canto V.
Key to Word and Letter F * tizzies.
The construction of other problems on the same model
will be found very interesting. Should the passage used be
less than the full number of sixtyfour words, a word may
here and there be cut in two, so as to occupy two squares, or
the superfluous squares may be left unoccupied. If the
quotation be too long, two words may be made to occupy a
single square, like " soon as " in the example given.
No. XV.— Hidden Proverbs. Solution.
The five proverbs are as follows :—
A rolling stone gathers no moss.
Too many cooks spoil the broth.
A live dog is more to be feared than a dead lion.
You cannot eat your cake and have it.
Peace hath her victories, no less renowned than war. .
To read them, first find the central letter, which is A,
This begins the first proverb. Immediately below this will
be found R, to the left of this 0, and above the two L's.
To the right of the last L are the letters I N. The G, com
pleting the word " rolling," comes next below the N, and
below this, S, the initial of the next word, " stone." From
the S, moving to the left, we have the remaining letters,
TONE, and so we read on, following the course of the
sun, round each square of letters in succession.
For greater clearness we exhibit separately the central
square and a few letters of the next square, showing the
commencement of the process.
LIN '
L A &
R S
E N T
No. XVI —The Five Arab Maxims. Solution.
The key to this puzzle consists in reading first the words
in the first and second lines alternately. Then those in the
first and third alternately. Then in the first and fourth, the
264 Puzzles Old and Neiv.
first and fifth, and first and sixth in succession. The maxims
will then be found to run as follows : —
"Never tell all you know ; for he who tells all he knows,
often tells more than he knows."
"Never attempt all you can do; for he who attempts all
he can do often attempts more than he can do."
" Never believe all you hear, for he who believes," etc.,
etc.
And so on, the words in the first line being common to
each maxim.
CHAPTER VI.
PUZZLES WITH COUNTERS.
Puzzles of this class are frequently propounded in more or
less fanciful forms, — e.g., a gardener is required to plant trees,
or an officer to place his men, in such manner as to answer
the conditions of the problem. From considerations of space,
we have thought it best to leave such fanciful elaborations
for the most part to the imagination of the reader. Should
he prefer to put the question in such a shape, he will have
little difficulty in inventing an appropriate legend.
No. I.
Required, to arrange eleven counters in such manner
that they shall form twelve rows, with three counters in
each row.
No. II,
Required, to arrange nine counters in such manner that
they shall form ten rows, with three counters in each row.
No. III.
Required, to arrange twentyseven counters in such man
ner as to form nine rows, with six counters in each row.
No. IV.
Required, to arrange ten counters in such manner that
they shall form five rows, with four counters in each row.
No. V.
Required, to arrange twelve counters in such manner that
they shall make six rows of four counters each.
No. VI.
Required, to arrange nineteen counters in such manner
that they shall form nine rows of five counters each.
265
266
Puzzles Old and New.
No. VII.
Required, to arrange sixteen counters so as to form ten
rows, with four counters in each row.
No. VIII.
Required, to arrange twelve counters in such manner that
they shall count four in a straight line in seven different
directions.
No. IX.
Required, to arrange nine white and nine red counters in
such manner that there shall be ten rows, of three counters
each, white, and eight rows of three each red.
No. X.
Given, a square, divided into nine smaller squares. Re
quired, to arrange counters in the eight outer squares in
such manner that there shall always be nine on each side of
the square, though the total be repeatedly varied, being
24, 20, 28, 32, and 36 in succession.
This is a very ancient problem. It is usually propounded
after the fashion following: A blind abbot was at the head
3 3 3
3 A 3
3 3 3
Fig. 380.
of a monastery of twentyfour monks, who were domiciled
three in a cell in eight cells, occupying the four sides of a
square, while the abbot himself occupied a cell in the
centre. To assure himself that all were duly housed for the
night, he was in the habit of visiting the cells at frequent
intervals, and counting the occupants, reckoning that if he
found nine monks in each row of three cells (see Fig. 380),
the tale was complete.
But the brethren succeeded in dueling his vigilance. First
Puzzles with Counters.
267
four of them absented themselves (reducing the number to
twenty), but still the abbot counted and found nine in a
row. Then these four returned, bringing four friends with
them, thus making twenty eight persons, and yet the normal
nine in a row was not increased. Presently four more out
siders came in, making thirty two. The result was the same.
Again, four more visitors arrived, making a total of thirty
six, but the abbot, going his rounds, found nine persons in
each row as before.
How was this managed ?
No. XI.
Required, so to place ten counters that they shall count
four in a row in eight different directions.
No. XII.
Required, so to place thirteen counters that they shall
count five in a straight line in twelve different directions.
No. XIII.
This is a puzzle of a different character.
Given, an eightpointed star, as shown in Fig. 381, and
seven counters. You are required to place the counters on
Fig. 381.
seven of the points of the star, in so doing strictly following
the rule following r viz., Each counter is to be drawn from a
vacant point along the corresponding line to another vacant
point, and there left. You then start from another vacant
point, and proceed in like manner till the seven points are
covered.
268
Puzzles Old and New.
No. XIV.— The "Okto" Puzzle.
The puzzle brought out under this name is a variation on
the foregoing. Each point of the star terminates in a small
circle, on which is printed the name of a given colour — black,
yellow, carmine, and so on. The counters, eight in number,
are of corresponding colours. The aspirant is required to
cover seven of the circles, according to the rule laid down for
the last puzzle, each with the counter of its proper colour.
No. XV. '
Given, the figure described in Fig. 382.
Fig. 382.
Required, to place at the intersections of its various lines
twentyone counters, in such manner as to form thirty rows
of three counters each, each group of three being united by
one of the lines.
No. XVI.
This is sometimes known as the " crowning" puzzle.
The reader will remember that at the game of draughts a
man reaching the opposite side of the board becomes a king,
and is " crowned " by having a second man placed on the
top of it. In the case of the puzzle we are about to describe
ten counters, or men, are placed in a row, and the player is
required to "crown" five of them after the following fashion.
He is to take up one counter, pass it to right or left over
Puzzles with Counters.
269
two others, and crown the one next in order, proceeding in
like manner till the whole are crowned.
A king, it should be stated, is still regarded as being two
counters.
No. XVILThe "Right and Left" Puzzle.
This is a very excellent puzzle, and has the special recom
mendation of being very little known. Rule on cardboard
a rectangular figure consisting of seven equal spaces, each
one inch square (see Fig. 383) . In the three spaces to the
left place three red, and in the three spaces to the right
three white counters, the space in the middle being left
unoccupied.
3
Fig. 383.
The puzzle is to transpose the red and white counters, so
that the three white shall be in the left hand, and the three
red in the right hand spaces. This is to be done in accord
ance with the following conditions, viz. : —
1. Each counter can only be moved one space at a time.
2. If a counter is divided from a vacant space by a single
counter only, it may pass over it into such vacant space.
3. Counters may only be moved in a forward direction —
i.e., red to the right, and white to the left. A move once
made cannot bo retracted.
No. XVIII.
This is a further development of the same problem. Rule
a sheet of paper into squares so that each horizontal row
shall contain seven, and each vertical row five, and upon
them place red and white counters (17 of each colour), as
shown in the diagram* (Fig. 384), the central space (No. 18)
being left vacant.
* If preferred, one corner of a draught or gobangboard may be used,
in place of the ruled paper.
270
Puzzles Old and New,
lb
29
23
30
10
17
24
31
18
25
32
26
33
13
14
20
2T
34
_ 21
®
28
35
Fig. 384.
You are required, under the same conditions as in the last
case, to transpose the red and white counters.
No. XIX.— The "Four and Four" Puzzle.
This is in general idea very similar to the two puzzles last
described, but it is wholly different in working.
10
Fig. 385.
Rule on cardboard a rectangular figure, divided into ten
squares, as Fig. 385, and in the first eight spaces, beginning
from the left hand, dispose eight counters, red and black
alternately.
The puzzle is, moving them two at a time, to get the four
7 2 3 4 5 6 7 8 9 10
m.
Fig. 386.
red and the four black counters grouped each colour together
without any interval, and this must be done in four moves
Ptizzles with Counters.
271
only. At the close of the operation the eight counters should
be as shown in Fig. 386.
They are then to be worked back again, after the same
fashion, to their original positions.
r

fcN
; ;
jM
NHMMMMt
w
Fig. 387.
This puzzle is sometimes arranged in the form of a slip of
wood, seven inches by two, with its central portion cut out
with a fretsaw, as shown in Fig. 387. The counters are
here replaced by little blocks of wood, each in the shape of a
collarstud. They may be shifted backwards and forwards
from gap to gap with great ease, but cannot be detached
from the board, and are therefore always available for use.
No. XX.— The "Five and Five" Puzzle.
This is the same as that last described, save that ten
counters, five of each colour, are used, and that the desired
transposition is to be effected in five moves.
No. XXI.— The "Six and Six" Puzzle.
This is again the same problem, but with twelve counters,
six of each colour, the transposition to be effected in six
moves.
No. XXII.— The Thirtysix Puzzle.
Thirtysix counters are arranged in the form of a square,
six rows of six each.
Required, to remove six counters in such manner that
the remaining counters shall still have an even number in
each row, horizontal and vertical.
272
Puzzles Old and New,
No. XXIII.— The "Five to Four" Puzzle.
Twentyfive counters are arranged in the form of a square,
five rows of five each.
Required, to remove five counters in such manner that
the remainder shall be four in a row, horizontally and
vertically.
No. XXIV.— No* Two in a Row.
With an ordinary draughtboard, and eight draughtsmen
or counters.
Required, so to dispose the eight men upon various
squares of the board that no two shall be in the same line,
either vertically, horizontally, or diagonally.
No. XXV.— The "Simple" Puzzle
A further and very interesting development of the last
mentioned ''poser" has been brought out, under the name
^_ ~,
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r~r
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11
...
Ji
, \ ;■ , ;„
r . ' . , ,■.  . ■■" ..;
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Fig. 388.
of the "Simple" Puzzle, by Messrs. Feltham. Simple as it
may be, in a sense, we have known much keen thought
expended upon it before the coveted solution was obtained.
Puzzles with Counters.
A special board of 67 squares is used, arranged as shown in
Fig. 388, and the problem is to place nine counters upon it,
in accordance with the conditions of the last puzzle.
N.B. — It should be stated that the indentations on either
side of the board do not affect the conditions. Thus, two
counters placed respectively in the left hand top and bottom
corner squares would be regarded as being in the same
line, notwithstanding that there is a break of continuity
between them.
No. XXVI.— The " English Sixteen " Puzzle.
A clever puzzle, under the above title, is issued by
Messrs. Hey wood, of Manchester. In the result to be at
tained it is almost identical with No. XVIII. (p. 269), but
the conditions are in this case somewhat different, and the
puzzle considerably more difficult.
A board, as illustrated in Fig. 389, is used, with eight
Fig. 389.
white and eight red counters. These are arranged on the
black squares, the red to the right, the white to the left, the
central square, No. 9 in the figure, being left vacant. The
problem, as in the puzzle abovementioned, is to transpose
the red and white counters, the men to be moved accor
ding to "draughts" rules — i.e., forward only; the whites
towards the spaces occupied by the reds, and the reds
towards the spaces occupied by the whites. The men move
only on the black squares, and therefore diagonally. A
T
274 Puzzles Old and New,
white man can pass over a black man, or a black man over
a white man, provided that the space next beyond is vacant.
No. XXVII.— The Twenty Counters.
Required, (a) so to arrange twenty counters as to form
therewith thirteen different squares.
(b) To remove six counters only from the figure formed
as above, so that no single square shall remain.
KEY TO CHAPTER VI.
PUZZLES WITH COUNTERS.
N.B. — It must not be taken for granted, in the case of
puzzles which demand a particular arrangement of counters,
that the solution given is the only one possible, as there
may frequently be two or more modes of arrangement which
will equally answer the conditions of the problem.
No. I.— Solution.
The eleven counters are arranged as shown in Fig. 390.
Fig. 390.
The five at bottom, count as two rows of three, the counter
in the middle being common to both.
No. II.— Solution.
\ \ / j
C©V W
& # H®
Fig. 391.
Arrange the nine counters as shown in Fig. 391,
275
276 Puzzles Old and New.
Ho. III.— Solution.
There are several different arrangements which will
#—#—# h—kt'—®
/ / /
(§>.— @..@: ®.....^~ .®
Fig. 392.
answer the conditions of this problem. Figs. 392 and 393
represent two of such arrangements.
!»—«>'■©!>.—
#— 0— (9
Fig. 393.
Key to Puzzles with Counters. 277
No. IV.— Solution.
This again may be solved in various ways. Either of
\ w
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Fig. 394.
Fig. 395.
the subjoined arrangements (Figs. 394 and 395) will answer
the required conditions.
\ /
No. V.— Solution.
©
& @ ®—
/'¥
A
Fig. 396,
Fig. 397.
The counters may be arranged as shown in Fig. 396 or
Fig. 397,
278
Puzzles Old and New.
No, VI.— Solution.
fe»lo»
'M
mm—m
Fig. 398.
The nineteen counters should be arranged as shown in
Fig. 398.
No. VII— Solution.
This is a puzzle of a class that sometimes perplexes by
reason of its very simplicity, the experimenter seeking for
some abstruse method of solution, while the real mode lies
close to his hand. All that is needed in this case is to
arrange the sixteen counters in four rows of four each,
forming a square. There are thus four vertical and four
horizontal rows, while the diagonals from corner to corner
supply the two additional rows required by the problem.
No. VIII.— Solution.
Lay out nine counters in three rows of three each, so as
to form a square. This done, distribute the remaining three
as follows : — place one counter on the first of the first row,
another on the second of the second row, and the third on
the last of the third row.
Key to Puzzles with Counters. 279.
No. IX.— Solution.
The counters must be arranged as in the subjoined dia
¥•#■
*■'»*
i s
m
Fig. 399.
gram (Fig. 399). Note that the centre counter is duplica
ted, a red counter lying half over a white counter, or vice
versa, so as to do double duty.
No. X.— Solution.
The secret lies in increasing or diminishing, as the case
may require, the number of persons in the corner cell?,
4 14
J
1 1
J
4 14
2 5 2
I
5 5
2 5 2
Fig. 403.
Fig. 401.
each of which counts twice over, and so, to a person as
doddering as the abbot must be assumed to have been,
seems at first sight to increase the general total. Thus
28o
Puzzles Old and New.
when the four monks absented themselves, the remaining
twenty were rearranged as in Fig. 400; and when they
returned with four other persons, the twentyeight were dis
17 1
7 7
1 7 1
9
9
9
9
Fig. 402.
Fig. 403.
posed as in Fig. 401. When four more visitors arrived, the
thirtytwo were distributed as in Fig. 402 ; and when the
final four arrived, the party, now numbering thirtysix, were
arranged as in Fig. 403.
No. XL— Solution.
This is a quibble dependent on the special wording of the
problem. You begin by distributing the counters in three
rows of three each, forming a square, and then place the
remaining counter on the centre one. You have now four
rows of four each ; but as each row can be counted in two
different directions— i.e., from right to left or left to right,
and vertical rows upwards or downwards — you are enabled
to count four in eight different directions, as required by the
problem.
No. XII.— Solution.
This is effected on a similar principle. You arrange nine of
the counters in three rows of three each, forming a square
as above described, and place the remaining four one on
each of the four " corner " counters. This gives you six
rows of five each, enabling you to count five in twelve
different directions.
No. XIII.— Solution.
The secret lies in working backwards throughout, each
time covering the point from which you last started. Thus,
placing a counter on a, draw it along the line a d, and leave
Key to Puzzles ivith Counters.
281
it on d, a is now the next point to be covered, and there
is only one vacant line, / a, which leads to it. Place, there
fore, your second connter on /, draw it along / a, and leave
it on a. The third counter must be placed on c, drawn
along c/, and left on/. The next placed on h, arid left on c.
The fifth is placed on e, and left on h. The sixth is placed
on 6, and left on e : and the seventh placed on g, and left
on b.
You now have the whole seven counters duly placed, and
only one point, g, left uncovered.
ls T o. XIV.— The "Okto" Puzzle. Solution.
The requirement in this case that each circle shall be
covered with a counter of a given colour does not in reality
add anything to the difficulty of the puzzle, though it
appears to do so to any one attempting it for the first time.
The experimenter has only to proceed as indicated in the
last solution, taking care in each case to use the counter
corresponding in colour with the circle on which he pro
poses to leave it,
No. XV.— Solution.
Fig. 404.
The counters must be placed as shown in Fig. 404,
282 Puzzles Old and New.
No. XVI.— Solution.
Supposing the row of counters to be indicated from left to
right by numbers as under —
1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Proceed as follows : —
Place 4 on 1, 6 on 9, 8 on 3, 2 on 5, and 10 on 7.
or
Place 4 on 1, 7 on 3, 5 on 9, 2 on 6, and 10 on 8.
