150

Challenging and Instructive

Puzzles

Pierre Berloquin
TRANSLATED BY
Charles Scribner, Jr.

BARNES
&.NOBLE
8 0 0 K S
NE W YORK
 Copyright © 1981 by Bordes, Paris, France.
Translation copyright © 1985 by Charles Scribner's Sons.

English translation by Charles Scribner, Jr.

This edition published by Barnes & Noble, Inc.

by arrangement with Pierre Berloquin.

Reprinted by arrangement with Scribner,

an imprint of Simon & Schuster.

All rights reserved. No part of this book may

be used or reproduced in any manner whatsoever
without the written permission of the Publisher.

1996 Barnes & Noble Books

ISBN 0-76070-116-4

Printed and bound in the United States of America.

� 9 8 7 6 543 2 1
FG
 For Dominique
who so often knows why
Problems
1. Water in the Desert 3
2. Pure Reason 3
3. The Age of His Digits 4
4. The Exchange of Knights 4
S. Squaring the Maltese Cross 5
6. Reconstructing Octagons 6
7. Basic Multiplication 7
8. A World of Fours 8
9. Circular Uncertainties 9
10. The Random Chord 10
11. What Coincidences? 10
12. Meetings on the Dot 10
13. Steer Metabolism 11
14. Lightning Calculators 11
15. Heterogeneous Squares 11
16. Antimagic 12
17. Strategic Dates 12
18. Manual Tactics 13
19. Two-Handed Confrontations 13
20. The Best Bridge 14
21. The Best Place for the Bridge 15
22. Thrift in Mirrors 15
23. Multiple Images 16
24. A False Proof 16
25. Visualizing Tetrahedrons 17
26. Truth and Brotherhood 17
27. Doubts About Euclid 18
28. The Sum of the Angles 19
x· CONTEN T S

29. Jet Psychology 20

30. The Parson and His Sexton 21
31. Revealing Ambiguities 21
32. An illegitimate Equilateral Triangle 22
33. Division by Geometry 23
34. Dismembering One Thousand 23
35. How Many Elevators? 24
36. The Queen Takes a Walk 24
37. An Attractive Number 25
38. If and Only If 25
39. Curvilinear Mystery 26
40. Less Water in the Desert 26
41. Superfluous Digits 27
42. A Difficult Juncture 27
43. Using Your Imagination 28
44. Father, Son, and Horse 28
45. Archimedes Trisects the Triangle 29
46. Predictable Pain 29
47. Watch Out for the Train 30
48. Knight Moves 30
49. Playing with Blocks 31
50. IQ Logic 31
51. Magic Cubes 32
52. Three Strokes of the Compass 33
53. To Be Continued 33
54. Five Folds 33
55. Nine Lines Make Twenty-One Triangles 34
56. A Cube of Bricks 34
57. Kicking a Goal 35
58. Other People's Husbands 35
59. Sequences and Gaps 36
60. Trick Squares 36
61. Circles and Square Roots of Seven 37
62. An Adequate Supply of Digits 37
63. Pawn Parity 38
64. Trick and Countertrick 39
65. Squaring a Star 39
66. Matchmaking 40
67. Prime Magic 41
 CONTENT S ' xi

68. Constructive Folds 41

69. Three and Five Make Four 42
70. The Price Is Right 42
71. Newton's Steers 42
72. Sixteen Equals Four Times Fifteen 43
73. Truncated Cubes 44
74. Leaping Counters 45
75. Trickery and Magic 45
76. Black and White Dots 46
77. Christmas Dinner 46
78. Flawless Cubes 47
79. What Times Are They? 47
SO. Beyond Thirst 48
81. Kings, Queens, and Knaves 48
82. Folding, Folding 49
83. Whodunit? 50
84. A False Unknown 50
85. Practical Trisection 51
86. The Bandbox 52
87. A Multicolored Chessboard 52
88. Breaking the Chain 53
89. What's It Good For? 53
90. Reversible Magic 54
91. A Cross and a Square 54
92. The Finishing Touch 55
93. Back to Trisection 55
94. The Symmetrical Cross 56
95. Century Sundays 56
96. To Have or Not to Have 56
97. The Area of the Slice 57
98. Integral Triangles 57
99. Vases 58
100. Writer's Cramp 58
101. Lining Up Blacks and Whites 58
102. More Alignments 59
103. Trisection in the Enlightenment 59
104. The Odds on the White Card 60
105. Pi in an Unfamiliar Context 60
106. Toward Unity 61
xii . CO NTENTS

107. An Accidental Ellipse 61

108. Your Basic Boat 62
109. Points and Lines 62
110. Squaring the Square 63
111. The Missing Token 63
112. Quadrature and Dissection 64
113. An Uneasy Cryptogram 64
114. Roots 65
115. The Triangulation of the Triangle 65
116. Travels on the Hexagon 66
117. A Bailout Fee 66
118. A Prize Hamster 66
119. The Women Man the Oars 67
120. Breaking Up a Year 67
121. Protagoras at the Bar 68
122. Squaring Polygons 69
123. Drawing the Line 70
124. A Band of Five Squares Creates Another Square 70
125. The Fifth Power 71
126. A Line to Follow 71
127. A Tricky Division 72
128. Unlisted Numbers 72
129. No Way to Make a Square 73
130. A Primary Law 73
131. Ten Digits for One 73
132. Inside Out 74
133. Raindrops Are Falling . . . 74
134. Remarkable Quadratures 75
135. Mobius and His Better Half 75
136. Quadruple Quadrature 76
137. Something New About Nines 76
138. The Trisection of Pappus 77
139. Archimedes Cuts It Up 78
140. Tails I Win 78
141. The Barrier of Thirst 79
142. Regrouping the Pieces 79
143. A Square for a Compass 79
144. Words and Numbers 80
145. Words, Numbers, and Sentences 80
 CO NTEN T S ' xiii

146. A Triangle in a Square 81

147. Mysterious Powers 81
148. Superior Antirnagic 81
149. Coupled Couples 82
ISO. No Holds Barred 82

Solutions begin on page 83.

 These mathematical puzzles have appeared in the newspaper
Le Monde during the past few years. A number of readers have
responded to the challenge by offering not only original solu­
tions and modes of attack but also by stating new problems
related to those proposed. A dialogue was thus initiated in
which my problems or those I reproduced from the literature
generated a new set by stimulating the minds of my corre­
spondents. This book presents a collection resulting from the
fruitful exchange.
 If one is being beaten over the head, it is a delight when the
blows come to an end, and so it is with brain teasers when
they solve themselves. I said: when they do.
But there is also a kind of gentle creaking sound in the act
of thought when the clue eludes us and a pleasing sort of fury
when the last hope turns out wrong, after which the mind goes
into a panic as it turns and turns in a vicious cycle.
Yet anxiety can prove delightful, too, while one catches on
to the threat of a paradox, even though contradictory results
in computation darken one's understanding. Puzzles, para­
doxes, and mental mazes form a grand trinity presiding over
the play of mind, and I have long partaken of the bitter brew
supplied by enigmas and conundrums.
All these mysteries have opened my eyes to the nature of
the world of play. I have learned to suspect the trap concealed
under some innocent statement and the dead end around the
comer from common sense. Never again shall I take the world
for a simple business. I have left the common ground for the
Garden of the Sphinx where, remote from the deceitful plati­
tudes of everyday life, I shall forever be found seeking the
solution.
 PROBLEMS· 3

1. Water in the Desert

The desert lies before you. Your mission is to plant a flag
a four days' march into the interior. You do not have any
special equipment and must rely on your own powers,
although you may enlist the help of one or more com­
panions.
Carrying food and the flag itself is not a problem. The
only limitation has to do with water: each person can
carry only a five-day supply of water. Thus, if you went
by yourself, the amount of water you could bring with
you would take you only two and a half days toward your
destination and back to your starting point. Given these
conditions, how can you fulfill your mission without using
more than a twenty-day supply of water and without
enlisting more than three companions?