There are several formulas which will answer the pro
blem, but a person trying the puzzle for the first time will
find some difficulty in hitting upon one of them.
No. xvil.— The "Right and Left" Puzzle, Solution.
The key to this puzzle lies in the observance of the
following rules : —
1. After having moved a counter, one of the opposite
colour must invariably be passed over it.
2. After having passed one counter over another, the
next advance will be with the same colour as such first
mentioned counter. The position will guide you whether
to move or to pass over, only one of such alternatives being
usually open to you.
The above rules, however, only apply up to a certain
point. After the ninth move you will find that the next
should be with a counter of the same colour ; but none such
is available. By this time, however, the puzzle is practically
solved. The counters are white and red alternately, with
the space to the extreme left vacant, and two or three ob
vious moves place the counters so as to answer the con
ditions of the problem.
Thus, if we begin with the white counters, the moves
will be as under (see Fig. 383), the spaces being designated
by the numbers, and the counters by the letters : —
1. D moves into space 4.
2. passes over D into spa.ce 5.
3. B moves into space 3.
4. D passes over B into space 2.
5. E passes over C into space 4.
6. .F moves into space 6.
7. G passes over F into space 7.
Key to Puzzles with Counters. 283
8. B passes over F into space 5.
9. A passes over D into space 3.
Here occurs the state of things to which we have referred
the position baing as in Fig. 405.
Fig. 405.
The next move should, according to the rule, be with a red
counter ; but there is only one counter, and that a white one,
D, which is capable of being moved in a forward direction,
and that only into 1. This move is made accordingly, and
the solution proceeds as follows, the remaining moves being
almost a matter of course : —
10. D moves into space 1.
11. F passes over A into space 2.
12. F passes over B into space 4.
13. B moves into space 6.
14. A passes over F into space 5.
15. F moves into space 3, and the trick is done.
If the operator prefers to begin with the red counters, the
moves will be as follows : —
1. G moves into 4.
2. J) passes over into 3.
3. E moves into 5.
4. G passes over into 6.
5. B passes over into 4.
6. A moves into 2.
7. D passes over into 1.
8. F passes over into 3.
9. F passes over into 5.
(From this point, as before, the rule ceases to apply.)
10. G moves into 7.
11. B passes over into 6.
12. A passes over into 4.
13. F moves into 2.
«14. F passes over into 3.
15. A moves into 5.
284 Puzzles Old and Neiv.
When the principle is once fairly mastered, the movements
can be executed with great rapidity, and with little fear of
any onlooker being able to repeat them from recollection.
The solution we have given is equally applicable to any
larger (even) number of counters, so long as the number of
spaces be one greater, and a vacant space be left in the
middle.
No. XVIII.— Solution.
You first deal with the middle row (15 to 21) after the
manner described in the last solution. You then move the
white counter now occupying space 25 into the central space
(18), and deal in like manner with the fourth row (22 to 28),
leaving space 25 vacant. Pass the counter occupying 11
into this space, and you are then in a position to deal with
the second row (8 to 14). When space 11 is again vacant,
move the counter occupying space 4 into it, and you are
then enabled to deal with the uppermost line (1 to 7).
Pass the counter occupying space 18 into space 4, and
that occupying space 32 into space 18. You are now in a
position to rearrange the last row (29 to 35). You have then
a vacant space (32) in the centre of the bottom row. Move
the counter occupying space 25 into this space, then pass
that occupying 11 into 25, and finally move the counter now
in 18 into 11.
No. XIX.— The "Four and Four" Puzzle, Solution.
The necessary transpositions are as follows : —
Shift the counters occupying spaces 2 and 3 to 9 and 10 ;
„ „ „ ,, 5 and 6 „ 2 and 3;
„ „ ,, ,, 8 and 9 „ 5 and 6;
„ „ „ „ 1 and 2 „ 8 and 9.
To work the counters back again, you have merely to
reverse the process, but to do this from memory is rather
more difficult than the original puzzle, and some amount of
practice is necessary before it can be done with facility.
No. XX.— The "Five and Five" Puzzle. Solution.
For the sake of brevity, we will distinguish the red and
black counters by the letters r and b respectively. They
will then stand at the outset as under : —
Key to Puzzles with Cottnters. 285
brbrbrbrhr . .
Position
after
1st move :
be . rbrbrbrr'b
55
55
2nd „
b b r r b r . . b r r b
55
55
3rd „
b b r . . r r b b r r b
55
55
4th „
bbrrrrrbb. . b
55
55
5th „
. . rrrrrbbbbb
N.B. — Where the number of pairs is odd, the second move
should be with the pair next following that in the centre.
No. XXI.— The "Six and Six" Puzzle. Solution.
Distinguishing the counters as before, we have at the
outset : —
brbrbrbr
b . . r b
b b
Position after 1st move
2nd
3rd
4th
5th
6th
55
5)
55
55
55
b b
b b
b b
r r .
r r r
r r r
r r r
r r r
r b
. b
b b
r
r
r
. . r
r r r
r r r
r r
r r
r r
r r
b b
No. XXII.— The ThirtySix Puzzle Solution.
\& d)) @) • ®) *
Fig. 406.
The six counters are so removed as to leave the re
mainder as under (Fig. 406).
No. XXIII.— The "Five to Four" Puzzle. Solution.
Remove the five counters of either diagonal, when those
remaining will be found to answer the conditions of the
problem.
286
Puzzles Old and New.
No. XXIV.— No Two in a Row. Solution.
Fig. 407.
Arrange the eight draughtsmen, or counters, as shown in
Fig. 407. '
No. XXV.— The " Simple " Puzzle. Solution.
L_
 : :
■
,/
w
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_/_v_
•
.
:
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J
Fig. 403.
The arrangement of the nine counters is as shown in
Fig. 403.
Key to Puzzles with Counters.
287
No. XXVI.— The "English Sixteen" Puzzle. Solution.
We know of no rule for working this puzzle. There are
several possible solutions. Among others, moving the men
in the following order will be found to answer the conditions
of the problem. The man to be moved is in each case indi
cated by the number of the square (see Fig. 389). It is not
necessary to specify the square to which it is to be moved.
As there is never more than one square vacant, the experi
menter cannot well go wrong in this particular.
It will be observed that the number of moves is 52, which
we believe to be the smallest which will suffice to transfer
the whole of the men.
11, 7, 9, 8, 10, 13, 11, 14, 9, 6, 8, 5,
7,
11, 9, 10, 8, 2, 1, 6, 3, 5, 7, 4, 9, 12,
15, 17, 14, 16, 13, 15, 11, 7, 9, 14, 11, 13, 10,
8, 9, 6, 8, 2, 5, 7, 11, 9, 12, 10, 8, 9.
No. XXVII.— "The Twenty Counters. 5 ' Solution.
(a) If the counters be arranged as shown in Fig. 409, it
will be found that they form seventeen perfect squares.
■—m m : ^
© © <§ ©
©•<D
Fig. 409.
Fig. 410.
(b) Fig. 410 shows the same arrangement of counters, less
six removed. Not a single square now remains.
CHAPTER VIL
PUZZLES WITH LUCIFER MATCHES.
There are many puzzles, of various degrees of merit, per
formed with the aid of lucifer matches. We ajDpend a brief
selection : —
No. I.
With eleven matches, Required, so to place them as to
make nine of them.
No. II.
With nine matches. Required, so to place them as to
make three dozen of them.
No. IIL
With nine matches. Required, so to place them as to
make three and a half dozen of them.
No. IV.
With three matches. Required, so to lay them on the
table as to make four of them.
No. V.
With three matches. Required, so to lay them on the
table as to make six of them.
No. VI.
With three wine glasses and three matches. Required,
with the three matches to form a bridge between the three
wineglasses, strong enough to support a fourth wineglass.
N.B. — Each match must rest on one glass only, and touch
such glass only at a single point.
2.8
Puzzles ivith Lucifer Matches.
289
No. VII.
With four wineglasses and four lucifer matches. Re
quired, so to place them as to form a bridge to support a
fourth wineglass, under the same conditions as in the last
puzzle.
No. VIII.
Four and twenty matches being arranged on the table so
=*'<=
©=
4
4
=#
Fig. 411,
as to form nine squares as in Fig. 411, required, to take
away eight matches and leave two squares only.
No. IX.
Having formed the two squares, as required by Puzzle
No. VIII., to form, with two matches only, a bridge from the
one to the other.
No. X.
Seventeen matches being laid on the table so as to form
f, :
&=
=8
^3«E=
dS
Fig. 412.
six equal squares as in Fig. 412, required, by taking away
five matches, to leave three squares only,
290
Puzzles Old and New.
No. XL
Seventeen matches being laid on the table so as to form
six equal squares (see last figure), required, by leaking
away six matches to leave two squares only.
No. XII.
Twelve matches being laid on the table so as to form four
equal squares (see Fig. 413), required, to remove and
Fig. 413.
replace four matches so as to form three squares only, of the
same size as at first.
No. XIII.
 Fifteen matches being laid on the table so as to form five
£ . B,
Fig. 414.
equal squares, as in Fig. 414, required, to remove three
matches so as to leave three such squares only.
No. XIV.
With five matches, to form two equilateral triangles.
No. XV.
With six matches, to form four triangles of equal size.
Pttzzles with Lticifer Matches. 291
No. XVI.
Four matches are here used. With a sharp penknife split
the upper end of one of them so as to form a notch, and
pare the end of another to a Avedge shape. Insert the wedge
into the notch, so that the two matches shall form an angle
Fig. 415.
of about 60°. With these two and a third match, placed so
as to lean against the point of juncture, form a tripod on the
table, as shown in Fig. 415.
The puzzle is to lift these three simultaneously with the
end of the fourth match. One hand only must be used.
No. XVII.
Ten matches are here used. You are required so to arrange
them as to lift nine of them with the tenth, using one hand
only.
No. XVIII.— The Magnetised Matches.
Break eight or ten matches in half. Fill a saucer with
water, and when the liquid has had time to become quite
still, lay the nonphosphoric ends on its surface in a circle,
like the spokes of a wheel, with a vacant space of about an
inch diameter in the centre.
Required: to compel the pieces of wood, without touch
ing them, to move towards the common centre, and when
they have done so, to make them again separate, and move
outwards towards the edges of the saucer.
292 Puzzles Old and New.
No. XIX.— The Fifteen Matches Puzzle.
This is in form a game, but ifc may also be presented as a
puzzle.
Fifteen lucifer matches (or counters) are laid side by side
on the table. One player takes one end of the row, the
other player the other. Each takes it in turn to remove as
many as he pleases, not exceeding three.
The object of each player is to avoid being the one to
remove the last match. To all appearance it is an equal
chance which shall do so, but a player who knows the secret,
if opposed to a novice, can always compel his adversary to
take it.
How Is it done ?
KEY TO CHAPTER VII.
PUZZLES WITH LUCIFEE MATCHES.
No. I.— Solution.
The matches are placed so as to form the word NINE.
No. II.— Solution.
The matches are placed so as to form the number XXXVI,
the Roman equivalent for 36.
No. III.— Solution.
This puzzle is solved by placing three matches in one
heap, and six in another. Total, three and a halfdozen.
No. iV.Solution.
The matches are so placed as to form the number IV.
No. V.— Solution.
This is the same as the last, save that the V is in this
case placed first, making the number VI (six).
Two matches may be made five (V) in like manner.
No. VI.— Solution.
The three matches are interlaced as shown in Fig. 416,
one resting on the. brim of each wineglass. The superin
cumbent weight binds them together, so that they will sustain
a fourth wineglass without difficulty.
293
294
Puzzles Old and New.
Tobacco pipes (long clays) are sometimes used instead of
Fig. 416.
lucifer matches, and made to support a tankard, with, even
better effect.
No. Vii.Solution.
'1
e^
Fig. 417.
This is on the same principle as No. VI., the arrange
ment being as shown in Fig. 417.
Key to Puzzles with Lucifer Matches. 295
No, YIIL— Solution.
Take away the matches forming the inner sides of the
*F=
y_;
o „■ ) ,
=0
a©
Fig. 418.
four corner squares, when you will have left two squares
only, the one in the centre of the other, as in Fig. 418.
No, IX.— Solution,
It will be found that the distance between the external
and inner squares is too great to be bridged by the length of
Fig. 419.
a single match. To construct the bridge, place one match
across one of the corners of the larger square, and a second
at right angles to it, resting on the corresponding corner of
the smaller square, as in Fig. 419.
296
Ptizzles Old and New.
No. X.— Solution.
Take away the two matches forming each of the upper
corners, and the centre match of the lower side. This will
leave three squares only, as in Fig. 420.
No. XI.— Solution.
Take away the four matches forming the inner sides of
the four squares to the left, and the two matches forming
Fig. 421.
the outer sides of the lower square to the right. You will
then have only two squares leffc ? a larger and a smaller, as
in Fig. 421.
No. XII.— Solution.
Take away the two matches forming the outer sides of
the upper right hand square, and the two forming the outer
sides of the lower left hand square. You have then lef(
two squares, lying diagonally. With the four matches yon
have removed, form a third, square in continuation of the
Key to Pussies zvitk Lttcifer Matches. 297
Fig. 422.
diagonal line, when you will have three squares, as shown in
Fig. 422.
No. XHT.Solution.
You remove the centre match of the upper side of the
Fig. 423.
figure, and the two matches forming the outer sides of the
lower square to the right, as in Fig. 423.
No. XIV.— Solution.
Arrange the five matches as shown in Fig. 424.*
* This is all but selfevident, and the puzzle would not have been
worth insertion but for the fact that it forms an appropriate "lead" to
the really clever puzzle which next follows.
TI13 illustration is not quite exact. The ends of the two matches on
either side should be in contact with those of the one laid transversely.
298
Puzzles Old and New.
Fig. 424.
No. XV.— Solution.
Place three of the matches on the table in the form of
a triangle, and hold the remaining three above them so as to
Fig. 425.
form a triangular pyramid, as shown in Fig. 425, a b c
representing the base, and d the apex.
No: XVI.— Solution.
Fig. 42G.
Take the fonrth match in yonr hand, and with its point
gently raise the two joined matches to a slightly more
Key to Puzzles with Lucifer Matches. 299
vertical position, so that the upper end of the third match
shall fall forward into the angle of the other two, as shown in
Fig. 426. By slightly raising the fourth match you lock all
three together, and they may be lifted without difficulty.
No. XVII.— Solution.
Lay one match, which we will call a, on the table, and
eight others across it, on alternate sides, with the heads
inwards, as shown in Fig. 427. Lay the last match, &, in
the furrow formed by the intersection of the eight crossed
matches. Now take hold of the end of a, and you may lift
the whole, as shown in Fig. 428, the one last placed form
Fig. 427.
Fig. 428.
ing a wedge between the upper ends of the eight suspended
matches, and so holding them together.
The matches used should be of wood, not wax, and fairly
large. It will be found a good plan to break off the head of
b, which otherwise projects inconveniently to the one side
or the other, and is now and then found to tip up, and
disturb the operation.
No. XVIII.— The Magnetised Matches. Solution.
This is a puzzle of a scientific character. To make the
matches gather in the centre, take a slip of blottingpaper,
three inches wide by twelve or fifteen long\ and roll it
loosely into a solid cylinder, in diameter about as large as
a sixpence. Hold this vertically aboA r e the saucer, letting
its lower end just touch the surface of the water. As
300 Puzzles Old and New.
the water rises by capillary attraction in the roll of
blottingpaper, a minute current is created moving from
the centre towards it, and the little pieces of lucifer match
consequently begin also to move in that direction. A good
sized lump of sugar may be substituted for the roll of
blottingpaper.
To cause the matches to move away from the centre, you
have only to proceed in the same way with a piece of soap
cut into a cylindrical shape, letting one end touch the water
in the centre of the group of matches, when they will forth
with scatter in all directions.
This new phenomenon, which has quite a magical appear
ance, depends upon what is known as the " surface tension "
of the water, which is disturbed by the introduction of the
soap.
"We have not space to go more minutely into the rationale
of this very curious experiment, but any reader who is in
terested in the subject will find full information in the new
edition of Chambers Encyclopedia. Titles — Capillarity and
Surface Tension.
ffo. XIX.— The Fifteen Matches Puzzle. Solution.
Victory will always lie with the player who removes the
tenth match, leaving five on the table. Thus suppose A and
B to be the players, and B to have the move, five matches
being left. If B now removes
1, A removes 3 ;
2,^1 „ 2;
3, A \ „ 1 ;
in each case leaving the last to be removed by B.
After one or two trials, the novice will probably perceive
that five is the critical remainder, and will endeavour to leave
that number. To prevent his doing so, his adversary must
so play in the earlier stages of the game as to leave nine
matches, when it will be equally impossible for the novice
to leave five, for, again
if B play 1, A will play 3 ;
if B play 2, A will play 2 ;
if B play 3, A will play 1 ;
and five will be left, with B to play.
Key to Puzzles with Lucifer Matches. 30 1
In like manner, to make sure of leaving nine, the adept
plays in the first instance so as to leave thirteen— i.e., if he is
the first to play, he removes two.
As between two players both of whom know the secret,
the first mnst necessarily win,
CHAPTER VIII,
WIRE PUZZLES.
One of the simplest of these is :—
No. I.— The United Hearts.
We have here (see Fig. 429) tlje representation, in
Fig. 429.
eopper wire, of a couple of hearts, the one interlaced within
the other. The problem is to separate them.
No. II.— The Triangle.
The wire is here bent into a triangle, or rather succession
of triangles, the one within the other, and terminates on the
outside in a ring passing round the next adjacent portion of
the wire (see Fig. 430). From the triangle hangs a long
wire loop, like a lady's hairpin, save that it is secured by a
sort of crossbar at the opposite end.
The puzzle is to detach the loop from the triangle.
Wire Ptizzles.
;o3
Fig. 430.
No. III.— The Snake and Ring.
The puzzle to which, this name is given consists of a
spiral coil of wire securely fastened off at the two ends.
Threaded on the wire is a brass ring (see Fig. 431), and the
problem is to disengage the ring from the spiral.
Fig, 431.
This is one of the simplest, and at the same time cleverest
puzzles of the wire series. To any one taking up the puzzle
for the first time its solution seems an impossibility, and yet
any one in the secret can disengage the ring in a single
second.
304
Puzzles Old and New.
No. iv. —The Hieroglyph.
The puzzle to which,, on account of its quaint shape, we
Fig. 432.
have given the above name is as depicted in Fig, 432.
The problem :j^to disengage the ring.
No, V.— The Interlaced Triangles,
Fig. 433.
The wire is in this case so manipulated as to form five
Wire Puzzles.
305
triangles, four of them in pair3, lying one above the other ;
the fifth of smaller size, connected with the larger triangles
by a tiny ring, and forming a "stop " to a larger ring, which
it is the " crux " of the puzzle to disengage (see Fig. 433).
No. VI.— The Double Bow and Ring.
This is a puzzle of especially simple appearance, consist
ing merely of two bows of wire united in such manner as to
form the shape of an hourglass, with a ring encircling its
narrower portion, (See Fig. 434=)
Fig. 434.
Although so simple in its elements, the usual problem
(the removal of the ring) will be found by no means easy
of solution until the secret is known.
No. VII.— The Egyptian Mystery.
Fig. 435.
The puzzle to which this highsounding title is given is of
,o6
Puzzles Old and New.
the same class, and a very good one of its kind,
ance it is as depicted in Fig. 435.
The problem is to disengage the ring.
In appear
No. VIII. —The Ball and Spiral.
This, shown in Fig. 436, is another very clever puzzle
Fig. 436.
The ball is permanently attached to the ring, and the latter
threaded on the spiral. The experimenter is required to
detach the ring from the spiral.
No. IX.— The Unionist Puzzle.
The puzzle to which this name is given consists of two
pieces of stout iron or brass wire, about jthree inches in
length and 5 of an inch in diameter, bent into horse shoe
Fig. 437.
shape, the ends being bent back on themselves with the
points slightly converging. These are interlaced as shown
in Fig. 437, and the puzzle is to disengage them. To do so
looks as it were the simplest thing in the world, but the
experimenter moves the two horseshoes this' way and that
Wire Puzzles. 307
way to no purpose, till suddenly, just as he is about to
"give it up," he hits by a happy accident on the right
position, and they come apart in his hands. He unites them
again readily enough, for it is a peculiarity of this class of
puzzles that their parts unite with delusive ease, their proper
position for this purpose being generally obvious. Again he
tries to separate them, flattering himself that this time he
will have little trouble in doing so, but, to his disgust, he
finds himself as far off as ever ; one position being, to the
uninstructed eye, pretty much the same as another, while
one only, of a score of possible positions, will admit of the
iesired release.
It is, of course, an understood thing that no force is to be
used.
No. X.— The Eastern Question.
Two pieces of stout wire are bent as shown in Fig. 438,
each forming an open ring, or segment of a spiral, with a
Fig. 438.
short straight piece by way of handle. They are interlaced
like the horseshoes, and the puzzle is to separate them.
No, XI.— The Handcuff Puzzle.
We have here (see Fig. 439) four rings, each forming a
circular segment of a closecoiled spiral, interlaced the one
with another.
:oS
Puzzles Old and New.
To solve the puzzle, all four must be disconnected.
Fig. 439.
No. XII.— The Stanley Puzzle.
The Stanley Puzzle (Perry & Co.), though not consisting
entirely of wire, belongs to this class. We have, in the first
place, a little medallion of stamped brass, bearing the pre
Fig. 440,
sentment of the celebrated explorer. From this (see Fig.
440) depends a wire loop, and from this again another piece
of stamped brass, narrow at the top, but widening towards
its lower end. Over this hangs a ring, which it is the prob
lem of the puzzle to remove.
KEY TO CHAPTER VIII.
WIRE PUZZLES.
No. I.— The United Hearts. Solution.
Pass the loop a of the lefthand heart (see Fig. 429)
through the ring* b of the righthand heart, and over the
ring c, in the direction shown by the arrow, when the two
will come apart. They may be joined again by reversing
the process,
No. II.— The Triangle. Solution.
Push the point of the long loop through the ring a (see Fig.
Fig. 441.
441), then pass it over the angles h, c, and d, in succession ;
this done, it may be drawn out clear through a. To re
engage it, reverse the process. 
No. III.— The Snake and Ring*. Solution.
Give the ring a quarterturn from left to right (it will be
found that it will only twist in one direction) ; and though
309
IO
Ptizzles Old and New.
its relation to the spiral is apparently unaltered, it will bo
found that it is now quite free, the condition of things being
as if the spiral had been simply passed through the ring,
and the latter then allowed to drop into its folds.
To again secure it, give it a turn in the opposite direc
tion.
Why so simple a movement should produce so curious a
result is, we confess, a mystery to ourselves. Perhaps some
of our readers will be more successful in finding a scientific
reason for it.
No. IV.— The Hieroglyph. Solution.
Hold the puzzle by the diamondshaped handle, and
with the opposite hand raise what, for want of a better
name, we may call the " triangle " up the perpendicular
loop, as shown in Fig. 442. It will be found that the
triangle does not work rigidly on the loop, but that either
side of it may in turn be brought close to the loop. Avail
Fig. 442.
ing yourself of this fact, work the ring down the lefthand
sides of the triangle and loop simultaneously in the direc
tion shown by the arrow. When it reaches the bottom of the
loop, push the triangle over to the left, and thencework the
ring in like manner up their right sides. When it reaches
the top of the loop on the righthand side it will be free.
No. y._ The Interlaced Triangles. Solution.
Hold the puzzle by the corner a (Fig. 433), and fold back
the upper part behind the lower. Push the movable
triangle (with the ring upon it) to the top of the upright
from which it hangs, and then pass the ring along down the
Key to Wire Puzzles.
1 1
two uprights now folded together, and up the two slanting
portions to a. Again open the puzzle, and push the ring
along the two horizontal wires to b. ]?old the puzzle again,
and slide the ring down along the double wires to c, then
upwards in a perpendicular direction, and it will be free.
To put it on again, repeat the process in the opposite
direction.
No. VI.— The Double Bow and Ring*. Solution.
Bend the two bows together, so that the one shall partially
cross the other, as shown in Fig. 443. Get the ring be
Fig. 413.
tween the two joints, pull it downwards over the two inner
wires, and the trick is done.
No. vil.— The Egyptian Mystery. Solution.
The secret here lies in the proper manipulation of the move
Fig. 444.
able loop a (see Fig. 435). Get this into the position shown
in Fio\ 444. Then fold the puzzle in half; lift the ring so
312 Puzzles Old and New.
as to bring 1 it over the two curved ends on the opposite
side ; work it up to the centre, down the opposite curve and
up again to the top, a in its present position offering no
obstacle to your doing so. When the ring reaches the top
at the lefthand side, it is free,
No. VllL— The Ball and Spiral. Solution.
Twist the ball up to the top of the spiral. When it
reaches this point, turn it over, so that the ring is no
longer engaged in the turns of the spiral, but lies between
them, encircling the upright wire in the centre. Now work
it downwards round and round in the opposite direction to
that by which you brought it up. When you reach the
bottom, pass the ring over the loop of the upright wire, and
the ball will be free.
No. IX— The Two Horseshoes. Solution.
To separate the horseshoes, hold them between the finger
and thumb of the right hand in manner shown in Fig. 445.
To get the right positions, note that both horseshoes have
the loop end downwards, but that the hook ends of the upper
one are turned towards the experimenter,, while those
Fig. 415. Fig. 446.
of the lower are turned away from him, and that the dis
engfag'ed arm of the lower horseshoe is to the left. With the
forefinger and thumb of the left hand take hold of the loop
of the lower horseshoe, slide it up the arm of the upper
horseshoe, at the same time turning it over till it assumes
the position shown in Fig. 446. Now move it .vertically
downward, and the two horseshoes are free.
Key to Wire Puzzles. 31
o L o
To reunite them, you have only to pass the points of the
one horseshoe between the points of the other, give them
a shake, so as to make them assume a casual position, and
the difficulty of separation is as great as ever.
No. X.— The Eastern Question Puzzle. Solution.
To separate the rings, get them into the position de
picted in Fig. 447, when they can be drawn apart with the
Fig. 447.
greatest ease, the one curved end sliding smoothly past the
other. To reunite them, hold them in the same position
and reverse the movement.
No. XI.— The Handcuff Puzzle. Solution.
To separate the first or outer pair, get them with the cut
ends in the same straight line, when they may be drawn apart
with ease. The other links are united on the same principle
as those of the Eastern Question Puzzle, last described. By
studying the instructions given in that case, and adapting
them to the present, success should be merely a question of
a little perseverance.
No. XII.— The Stanley Puzzle. Solution.
To get the ring off, fold the loop backwards against the
Stanley medallion, as shown in Fig. 448. Then push the
ring upward into the horseshoeshaped space, and then work
the outer portion of its circumference round the medallion
from left to right, till it reaches the position shown in the
diagram, When it reaches this point, it is free.
14
Puzzles Old arid New.
Fig. 448.
The solution seems almost childishly simple when known,
but the puzzle will, notwithstanding, give a good deal of
trouble to any one trying it for the first time.
CHAPTER IX,
"QUIJ3£^e" OR "catch" puzzles.
The present Chapter will be devoted to puzzles which de
pend upon some double meaning or nonnatural interpreta
tion of the question, whereby it assumes a second and less
obvious signification.
No. I.— A Remarkable Division.
A gentleman divided seven and sixpence between two
fathers and two sons, each father and each son receiving
half a crown. How did he manage it ?
No. II.— Subtraction Extraordinary.
Required, to take one from nineteen and leave twenty.
How is it to be done ?
No. ill.— Two Halves Greater than the Whole.
Prove that seven is the half of twelve.
No. IV.— A Distinction and a Difference.
What is the difference between twice twentyfive and
twice five and twenty ?
No. Y.— The Family Party.
A family gathering included 1 grandfather, 1 grandmother,
2 fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother,
2 sisters, 2 sons, 2 daughters, 1 fatherinlaw, 1 motherin
law, and 1 daughterinlaw, and yet there were only seven
persons present.
How can the two statements be reconciled ?
3 t 6 Puzzles Old and Nezv.
No. VI.— A Sum in Subtraction.
What is the difference between six dozen dozen and half a
dozen dozen ?
No. VII.— Another Sum in Subtraction.
What is the difference in capacity between twenty four
quart bottles and fourandtwenty quart bottles ?
No. VIII.— Three Times Six.
Place three sixes together so as to make seven.
o
No. IX.— A New Way of Writing* 100.
Required, to express 100 by repetition of the same figuro
six times over.
No. X.— A Seeming" Impossibility*
Required, to find six times thirteen in twelve.
No. XI.— Multiplication Extraordinary.
What three figures, multiplied by five, will make six ?
No. XII.— A Question in Notation.
How would you write in figures twelve thousand twelve
hundred and twelve ?
No. XIII.— The Miraculous Herrings.
Five herrings were divided among five persons. Each
had a herring, and yet one remained in the dish.
How was this managed ?
o
No. XIV.— Two Evens make an Odd.
Prove that two sixes make eleven.
"Quibble" or "Catch" Puzzles. 317
No. XV.— Six made Three.
Out of six chalk or pencil strokes — thus, 
to make three, without striking out or rubbing out any,
No. XVI.— A Singular Subtraction.
Required, to take ten from ten so that ten shall remain,
No. XVII.— A Sum in Addition.
Required, to add 2 to 191 and make the total less than 20.
No. XVIII.— The Flying* Sixpence.
A sixpence being placed in each hand, and the arms ex
tended shoulder high, required, to bring both coins into,
one hand without allowing the arms to approach each other.
No. XIX.— The Last Thing Out.
You undertake to show another person something which
you never saw before, which he never saw before, and
which, after you both have seen it, no one else will ever see
again.
How is it to be done ?
No. XX.— The Three Gingerbread Nuts.
This is propounded in the shape of a conjuring trick,
usually after two or three bondfide tricks have been per
formed. You place three gingerbread nuts* on the table,
and cover each with a borrowed hat. You make a great
point of having nothing concealed in your hands, and pro
fess your willingness to allow the audience, if they please, to
mark the three articles, so that there can be no question of
substitution.
You then take up each hat in succession^ pick up the nut
beneath it, and gravely eat it, replacing the hat mouth down
ward on the table. Any  one is at liberty to see that there
is nothing left under either hat. You then undertake to
* Tn default of gingerbread nuts, three almonds, raisins, or any other
small eatable articles may be substituted. '_
o
1 8 Puzzles Old and New.
bring the three nuts under whichever of the three hats the
company may select; and the choice being made, you at
once do so.
How is it to be done ?
No. XXI.— The Mysterious Obstacle.
You undertake to clasp a person's hands in such manner
that he cannot leave the room without unclasping them.
How is it to be done ?
No. XXIL— The Bewitched Right Hand.
You undertake to put something into a person's left hand
which he cannot possibly take in his right.
How is it to be done ?
No. XXIIL— The Invisible Candle.
You undertake to place a lighted candle in such a position
that it shall be visible to every person save one ; such person
not to be blindfolded, or prevented from turning about in any
manner he pleases.
How is it to be done ?
No. xxiv.— The Draper's Puzzle.
A draper, dividing a piece of cloth into yard lengths,
found that he cut off one yard per second. The piece of cloth
was 60 yards in length.
How long did it take him to cut up the whole ?
No. XXV.— The Portrait.
A portrait hung in a gentleman's library. He was asked
whom it represented. He replied, —
"Uncles and brothers have I none,
But that man's father is my father's son."
"What relation was the subject of the portrait to the
speaker ?
20 Puzzles Old and New.
No. XXXII.— The Mysterious Addition,
1. I add one to five, and make it four.
How can that be P
2. What must I add to nine to make it six ?
No. XXXIII.— Arithmetical Enigma.
From a number that's odd cut off its head.
It then will even be ;
Its tail, I pray, next take away,
Your mother then you'll see.
Ho. XXXIV.— A New Valuation.
If five times four are thirty three,
What will the fourth of twenty be ?
No. XXXV.— Easy, when You Know It.
What two numbers multiplied together will produce
seven ?
No. XXXVI.— Necessity the Mother of Invention.
I have a bottle of wine, corked in the ordinary way.
Unfortunately, I have no corkscrew.
How can I get the wine out, without breaking the glass, or
making a hole in the cork ?
No. XXXVII.— A Singular Subtraction.
From six, take nine, from nine take ten, from forty take
fifty, and yet have six left.
How is it to be done ?
No. XXXVIII.— A Vanishing Number.
There is a number of ihree figures, in value not very far
short of a thousand, but when halved its value is nothing.
What is it ?
No. XXXI.— A Short Year.
The year 1892 was one of the shortest on record.
How do you prove it ?
"Quibble" or "Catch" Puzzles. 31
No. xxvi.— The Charmed Circle.
You invite a gentleman to stand in the middle of th
room. Taking a piece of chalk, you undertake to draw round
him a circle which he cannot jump out of.
How is it to be done ?
No. XXVII— The Egg and the Cannonball.
Exhibiting an egg and a cannonball, you hold forth
learnedly on the extraordinary strength of a perfect arch,
and, still more, of a perfect dome, remarking that few
people know how strong even the shell of an egg is, if it is
placed in a proper position. In proof of your assertion, you
Undertake to place the egg, without covering it in any way,
in such a position that no one present can break it with the
cannonball.
How is it to be done ? \
No. xxviii.— A Curious Window,
A window in a certain house has recently been made
twice its original size, but without increasing either its
height or width.
How can that be ?
No. xxix— A Queer Calculation.