2. Pure Reason
At the beginning of this century, long before groups and
relations were in vogue, algebra itself was not well re­
ceived in educational circles. There was some doubt as to
its pedagogical value. Logical arguments were considered
preferable to unknowns and equations. Here we have an
opportunity to return to that era and rediscover the utility
of "pure reason."
It takes a train 7 seconds to pass by a stationary observer
and 26 seconds to traverse the length of a station 380
meters long.
What is the speed of the train and its length?
4 THE GARDEN OF THE SPHINX
.

3 . The Age of His Digits

Let us remain pure. Here again, as in the case of the
enigma in problem 2, unknowns and equations are not
permitted. How old was a person in 1898 if the age in
question is the sum of the digits of the year of his birth?

4. The Exchange of Knights

This problem can be solved directly by prolonged trial
and error, but it is more interesting to solve it more sys­
tematically by means of a shortcut.
On a chessboard of three files and four ranks, three
black nights (black stars) are face-to-face with three white
knights (white stars). How many moves does it take for
the black knights to change places with the white ones?

***

* **
 PROBLEMS· 5

5 . Squaring the Maltese Cross

In spite of the fact that the shape o f the Maltese cross is
radically different from that of the Greek cross, it is pos­
sible to transform it into a Greek cross. Several cuts with
a pair of scissors and reassembling of the pieces correctly
are all it takes. Can you find the solution with eight pieces?

\ /
� 1/
"'� \ J .e!'"
..... N {.."

r-;iI"�"'7
� I � ..... ",,
......
-

/I 1\
L \.
6 THE GARDEN
. OF THE SPHINX

6. Reconstructing Octagons
A decoupage problem consists of cutting up one figure
in order to reassemble its pieces in the shape of another
figure. A closely related problem is one of assembling­
starting with several figures of identical shape and then
cutting them into pieces as economically as possible in
order to reconstruct a similar figure of greater size.
Are you able to cut these three little octagons into a
total of ten pieces and then reconstruct a single octagon
with three times the area?
 PROBLEMS' 7

7. Basic Multiplication
There is a crude way of performing multiplications that
requires nothing more than the ability to double and to
halve (Le., to multiply and to divide by 2). For instance,
suppose we want to multiply 43 by 81 . The two numbers
are written at the top of two columns. In one column, a
number is successively divided by two, disregarding re­
mainders. In the other column the second number is suc­
cessively multiplied by two until both columns have the
same number of lines. The numbers on the right corre­
sponding to any even numbers on the left are bracketed.
The remaining ones are added up to give the correct prod­
uct. Can you explain why?

43 81
21 162
10 324( )
S 648
2 1 296 ( )
1 2 592
3 483
8 . THE GARDEN OF THE S PHINX

8. A World of Fours
The expressions that employ as digits only three 4s and
the conventional mathematical symbols are endlessly var­
ied.
4 x 4!
4
4 x 44
4 x 4 x 4!!
A little research shows that one can represent most values
of the positive integers, assuming a resourceful use of
arithmetic operations.
But is it possible to construct a universal formula adapt­
able to the expression of any positive integer n?
If necessary, one can use an unlimited number of op­
erations.
 PROBLEMS· 9

9. Circular Uncertainties
A chord is drawn in a circle in a random way. What is
the probability that its length is greater than the side of
an inscribed equilateral triangle?
One way to calculate that would be to choose, for rea­
sons of symmetry, the direction of the chord and to con­
sider its intersection with the perpendicular diameter. The
chord is of maximum length when it intersects the di­
ameter in the middle of the radius. The probability in
question is therefore V2.
Another method would be to consider the midpoint of
the chord. This midpoint must lie within a concentric
circle of half the radius. Since the area of the new circle
is one quarter as large, the probability in question is 1/.4.
Joseph Bertrand proposed this problem half a century
ago as a means of criticizing continuous probabilities and
added a third method of calculation that leads to a result
contradicting the two preceding results.
What is this method?
10 • THE GARDEN O F THE SPHINX

10. The Random Chord

If a chord is drawn on a circle "at random," the probability
that it will be greater than the side of an inscribed equi­
lateral triangle depends on the point of view of the cal­
culator (see problem 9) and may be %, %, or Y4.
A reader of Le Monde has shown that if one adheres to
the ambiguous reasoning that has produced the paradox,
one can contrive to "vary" the probability in a continuous
fashion between 0 and 1 . By what construction can that
variable probability be produced?

11. What Coincidences?

Is it possible that one could suddenly wake up, look at
the alarm clock, and say, "That's strange! It has only one
hand." At that instant, the three hands of the clock co­
incided exactly.
The three hands of the clock, which indicate the hour,
the minute, and the second, are in exact alignment at
twelve o'clock. Can they coincide at any other time in
twelve hours?

12. Meetings on the Dot

The hour, minute, and second hands of an ordinary alarm
clock cannot coincide exactly (except at twelve. 0' clock).
At what other moment in the day are the three hands
closest to perfect alignment, or, in other words, lying
within the smallest acute angle?
 PROBLEMS ' 11

13. Steer Metabolism

A diet with precise goals is not always as simple as it
seems. Have you any idea what it entails? A steer weigh­
ing 630 kilograms requires 13,500 calories a day for its
"maintenance fodder," which keeps it as it is, in good
health and without getting fatter. That amount of food
turns out to be proportional to its external surface. How
many calories does a steer of 420 kilograms require for its
maintenance fodder?

14. Lightning Calculators

"Bona fide" calculating prodigies in reality combine ex­
ceptional abilities with certain simple calculations in order
to conserve their energies. Suppose that one of these should
undertake to calculate in his head the sixty-fourth root of
a twenty-digit number. If he knows in advance that the
result is an integer, does he need to know all the digits
of the first number?

15. Heterogeneous Squares

Oassically, a square containing the first of n2 integers is
"magic" when its lines, columns, and diagonals all add
up to the same sum. Can one imagine a property opposite
to magic? Is there such a thing as nonmagic? A first step
consists of defining "heterogeneous squares." In such a
square, each of the sums of its lines, columns, and di­
agonals would be different. Why is there no heteroge­
neous square of order 2 (containing 1, 2, 3, 4)? Can you
construct one for order 3?
12 • THE GARDEN OF THE SPHINX

16. Antimagic
In an IIantimagic" square the negation is more systematic
than in a heterogeneous square (see problem 15) . The
sums of the rows, columns, and diagonals are not just
different: they must form a sequence. There is no anti­
magic square of order 3, and the existence of one appears
to be impossible. Can you construct one of order 4 (there
are twenty known examples) and one of order 5?

17. Strategic Dates

Two contestants play at exchanging dates. (They do not
take the year into account. ) The first must name a day in
January, for example, January 14. Then each player in
tum picks a later day, retaining either the month or the
date just named by his adversary.
For example, after January 14, it is possible to name
January 16 or January 20, April 14 or October 14. One
contest might begin as follows: January 14, April 14, April
27, April 30, June 30 . . .
The winner is the first to reply December 31 . Which of
the players is able to clinch the victory by an unbeatable
strategy and how?
 PROBLEMS ' 13

18. Manual Tactics

Two players compete with one another by exchanging
prime numbers by signs. Each player, in turn, holds up
a number of fingers on one hand. (Zero is not allowed . )
The cumulative total must remain a prime number. One
match might run as follows:
A: 1
B: 1 (+ 1 = 2)
A: 3 (+ 2 = 5)
B: 2 (+ 5 = 7)
What unbeatable strategy can the first player adopt?

19. Two-Handed Confrontations

Two contestants compete by exchanging prime numbers
by digital signs as in problem 18. But this time each player
in turn can use the fingers of one or both hands, zero
again being excluded.
What is the winning strategy available to the one who
plays first?
14 . THE GARDEN OF THE SPHINX

20. The Best Bridge

,
,
fa
,
I
I
I
I
I
I
I

-----... b
�
... -
--
-

Two towns are situated on the same shore of a river. The

bank of the river forms a straight li ne. There is a project
to build a road that extends in a straight line from one
town to a bridge across the river and then extends in a
straight li ne to the second town. In this way the same
bridge can serv e both towns as a means of crossing the
river.
What is the sim plest way to find the best place to build
the bridge, the goal being to minimiz e the length of the
road from one town to the other?
 PROBLEMS ' 15

21. The Best Place for the Bridge

Th e towns a and b are situated on opposite sides of a
canal, the banks of which form two parallel lines. In this
case, as distinguished from problem 20, the width of the
canal must be taken into account. The object is to plot a
road in a straight line from town a to a point c' on the
canal, where there is a bridge perpendicular to the banks
of the canal, across the bridge to the point d on the other
side and then proceed in a straight line to town b . Where
should the bridge c'd lie to provide the shortest path from
town a to town b ?