A hundred and one by fifty divide,
And next let a cipher be duly applied ;
Then, if the result you should rightly divine,
You'll find that the whole makes but one out of nine.
No. XXX.— Arithmetical Enigma.
Write down a cipher, prefix fifty, to the right place five,
and to the whole add onefifth of eight. The result will
give you the most important factor in human happiness.
i
V
"Quibble" or '" Catch" Ptizzles. %2\
No. XXXIX.— A Queer Query.
Twice ten are six of us,
Six are but three of us,
Nine are but four of us ;
What can we possibly be ?
Would you know more of us,
Twelve are but six of us,
Five are but four, do you see ?
No. XL.— The Mouse.
A mouse found in a box a number of ears of corn, and
set to work to carry them off to his hole. He brought out
with him three ears at each journey, and it took him nine
journeys to remove the whole.
How many ears of corn were there in the box ?
No. XLL— The Fasting Man.
How many hardboiled eggs can a hungry man eat on an
empty stomach ?
No. XLIL— The Family Party,
An old gentleman was asked who dined with hirn on
Christmas Day. " Well, we were quite a family party," he re
plied ; " there was my father's brotherinlaw, my brother's
fatherinlaw, my fatherinlaw's brotherinlaw, and my
brotherinlaw's fatherinlaw."
It afterwards transpired that he had dined alone, and yet
his statement was correct.
How could that be ?
No. XLIIL— A Reversible Fraction.
Required, to find a fraction whose numerator is less than
its denominator, but which, reversed, shall remain of the
same value.
Y
322 Puzzles Old and New.
No. XLIV.— The Three Counters.
Three counters are laid in a row on the table.
Required, to take the middle one away from the middle
without touching it.
No. xlv.— Magic Made Easy.
Borrow a halfcrown and a penny, and hold them one in
each hand, with the hands open, in front of you, the hands
being about two feet apart. Now close the hands, and
announce that you will make the coins change places with
out again opening your hands, which you proceed to do
accordingly.
How is it done ?
KEY TO CHAPTER IX.
" QUIBBLE " OR " CATCH " PUZZLES.
No. I.— A Remarkable Division. Solution.
There were only three persons who shared in the gift",
related to each other as son, father, and grandfather. Each
is necessarily a son (of somebody), while the two elder are
fathers also.
No. II.— Subtraction Extraordinary. Solution.
Write nineteen in Roman numerals — XIX. Remove the
I., and you have XX.
No. III.— Two Halves Greater than the V/hole.
Solution.
Write twelve in Roman numerals — XII. Halve the num
ber by drawing a line horizontally across its centre, and the
upper half is VII.
No. IV.— A Distinction and a Difference. Solution.
There is a difference of twenty ; twice twentyfive being
fifty, while twice five, and twenty, make thirty only.
No. V.— The Family Party. Solution.
The party consisted of three children (two girls and a boy),
eir father and mother, and their father's father and mother.
will be found that these, in their various relations to each
:her, fill all the characters named. Thus the father, in
elation to his own father, is also a son, and so on.
323
324 Puzzles Old and New.
No. VI.— A Sum in Subtraction, Solution.
The former are six gross, the latter six dozen only. The
difference is therefore 864 — 72 = 792.
No. VII.— Another Sum in Subtraction. Solution.
56 quarts : twenty fourquart bottles holding 80 quarts,
while four and twenty quart bottles hold 24 quarts only.
No. VIII.— Three times Six. Solution.
No. IX.— A New Way of Writing 100. Solution.
99f( = 100).
No. X.— A Seeming Impossibility. Solution.
If you write down the numbers 1 to 12 inclusive, taking in
pairs the first and last, the second 'and last but one, and so
on, you have —
1 and 12 mate 13
2 and 11 ,,
13
3 and 10 ,,
13
4 and 9 ,,
13
5 and 8 ,,
13
6 and 7 .;
13
Your six thirteens are thus accounted for.
No. XI.— Multiplication Extraordinary, Solution.
Answer, 1. 1} x 5 = 6.
No. XII.— A Question in Notation. Solution.
Answer, 13212.
No. XIII.— The Miraculous Herrings. Solution.
The last of the five received his herring in the dish.
Key to "Quibble" or "Catch" Puzzles. 325
No. XIV.— Two Evens make an Odd. Solution.
This is another of the "catches" dependent upon the use
of Roman numerals. One six (VI.) is placed above another
six, but the latter in an inverted position (Al)? the combina
tion making XI.
No. XV.— Six made Three. Solution.
Add the necessary lines to complete the word " three," thus
THREE.
No. XVI.— A Sum in Subtraction. Solution.
The propounder of this puzzle should be wearing gloves,
and the problem is solved by taking them off. The ten fin
gers of the gloves are taken from the ten fingers of the hand,
and the latter still remain.
No. XVII.— A Sum in Addition. Solution.
Draw a line nnder the final 1, and place the 2 under it,
when the result will be 19J.
No. XVIII— The Flying* Sixpence. Solution.
Place yourself so as to bring one hand just over the man
telpiece, and drop the coin contained in such hand upon
the latter. Then, keeping the arms still extended, tuim the
body round till the other hand comes over the coin. Pick it
up, and you have solved the puzzle, both coins being now in
one hand.
No. XIX.— The Last Thing" Out. Solution.
^he puzzle is solved by cracking a nut, showing your
locutorthe kernel, and then eating ifc.
C3
). XX.— The Three Gingerbread Nuts. Solution.
This is a very ancient " sell," but it still finds victims.
. ie performer's undertaking, is performed by simply putting
.1 the hat selected. No one can deny that the three nuts
,re thereby brought under the hat.
326 Puzzles Old and New.
No. XXI.— The Mysterious Obstacle. Solution.
You perform your undertaking by clasping the person's
hands round the leg of a loo table, a piano, or other object
too bulky to be dragged through the doorway.
No. XXII.— The Bewitched Right Hand. Solution.
You place in the person's left hand his own right ellcw,
which, obviously, he cannot take in his right hand.
No. XXIII.— The Invisible Candle. Solution.
You place the candlestick upon the head of the person who
is not to see it.
No. XXIV.— The Draper's Puzzle. Solution.
It took him 59 seconds. Most people are apt to say 60, for
getting that the 59 bh cut separates the last two lengths, and
that, therefore, a 60th is unnecessary.
No. XXV.— The Portrait. Solution.
The portrait represented the speaker's son, as will be seen
after a moment's consideration. The speaker says in effect,
"The father of that man is my father's son" ; in which case
the father of the subject must be either a brother of the
speaker, or himself. He has already told us that he has no
brother. He himself must therefore be the father, and the
portrait represents his son.*
No. XXVI.— The Charmed Circle. Solution.
The circle is drawn on the clothes of the victim, round the
w r aist.
* This venerable puzzle forms the subject of a humorous article, e. i
titled " Prove It," in a recent number of the Idler. Its most amusing
feature is that the writer has himself gone astray, the story proceeding
on the assumption that the speaker himself is the subject of the portrait,
and being based on the (by no means imaginary) difficulty of demon
strating that fact to other people.
Key to "Quibble" or "Catch" Puzzles. 327
No. XXVII.— The Egg and the Cannon Ball.
Solution.
Yon place the egg on the floor, in one corner of the room,
in which position the walls on either side make it impossible
to touch it with the cannon ball.
No. XXVIII.— A Curious Window. Solution.
The window was diamondshaped. By enlarging it to a
square its area is exactly doubled, without increasing either
its height or width.
A window shaped as an isosceles or rightangled triangle
will equally answer the conditions of the puzzle.
No. XXIX— Queer Calculations. Solution.
Take the Roman equivalent for 101 (CI), and divide the
two letters by inserting between them the equivalent for 50
(L). Add 0, and you have CLIO, one of the nine Muses.
No. XXX.— An Arithmetical Enigma. Solution.
This is on the same principle. L stands for 50, the cipher
for the letter 0, and V for 5, while E is onefifth of eight
(eig h t), the whole forming the word LOVE.
No. XXXI.— A Short Year. Solution.
It began on a Friday, and ended on Saturday.
No. XXXII.— The Mysterious Addition. Solution.
1. Write five in Roman characters (V) ; add I, and it
becomes IV.
2. The letter S, which makes IX SIX.
No. XXXIII.— An Arithmetical Enigma* Solution.
Seven — Even — Eve.
No. XXXIV.— A New Valuation. Solution.
8J.
328 P tizzies Old and New.
No. XXXV.— Easy, when You Know It. Solution.
Seven and one.
No. XXXVI.— Necessity the Mother of Invention.
Solution.
Push the cork in.
No. XXXVII.— A Singular Subtraction. Solution.
SIX IX XL
IX X L
SIX
No. XXXVIII.— The Vanishing 1 Number. Solution.
The number is 888. When halved it becomes §£§ = 0.
No. XXXIX.— A Queer Query. Solution.
This is a mere " sell." The answer is " Letters." In the
word "twenty" there are six letters, in the word "six"
three, and so on.
No. XL —The Mouse. Solution.
There were nine ears of corn in the box. The mouse
brought out three ears at each journey, but two of them
were his own.
No. XLL— The Fasting Man. Solution.
One only; for after eating one his stomach would no longer
be empty.
No. XLII.^The Family Party. Solution.
The very peculiar state of things described is accounted
for as follows. The old gentleman was a widower, with
a daughter and sister. The old gentleman and his father
(who was also a widower) married two sisters (the wife of
the old gentleman having a daughter by a former husband) ;
Key to " Quibble " or " Catch " Puzzles. 329
the old gentleman thus became his father's brotherinlaw.
The old gentleman's brother married the old gentleman's
stepdaughter ; thus the old gentleman became his brother's
fatherinlaw. The old gentleman's fatherinlaw married
the old gentleman's sister, and the old gentleman thus be
came his fatherinlaw's brotherinlaw. The old gentle
man's brotherinlaw married the old gentleman's daughter,
whereby the old gentleman became his brotherinlaw's
fatherinlaw. He therefore himself filled all the four
characters mentioned.
No. XLIII— A Reversible Fraction. Solution.
q. Turn the paper upside down, so as to bring the de
nominator into the place of the numerator, and vice versa.
The fraction will still be q.
No. XLIY.— The Three Counters. Solution.
Remove one of the end counters and transfer it to the
opposite end. You have not touched the middle counter,
but it is no longer in the middle,
No. XLV.— Magic Made Easy.
This puzzle, like that last described, depends on a double
meaning. The spectators naturally prepare themselves for
some more or less adroit feat of jugglery, but you perform
your undertaking by simply crossing the closed hands. The
right hand (and the coin in it) is now where the left was
previously, and vice versa.
CHAPTER X.
MISCELLANEOUS PUZZLES.
No. I.— The John Bull Political Puzzle.
The appliances for this puzzle, brought out by Messrs.
Jaques & Son, consist of a cardboard box, on the bottom of
which, are described three stronglymarked concentric
circles, with other fainter lines connecting tliem (see Figs.
449, 450), and nine small counters, three white, three red,
and three blue. Each group of three bears the letters, C, U,
Fig. 449.
Fig. 450.
and L, standing for Conservative, Unionist, and Liberal, re
spectively. Wherever on the board a thin line intersects a
thick one, the point of juncture is marked by a circular
" spot," a quarter of an inch in diameter. These are nine in
number, forming straight radial lines of three each. At one
corner of the board is a tenth spot (marked 10 in the figure),
connected with the main diagram by a faint curved line.
The nine counters are to be placed at the outset promis
cuously upon the nine spots, the outer one, 10, being at
starting unoccupied. The counters are then to be moved, one
at a time, along the thick or thin lines, into the spot which
330
Miscellaneous Puzzles.
1 > T
00 l
happens for the time being to be vacant, until all three
colours and all three letters are found in each circle and in
each row of spots.
The first move is, as a matter of course, to shift the nearest
counter into the 10 spot, thereby giving room to manipulate
the others. The last move will be to replace this counter in
its original position, leaving the 10 vacant as at first.
No. II.— The Pig in Sty.
This is a small board, 4 inches square, and marked as
shown in Fig. 451, each of the twentyfive circles represent
ing an opening a quarter of an inch deep. Of these, all save
the centre square are at the outset occupied by cylindrical
wooden pegs, of which the eight occupying the middle of the
Fig. 451.
board are lettered as shown, the various letters forming, as
will be observed, the words Pig in Sty. The peg occupying
the second hole in the bottom row bears the effigy of a pig,
the others being left blank.
The object to be achieved is to get the "Pig " into the
middle hole, representing the " Sty," and this is to be done
according to the rules following : A peg can only be moved
into the hole for the time being vacant. If moved along the
solid black lines, it must overleap two other pegs. If moved
along the dotted diagonals, it can only be moved into an
adjoining space, without overleaping. Subject to these two
rules, a peg may be moved in any direction.
When the puzzle is solved, the lettered pegs must again
332
Puzzles Old and New.
occupy their original positions. The pig must be in the sty,
and the hole he previously occupied must be vacant.
The necessary moves should not exceed twelve in number.
No. IILHide and Seek.*
This is a puzzle of a very novel and ingenious kind ; in
deed, we believe it to be unique. At any rate, we have come
across no other upon precisely the same principle.
It consists of a cardboard box, four inches square, with
the top and bottom of glass. The intermediate space is oc
cupied by a metal plate, divided by upright partitions (just
high enough to touch the glass) into a number of different
compartments, somewhat after the fashion of a maze (see
Fig. 452) . On turning the box over, we find that the under
Fig. 452.
side of the plate is divided after a similar fashion, but that
the shape of the compartments is in this case different, the
partitions on this side running in different directions. There
is no direct communication between the compartments of the
top, or between those of the bottom, each being fully en
closed on all sides; but the intermediate "floor" is perforated
with a number of holes, (in most cases two to each com
Perry & Co., Limited.
Miscellaneous Puzzles.
333
part ment), which form a means of communication between
the npper and the lower compartments, and therefore in
directly between compartments on the same side.
In the front of the box (see the diagram) will be seen a
small round hole. A little leaden ball, a, is introduced at this
point, and allowed to drop through the hole immediately in
front of the opening. It is then to be made to travel, up one
hole and down another, from compartment to compartment,
till it comes out again at the hole to the right of the open
ing, the box being turned over at each stage so as to enable
it to fall in the desired direction.
No. IV.— The Brahmin's Puzzle."
This very clever puzzle is professedly based on a Hindu
legend, to the following effect : —
At the beginning of the world, Brahma set up in the
great Temple of Benares three diamond pyramids. Round
the first of them he hung sixty four rings, made of purest
gold, and arranged in regular order, the largest ring en
Fig. 453.
circling the foot of the pyramid and the smallest its top.
And Brahma said unto the priests, " Transfer these sixty
four rings from the first pyramid to the third, transposing
one ring at a time only, and putting it either on a vacant
pyramid or on a larger ring. By the time jou have executed
this task the end of the world will be near."
As few persons would care to attempt a puzzle which pro
fessedly takes some thousands of years to solve, it has been
found necessary to modify the conditions of the problem, the
number of rings to be transposed being reduced from 64 to
* This puzzle also is published by Messrs. Ferry & Co. , Limited.
334 Puzzles Old and New.
8. Instead of gold, they are in this case discs of cardboard,
coloured alternately orange and black. The three diamond
cones are represented by three little wooden slabs (see Fig.
453), each with a cylindrical peg standing up in its centre.
The method of transposition is the same as laid down in the
legend.
No. V Cardan's Ring's.
We take this puzzle next in order, as having a close
affinity in principle with the problem of Brahma, which
precedes it. It is one of the oldest of known puzzles, having
been learnedly discussed by the mathematician, Jerome
Cardan, as early as the sixteenth century. Whether it was
his own invention is doubtful, but it was for many genera
tions associated with his name. At the present day people
have forgotten all about Cardan, and the problem is now
more frequently referred to by the less distinctive title of
li The Puzzling Rings." In French it is known as La
Baguenaudier, and it is said to be now and then found on an
enlarged scale in English rural districts, forged in iron, and
appropriately called " The Tiring Irons." It has more than
once been deemed worthy of notice by mathematicians, the
learned Savilian professor, Dr. Wallis, devoting to it a spe*
cial section of his treatise on Algebra (1685),' under the
title Be Gomjolicatis Annul is.
The apparatus consists of four parts : —
(1) A wire bow or shuttle (sometimes provided with a
handle at one end).
(2) A flat bar of wood, metal, or bone, a trifle larger than
the bow, with holes through it at regular intervals, corre
sponding in number with the rings.
(3) A number (six to twelve, as the case may be), of rings,
which should in internal diameter be just double the ex
ternal width, and in thickness onethird of the internal
width of the bow.
(4) A series of short wires, corresponding in numb or with
that of the rings.
One end of each wire passes through one of the holes in
the bar, and is rivetted on the opposite side, though the hole
is of such a size as to allow it free play. It thence passes
through the bow, and through one of the rings, and its oppo
Miscellaneous Puzzles. 335
site end is then bent round another ring, the result being
as shown in Fig. 454 (representing the 10ring form of the
puzzle). The rings are all threaded on the bow, each (with
one exception) passing around the wire of its righthand
Fig. 451.
neighbour. The exception is the ring to the extreme right,
which, having no neighbour on its outer side, enjoys a degree
of liberty not shared by the remaining rings. The puzzle
is to get the rings off the bow.
It should be mentioned that, as in the case of the Brahmin
Puzzle, each additional ring doubles the time occupied in the
solution. With seven rings, the puzzle requires 85 moves to
solve it; with eight rings, 170; with nine, 341; and with ten,
682. With eleven it would require 1,365 movements, and
with twelve, 2,730. Ten rings are the popular limit, and
we have therefore selected the puzzle in this form for illus
tration.*
No. VI.— The Knight's Tour.
Chess problems, in the ordinary sense, are interesting only
to the chessplayer, and would, therefore, be out of place in
the present pages, which are designed for all and sundry.
But there is one particular chesspuzzle, the socalled
Knight's Tour, which requires no knowledge of chess, and
may be attempted with success even by a person quite un
acquainted with the game.
We may take it for granted that every reader knows that
the chessboard consists of 64 squares (eight rows of eight
squares each, black and white alternately). Some readers
may, however, not be aware of the nature of the knight's
move. The knight at chess moves in a rather peculiar way
* The solution given in the Key will equally apply to the case of any
smaller number of rings, the only difference being that the process will
be cut short at an earlier stage.
Puzzles Old and New.
— viz., one square straight (either forward, backward, or
sideways), and one square diagonally (to right or left) from
the square thus reached, forming a sort of zigzag. Thus,
assuming the knight to be placed on the square marked K in
Fig. 455, he might be moved to either of those indicated by
an asterisk.
The problem known by the name of The Knight's Tour
is to move the knight from square to square of the boarjl in
such manner that he shall, in the course of 64 moves, have
rested (once and once only) on every square.
The experimenter is sometimes permitted to choose for
himself from which square he will start. Under stricter
conditions, he is required to start from a given square.
1
'S
*
w
K
#
1
#
*
Fia. 455.
It is sometimes also made a condition that he should finish
within a single move of the square from which he started.
Indeed, no solution is now regarded as " perfect" which does
not fulfil this requirement.
Before attempting to solve the puzzle, the reader is recom
mended, if using the actual chessboard, to provide himself
with a supply of small counters, and to place one by way^of
" mark " on each square to which he moves the knight, so
that there may be no doubt as to which squares have or have
not been visited.
Another method, preferred by some, is to use, instead of
the actual board, a piece of paper ruled in sixtyfour squares,
to represent a chessboard. Each move may then be noted
by simply drawing with a pencil a straight line from the
one square to the other.
Miscellaneous Puzzles. 337
No, VII.— The Knotted Handkerchief.
Required, to take a handkerchief, twisted ropewise, by its
opposite ends, and, without letting go of either end, to tie a
knot in the middle.
No. VIII.— Crossette,
Arrange in the form of a circle ten smaller circles (say
coins or counters), as shown in Fig. 456.
Starting from any circle you please, and calling such circle
1, the next 2, and so on, strike out the fourth. Then start
#
*
Fig. 456.
again from any circle you please, count 1, 2, 3, 4, and strike
out the fourth. Proceed as above until all but one have been
struck out.
You may count either backwards or forwards. Circles
already struck out are to be reckoned in counting, but the
count of "four " must in each case fall upon a circle not al
ready struck out.
This puzzle may be most conveniently worked with the aid
of "reversi" counters, which, as the reader is probably aware,
are red on the one side and black on the other. Ten of these
are arranged in a circle, with the red side uppermost, and as
each is " struck out " it is turned over, so as to bring the
black side uppermost.
,38
Puzzles Old and New.
No. IX.— SingleStroke Figures.
A good deal of ingenuity may be exercised in the attempt
to describe geometrical figures without taking, off the pencil,
or passing over any line for the second time.
C
Fig. 457.
Fig. 458.
Figs. 457 and 458 may be tried by way of examples.
From these other figures may be constructed, and tested in
like manner.
Fig. 459.
Fig. 459, the double crescent, or socalled Seal of Mahomet,
is another pretty example, the legend being that the pro
phet was accustomed to describe it with one stroke of his
scimitar, by no means a difficult feat, notwithstanding its
apparent complexity.
No. X— The Balanced Egg*. Another Method.
We have already (p. 72) described a method of performing
this feat with a mechanical egg, constructed for the purpose ;
but it is also quite possible to perform it with an ordinary
Miscellaneous Puzzles.
'>9
JO"
egg, and without having recourse to the somewhat heroic
expedient of the great navigator.
How is it to be done ?
No. XI.— Solitaire Problems.
Solitaire, though commonly referred to as a " game,"
belongs rather to the category of puzzles, the problems which
it affords being numerous and interesting. It is played
with a circular board, as shown in Tigs. 460, 461, with thirty
seven hemispherical depressions,* . in each of which rests a
small marble, or glass ball. One of these being removed from
the board, another is moved into the vacant space thus created,
but in so doing it must pass over one intervening ball, lying
Fig. 460.
Fig. 461.
in a straight line (not diagonally) between it and the hole.
This intervening ball is removed from the board (just as a
man is "taken "at draughts), and another move is then
made, after the same fashion, into one or other of the two
holes now left vacant, a fresh man being removed from the
board at each move.
For the third move there will be three holes vacant, for the
next four, and so on, one ball being removed from the board
at each move. The puzzle, in its simplest form, is to re
move all the balls save one, which last cannot be removed,
inasmuch as it has no second ball to pass over.
* Some Solitaire boards are made without the four corner holes, Nos.
4, 8, 30, and 34 in the diagram. In this case 33 balls only are used.
340
Puzzles Old and 'New.
Sometimes it is left optional in what part of the board
such last ball shall remain. In more elaborate forms of the
puzzle the player is required to leave one or more balls in
a particular hole or holes, previously determined.
Each alteration of the startingpoint makes a fresh pro
blem. We have selected three examples, viz : —
1. Starting with No. 1 as the vacant hole, to leave the
last ball in No. 37.
2. Starting with No. 19 (the centre hole) vacant, to leave
all the outer holes (1, 2, 3, 8, 15, 22, and so on) occupied, and
the last ball in the centre hole. (This is sometimes known
as " The Curate and his Flock.") To be done in 19 moves.
3. (The Triplets.) Starting with the centre hole vacant,
to leave holes 1, 3, 6, 9, 12, 15, 17, 18, 20, 21, 23, 26, 29, 32,
35, and 37 occupied, forming a geometrical figure. (To be
done in 20 moves).
By devising other geometrical figures, and endeavouring to
produce them according to the conditions laid down, a very
interesting series of puzzles may be produced.
No. XII.— Skihi.
This also is sold as a game, but comes more properly with
in the category of puzzles. It is a patent, and the property
of the Skihi Novelty Company, London, W.C.
The set consists of 48 square cards, 2 inches each way,
and of various colours. Each card has four slots cut in it,
Fig. 462.
Fig. 463.
as shown in Eig. 462. There are also 10 circular cards, each
with three slots, as shown in Eig. 463.
These cards may be utilized to form an almost unlimited
number of fanciful designs. We subjoin a few examples
(see Eigs. 464467), which will give some idea of the very
Miscellaneous Puzzles.
34*
Fig. 464.
Fig. 465.
Fig. 466.
FiGc 467.
342 Puzzles Old and New.
wide capabilities of tins clever toy. All of these may be
constructed with a single set. By using three or four sets
in conjunction, very much more ambitious designs may be
executed.
No. XIII.— A Card Puzzle.
Taking the four " fives " from a pack of cards, you are re
quired to arrange them, face upwards, in such manner that
only four pips of each shall be visible.
No. XIV.— Another Card Puzzle.
The aces and court cards being removed from the pack,
required to arrange them in four rows, in such manner that
neither horizontally nor perpendicularly shall there be two of
same rank or same suit in any one row.
No. XV.— The Floating Corks.
The requirements for this puzzle are seven wine corks (not
tapering but of cylindrical form) and a basin of water.
The experimenter is required, without weighting them in
any way, to make the seven corks float upright in the water.
No. xvi.— The Obstinate Cork.
For this puzzle a winebottle is required, and a cork a size
or two too small for the neck, so that if inserted in the ordi
nary way it would fall into the bottle.
The bottle being held horizontally, with the cork resting
just within the neck, the experimenter is invited to try
whether he can, by blowing, force it into the bottle. If he
does not know the secret, he will pretty certainly fail.
No. XVII.— Fixing the Ring.
This is described, with a touch of poetry, as a " matri
monial " puzzle. It is, in fact, a puzzle for two persons, who,
to give it the proper touch of sentiment, should be a lady
and a gentleman.
The appliances consist of a silken cord and a plain gold
ring. The lady holds the cord, and the gentleman the ring',
Miscellaneous Puzzles.
o<¥3
and they are required, by their joint efforts, to tie the cord in
a knot round the ring, each using one hand only.
No. XVIII.— The Treasure at Medinet.
This puzzle comes from Germany, but is said to be of
Oriental origin. The legend accompanying it is to the effect
that an Eastern prince, Haroun al Elim, in far back times,
ruled over a range of country with eight sugarloaf hills, on
each of which was erected a fortress. Each fortress, with
the surrounding district, was under the command of a
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Fig. 468.
governor, but the jealousies of the eight governors and their
respective under'iigs led to affrays and bloodshed whenever
they chanced to meet. To lessen the chance of such meet
ings, Haroun made a number of roads, eight crossing his
kingdom in one direction, eight more at right angles to them,
and others crossing diagonally. These were so arranged with
reference to the castles that the occupants of each castle
had a clear road in each direction through and out of the
prince's territory without passing any other castle.
The castles, says the legend, are now in ruins, and the roads
344 Puzzles Old and New.
no longer traceable ; but a plan of them is still preserved
among the archives of the Mosqne Al Redin, at Medinet, on
the coast of the Red Sea. Unfortunately, the plan, which
was folded in four, has been worn by age into four separate
fragments, and the ntmost skill of the Cadi of the mosque
has failed to discover their proper relative positions. He has
therefore offered a reward — a treasure of ancient jewellery,
preserved at Medinet — to anyone who may succeed in placing
the four fragments in their original positions — viz., with no
two castles on either road, either horizontal, perpendicular,
or diagonal.
For the use of the Infidel, the severed map has been re
produced on four separate cards, as A, B, C, and D, in Fig.
468.
No. XIX.— The Four WineGlasses.
Given, four wineglasses, of same shape and size.
Required, so to arrange them that the centre of the foot
of any one of them shall be equi distant from all the rest.
No. XX.— One Peg* to Fit Three Holes.
A brass plate (Fig. 469) has three openings, one circular,
one square, and one triangular. The experimenter is handed
a knife, and a cork which just passes through the circular
Fig. 469.
hole. He is required so to cut the cork that it shall exactly
fill any one of the three openings.
For lack of the brass plate, a piece of stiff cardboard may
be cut so as to answer the same purpose.
No. XXL— The Balanced Pencil.
Given, a lead pencil, and a penknife, with which you
sharpen the pencil to the finest possible point.
Required, to bilanoe the pencil in an upright, or nearly
upright, position on the tip of the forefinger.
Miscellaneous Pitzzles. 345
No. XXII.— To Balance an Egg* on the Point of a
WalkingStick.
We have already described a puzzle (No. X) in which,
an ordinary egg is made to stand upright, but in this case the
difficulty of the feat is enhanced by the fact that the egg is
to be balanced on the end of a walkingstick.
The articles employed are an egg, a cork, and a couple of
dinner forks.
.Required, to balance the egg, by the aid of the other three
articles, on the smaller end of the stick.
No. XXIII.— The Ashantee Horseshoe.
The requirements for this puzzle are a miniature horse
shoe of wood or cardboard, and a couple of pieces of stout
brass wire, of the same length as the horseshoe. By the aid
of one of these the horseshoe is propped up in a slanting
position,* as shown in Fig. 470.
Fig. 470.
The experimenter is required (without touching either with
the hands) to lift both wire and horseshoe simultaneously
with the second piece of wire.
No. XXIV.— A Feat of Dexterity.
Fill a wineglass to the brim with water, and place it on
the corner of a tablenapkin or pockethandkerchief spread
* The horseshoe should have a minute notch or depression on its
binder side, to receive the upper end of the wire, and so prevent
slipping.
346 Puzzles Old and New.
over the edge of a table,* the remaining part hanging down,
and being kept from falling by the weight of the glass.
Puzzle : to remove the handkerchief without touching
the glass or spilling any of the water.
No. XXV.— The Divided Square.
Given, a square of cardboard of, say, two inches each way.
Required, to divide it into five equal squares.
No. XXVLThe "Oval" Problem.
Given, a sheet of drawingpaper and a pair of ordinary
drawingcompasses.
Required, without any other aid, to describe an oval on
the paper.
No. xxvil.— The Floating Ball.
This is more of a game than a puzzle, though it partakes
of the nature of the latter.
A hollow rubberball, two inches in diameter, is set afloat
in a tub or basin of water, and the players are challenged to
take it out, using the mouth only.
Any one not acquainted with the secret will make a great
many attempts before he finally succeeds.
No. XXVIII.— The Cut PlayingCard.
Given, a playingcard or an oblong piece of cardboard of
corresponding size.
Required, so to cut it, still keeping it in on3 piece, that
a person of ordinary stature may be able to pass through it.
N.B.A bit of roan or morocco leather makes a very good
substitute for the cardboard.
* The napkin or handkerchief should be in direct contact with the
polished surface of the table, no second cloth intervening.
Miscellaneous Puzzles.
347
No. XXIX.— The Mitre Puzzle.
Given, a piece of paper" or cardboard shaped as Fig. 471
(a distant likeness of a bishop's mitre).
vvzA
l'iG. 471.
Required, to divide it into four parts, all of the same shape
and size.
No. XXX.— The Five Straws.
Given, five straws, each three to four inches in length,
and a shilling.
Required, by holding the end of one straw only, to lift all
the remainder.
No. XXXI.— The Three Fountains,
A
HIK
/1\
E
F
G
Fig. 472..
A, B, C, J) (Fig. 472) represents a walled space; E, F, and
343
Puzzles Old and New.
G, three houses, and II, I, K, three fountains. It is re
quired to lay pipes in such manner as to bring water from
I to (?, from H to F, and from K to E ; but the pipes must
not cross each other, nor must they pass outside the enclo
sure.
How is it to be done ?
No. XXXILThe Two Dog's.
The two dogs depicted in Fig. 473 are obviously dead.
Fig. 473.
Required, by the addition of four more lines, to restore
them to life aoain.
How is it to be done ?
No. XXXTIL— Water Bewitched.
Required, to place a glass of water in such a position that
the glass cannot be lifted without sjDilling the whole of the
water.
No. XXXIV.— The Balanced Halfpenny.
The requirements for this puzzle are an ordinary hairpin,
a long steel pin, such as ladies use to keep their bonnets in
position, a halfpenny, and a fingerring, about equal to it in
weight.
You are required, by the aid of the other two articles, to
balance the halfpenny on the point of the bonnetpin.
Miscellaneous Pussies. 349
No. XXXV.— The Balanced Sixpence.
It would stagger most people to be invited to balance a
sixpence .on edge on the point of a needle, and yet, if you
know how to do it, the feat is not only possible, but easy.
The requirements for the trick are to be found in any
household. They are a corked winebottle, a second cork,
of somewhat smaller size, a needle, and a couple of dessert
forks of equal size and weight. Last, but not least in im
portance, the sixpence.*
Having provided himself with these aids, the reader is
invited to try whether he can solve the puzzle.
m. xxxvi.— Silken Fetters.
This is a puzzle for two persons, preferably a lady and
gentleman. Two pieces of ribbon, each, say, a yard and a
half in length, are required. One end of the first ribbon is
Fig. 474.
to be tied round each of the lady's wrists, and the second
ribbon is then in like manner secured to the gentleman's
wrists, one end of it, however, being first passed inside the
lady's ribbon, so that the pair are held captive, after the
manner illustrated in Fig. 474.
* Should there be any difficulty in procuring a sixpence, a half
sovereign will do equally well. Two penknives, of equal size and weight,
may be substituted for the forks.
35o
Puzzles Old and New.
The puzzle is to disconnect tliem, but without untying
either of the knots.
How is it to be done ?
No. XXXVll.The Orchard Puzzle.
A farmer had an orchard, wherein were twelve fruittrees,
in the positions shown in Fig. 475. On his decease he
directed that the orchard should be equally divided between
Fig. 475.
his four sons, with the proviso that the portion taken by
each was to be of the same size and shape, and to contain
three of the twelve fruittrees.
How was it done ?
No. XXXVIII— The Gook in a Difficulty (La Question
cle la Marmite).