.b
\
\
\
\
\
\
\
\
\
\

"\
\

- - --
-- - -
d C -.

22. Thrift in Mirrors

It would seem that we live in an excessively luxurious
world. E ven our mirrors are larger than they need to be.
If I am five feet six inches tall, how high a vertical m irr or
do I need to see all of me at once if I stand in front of it
and do not move m y head?
16 • THE GARDEN OF THE SPHINX

23 . Multiple Images
I am at a point in front of two mirrors that rotate about
a vertical axi s lying in the plane that bisects them.
Dependin g on the angle between the mirr ors, how many
images of myself will I see?

24. A False Proof

Yes, your memory serves you well. There is only one
perpendicular from a point to a plane . In that case, what
is wrong wi th the following demonstrati on, whi ch " proves"
that there is an infinity of perpendiculars from an external
point to a given plane? Let p be the external point and a
and b be two arbitrary points on the plane. Two spheres
with diameters pa and pb cut the plane in two circles that
intersect at c and d. Since c is on the circles of diameters
pa and pb (the great circles of the spheres), pc is orthogonal
to ca and cb, and therefore to the plane . Likewise, d is
certainly on the circles of diameters pa and pb, and po is
orthogonal to da and dp, and accordingly, orthogonal to
the plane. Whatever the points a and b may be, c and d
can have an infinite number of values, and thus there are
an infinite number of perpendiculars .
 PROBLEMS ' 17

25. Visualizing Tetrahedrons

How good is your geometric imagination? Are you able
to visualize in three-dimensional space? Are you able to
construct eight tetrahedrons (not necessarily regular), in
pairs, each having a portion of a side in common? This
common portion may not be reduced to a line or point,
nor can it be shared by more than two of the tetrahedrons.

26. Truth and Brotherhood

There are two brothers who are both scrupulously truth­
ful, with only one exception: each one lies about his birth­
day on the day of his birthday.
If you ask them on New Year's Eve what their birthdays
are, one will say "yesterday" and the other will say "to­
morrow. " On New Year's Day, they both give the same
answers. What is the birthday of each?
18 • THE GARDEN OF THE SPHINX

27. Doubts About Euclid

Is the parallel axiom really necessary? Here is a proof that
does not employ it in order to establish a theorem that
could be substituted for the famous axiom.
Let us "show" that if two parallel lines are cut by a
transversal, the sum of the interior angles on the same
side of the transversal is equal to two right angles. Let a,
b, c, and d be the internal angles determined by the trans­
versal. Three hypotheses are possible, depending on
whether the sum of the angles on the same side are (1)
greater than two right angles; (2) less than two right an­
gles; (3) equal to two right angles.
The first hypothesis entails that a + d > 180°,
b + c> 180°, and therefore a + b + C + d > 360°. But
a + b = c + d = 180°, from which it follows that
a + b + c + d = 360°.
The hypothesis is contradictory and must be aban­
doned; the same thing occurs with the second by revers­
ing the inequalities. There is only one remaining possi­
bility: the sum of the internal angles is equal to two right
angles.
Where is the error in this line of reasoning?
 PROBLEMS ' 19

28. The Sum of the Angles

As a further doubt about the validity of the parallel pos­
tulate (see problem 27), one can appear to prove directly
that the sum of the angles of any triangle is equal to 180°.
Take any triangle and divide it arbitrarily into two tri­
angles. The internal angles of the figure are a, b, c, d, e,
and f. Let us designate as x the sum of the angles of a
triangle-what we are seeking to determine.
We have a + b + f = x = c + d + e and a + b
+ c + d + e + f = 2x. But e and f, being adjacent and
supplementary, have the sum 180°. Therefore:
a + b + c + d + 180° = 2x
x + 180° = 2x
x = 180°
Where is the error?
20 • TH E G A R D E N OF THE S P H I N X

29 . Jet Psychology
A cylindrical container 4 meters high isfu ll of water. Three
holes are made in it in a vertical row, 1, 2, and 3 meters
from the ground, respectively. How will the drops of
water land that spurt from the container at the beginning
of the experiment? Will they strike the ground at the same
place, and, if not, how will their points of fall be distrib­
uted in relation to the container?
 PRO BLEMS ' 21

30. The Parson and His Sexton

A parson once said to his sexton, "Today I saw three of
our parishioners. The product of their ages was 2,450.
Can you tell me what their ages were?"
Sexton: No.
Parson: If I add that the sum of their ages is twice
yours, can you give me the answer?
Sexton: Not yet.
Parson: I'll also tell you that the oldest is older than
I am.
Sexton: Now I have all the information I need.
If we assume that the parson and his sexton are both
able mathematicians, what are the ages of the three pa­
rishioners?

31. Revealing Ambiguities

Simon and Paul are two friends; x andy are two integers
between 2 and 99, inclusive. The only thing that Simon
knows about the two integers is their sum, x + y, while
the only thing Paul knows about them is their product,
xy. In this situation, each of them attempts to identify x
andy.
Simon says to Paul: "I do not have enough informa-
tion."
Paul replies: "Me, neither."
At that point Simon says, "Now I do."
Then Paul says, "So do I. "
Can you determine a pair of numbers consistent with
their dialogue?
22 • THE GARDEN OF THE SPHINX

32. An Illegitimate Equilateral

Triangle
Inasmuch as it is impossible to trisect an angle using only
a compass and straightedge, classical geometry takes very
little interest in the triangle formed by the trisectors of an
arbitrary triangle. It is, in fact equilateral, as is usually
established by trigonometry. Nevertheless, it is possible
to prove it geometrically. Can you demonstrate such a
theorem?
 PROBLEMS ' 23

33. Division by Geometry

Is 54 divisible by 3? In geometry that is not always evident.
Let us test it by experiment.
In this rectangle of 56 squares, two squares have been
removed, leaving 54-a number divisible by 3. Is it pos­
sible to cover all the squares of the remaining figure with
triple dominoes (i.e. units made up of three squares in a
row)?

§
34. Dismembering One Thousand
The number 1,000 can be expressed in a great many ways
as the sum of four even positive integers (zero excluded).
For example, 1,000 = 2 + 4 + 66 + 928. It can also be
expressed in a great many ways as the sum of four odd
integers, such as 1 + 3 + 5 + 991 = 1,000. Of these two
ways of breaking up 1,000, which are the more numerous,
the sums of even numbers or of odd numbers?
24 • THE GARDEN OF THE SPHINX

35 . How Many Elevators?

An apartment house has seven floors (above the ground
floor) and a number of elevators. Each elevator travels
between the ground floor and the seventh floor. But in
order to conserve energy, each elevator stops only at three
of the six intermediate floors. Its stops at the other three
are omitted. Since it is desirable to be able to get from
any floor to any other floor without having to change
elevators, how many elevators must be installed, and what
floors do they serve?

36. The Queen Takes a Walk

The queen in chess is able to cover every square in the
standard 8 x 8 chessboard and return to her starting point
in sixteen moves. This kind of closed circuit ends up at
the starting point, but if one permits her to traverse a
single square several times, she can accomplish her trip
in only fourteen moves. How can one plot that trip (which
is the shortest one achieved to date)?

- - - - - - -.
,,- - - -

....
r -,- - - - -
,

I ,

I r- - � ....
I '
, ,
• ->

__ .J

L _ _ __ _ _ _ _ �
 PROBLEMS · 25

37. An Attractive Number

The number 495 has a strange attraction for three-digit
numbers. For example, take any three-digit number (in
which the digits are not all the same); for this case, let us
pick 265. Rearrange its digits in decreasing order to pro­
duce 652. Now reverse the order of the digits to 256 and
subtract the last number from the next-to-Iast one:
652 - 256 = 396. Perform the same operation on 396:
963 - 369 = 594. Do the same for 594: 954 - 459 = 495.
As Kaprekar discovered, no matter what number you start
with, it will invariably lead to 495. Why is that?