This is a puzzle of French origin. Our illustration (Fig.
476) represents a fire in the open, and a stewpan, which is
Miscellaneous Puzzles. 351
to be suspended over it. As will be seen, the four uprights
are too far apart to support it, and, being of wood, the cook
dares not drive them in any nearer to the fire. The only
appliances at his command are four pieces of hoopiron, as
shown, in length a little less than the distance from upright
to upright. With these he is required to form a support for
his stewpan.
How does he manage it ?
No. XXXIX.— The Devil's Bridge (Le Pont du
Didble).
This is a puzzle on the same principle.
Given, three uprights (see Fig. 477), representing three
hills, or towers, and three flat pieces, representing three
Fig. 477.
planks or girders, not long enough to reach from upright
to upright.*
Required, with these materials to construct a bridge from
point to point.
No. XL— The Two Corks.
Take two winebottle corks, and hold them as shown in
Fig. 478 — viz., each laid transversely across the fork of the
thumb. Now with the thumb and second finger of the right
hand (one on each end) take hold of the cork in the left
hand, and, at the same time, with the thumb and second
* If the specially manufactured form of the puzzle is not available
three wineglasses may supply the place of the three uprights, and three
tableknives that of the cross pieces.
352
Puzzles Old and New.
finger of the left hand take hold of the cork in the right
hand, and draw them apart.
Fm. 478.
The above sounds simple enough, but the neophyte will
find that the corks are brought crosswise, as shown in Fi«\
Fig. 479.
479. The puzzle is to avoid this, and enable them to part
freely.
No. XLI.— The Divided Farm.
This is a puzzle of the same class as No. XXXVII. (see p.
350). A house, IT, (see Fig. 480) stands in a square enclosure.
Fig. 480.
Within the same enclosure are ten trees, placed as shown in
the figure. The owner (one of those eccentric persons
Miscellaneotis Puzzles.
*> <r "»
whom we only hear of in connection with puzzles) made a
will whereby he directed that the house should be occupied
by his five sons jointly, but that the land was to be divided
between them, each to have a piece of the same size and
shape, and each piece to enclose two of the ten trees.
How must the land be divided, in order to carry out the
testator's intentions ?
No. XLIT.— The Conjurer's Medal.
This is as represented in Fig. 4S1. The medal has five
holes in it, and the puzzle is to work the ring (which, as will
Fig. 481.
be observed, has a gap in it) from hole to hole until it is
finally detached from the medal.
No. XLIII.— The Maze Medal.
This (see Fig. 482) is a puzzle of the same kind, but some
what more complicated, and different as regards the object
aimed at, the experimenter in this case being required to
start with the ring detached from the medal, and to work it
into the hole marked Home,
AA
354
Puzzles Old and New.
Fig. 482.
No. XLIV.— The Puzzle KeyRing.
We have here a keyring in the form of a horseshoe, with
the space between its arms closed by a smaller horseshoe (see
Fig. 483.
Fig. 483). The ring is not " split," but the keys are put on
and taken oh by a little secret process, the nature of which
the experimenter is invited to discover.
Miscellaneous Puzzles.
355
No. XLV.— The Singular Shilling".
A handkerchief being spread out squarely upon the table,
and a shilling laid on its centre, required, so to pick up the
handkerchief as to bring it into a vertical position, the
shilling still remaining in the centre, supported by the hand
kerchief only.
No. XLVL— The Entangled Scissors.
Pass the loop end of a piece of doubled string through one
of the bows of a pair of scissors, then pass the opposite ends
through the loop, thence through the second bow, and
finally tie them round a walkingstick or ruler, as shown in
Fig. 484.
Fig. 484.
The puzzle is to disengage the scissors without untying
the cord, or slipping the string off the stick.
If preferred, the ends of the cord may be held by a second
person, in which case the use of the stick will be unneces
sary.
356 Puzzles Old and Nezv.
No. XLVIL— The Penetrative Penny.
In a piece of stout paper cut a circular hole the size of a
shilling — i.e., about T \ less than an inch. Invite any one to
pass a penny through the hole without touching the coin or
tearing the paper. He will naturally tell you that it can't
be done, the diameter of a penny being 1 inch, or y\ inch
larger than the hole. And yet the thing can be clone —
easily done ; and the reader is invited to find out how to
do it.
rTo. XLYIII.— The Packer's Secret.
This is a very ingenious little puzzle, of French origin.
It consists of a shallow cardboard box, 3 inches in diameter,
containing twelve boxwood discs, each three quarters of an
inch in diameter and rather more than an eighth of an inch
in thickness. The puzzle is so to arrange these in the box
(whose area they only about three parts cover) that it may
be turned upside down without their falling out.
KEY TO CHAPTER X.
MISCELLANEOUS PUZZLES.
No. I.— The John Bull Political Puzzle. Solution.
ISTo absolute rule can be given for the solution of this
puzzle, as the number and direction of the moves will
necessarily vary according to the position in which the nine
counters happen to be placed at the outset. We propose to
take a couple of specimen positions, and if the reader duly
studies the modus operandi in these two cases, he should
have little difficulty in finding appropriate solutions for
other positions.
Distinguishing the counters for facility of reference as
follows : —
Red Conservative
White
Blue ,,
Red Liberal
White „
Blue ,,
Red Unionist
White „
Blue „
as R.
W
5? B.
„ R.
n W.
„ B.
., R.
C.
c.
c.
L.
L.
L,
IT.
W.U.
B. U.
we will suppose that the original (haphazard) position is
as follows : —
(see Fig. 449)
On spot 1
„ 2
„ 3
„ 4
„ 5
„ 6
„ 7
• „ 8
„ 9
B.
B.
W.
R.
W.
W.
B.
R.
R.
C.
u.
a
L.
L.
U.
L.
U.
C.
357
35o Jruzz
les Uu
x a
na i\i
ew.
The puzzle may then be solved
as foil
ows: —
Move R. 0.
from 9 to 10.
>>
R. U.
55
8 „
9.
55
W. U.
55
6 „
8.
55
B. 0.
55
1 „
6.
55
W. C.
55
3 „
1.
55
B. IT.
55
2 „
3.
55
R. L.
55
4 „
2.
55
B. G.
55
6 „
4.
55
W. L.
55
5 „
6.
55
B. 0.
55
4 „
5.
55
R. U.
55
9 „
4.
55
R. C.
55
10 „
9.
leaving the counters in this case as shown in Fig. 450 ;
though, of course, many other final arrangements would
equally answer the conditions of the puzzle.
Again, suppose the original position to be as follows : —
R. L. on 7.
W. 0.
B. U.
B. C.
R. U.
W. L.
R. C.
B. L.
W. IT.
The puzzle may then be solved as follows :—
Move W. U. from 9 to 10.
J5
55
55
55
55
55
35
3?
R. L.
W. 0.
B. L.
R. TJ.
R. L.
W. U.
4
6
5
7
9
10
5?
55
5 5
•>■>
9.
4.
6.
5.
7.
9.
No. II.— The Pig in Sty. Solution.
For convenience of reference, it will be desirable to dis
tinguish the various holes by numbers, as under (Fig. 485),
each square representing a hole.
Key to Miscellaneous Puzzles.
359
The "pig" at the outset is placed in No. 22, and No. 13
(the centre hole) is vacant.
1
2
3
4
5
6
7 :
8
9
10
11
12
13
14
15
16
17
18
19 20
21
22
23 24
25
Fig. 485.
First Solution. In Twelve Moves.
Move 9 to 13 ; 3 to 9 ; 7 to 3 ; 22 to 7 ; 18 to 22 ; 24 to 18;
9 to 24 ; 13 to 9 ; 7 to 13 ; 3 to 7 ; 18 to 3 ; 22 to 18.
Second Solution. In Eleven Moves.
Move 19 to 13 ; 4 to 19 ; 10 to 4 ; 7 to 10 ; 22 to 7 ; 25 to
22 ; 19 to 25 ; 13 to 19 ; 7 to 13 ; 10 to 7 ; 22 to 10.
No. III.— Hide and Seek. Solution.
Figs. 486 and 487, the first representing the upper, and
the second the under side of the puzzle, show the course
which the ball must travel in order to fulfil the conditions.
The letter A indicates the point of entrance and exit, and
the numbers show the order in which the various holes are
to be passed.
A dot beneath a number indicates that the ball (relatively
to the side represented bj the diagram) passes downivards
through the hole in question. A cross, that the ball is to be
brought through that hole from the opposite side.
3 Co
Puzzles Old and A T ew.
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Fig. 487.— Under Side.
Key to Miscellaneous Puzzles.
;6i
No. IV.— The Brahmin's Rings. Solution.
For facility of reference it will be desirable to distinguish
die rings or discs by numbers, calling the smallest No. 1,
the next larger No. 2, and so on up to the largest, which
will be No. 8. The three "cones," which we will assume
to be placed in a row before us, we will distinguish (see
Fig. 453) by the letters L, C, and R, respectively equivalent
to Left, Centre, and Right; and we will suppose that the
rings are arranged at the outset in proper order (No. 1
uppermost, and so on) upon the lefthand cone, L. We then
proceed to move as follows : —
(1) 1 on C : (2) 2 on R : (3) 1 on II.
We have now transferred two of the rings, and it will be
observed that it has taken three moves to do it. A rash
experimenter might hence conclude that it would only require
eight times three moves to transfer all eight rings, but he
would be vastly mistaken, for before each f olio win g ring 1
can be brought to the base of the new heap a constantly in
creasing number of transpositions is necessary.
We proceed : —
(4) 3 on : (5) 1 on L : (6) 2 on C : (7) 1 on 0.
We have now transferred three of the rings.
(8) 4 on R : (9) 1 on R : (10) 2 on L :
(ll)lonL: (12) 3 on R: (13) 1 on C :
(14) 2 on R : (15) 1 on R.
Four rings are now transferred. Continuing : —
(16) 5onC: (17) 1 onL
(19) 1 on C : (20) 3 on L
(22) 2 on L : (23) 1 on L
(25) 1 on C : (26) 2 on R
(28) 3 on C : (29) 1 on L
(31) 1 on C.
Five rings are now transferred, and it has taken 31 moves
to transfer them.
(32) 6 on R : (33) 1 on R :
(35) 1 on L : (36) 3 on R :
(38) 2onR: (39) 1 on R :
(41) 1 on L : (42) 2 on C :
(44) 3 on L : (45) 1 on R :
(18) 2 on C:
(21) 1 on R :
(24) 4 on C :
(27) 1 on R :
(30) 2 on C :
(34) 2 on L :
(37) 1 on C :
(40) 4 on L :
(43) 1 on C :
(46) 2 onL:
,62
Puzzles Old mid New.
(47) 1 on L
(48) 5 on R :
(49) 1 on C :
(50) 2 on R
(51) 1 on R :
(52) 3 on C :
(53) 1 on L
(54) 2 on O :
(55) 1 on C :
(56) 4 on R
(57) 1 on R :
(58) 2 on L :
(59) 1 on L
: (60) 3 on R:
(61) lonC:
(62) 2 on R
(63) 1 on R.
Six rings have now been transferred
.
(64) 7 on C .
(65) 1 on L :
(66) 2 on C :
(67) 1 on C :
(68) 3 on L :
(69) 1 on R :
(70) 2 on L :
(71) 1 onL:
(72) 4 on C :
(73) 1 on C :
(74) 2 onR:
(75) lonR:
(76) 3 on C :
(77) 1 on L :
(78) 2 on C :
(79) 1 on C :
(80) 5 on L :
(81) 1 onR:
(82) 2 on L :
(83) 1 on L :
(84) 3 on R :
(85) 1 on C :
(86) 2 on R :
(87) 1 onR:
(88) 4 on L
(89) 1 on L :
(90) 2 on C :
(91) 1 on C :
(92) 3 on L :
(93) 1 onR:
(94) 2 on L
(95) 1 on L :
(96) 6 on C :
(97) 1 on C :
(98) 2 on R :
(99) 1 on R :
(100) 3 on C
(101) 1 on L :
(102) 2 on C :
(103) 1 on C
(104) 4 on R:
(105) 1 on R :
(106) 2 on R
(107) 1 on L :
(108) 3 onR:
(109) 1 on C :
(110) 2 onR:
(111) 1 on R:
(112) 5 on C :
(113) 1 on L :
(114) 2 on C :
(115) 1 on C
(116) 3 on L :
(117) lonR:
(118) 2 on L
(119) 1 onL:
(120) 4 on C :
(121) 1 on C :
(122) 2 onR:
(123) 1 onR:
(124) 3onC:
(125) lonL:
(126) 2 on C :
(127) 1 on C.
We have now transferred seven of the eight rings. We
have made no less than 127 moves, and have, as a matter
fact, completed just half our task. We proceed : —
of
(128) 8 on R :
(129) 1 on R :
(130) 2 on L :
(131) 1 on L :
(132) 3 on R :
(133) 1 on C :
(134) 2 on R :
(135) 1 on R :
(136) 4 on L :
(137) 1 onL:
(138) 2 on C :
(139) 1 on C :
(140) 3 on L :
(141) 1 onR:
(142) 2 on L :
(143) 1 on L :
(144) 5 onR:
(145) 1 on C :
(146) 2 onR:
(147) 1 on R :
(148) 3 on C :
(149) 1 onL:
(150) 2 on C :
(151) 1 on C :
(152) 4 on R:
(153) 1 on R :
(154) 2 onL:
(155) 1 on L :
(156) 3 on R :
(157) 1 onC:
Key to Miscellaneous Puzzles. 363
(158) 2 on R :
(159) 1 on R :
(160) 6 on L :
(161) 1 onL:
(162) 2onC:
(163) 1 on C :
(164) 3 on L :
(165) 1 on R :
(166) 2 on L :
(167) 1 on L :
(168) 4 on C :
(169) 1 on C:
(170) 2 on R:
(171) 1 onR:
(172) 3 on C :
(173) lonL:
(174) 2onC:
(175) 1 on C :
(176) 5onL:
(177) 1 onR:
(178) 2 on L :
(179) lonL:
(180) 3 onR:
(181) lonC:
(182) 2 on R :
(183)1 onR:
(184) 4 on L :
(185) lonL:
(186) 2 on C:
(187) 1 on C :
(188) 3 on L :
(189) 1 on R :
(190) 2 on L :
(191) 1 on L :
(192) 7 onR:
(193) 1 on C :
(194) 2 on R :
(195) 1 onR:
(196) 3 on C :
(197) lonL:
(198) 2onC:
(199) 1 on C :
(200) 4 on R :
(201) 1 onR:
(202) 2onL:.
(203) 1 on L :
(204) 3 on R :
(205) 1 on C :
(206) 2 on R :
(207) 1 on R :
(208) 5 on C :
(209) 1 on L :
(210) 2 on C :
(211) 1 on C :
(212) 3 on L :
(213) 1 on R :
(214) 2 on L :
(215) 1 on L :
(216) 4 on C :
(217) 6 on C :
(218) 2onR:
(219) 1 on R :
(220) 3 on C :
(221) lonL:
(222) 2 on C :
(223) 1 on C :
(224) 6 on R:
(225) 1 onR:
(226) 2onL:
(227) 1 on L :
(228) 3 on R :
(229) 1 on C :
(230) 2 on R :
(231) 1 on R :
(232) 4 on L :
(233) 1 on L :
(234) 2 on C :
(235) 1 on C :
(236) 3 on L :
(237) 1 on R :
(238) 2 on L :
(239) 1 on L :
(240) 5 on R :
(241) 1 on C :
(242) 2 on R :
(243) 1 onR:
(244) 3 on C :
(245) 1 on L :
(246) 2 on C :
(247) 1 on C :
(248) 4 on R :
(249) 1 on R :
(250) 2 on L :
(251) 1 on L :
(252) 3 on R :
(253) 1 on C :
(254) 2 on R :
(255) 1 on R.
The whole of the eight rings are now transferred. It will
be observed, on studying the steps of the process, that each
additional ring doubles the number of moves necessary to
the transposition, with one move in addition. The number
of moves, therefore, which would be needed to transfer 64
rings, according to the legendary shape of the original prob
lem, would be considerably higher (by reason of the unit
added at each stage of the progression) than the number of
grains of corn in the familiar "Chessboard " problem (1 for
364 Fttzzles Old and New.
the first square, 2 for the second, 4 for the third, and so on
throughout the sixtyfour squares of the board). As the
total number of grains in the lastmentioned case is 18,446,
774,073,709,551,615: and as it is computed that the count
ing of a single billion (1,000,000,000,000), at the rate of 100
a minute, would occupy 19,024 years, it may well be ima
gined that the assurance of Brahma, that the end of the
world would arrive before the priests' task was complete,
was really a mild way of putting the matter.*
No. V.— Cardan's Rings.