38. If and Only If

A businessman wishes to give salary increases to two of
his three employees: Tom, Dick, and Harry. He has ar­
rived at a statement that summarizes the logical problem
presented by the choice he must make. If and only if I
give Tom or Dick or both a raise and do not raise Harry,
then I shall raise Tom; and if I raise Dick, I shall raise
Harry.
In light of this condition, who will get a raise?
26 • THE GARDEN OF THE SPHINX

39. Curvilinear Mystery

Curvilinear figures are more mysterious than rectilinear
ones. Can you calculate the area of the curvilinear square
produced by the four quarter circles centered on the apexes
of the big square?

40. Less Water in the Desert

Nothing is more stimulating to the intellect than a desert
trip. Following the publication of problem 1 and its so­
lution, I received numerous letters from readers who suc­
ceeded in planting the flag in the required spot and at
the same time using less than fifteen days' worth of water
and also with a reduced crew. Can you figure out how
that can be done?
 PROBLEMS ' 27

41. Superfluous Digits

This example of long division is uniquely determined by
the single digit appearing in the quotient. All the missing
digits, which are indicated by dots, can be deduced. See
if you can recover them.

. . 8. .

42. A Difficult Juncture

What happens when two cylinders intersect? Take two
cylinders of radius 1 whose axes of symmetry are per­
pendicular to one another in the same plane. What is the
volume of their intersection?

R
O=7�
U
28 . THE GARDEN OF THE SPHINX

43 . Using Your Imagination

You are given a three-digit number, the last, middle, and
first digits of which are the 100th, 200th, and 300th digits,
respectively, in the decimal expansion of 1i'. Square the
number, double the result, and raise the double of the
square to the fifth power. Now subtract the next-to-Iast
result from the last. Without using mathematical tables,
calculators, or extensive computation, can you determine
the units digit of the number resulting from the final
subtraction?

44. Father, Son, and Horse

A man and his son must make a 60-kilometer journey.
They have a horse for the trip that is able to average 12
kilometers an hour. The problem is that the horse can
carry only one person at a time. If one person rides, the
other must walk. Assuming that the father walks at a rate
of 6 kilometers an hour and his son 8, how many hours
will the trip take if they arrive at their destination at the
same time?
 PROBLEMS · 29

45 . Archimedes Trisects the

Triangle
One of the most famous insoluble problems of geometry
is to find a general method of trisecting an angle by means
of a straightedge and compass only. In this case, "insol­
uble" does not mean that the correct construction has
eluded mathematicians up to now; rather, that it has been
demonstrated to be impossible in principle, since the prob­
lem is equivalent to finding a general solution to the equa­
tion of the third degree without any irrational number
other than 0.
On the other hand, if one liberalizes the "rules" of the
problem to permit devices in addition to the straightedge
and the compass, it can be readily solved. One method,
attributed to Archimedes, depends on inscribing on a sep­
arate piece of paper (or on the straightedge itself ) marks
representing the distance between the endpoints of a line
segment. Can you reconstruct that method?

46. Predictable Pain

A logician once observed that one of his eyelashes was
growing into his eye and was irritating the cornea. So he
plucked it out. When the complaint recurred, he decided
to study the problem. In this way, he noticed that the
irritation flared up two days later, and two days after
that, and then at successive intervals of 5, 3, 1, 3, 4, 3,
3, 2, 2, 6, 1, 2, etc. Has our logician grounds for continuing
to believe that the eyelash follows a regular and predict­
able growth pattern? Is he able to deduce a precise law
governing the phenomenon?
30 • THE GARDEN OF THE SPHINX

47. Watch Out for the Train

A man is walking along a one-way railway bridge. He is
two-thirds of the way over it when he sees a train ap­
proaching him at the rate of 45 miles per hour. In this
situation he is able to escape in the nick of time by running
at a constant speed. The interesting thing is that it makes
no difference in which direction he runs. Without any
algebra or any equation, can you calculate his speed?

48. Knight Moves

On this chessboard of 3 x 3 squares, a knight can make
sixteen different moves (two moves from each of the outer
squares).
How many moves can it make from a chessboard 7 x 9,
or more generally, p x q?

�\
\

'.

•
\
 PROBLEMS ' 31

49. Playing with Blocks

A set of children's blocks has exactly the right number of
pieces to construct two separate squares or a cube, the
side of which is equal to the difference between the sides
of the two squares. What is the smallest number of blocks
for such a set, and what are the two constructions?

---------.
" , '.
,
.
"
,
,' .
f,---- - -,.-,;;;.
�-+_
, I
.....
-
-
•
I
•
I
I
I

't:=�:::::::11

50. IQ Logic
I n this puzzle, let us grapple with the sort of problem in
arithmetic reasoning that has become such a fixture in
intelligence tests. We start with the sequence
1 1 1 1 1
:5 45 117 221 357'·'
I f the terms of this sequence follow a logical pattern,
what is the next fraction? What is the sum of the first
thirteen terms?
32 . THE GARDEN OF THE SPHINX

51 . Magic Cubes
Filling i n a magic square takes care and time. You have
to distribute the first n2 whole numbers in conformity with
2n + 2 constraints: the sums of the lines, columns, and
diagonals have to be equal to one another. B ut to fill in
a magic cube calls for the next higher degree of patience,
for you must distribute n3 whole numbers in conformity
with 6n + 4 constraints: to create this equality along the
edges and the four long diagonals. Can you fill a 3 x 3 x 3
cube with the first 27 whole numbers?

.
: ;).:.}\).; :
.

3 :. :;:;:.:. ::. ::::.;..:;::::= .: .:;:.::

. .
. .
.

. ..
.

>}!/ .\;:
. . .
.
:;
. ... . ... ... . .. .. . . :. ..
.
. .
. .
.
. .
.

.
 PROBLEMS ' 33

52. Three Strokes of the Compass

Th e construction of the mean proportional is one of the
most standard of all geometric operations, provided one
assumes the right to draw parallel lines . But do you know
the simplest way of performing the construction? If lengths
a and b are lying suitably on a straight line, their mean
proportional can be determined by three applications of
a compass . H ow?

53. To Be Continued
I, 2, 3, 4, 5 . . .
What is the next number? F acing such a question on
an intelligence test is enough to daunt the most inventive
minds, which are often able to come up with a number
of answers, in addition to 6. A handbook of integral se­
quences contains no fewer than twenty-two integral se­
quences, all starting with I, 2, 3, 4, 5. Some of them go
beyond 10 in their similarity to the sequence of natural
numbers. Can you find at least six such sequences?

54. Five Folds

Given a piece of paper in the shape of a square, without
using any instrument, how can one produce a smaller
square the area of which is three-quarters that of the
original square?
34 . THE GARDEN OF THE SPHINX

55 . Nine Lines Make Twenty-One

Triangles
Arranged symmetrically in this particular way, nine l ines
produce only nine triangles, provided one does not count
the triangles that overlap one another. But it is possible
to improve this score considerably. In fact, it is possible
that nine lines can produce at least twenty-one non­
overlapping triangles.

56. A Cube of Bricks

We have on hand twenty-seven parallelepiped bricks
measuring 2 x 1 X V2. H ow can one assemble these to
produce a cube of dimensions 3 x 3 x 3?
 PROBLEMS ' 35

57. Kicking a Goal

A soccer player in possession of the ball is running down
a sideline toward the opposing goalposts. Along that line,
what is the best point for him to shoot a goal? In other
words, what is the point when the goalposts can be seen
to offer the widest angle? One supposes that the field is
100 meters long and 70 meters wide. The goalposts are
7.32 meters apart.

58. Other People's Husbands

In a situation that has become a classic one in puzzle
literature, n married couples find themselves on the bank
of a river they wish to cross. They have at their disposal
a boat that can hold only (n - 1) persons at one time.
The logic of the situation demands that no wife be on the
bank or in the boat without her husband if other husbands
are there, even if the other husbands are accompanied by
their wives. Supposing that n is equal to or greater than
7, how many crossings are necessary?
36 • THE GARDEN OF THE SPHINX

59 . Sequences and Gaps

There are a great many plausible sequences of integers
starting 1, 2, 3, 4, 5 that are otherwise different. This was
shown to be so in problem 53. But it is possible to go
further in that direction. In fact, it is possible to find an
infinite number of sequences that begin with 1, 2, 3, 4, 5
and continue beyond 10, 100, or any number you may
choose. Yet each one of these will contain a gap some­
where beyond the last number specified, no matter how
large that number may be. How can a simple formula
define such a sequence? Obviously, a formula like "all
the whole numbers except 53" is unacceptable. For ex­
ample, what sequence starts with 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12 and does not contain 13?

60. Trick Squares

Given a checkerboard, for example, one with sixteen squares
(4 x 4), how does one fill the squares with numbers in
compliance with the following three rules:
I} Each square contains a positive integer greater than
1.
2} The integers in two squares meeting along a line
have a common factor.
3} The integers in two squares meeting at a single point
have no common factor.
It is possible to fill this kind of square, which we shall
call a "trick square" (since the name "magic" is used
elsewhere). The problem is What are the smallest numbers
that will work?
 PROBLEMS · 37

61. Circles and Square Roots of

Seven
In geometry at least, the circle is all-important and de­
serves admiration, if not veneration. For geometers ob­
sessed with the beauty of the circle, Quimper de Lanascol
even invented a word, "cyclolatry," in his voluminous
work on the geometry of the compass in which he com­
piled the constructions possible with the use of a compass
alone. One of these beautiful constructions is that of the
length of the square root of seven, given a unit length.
Do you know how to represent that with only four arcs
of a circle and without a straightedge?

62. An Adequate Supply of

Digits
In connection with this puzzle, let us deal with the con­
cerns of the now virtually obsolete profession of com­
positors.
It requires eleven digits to print all the numbers from
1 to 10 and one hundred ninety-two digits to print all the
numbers from 1 to 100.
Now let us approach the question in a more general
way. How many digits does one need to print all the
numbers from 1 to N if N has m digits?
38 . THE GARDEN OF THE SPHINX

63. Pawn Parity

On a checkerboard of n dimensions on which the lower
left-hand square is black (as on a regular chessboard), one
places pawns in such a way that there is exactly one pawn
assigned to any row and column.
Is the number of pawns on the white squares always
odd or even, or can it vary?
 PROBLEMS ' 39

64. Trick and Countertrick

On the basis of trick squares (see problem 16), a puzzle
fan has proposed the construction of antitrick squares in
which the basic relations have been inverted to require a
common factor at each angl e and relatively prim e numbers
for adjacent columns.
H is smallest solution is:

115 52 33 56

26 15 28 99

51 14 45 76

112 156 38 135

Is there a small er one with its highest number less than
156?

65. Squaring a Star

Is it possible to cu t this star into eight pieces that can be
reassembled into a square? What is the most symmetrical
solution?
40 .
THE GARDEN OF THE SPHINX

66. Matchmaking
The art of linking matches provides interesting analytical
problems that require care and thought. For a given num­
ber of matches, one must find the two-dimensional dia­
grams in which each match touches at least one other at
an endpoint. Thus, there are only three ways of con­
necting three matches and five ways of connecting four.
In how many ways can one connect five matches or six
or seven or eight . ?. .
 PROBLEMS · 41

67. Prime Magic

8 1 6

3 5 7

4 9 2

This magic square is the smallest one formed by nine

consecutive integers. For its size, it is unique and nearly
symmetrical in all its internal relations . Its columns, rows,
and diagonals all have the same sum: 15. If we should
decide to make such a square using prime numbers only,
what is the smallest possible square?

68. Constructive Folds

A piece of paper not only gives a mathematician some­
thing to write on; it also gives a tool for the study of plane
geometry. Its folds are automatically rectilinear and de­
termine straight lines . Various folds also provide certain
relations, certain bisectors, etc. By folding a square piece
of paper, is it possible to construct a regular hexagon?
42 . THE GARDEN OF THE SPHINX

69. Three and Five Make Four

A well-known problem consists in taking an B -pint con­
tainer filled with water, plus two empty containers of 5
and 3 pints, and then using these to measure out exactly
4 pints in the simplest way.
Can you avoid trial and error and come up w ith a
systematic and elegant solution to this problem?

70. The Price Is Right

E very morning a radio station gives its listeners the chance
to win an expensive prize, which is described in detail
except for its brand . Ten listeners attempt successively to
name the exact retail price (a whole number of dollars) .
For each of their guesses the MC indicates whether it is
correct, too high, or too low. The first listener to gi ve the
right price wins the prize. There is a strategy that will
ensure a win if one person were able to make all ten
guesses or if ten listeners were able to follow a single
strategy by teaming up. Can you describe such a strategy?

71 . Newton's Steers
Tradition attributes to Newton a curious problem that can
be solved without using calculus.
In twelve days, seventy-five steers crop a 60-acre field,
while eighty-one steers crop a 72 -acre field in fifteen days.
H ow many steers would it take to crop a 96-acre field in
eighteen days? (It is stipulated that the grass grows uni­
formly in all the fields and is the same height in all of
them at the beginning of the problem . )
 PROBLEMS ' 43

72. Sixteen Equals Four Times

Fifteen
Following up on several problems of aligning points, a
puzzle fan came up with a way of distributing sixteen
points on a plane that will simultaneously create fifteen
alignments of four points . H ere is one of the solutions:

Can you find another?

44 • THE GARDEN OF THE SPHINX

73 . Truncated Cubes
Are you a s well acquainted a s you should be w ith the
properties of polyhedrons-those special shapes that play
such important roles in the organization of space itself?
The cube, in spite of its apparent simplicity, is a
polyhedron rich in various symmetries. It holds in addi­
tion to its planes of symmetry six axes of order 2, four
axes of order 3, and three axes of order 4. We shall not
provide a diagram here. It is a tricky task to visualize the
transformations of these axes as the result of various trun­
cations of the cube. What will they become if one lops
off
1) an apex in order to create an equilateral triangle?
2) two apexes diametrically opposed in the same way?
3) four apexes that are not adj acent?
 PROBLEMS ' 45

74. Leaping Counters

Twelve numbered positions are arranged in a circle. One
starts four counters marked a, h, c, d placed on the num­
bers I, 2, 3, 4. A counter moves from one position to
another by jumping four places (empty or filled) in clock­
wise or counterclockwise direction in order to land in a
fifth position, which must be empty. After a certain num­
ber of such displacements, the four counters will end up
in the original four places, the order of the counters being
either the original one or a new one .
In how many different sequences can the four counters
end up in the first four positions?

a
b
1
12 2
3c
11

10 • 4d
9 5
8 6
7

75. Trickery and Magic

Does magic adapt itself to trickery as it permits "antitrick­
ery" as defined in problem 60? There exists at least one
square 4 x 4 filled with whole numbers such that all the
rows, columns, and the two diagonals add up to the same
sum. The numbers that are in vertical or horizontal contact
with one another are relatively prime. Those that are at
an angle to one another have at least one common divisor.
This is what we have decided to call a supertrick square.
Can you construct it?
46 . THE GARDEN OF THE SPHINX

76. Black and White Dots

H ere is one of the ways of arranging ten dots in five
alignments of two black dots and two white dots. H ow
would you arrange twenty-one dots in fourteen align­
ments of two black dots and two white dots each?

77. Chri s tma s Dinner

For six brothers of a certain family, Christmas and N ew
Year's Day are the two traditional occasions to dine to­
gether with their six wives. The meals take place at a
single round table where the brothers sit according to their
ages, always in the same chairs. They leave between them
empty places for their wives, none of whom sits beside
her husband.
If the wives never sit in the same positions twice, how
many years will it take for them to exhaust all the possible
seating arr angements?
 PROBLEMS·47

78. Flawless Cubes

Constructing cubes constitutes an unlimited universe of
puzzles, some of which are as difficult as anyone could
ask for. In a "simple" puzzle, one is set the problem of
constructing all the 3 x 3 x 3 cubes that can be made
out of L-shaped tricubes (i. e . , a solid figure in which three
cu bes are joined at right angl es). Th e diagram below shows
one solution. In the lower left-hand comer of each layer,
a dot indicates that the tricube extends upward from the
page. But this assemblage has a " flaw" inasmuch as cer­
tain pairs of tricubes overlap each other in such a way
that both are in the same vertical or horizontal layer. Is
it possible to ov ercome this defect?