We quote the following instructions, the clearest and most
practical that we have yet seen for the solution for this
puzzle, from an anonymous American writer.f
"Take the bow in your left hand, holding it at the end, B,
and consider the rings as being numbered 1 to 10. The first
will be the extreme ring to the right, and the tenth the
nearest to your left hand. (See Fig. 454.)
"It will be seen that the difficulty arises from each ring
passing round the wire of its righthand neighbour. The
extreme ring at the right hand, of course, being unconnected
with any other wire than its own, may at any time be drawn
off the end of the bow at A, raised up, dropped through the
bow, and finally released. After you have done this, try to
pass the second ring in the same way, and you will not
succeed, as it is obstructed by the wire of the first ring ; but
if you bring the first ring on again, by reversing the process
by which you took it off, — viz., by putting it up through the
bow and on to the end of it, — you will then find that by taking
the first and second rings together they will both draw off,
lift up, and drop through the bow. Having done this, try
to pass the third ring off, and you will not be able, because
it is fastened on one side to its own wire, which is within
* As a somewhat similar example of an impossible task, a person may
be invited to try in how many ways he can arrange the 28 "cards " of a
set of dominoes according to domino rules — i.e., 1 against 1, 2 against 2,
and so on. A German mathematician, Dr. Beiss, has computed that the
possible number of such combinations is 15,918,459,863,040 ; and that
supposing two minutes to be occupied in making each combination, the
time occupied for the whole would be something over C0,000 years.
f For another very clever explanation of this puzzle seethe Ency
clopedic Metlwdique des Jeux, pp. 424 et seq.
Key to Miscellaneous Puzzles. ^65
the bow, and on the other side to the second ring, which is
without the bow.
" Therefore, leaving the third ring for the present, try the
fourth ring, which is now at the end all but one, and both
of the wires which affect it being within the bow, you will
draw it off without obstruction. In doing this, you will have
to slip the third ring off, which will not drop through for
the reasons before given ; so, having dropped the fourth ring
through, you can only slip the third ring on again. You
will now comprehend that (with the exception of the first
ring) the only ring which can at any time be released is that
which happens to be second on the bow, at the righthand
end ; because both the wires which affect it being within the
bow, there will be no impediment to its dropping through.
" You have now the first and second rings released, and
the fourth also, the third still fixed, to release which we
must make it last bat one on the bow. To effect this, pass
the first and second rings together through the bow, and on
to it ; then release the first ring again by slipping it off and
dropping it through, and the third ring will stand as second
on the bow, in its proper position for releasing, by drawing
the second and third off altogether, dropping the third
through, and slipping the second on again. Now, to release
the second, put the first up, through and on the bow, then
clip the two together off, raise them up, and drop them
through.
" The sixth will now stand second, consequently in its
proper position for releasing, therefore draw it towards the
end, A; slip the fifth off, then the sixth, and drop it through ;
after which, replace the fifth, as you cannot release it until
it stands in the position of a second ring. In order to effect
this, you must bring the first and second rings together,
through and on to the boAv ; then, in order to get the third
on, slip the first off, and down through the bow ; then bring
the third up, through and on to the bow, then bring the
first ring up and on again, and releasing the first and second
together, bring the fourth through and on to the bow, re
placing the third. Then bring the first and second together
on, drop the first off and through, then the third the same ;
replace the first on the bow, take off the first and second
together, and the fifth will then stand second, as you desired ;
draw it towards the end, slip it off and through, replace the
fourth, bring the first and second together up and on again ;
^66 Puzzles Old and New.
o
release the first, bring on the third, passing the second
ring on to the bow again ; replace the first, in order to
release the fir^t and second together, then bring the fourth
toward the end, slipping it off and through; replace the
third, bring the first and second together up and on again,
release the first, then the third, replacing the second, bring
the first up and on, in order to release the first and second
together, which having done, your eighth ring will then
stand second ; consequently you can release it, slipping the
seventh on again.
" To release the seventh, you must begin by putting the
first and second up and on together, and going through, the
movements in the same succession as before, until you find
you have only the tenth and ninth on the bow ; then slip the
tenth off and through the bow, and replace the ninth. This
dropping of the tenth ring is the first effectual movement
toward getting the rings off, as all the changes you have
gone through were only to enable you to get at the tenth
ring.
" You will now find that you have only the ninth left on
the bow, and you must not be discouraged on learning that
in order to get that ring off, all the others to the right hand
must be put on again, beginning by putting the first and
second together, and working as before, until you find that
the ninth stands as second on the bow, at which time you
can release it. You will then have only the eighth left on the
bow; you must again put on all the rings to the right hand,
beginning by putting up the first and second together, till
you find the eighth, standing as second on the bow, or in its
proper position for releasing, and so you proceed until you
find all the rings finally released.
"As you commence your operations with all the rings
ready fixed on the bow, you will release the tenth ring in
170 moves; but as you then have only the ninth on, and as
it is necessary to bring on again all the rings up to the ninth,
in order to release the ninth (which requires sixteen moves
more), you will, consequently, release the ninth ring in 256
moves. For your encouragement, your labour will diminish by
one half with each following ring which is finally released.
The eighth comes off in 128 moves. The seventh in 64
moves, and so on, until you arrive at the second and first
rings, whicli come off together, making 681 moves, which are
necessary to take off all the rings.
Key to Miscellaneous Puzzles.
,67
" With the experience yoa will by this time have acquired,
it is only necessary to say that, to replace the rings, you
begin by putting up the first and second together, and follow
precisely the same system as before, in reverse order. "
No. VI.— The Knight's Tour. Solutions.
This puzzle, like the foregoing, has repeatedly engaged
the attention of mathematicians, and a great many solutions
have been placed on record. Of these we select a few, of
various character, by way of specimens. Figs. 488 and 489
43
50
4i
40
21
4G
7
Gl
10
41
53
■
55
45
22
8
30
20
12
43
GO
57
G
47
6S
Puzzles Old and New.
Key to Miscellaneous Pttzzles.
3^9
finish within one move of the square from which he started.
If this condition be fulfilled, it is clear that the beginning
may be made from any square, the series of moves forming an
endless chain, which may be broken at any point without
otherwise interfering with its regular sequence, Euler begins
at the lefthand bottom corner square, and finishes on the
square next adjoining. Du Malabare begins nearly in the
centre of the board, and also finishes on the square next
adjoining.
1
6
51
S
11
60
57
54
50
13
2
61
52
55
10
59
5
64
7
12
9
58
53
56
14
49
62
3
16
47
36
31
63
4
15
48
35
30
17
46
24
21
26
41
44
29
39
32
37
27
43
23
20
34
45
18
22
25
28
43
40
19
3S
33
Fig. 491, (Monneron.)
Figs. 492, 493, and 494 represent other " perfect " solu
tions. It will be observed that the diagram formed by the
indicating lines now and then falls into the shape of a more
or less symmetrical figure, and the reader will find it an
entertaining exercise to describe either the centre or marginal
portion, or both, of such a figure on his paper chessboard,
and then try whether he can complete the " tour," taking the
lines already laid down as part of the route to be travelled.
The feasibility of such an attempt will, of course, depend
upon the figure selected, but can only be determined by
actual experiment.
Meanwhile, the less ambitious reader, who does not aim at
B B
7°
Puzzles Old and New.
Fig. 492.
Fig. 493.
Key to Miscellaneous Puzzles. 371
these higher flights, but merely desires to perform the feat
in the easiest and simplest way, may do so by adopting the
following method, the invention of a scientific German writer,
named Warnsdorf : " Place the knight on any square you
like, and begin by moving him to that square from which he
would command the fewest points of attach ; observing that if
his power would be equal on any two squares, you may play
him to either ; and that when a square is once covered, it
is not to be reckoned among those which he commands.
Continue moving him on this principle, and he will traverse
the 64 squares in as many moves."*
Fig, 494.
As an aid to the observance of this rule, it should be noted
that the knight, from either of the corner squares, commands
two squares only. If placed on an outer, row of the board next
to a corner square, he commands three squares. From any
other square of an outer row he commands four squares.
From the square next diagonally adjoining a corner square
he commands four squares. From any other point in a
second row he commands six squares, and from either of the
George Walker : A Ncio Treatise on Chess, 1832.
372
Puzzles Old and New.
sixteen squares in the centre lie commands eight squares, as
suming, of course, that they have not been already visited. It
follows that the squares in the outer rows, and particularly the
corner squares, should be first disposed of, thence gradually
working round and round to the centre. We append (Fig. 495)
an example of the Knight's Tour worked on this principle.
We have selected as startingpoint the same square as in
Du Malabare's solution. In accordance with the rule given,
3
40
19
36
5
50
21
34
18
37
4
51
20
61
35
52
55
64
6
49
41
38
2
39
54
33
22
17
42
27
62
1
58
48
7
13
57
60
53
23
32
16
43
28
14
63
56
59
8
47
12
15
29
44
26
45
10
/^r
24
ii
30
25
46
9
Fig. 495.
we first make for the nearest corner. From this point up to
the thirteenth the moves are a matter of course. At the
fourteenth move we have a choice of two squares, one next
the startingpoint, the other diagonally adjoining the square
to which a twelfth move was made. But we find on com
parison, that at the former the knight w r ould command seven
vacant squares, while from the latter he commands six only,
and we therefore give the preference to this square. From
this point we work outwards again. We cannot move to a
corner square, but we can secure one next to a corner, and
commanding only two vacant squares, We now run on,
skirting the board as before, to the twentyseventh move,
Key to Miscellaneous Puzzles. 373
inclusive ; and then back again, still on the two outer rows,
and securing another corner on our way, to the thirtyeighth
move, inclusive. For the thirtyninth we are compelled
to move into the third row, but immediately get back into
the outer rows, and again run on, in these same rows, to
number fiftyone, inclusive At fiftytwo we are compelled
to move into the centre. We have at this point the choice
of three squares, but two of these command three squares
each, while the other commands one only. We therefore
decide for this one. Fiftythree is a forced move. For fifty
four we have a choice of two moves, both equally eligible,
and a like alternative for fifty five. Fifty  six demands
more caution. One possible square commands two others,
another one only. We therefore choose the latter. Fifty
seven and fiftyeight are forced moves. For fiftynine there
are again two squares available, one commanding two, the
other one square only. If we were to move to the former,
we should fail, even though on the very point of success ;
but, avoiding this pitfall, the remaining moves are found to
be compulsory, and to bring the problem to a successful
issue.
It will be observed that the final move, 64, has brought
the knight to within one move of the point from which he
started, according to the condition of the perfect tour.
When the startingpoint is a corner square, a little extra
precaution will be needed to fulfil this condition ; for, inas
much as a corner square commands two squares only, and
the second move occupies one of such two squares, the
other must be kept vacant to receive the knight at the sixty
fourth move.*
* The Knight's Tour is frequently performed as a show feat, by
means of a mnemonic formula. The squares of the board are in this ease
regarded as being numbered consecutively, 1 to 8 in the top row, 9 to
16 in the second row, and so on. The proper sequence of the squares
for the purpose of the " tour" is then committed to memory by means
of a doggrel verse, or prose sentence, each syllable of which is so ar
ranged as to recall a corresponding number. The performance of the
feat in this shape involves no intellectual process beyond a very trifling
effort of memory.
74 Puzzles Old and Nezv.
No. VII.— The Knotted Handkerchief. Solution.
The secret lies in the manner of taking hold of the hand
kerchief. This is laid, twisted ropewise, in a straight line
upon the table. The performer then folds his arms, the
fingers of the right hand coming out above the biceps of the
left arm, and the fingers of the left hand being passed beloiv
the biceps of the right arm. With the arms still in this po
sition, he bends forward and picks up the handkerchief, the
right hand seizing the end lying to the left, and the left hand
that which lies to the right. On drawing the arms apart,
it will be found that a knot is formed upon the centre of the
handkerchief.
No. VIII.— Crossette. Solution.
It will be found that unless the experimenter proceeds in
accordance with a regular system he will fail, the count of
four beginning very soon to fall upon circles already crossed
out, and so failing to meet the conditions.
To solve the puzzle, after striking out a given circle, miss
three before starting again. Thus, suppose the start to be
made from No. I (Fig. 496) :—
10  . 2
8 4
6
Fig. 496.
No. 4 will in such case be the first to be struck out.
Miss three, and start again at 8 ; No. 1 will then be the one
to be struck out. Begin again at 5, and strike out 8.
Key to Miscellaneous Puzzles.
75
Again at 2, and strike out 5,
Again at 9, and strike out 2.
Again at 6, and strike out 9.
Again at 3, and strike out 6.
Again at 10, and strike out 3.
Again at 7, and strike out 10.
You have tlius struck out nine of the ten circles.
It will be observed that you strike out at each round the
number with which you started at the previous round.
No. IX.— Single Stroke Figures. Solution.
Fig. 457 may be described, according to the conditions of
the puzzle, by starting at the point E, thence carrying the
pencil from E to A, from A to B, from B to G, from G to D,
and from D to E ; then from _E7 to B, B to D, D to A, and
A to G.
Fig. 458 may be described by tracing the lines from A to
B, B to G, GtoB,D to A, A to F, F to 0, G to E, E to F,
F to B, B to E, and E to D.
Fig. 459 may be described by starting at A, then passing
along the curve AGE, from JD along JDEB, from B along
BFG, and from G along GEA.
The solubility of a puzzle of this kind by no means depends
upon the intricacy of the figure. A square with diagonals
united, for instance (see Fig. 497), cannot possibly be de~
Fig. 497.
scribed without taking off the hand. The test is as follows :
If the points of intersection are all formed by even numbers
of lines, or if there are not more than two such points
involving an odd number of lines, the puzzle is soluble ;
if otherwise, it is insoluble.
76
Puzzles Old and New.
No. X.— The Balanced Egg*.
Solution.
Another Method.
The difficulty in balancing an egg arises from the fact that
the yolk, which is of greater specific gravity than the white,
lies in an unstable condition about the middle of its longer
axis, and so makes the egg, in familiar language, topheavy.
If the egg be well shaken, so as to break up the yolk and
mix it thoroughly with the white, the whole contents become
of uniform specific gravity, and the egg may then be balanced
on its larger end without difficulty.
No. XI.— Solitaire Problems. Solutions. No. 1.
Starting with hole No. 1 vacant, to leave the last ball in
hole No. 37.
The moves are as under: —
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
3 to 1.
12 ,
, 2.
13 ,
, 3.
15 ,
, 13.
4 ,
, 6.
18 ,
, 5.
1 ,
, 11.
31 ,
, 18.
18 ,
, 5.
20 ,
, 7.
3 ,
; 13.
33 ,
, 20.
20 ,
; 7.
9 ,
, n.
16 ,
, 18.
23 ,
, 25.
22 ,
, 20.
No. 18. From 29 to 27.
No. 35.
19. „ 18
20. „ 31
21. „ 34
22. „ 20
23. ^T 37
24. „ 5
25. „ 18
26. „ 20
27. „ 33
28. „ 2
29. „ 8
30. „ 6
31. „ 19
32. „ 36
33. „ 30
34. „ 26
From 35 to 37.
31.
33.
32.
33.
27.
18.
20.
33.
31.
12.
6.
19.
32.
26.
32.
36.
Key to Miscellaneous Puzzles.
377
No. 2. The Curate and His Flock.
The moves are in this case as under : —
No. 10. From 24 to 26.
27 „ 25.
33 „ 31.
25 „ 35.
29 „ 27.
14 ;, 28.
27 „ 29.
19 „ 21.
7 „ 20.
2.
55
3.
55
4.
55
5.
55
6.
55
7.
)5
8.
55
9.
55
6 t
4 ,
o 19.
, C.
18 ,
, 5.
6 ,
, 4.
9 ,
5 11.
24 ,
, 10.
11 ,
, 9.
26 ,
, 24.
35 ,
, 25.
No
. 19. Fn
11.
12.
13.
14.
15.
16.
17.
18.
From 21 to 19.
No. 3. The Triplets.
The moves are in this case as under :■ —
No. 1. From 6 to 19.
2.
, 10 ,
12.
3.
, 19 ,
• 6.
4 ,
, 2 ,
12.
5.
i 4 ,
6.
6.
5 17 ,
, 19,
7.
, 31 ,
18.
8.
5 19 ,
.17.
9.
, 16 .
18.
10.
, 30 ,
17.
No. 11. From 21 to 19.
12. ,
13. ,
14. ,.
15.
i6. ,;
17. , :
18.
19.
20.
7 ,
, 20
19 ,
; 2i
22 ,
, 20
8 ,
, 21
32 ,
, 19
28 ,
, 26
19 ,
, 32
36 ,
, 26
34 ,
, 32
No. XII.— Skihi. Solution.
For the formation of the Skihi designs nothing more is
needed than a modicum of the constructive faculty, supple
mented by a good stock of perseverance. The cards must be
coaxed, not forced, into position. If this caution be borne
in mind, and the diagram carefully studied beforehand, the
execution of the most elaborate design becomes a mere matter
of time and patience.
We will take, by way of example, the Maltese cube (Fig.
464). Though so simple in appearance, it is by no means
one of the easiest to construct, no less than eighteen cards
being employed, and some little skill being needed to join
them neatly together.
78
Puzzles Old and New.
First take tlie card which is to form the top of the central
or " solid " portion, and with this combine, by means of the
slots, the four cards which are to form the uprights of the
upper portion. Slot must be inserted into slot, and each
card pushed well home. We next add the four cards which
form the four sides of the cube (only the upper portions of
which are visible in the diagram). These duly in place, we
insert the four cards which form the horizontal platform in
centre ; and this is a more difficult task, for each such card
has to be worked into one of the angles of the cube. When
these four are in position, the next thing to be done is to join
together five cards as we did in the first instance (four round
one centre card) and when joined, unite these to the figure
already made. Our cube is now complete; and by following
the same method of construction it will be easy to put to
gether either of the other figures,*
No. XIII,— A Card Puzzle. Solution,
The four cards are arranged as shown in Fig. 498, one pip
of each being hidden by the overlapping corner of the next.
No. XIV.— Another Card Puzzle. Solution.
First take the four aces, and with them form one of the
diagonal rows, beginning from the top lefthand corner. We
* A little powdered French chalk rubbed into the slots will be found
very useful to diminish friction. The cards when new are apt to go
somewhat stiffly.
Kev to Miscellaneous Puzzles,
79
will suppose that their order is ace of spades, ace of clubs,
ace of hearts, ace of diamonds, Next take the kings, and
with them form the second diagonal, bearing in mind that
the two centre kings must be of the opposite suits to the two
centre aces, viz., in this case the king of spades and king of
diamonds. Next place the four queens, beginning with the
Ace of
Spades
Knave of
Hearts
Queen of
Diamonds
King of
Clubs
Knave of
Diamonds
Ace of
Clubs
King of
Spades
Queen of
Hearts
Qneen of
Clubs
King of
Diamonds
Ace of
Hearts
Knave of
Spades
King of
Hearts
Queen of
Spades
]
Knave of
Clubs
Ace of
Diamonds
Fig. 499.
two squares next the top righthand corner, and then filling
the corresponding squares next the lefthand bottom corner.
A very little consideration will show you which queen should
occupy each space. Then place the four knaves in the re
maining spaces, duly bearing in mind the conditions, and you
will finally have the arrangement shown in Fig. 499.
The arrangement will, of course, be subject to variation
according to the cards with which you elect to commence.
No. XV.— The Floating Corks. Solution.
If the corks be taken singly, the feat is impossible ; but if
they be gathered into a bundle, and in that position grasped
With both hands, held under water for a moment, so as to get
well wetted, and then brought slowly back to the surface, the
bundle will float upright, as shown in Fig. 500, the several
,So
Puzzles Old and New.
corks which compose it being held together by a sort of
capillary attraction.
Fig. 500.
No. XVI.— The Obstinate Cork. Solution.
On first trying this experiment the cork will be found to
be forced out of, instead of into the bottle, and the more
vigorous the "blow "the quicker will be its return. This
arises from the fact that the act of blowing drives a certain
amount of air round the cork into the bottle. This com
presses the air already contained in the bottle; the cork goes
in a little way, but the moment the compression ceases, .the
air expands, and in so doing forces the cork out again.
And yet, to any one in the secret, the feat is perfectly
simple. Take a quill or other small tube, and blow steadily
through this against the centre of the cork. . ^The quantity
of air in the bottle is now not increased, and the cork goes in
without the least resistance.
No, XVII— Fixing" the Ring 1 . Solution.
Let the lady hold the cord at about three inches from the
upper end. The gentleman passes the ring over it, then takes
hold of the upper end of the cord, and draws it through the
lady's fingers to within an inch or two of the opposite end.
The cord is held slack, the ring' hanging in the centre of the
bight. The gentleman then passes the end he holds round
the opposite portion of the bight, draws it through the loop
thus formed, and the trick is done.
Key to Miscellaneous Puzzles.
38i
No. xvii r.— The Treasure at Medinet. Solution.
See Fig. 501 for the proper arrangement of the cards.
to
\
\
/
X
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F.g. 501.
The position of the letters shows which way the original
top of each card (see Fig. 468) is to be turned.*
No. XIX.— The Four WineGlasses. Solution.
Place three of the wineglasses on the table so as to form
on equilateral triangle, each side being equal to the height
of a siDgle glass. Then place the fourth glass upside down
in the centre.
No. XX.— One Peg to Fit Three Holes. Solution.
It will be observed that one side of the " square " is just
equal in length to the diameter of the "circle." Cut the cork
to this length, as a in Fig. 502 ; and if inserted sideways, it
will then just fit the square hole. It already fits the circular
hole. To adapt it to fit the triangular space also, draw a
straight line across one end of it through the centre, and
from such line cut an equal section in a sloping direction
* For two other problems of a very similar character, see Puzzles
with Counters, Nos. XXIV. and XXV.
3 32
Puzzles Old and New,
down to each side of the circular base. The cork Avill then
Fig. 502.
assume the shape of b in the same figure, and will fit either
one of the three holes.
No. XXL— The Balanced Pencil. Solution.
Stick the blade of the penknife (which should be a small
and light one) into the pencil near the point, in the direction
of its longer axis. Then partially close the knife. The pre
cise angle must be ascertained by experiment, as it will vary
with the length and weight of the two articles. When you
have discovered it, the pencil may be balanced, as shown in
Fig. 503, on the tip of the finger.
No. XXII.— To Balance an Egg on the Point of a
WalkingStick. Solution. .
Have the egg boiled hard.* Thrust the tw T o forks into
* It is said that the feat is possible even with a raw egg, but it is in
this ease much more difficult, the contents of the egg being in a con
dition of unstable equilibrium,
Key to Miscellaneous Puzzles.
333
the cork, one on eacli side, so that they shall form an angle
of about 60° to each other. Hold the stick, ferule upwards,
firmly between the knees. Place the egg on end upon the
Fig, 501
ferule, and the cork on top of it, as shown in Fig. 504.
After a few trials you will be able to balance the egg in an
erect position.
No. XXIII.— The Ashantee Horseshoe. Solution.
Pass the second piece of wire between the first piece and
the horseshoe, and raise the latter so that the first wire shall
fall just inside the curve, the two viewed edgeways forming
an X, though with the urrper arms very much shorter than
the lower. Now raise the second piece of wire into the
angle thus formed, and all will be lifted together.*
* See Fig. 426, which shows the method of performing a similar feat
with four lucifer matches, the fourth lifting the other three.
3^4
Puzzles Old and New.
No. XXIV.— A Feat of Dexterity. Solution.
With the left hand take hold of the hanging portion of
the handkerchief and raise it to a horizontal position, in as
straight a line as possible. Then with the right hand make
a quick downward "chop " at the cloth at about six inches
distance from the table. The cloth will be drawn away with
a jerk, but the glass will remain undisturbed, the vis inertice
of the latter overcoming the very slight friction occasioned
by the removal of the cloth.
A neophyte, attempting to solve this puzzle, endeavours
to pull away the cloth by a fraction of an inch at a time.
In such case the glass inevitably comes with it.
rTo. XXV.— The Divided Square. Solution.
The converse of this problem has been dealt with under
the head of Dissected or Combination Puzzles (No. XXI.),
but many persons who are well acquainted with the method
of forming a single square out of five smaller squares would
be quite at a loss if they were asked to perform the opposite
process. It is, however, extremely simple when you know
how it's done.
Suppose A B G D (Fig. 505) to represent the square to be
divided. Find the centre of each side, represented by the
letters E F G H. Draw straight lines from Hto B, B to F,
A to G, and E to C. Cut the. cardboard through the lines
thus marked. This will give us nine segments, which we
G C
Fig. 505.
JS
6 1
1 5
6
3
4
Fig. 506.
will distinguish accordingly by the numbers 1 to 9 inclusive.
We proceed to rearrange the various segments as shown in
Fig. 506, and the problem is solved.
Key to Miscellaneous Puzzles.
385
No. XXVI.— The "Oval" Problem. Solution.
Roll the paper into the form of a cylinder, and then pro
ceed as though you desired to describe a circle upon it.
When the paper is unrolled, you will find that, instead of a
circle, you have an oval.
No. XXVII.— The Floating Ball. Solution.
Just as the lips touch the ball, inhale vigorously, and the
ball will be drawn towards them by exhaustion of the air.
Maintain the exhaustion till you have fairly lifted the ball,
and then let it fall from the mouth to the hand.
No. XXVIIL— The Cut Playing Card. Solution.
Fold the card down the centre, and cut through the line
a
Fig. 5C7.
thus made to within a quarter of an inch of each end. The
card will then be as a (Fig. 507). Xext, with a sharp pen
c c
3 86
Puzzles Old and New.
knife, cut through both thicknesses, alternately to right and
left, but each time stopping within a quarter of an inch of
the edge, as b. The cuts should be about an eighth of an
inch apart. "The card when opened will be as c. Open it
out still further, when it will form an endless strip, of such
a size as to pass easily over a person's body.
No. XXIX— The Mitre Puzzle. Solution.
Fig. 508.
Divide as indicated in Fig. 508.
No. XXX. The Five Straws. Solution.
Interlace the five straws after the manner shown in Fig.
509, the shilling forming a .sort of wedge, locking all to
Fig. 509.
gether. They may then be lifted by the end of one straw as
required by the puzzle.
Key to Miscellaneous Puzzles. 387
No. XXXI.— The Three Fountains. Solution.
This, though at first sight perplexing, is really a very
simple problem. Fig. 510 represents one method out of two
Fig. 510.
or three, either of which would answer the conditions of the
puzzle.
No. XXXII.— The Two Dog's. Solution.
Add lines as shown dotted in Fig. 511, and turn the pic
ture partially round, so that what was originally its side is
Fig. 511.
now its top. The dogs will now appear not only to be alive,
but to be running at full speed.
3 88
Puzzles Old and New,
No. XXXIIL— Water Bewitched.
Fill a wineglass with water to the brim. Lay a piece
of card flat upon it, and turn it over, keeping the card mean
while in place with the hand. When the glass is inverted
the card will not fall, though the hand be removed, neither
will the water run out, being kept in position by atmos
pheric pressure. Place the glass thus inverted on a smooth
wooden table, near the edge, and cautiously draw away the
card. The water will still not run out so long as the glass
is not moved, but the moment any one lifts it the whole
will be spilt.
No. xxxiv.— The Balanced Halfpenny. Solution.
The first step is to bend the hairpin into the form shown
Fig. 512.
at a (Fig. 512). Use the narrow loop of this as a clip to
Key to Miscellaneous Puzzles. 389
hold the halfpenny, bending the wire closer and closer till you
have the coin secure. Hang the ring on the hook at the
opposite end of the wire, and then proceed to balance it as
shown in our ill a st ration. A good many trials will probably
be necessary before you are able to ascertain the precise
point to which to apply the pin ; but, this once found, you
may even set the coin spinning (by gently blowing upon the
ring) without destroying its equilibrium.
No. XXXV.— The Balanced Sixpence.
The first step is to fix the needle, point upwards, in the
cork of the winebottle. The next, to cut a slit, a quarter of
an inch deep, across the top of the smaller cork, and to press
the sixpence as far as it will go into the cut so made.*
Holding the cork with the coin downwards, thrust the
two forks into it (one on either side) in an upward direction,
at an angle of about
Fig. 513.
30° to the centre
of the cork. Now
bring the edge of the coin carefully down upon the point of
the needle (see Fig. 513), and, if the forks are properly
adjusted, it will remain balanced, and the cork may even be
spun round at considerable speed with little fear of dis
placing it.
* The diagram is hardly accurate in this particular. The slit
in the cork should be deep enough to admit about one half the diameter
of the sixpence.
39°
Puzzles Old and New.
No. XXXVI.— Silken Fetters. Solution.
This is a very simple matter, though, like a good many of
the puzzles contained in this book, it is perplexing enough
till you know "how it's done."
Let the gentleman pass the bight of his own ribbon (from
the arm outwards) through the loop which encircles one of
the lady's wrists, over the hand, and back again, when it will
be found that they are freed from the link which united
them. Their individual bonds must, of course, be removed by
untying in the ordinary way.
No. XXXVI r.— The Orchard Puzzle. Solution.
Fig. 514.
The orchard was divided as shown in Eig. 5J4.
No. XXXVIII.— The Cook in a Difficulty* Solution,
Fm. 515.
The four pieces are arranged as shown in Fig. 515, when
they will mutually support each other. The small black
Key to Miscellaneous Puzzles.
39i
semicircles seen in the diagram represent in each case the
visible portion of the upper end of the supporting pillar.
No. XXXIX.— The Devil's Bridge. Solution.
Fm. 516.
The three pieces are arranged as shown in Fig. 516.
No. XL,
The secret lies
brought together,
the palms of both
sequence we have
palm of the rigid
outward, in the act
get in each other's
least difficulty.
—The Two Corks. Solution.
in the position of the hands as they are
The uninitiated brings them together with
turned towards the body, with the con
described. To solve the puzzle, turn the
hand inward, and that of the left hand
of seizing the corks. They will then not
way, but may be separated without the
No. XLL— The Divided Farm. Solution.
•
♦
# 11
f
f
*
*
#
Fig. 517.
The farm was divided as shown in Fig. 517.
,92
Puzzles Old and New.
No. XLIL— The Conjurer's Medal. Solution.
Starting with the ring in the hole a, turn it round till the
gap in it is next to the opening through which it passes, and
slide it down to the opening b (Fig 481.) Turn the ring half
round, and pass on the cut part to c. Thread the ring into
this hole, and out of b. When it is once fairly in c, you have
only to turn it once more half round, and it will be free.
To reinsert the ring in a, the process must be reversed.
No. XLIII.— The Maze Medal. Solution.
Insert the ring at the point A, and slide the open portion
along into the space marked 1. Now turn the ring round
till the opening comes outside the . medal ; pass it in
again at the point B, and slide it along till the opening
reaches the hole 2. Turn the opening back to 1, and
thence slide it into the hole marked 3. Turn the opening
back to 2, and thence slide the ring into hole 4. Turn
the opening back to 3, and slide it on to, 5. By turning the
opening back to 4, next turn the opening back to 5, work
it into 6, and thence out at A, to the outside of the medal.
Work the opening back to 6, and thence pass it, along the
dotted line, into the final hole — Home.
To extricate the ring, proceed in precisely reverse order.
No. XL1V.— The Puzzle KeyRing*. Solution.
To clearly explain the construction of the ring, it will be
Fig. 518.
Fig. 519.
needful to illustrate its two portions separately. See Fig.
518, in which a represents the larger horseshoe, or ring
Key to Miscellaneous Puzzles.
393
proper ; and c the smaller horseshoe, which fills up the
opening. This latter is grooved all round, and revolves
freely between the two arms, b b. To place a key on the ring, c
is turned round so as to bring its opening outward, as in Fig.
518. The bow of the key is placed within the opening, and c is
then again turned round, as in Fig. 483, when the key will
fall within the ring.
To remove the key, reverse the process.
No. XLV.— The Singular Shilling*.
Take hold of two diagonally opposite corners of the hand
kerchief, with the thumb on the upper surface of each, and
stretch vigorously. The handkerchief will be found to form
a tense fold, or " overlap," on either side of the coin ; and if
the handkerchief, still in this condition, be lifted into a per
pendicular position, the shilling will remain gripped in the
fold, and will not fall.
No, XLVL— The Entangled Scissors. Solution.
Pass the loop through the opposite bow, and over the ends
of the scissors, when they will be free.
No. XLVIL— The Penetrative Penny. Solution.
Fold the paper exactly across the centre of the hole ; then
lake it in both hands, and ask some one to drop the penny
Fig. 520.
into the fold. Let it rest just over the hole, its lower edge
projecting below. Bend the corners of the paper slightly
upwards, as indicated in Fig. 520. This elongates the
opening, and, if the movement be continued, the penny
will, after a second or two, fall through by force of its own
weight. The paper remains uninjured.
394
Puzzles Old and New.
No. XLVIIL— The Packer's Secret. Solution.
As the reader will doubtless have guessed, the secret lies
in a sort of starlike arrangement of the counters in the box.
Turn out all but seven of the counters, and of these place
one in the centre of the box, with the other six round it, as
Nos. 1, 2, 3, 4, 5, 0, 7 in the annexed diagram (Fig. 521).
Fig. 521.
Keeping these steady by pressing the fingers of the left hand
upon them, proceed to place the remaining counters as indi
cated by Nos. 8, 9, 10, 11, 12. But yon are still one counter
short. This you supply by removing the centre counter
(No. 1), and placing it in the spot marked 1A.
The whole are now securely" wedged together, and the open
box may be turned upside down without any fear of displacing
them.
THE END.
Eutler & Tanner, The Selwood Printing Woiks, Frome, and London.
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