79. What Times Are They?

With two ears, how many chiming clocks can you follow
at one time?
There are two such d ocks that are out of synchrony by
three seconds at most. They happen to chime the same
time at a certain hour. The strokes of the one occur every
five seconds. The other's strokes are every four seconds.
Your ears wi ll confuse two strokes if there is a smaller
interv al than a second between them. If one hears a total
of thirteen strokes, what time is it?
48 . THE GARDEN OF THE SPHINX

80 . Beyond Thirst
Let us make a determined-though purely theoretical­
assault on the problem of thirst. It is obligatory to plant
a flag in the desert a four-day march from the starting
point. The number of companions can be as many as one
chooses, but none of them can bring more than a five­
day supply of water. We have already had a solution
involving a twenty-day supply of water (problem 1) and
a fourteen-day supply (problem 40) . Can you cut this
supply further by a few days' worth? To do this, you may
have to overcome a slight inhibition.

81 . Kings, Queens, and Knaves

Three players divide up a pack of fifty-two playing cards
in a random way into three "hands," not all with equal
numbers of cards. Then each player takes one of these
hands, the third player having the smallest. In looking at
their cards, each of the first two players notices that if he
should draw at random two cards from his hand, there
is only one chance in two that neither of them will be a
face card (L e . , a jack, queen, or king) .
H ow many face cards are there in the third hand?
 PROBLEMS ' 49

82. Folding, Folding

The folding of paper offers numerous resources for geo­
metric constructions, even though it's possible only to
determine the bisector of an angle or the middle of a line
and the perpendicular to it (see problem 68) .
The figure below shows one way of making an octagon
appear on a square sheet of paper. The paper is folded
into four squares. Each square is folded into two isosceles
triangles. The bisectors of the two demi-right angles give
the sides of an octagon. The sides are clearly equal, and
one can verify that all the angles equal 1350•
Can you find another, completely different method of
determining an octagon by folding a square sheet of pa­
per?
50 • THE GARDEN O F THE SPHINX

83 . Whodunit?
Three suspects testify as follows:
A: B is guilty.
B: A just lied.
C: A is guilty.
A: The next statement of C will be true.
B: The last statement of A is false.
C: The two last statements of A are false.
Wh o is guilty?

84. A False Unknown

A B C X
D E C D
B B C D
C A e X

x x x x

In this addition of four numbers, each letter takes the

place of a different digi t, but an error has crept in. One
of the letters is incorrect. Can you tell which one?
 PROBLEMS ' 51

85 . Practical Trisection
In the nineteenth century, no one could have predicted
the technical arsenal we have at our disposal today for
drawing curves or constructing the most complex geo­
metric figures. But the mathematicians of the last century
were not wholly unequipped. They had the so-called jointed
devices. Their curious combinations of arms, joints, and
sliding parts were used to partition line segments and
angles, inverse circles, and extract the roots of quadratics.
Here, for example, is the rhomboid invented in 1871 by
Roberts for bisecting angles. The two longer limbs are of
equal length, as are the shorter ones. Can you rediscover
a similarly simple device for trisecting angles?
52 • THE GARDEN OF THE SPHINX

86. The Bandbox

Let us give the name "bandbox" to a solid parallelepiped
constructed by gluing together two cubes . We shall accept
only those constructions in which each bandbox contacts
the rest of the construction in at least one whole surface
(rectangular or square) . Given that definition, there are,
in fact, only two constructions that can be made with two
bandboxes . How many can be made with three?

( ---&.( _D
____ .-
87. A Multicolored Chessboard
If one sets out to color the squares of a chessboard, how
many different colors does it take to make it possible to
place the king anywhere on the board without his being
able to link two squares of the same color by one move?
 PROBLEMS · 53

88. Breaking the Chain

Consider a surveyor's chain made up of 23 links . How
many links must one open in order to have a set of seg­
ments that can be combined (by relinking) to form all
lengths between 1 and 23?

89. What's It Good For?

This apparatus can be used to solve a definite geometric
problem. How does one use it, and what is the geometric
construction involved? Let us specify that the curve is a
circle centering on c and that ab = be = cd . The apparatus
is as useful as the length of the vertical arm.
54 . THE GARDEN OF THE SPHINX

90. Reversible Magic

A numerical palindrome is a number that can be read
backward or forward without changing it (e. g . , 22, 747,
or 5473672763745) . Let us take the first twenty-five nu­
merical palindromes starting with 1 1 . Is it possible to con­
struct a magic square with these; that is, a 5 x 5 square
in which all the rows and columns, as well as both di­
agonals, add up to the same number?

91 . A Cross and a Square

This unique cross has an unusual property: it can be cut
into eight pieces that can be reassembled into a square.
Can you find a way to do that?
 PRO BLEMS · 55

92. The Finishing Touch

Two players confront each other over a ticktacktoe board
containing nine empty squares. Each player in turn fills
in a number of squares with x's, either along one row or
down one column, but not both. The squares he fills in
do not need to be contiguous. The winner is the one who
fills in the last square.
Given these rules, is it possible for one player to be
sure of winning, and if so, which player?

x X

93. Back to Trisection

Let us return to trisecting an angle. One of my readers
has noted that the apparatus described in problem 89
cannot be used for all angles, whatever the length of its
movable arm. Also, he proposes another device that will
work for all angles. It has seven arms and six hinges. Can
you reinvent the gadget?
56 • THE GARDEN OF THE SPHINX

94. The Symmetrical Cross

Let us go back to the cross in problem 91 with the same
aim of cutting it up to produce a square, but this time
with a different condition.
Instead of trying to find the smallest number of pieces,
we must try to make a square out of symmetrical pieces.
Can that be done?

95 . Century Sundays
January 1, 1978, fell on a Sunday. That being the case,
what day will it fall on at the start of the next century?
As a more general consideration, what is the probability
that the centuries of the Gregorian calendar commence
on Sunday?

96. To Have or Not to Have

You are playing bridge, and when the deal is made, each
player has thirteen of the fifty-two cards. Which is the
more probable of the following two distributions: (1) that
you and your partner hold all the red face cards and all
the lower cards that are multiples of 3, or (2) that you
and your partner hold none of these cards?
 PROBLEMS ' 57

97. The Area of the Slice

Is it very difficult to determine areas that are bounded by
curved lines? Can you calculate the area of this one, de­
fined by two quarter circles crossing each other as shown
in the diagram . They have the same radius and are cen­
tered on two sides of a square. The evaluation of the
shaded sections should not take more than a few lines of
calculation.

98. Integral Triangles

The right-angle triangle is familiar to every student, and
the ones that have solutions in whole numbers (3, 4, 5;
5, 12, 13; etc.) are a j oy to fledgling mathematicians, not
least because their areas can also be expressed by whole
numbers.
Is it possible to maintain these attractive properties
without the right angle? In doing so, we would maintain
the requirement of sides measurable by whole numbers
but add a new property: the sum of the sides must be
equal to the area of the triangle . Can this be done? If so,
show how to find some of these triangles.
58 . THE GARDEN O F THE SPHINX

99 . Vases
One has two vases with capacities o f 3 and 5 pints, re­
spectively, both filled with water. Without the help of
another container and by simply pouring water from one
vase to another, one wishes to end up with 4 pints of
water in a third vase of more than 4 pints that was empty
at the outset. What is the required capacity of the third
vase?

100. Writer's Cramp

A writer is beset by a curious difficulty. The closer he gets
to the end of his work, the slower he writes. When he
begins a piece of work, his daily output is proportional
to the number of pages remaining to be written. Thus,
for a certain book, it takes him ten days to write the first
page and fifty days to write the last.
What is the length of the book, and how many days
does it take for him to write it? (The remaining number
of pages is rounded out each time to the nearest integer.)

101 . Lining Up Blacks and

Whites
Let u s follow up o n problem 76, which distributes twenty­
one points in fourteen alignments of two white and two
black points each, but without symmetry. Can you find
a symmetrical solution?
 PROBLEMS ' 59

102. More Alignments

Let us go further with the alignment of black and white
points (problems 76 and 101). With only one additional
point-making twenty-two all told-see if you can create
twenty-one alignments of two white and two black points
each.

103. Trisection in the

Enlightenment
Among the preceding puzzles (89 and 93), several devices
for trisecting the angle have been described. Following
their publication, one puzzle fan wrote to me to point out
that the history of such inventions does not date back to
the nineteenth century, as recent publications on the sub­
ject would lead one to believe, but can be found as early
as the eighteenth century. The Marquis de l'Hopital found
a solution to this technical problem, which was published
posthumously in 1720 in a work entitled /IA Treatise on
the Analysis of Conic Sections and their Importance in
Solving Determinate and Indeterminate Problems." His
problem VI proposes to /Idivide any given angle into any
odd number of equal parts by means of a mathematical
instrument. " This involved a device that could be im­
mediately used for trisection. Next, with a slight modi­
fication, it could be used to divide the angle into five
parts, into seven parts, etc.
Can you reconstruct this device, which was constructed
out of jointed or sliding rods?
60 • THE GARDEN OF THE SPHINX

104. The Odds on the White Card

I shall place three cards in a hat. One is white on both
sides, the second is red on both sides, and the third is
white on one side and red on the other. I draw a card at
random, and without looking at it, place it on a table.
When I look at it now, I see that the side that is face up
is white. And I deduce from that that there is one chance
in two, or even money, that it is either the first card or
the third. I bet you four dollars to three dollars that the
other side of the card is white. Which of us has the better
bet?

105. Pi in an Unfamiliar Context

Study the three shaded lunes. A fan has trapped 'TT within
them. In fact, it is possible to cut up remaining portions
of the circle into six pieces that can be used to create a
regular hexagon that, in tum, in a familiar cutting process,
yields a square. It is equally possible to cut out six pieces
that can be reassembled into a rectangle in which the sides
are in the relation 1 V2Iv'3.
How can this be done?
 PROBLEMS ' 61

106. Toward Unity

0. 8461538461538 . . .

and
0. 1538461538461 . . .
are two repeating decimals in which the same series of
digits repeat indefinitely. If one adds up each correspond­
ing pair of digits, one obtains a series of 9s. Can one
deduce from this that the sum of the two numbers is
exactly I?

107. An Accidental Ellipse

In our everyday life we seldom encounter ellipses. It is
not a widespread geometric figure as compared to the
omnipresent circle . And if we do not see ellipses very
often, we draw them even less.
There exists, however, in many homes a simple device
(not used for cooking) that under certain circumstances
can produce an ellipse. Although that is by no means its
normal function, if this device breaks down through an
accident, almost all of its points will individually describe
a portion of an ellipse .
What is the device, and what is the accident?
62 • THE GARDEN OF THE SPHINX

108. Your Basic Boat

Ordinarily, a boat makes headway by acting against some­
thing else: against the water if it is propelled by oar or
by screw, against the wind and water if it uses a sail, or
against the ground if it is towed. Now, is it possible for
a boat to move ahead without acting on anything outside
itself? Think of yourself in a boat that is entirely enclosed
in still water with no current and no wind. Would you
be able to make it move?

109. Points and Lines

This figure represents the present record achievement to
date in the alignment of twenty-one points: eleven align­
ments of five points are represented. But one notices at
once a defect in the arrangement: there are four unused
intersections. They do not contain any point of a sort that
would complete the perfection of the whole. One could
surmise the possibility of doing better, and this, in fact,
has been achieved: a new plane figure of twenty-one points
has been discovered that shows two improvements: (1)
there are twelve alignments of five points, and (2) all the
intersections are employed.
What is that arrangement?
 PROBLEMS · 63

110. Squaring the Square

If the quadrature of circles must be discarded as a valid
mathematical problem, at least the problem of combining
squares is within our reach. In spite of one's first reac­
tions, it is literally possible to combine two squares to
produce a third. Can you show how to do it?

111. The Missing Token

A child who has been given ten tokens marked 0, I , 2,
3, 4, 5, 6, 7, 8, and 9 announces that he has lost one. His
father does not know the number of the missing token,
but he is an able logician and engages his son in the
following dialogue:
Father: Can you divide the remaining tokens into
three groups that add up to the same number?
Child: Yes.
Father: Can you also divide these into four groups
that add up to the same number?
Child: Yes.
Father: Then I can tell you the missing token.
Which token is missing?
64 . THE GARDEN OF THE SPHINX

112. Quadrature and Dissection

The task of juxtaposing two squares to form a third (pro­
posed in puzzle 1 10) was achieved there by arithmetic. In
giving the solution, I expressed a doubt as to the existence
of a geometric solution. But this did not discourage one
of my readers from finding one. This reader had a com­
pletely valid approach: "to juxtapose two squares to form
a figure susceptible to being cut up into pieces that would
then form a third square. " (The two squares of the j oined
figure here are of any dimensions. )
How many cuts are necessary to produce the pieces to
form a third square?

113 . An Uneasy Cryptogram

The following message was enciphered by a relatively
simple method, but the plaintext has an unusual feature
that may cause you trouble . See if you can decipher it.

XCWXI RQJWJ KRYQS ZIQMQ CXCTX

VDSDS ZHDQS WJHOX CYJDS AXBRD
KJXJW QTQYQ KqOQ XHQCT QMXIF
DUVHQ IIXCX JIBDK JWJWX IMQIY
QSZIU XHIJM QHCXC VDUTH KVQTT
XSJXD C.
 PROBLEMS ' 65

114. Roots
Pocket calculators are now ubiquitous in the office and
the home, although most are limited to the simplest arith­
metic operations . However, it is possible to extract square
roots quite quickly even with a calculator that does not
have a square-root function . How can you use that device
to find square roots with a minimum of trial and error?

115 . The Triangulation of the

Triangle
In reflecting upon the problem of the quadrature of the
square (cutting two juxtaposed squares to make one square
from the resulting pieces; see puzzle 1 12), one of my
readers discovered an intriguing related problem. The me­
dian of any triangle is cut to form two triangles of equal
areas. How can one be cut to obtain pieces that can re­
constitute the other?
66 . fHE GARDEN OF THE SPHINX

116. Travels on the Hexagon

Can you find the twelve ways of joining the apexes of a
hexagon by a continuous sequence of straight lines that
form a closed figure and do not pass through any apex
more than once?

117. A Bailout Fee

Because of a mistake, a well 20 yards deep has been
contaminated. Formerly empty, it is now filled with fuel
oil. A workman undertakes to empty the well by means
of a pail with a rope and pulley. In that way he is able
to lower the level of the oil 1 yard at a time. His charge
is $400. If he stops work halfway through (i . e . , at the end
of 10 yards), how much is he owed?

118. A Prize Hamster

A researcher in an intelligence-testing institute came in
to his laboratory one day with a hamster that had been
trained by his children. It shattered every record for ne­
gotiating a maze . He had no trouble in selling it to a
colleague for $425. The latter taught the hamster addi­
tional skills and sold it for $470 to a third party, who
proceeded to sell it to another for $535. That person sold
it for $594 to someone else, who sold it for $716 to some­
one else, who sold it for $802. If this last buyer intends
to pass it on to someone else, how much will he ask for
it?
 PROBLEMS · 67

119. The Women Man the Oars

Crossing a river is an old chestnut of recreational math­
ematics. In a general way, it involves crossing in a boat
that seats fewer people than want to cross. Another re­
quirement, which is more logically than socially appeal­
ing, is that none of the husbands is willing to let his wife
be in the company of another man, either in the boat or
on shore, without his being present.
After reading such a problem (58), a woman wrote in
to express the regret that (quite apart from the outmoded
j ealousy) the women were not deemed capable of man­
aging the boat themselves.
To meet this obj ection, here is a problem in which the
jealousy is still a factor but in which one woman, even
by herself, is able to row the boat across . Four couples
are on one bank. How many crossings will it take to
transport everyone to the other bank in a boat that cannot
hold more than three persons?

120. Breaking Up a Year

Let us start with 1980; 1979 x 1 = 1979 is the smallest
number one can find that is the product of two nonneg­
ative integers, the sum of which is 1980.
Turning this around, what is the largest number you
can obtain as the product of two whole numbers that add
up to 1980?
68 • THE GARDEN O F THE SPHINX

121 . Protagoras at the Bar

The paradox of Protagoras and his pupil is a classic of
logic. As a teacher of law, the great sophist once accepted
unusual terms of payment from one of his students: he
would waive payment until the student won his first suit.
As it turned out, the student did not practice law upon
completing his studies and therefore neither lost nor won
a suit. And he did not pay Protagoras for his lessons.
On these grounds Protagoras initiated a suit in order
to force the student to pay up, and in this legal action
each side took an impeccably logical position: Protagoras
argued that if the student lost the case, he would have
to pay, since that was the obj ect of the suit, but if the
student won, he would still have to pay, because that
was the requirement of his contract.
The student argued that if he lost the case he would
not have to pay, since those were the terms of the con­
tract, but if he won his case, he would not have to pay,
because that was the decision of the court.
What ought Protagoras to have done in order to get
paid?
 PROBLEMS ' 69

122 . Squaring Polygons

The art of cutting up geometric figures and then reassem­
bling the pieces into different figures is an ordinary source
of inspiration for puzzle lovers . The goal is the traditional
one of the mystique of geometry: to produce a square.
By now almost all the classical geometric figures have been
explored, and it remains to deal with less commonplace
ones. For example, take the figure in the diagram. Can
you cut it into eight pieces that can be assembled into a
square?
70 • THE GARDEN OF THE SPHINX

123 . Drawing the Line

To retrace a geometric form without lifting the pencil off
the page or crossing the same line twice is a well-known
diversion for the aficionado of geometric puzzles. Here is
a real stumper. It is possible to traverse this diagram not
only without breaking the two rules but also by ending
up at the starting point. The solution is systematic and is
applicable to all multiplications of the original figure. Can
you find it?

124. A Band of Five Squares

Creates Another Square
This band made up of five contiguous squares is easily
cut into five pieces that can be reassembled into a square.
The solution is given below.
The symmetry of this result is very appealing, but it
conceals an even better solution: to cut the band into as
few as four pieces that can be assembled into a square.
How is that done?
 PRO BLEMS ' 71

125. The Fifth Power

The ubiquity of calculators in our everyday life relegates
number theory increasingly to the researches of experts
or the diversions of amateurs . Let us enj oy the fact. Can
you demonstrate a property of numbers that is as sur­
prising to those who have forgotten it as to those learning
it for the first time: the difference between the fifth power
of a number and the number itself is divisible by five.
Can you prove in other terms that n5 - n is a multiple
of five?

126. A Line to Follow

Certain problems of recreational mathematics sum up the
nature of scientific thought in an astonishing way, when
one considers its conflicting opinions and its unexpected
solutions . That is certainly the case with problem 123,
Drawing the Line.
First phase: someone proposes a problem that involves
tracing a figure without raising the pencil from the paper
or doubling back and at the same time ending at the
starting point. Second phase: some readers think they can
show that this is impossible. Third phase: a solution is
published that fits the rules. Fourth phase: someone dem­
onstrates that the problem is childishly simple, since he
has solved it and now wishes to add the constraint of
never crossing a line.
Can you solve this amended version of the problem?
72 • THE GARDEN O F THE SPHINX

127. A Tricky Division

When one-quarter of a square is removed in the form of
a piece that is itself a square, what is left of the original
square can be divided into four equal and superimposable
regions . This is a well-known mathematical puzzle that
taxes the ingenuity of anyone encountering it for the first
time. But don't even try to solve it; the figure below gives
you the solution. Suppose, however, that the part re­
moved is in the shape of a triangle. In that case, how
would you divide what is left of the square into four equal
and superimposable regions?

128. Unlisted Numbers

A telephone company assigns to its customers all the
numbers composed of six digits, but in order to avoid a
certain type of mistake it excludes the numbers that con­
tain the digits 12 in any two consecutive positions. How
many possible telephone numbers will have to be elimi­
nated on this account?
 PRO BLEMS ' 73

129. No Way to Make a Square

11
111
1111
11111
Intuitively, w e feel that in the above sequence o f numbers
composed entirely of Is there are no squares of integers.
But can you prove that conj ecture?

130. A Primary Law

For every prime number p greater than 3, there exists a
whole number m such that p2 = 24m + 1 . Is this some­
thing profound or trivial? How is it accounted for?

131. Ten Digits for One

The jungle of integers can be explored by the amateur
mathematician . He may, as he chooses, either arm himself
with the weapons of logic or take pleasure in dealing with
phenomena beyond his comprehension by merely guess­
ing them and experimenting with them empirically. Here
is a situation that is well suited to either of those two
approaches.
= 57,429
9
6,381
Nine is expressed here as a fraction containing all the
integers from 1 to 9 taken one at a time . If one admits 0
in order to have all ten digits, it is possible to represent
9 by at least six fractions, although purists may find they
stop at three . How many of these expressions can you
find?
74 • THE GARD E N O F THE SPHINX

132. Inside Out

In connection with problem 123, which consists of tracing
a closed straight line without a crossing, an interesting
question can be raised: the closed curve of every solution
defines an inside area and an outside one. Under this
condition, is each of the forty-two triangles of the figure
always on the inside, is it on the outside, or does its
position vary with the particular solution?

133 . Raindrops Are Falling • • •

If you have to go somewhere on foot on a rainy day, will

you get less wet if you run, will you intercept more rain­
drops that way, or does it make no difference how fast
you go? I wonder about that question every time there's
a shower without attempting to deal with it mathemati­
cally. It's not a good time to cope with the problem, and
as soon as the sun comes out, I tend to forget about it.
One of my readers, however, has attacked the problem.
In his reasoning, he represents the human body by a
parallelepiped, which for practical purposes is an accept­
able approximation. What is the result?
 PROBLEMS · 75

134. Remarkable Quadratures

In problem 110 we juxtaposed the squares of whole num­
bers in order to produce other squares . Thus, 16 and 81
form 1681, which is the square of 41 . One of our colleagues
has gone further in this direction in exploring all possi­
bilities up to a billion and itemizing 119 solutions (ob­
tained with the aid of a programmable pocket calculator) .
He has discovered some extraordinary numbers, one of
which is 5,184, 729 = 22772, formed by 5184 = 722 and
729 = 272. Can you find another that is a square that can
be decomposed in two different ways in the juxtaposition
of squares? (X2 = A2B2 = C2D2)

135 . Mobius and His Better Half

Everyone is able to construct or to visualize a Mobius
band by cutting a uniform strip of paper and then giving
it a half tum before joining the extremities. It is easy to
cut this band into two equal parts, either along a line that
j oins the two sides and opens it or by a central cut that
leaves two bands of the same length intertwined .
But how does one obtain tw o parts o f equal area in a
continuous straight line along the paper that only touches
one edge once?
76 . THE GARDEN O F THE SPHINX

136. Quadruple Quadrature

"Algebraic" quadrature continues its evolution (see prob­
lem 134) . In this realm-seemingly not very mathemati­
cal-the aim is to j uxtapose the squares of integers in
order to create other such squares. We have already ob­
tained some startling results. Now we endeavor to go
further by finding four squares that will create a fifth
square if juxtaposed. Can you find the number x such
that x2 = A2B2C2D2, where the squares are juxtaposed but
not multiplied and where A2B2 and C202 are also squares.

137. Something New About

Nines
Nine is a remarkable number: it is the only integral square
composed of odd digits (digit) only. Every other square
(to base 1O) contains at least one even digit. Can you find
a proof for the last statement? Is it possible to prove it by
generalizing the proof for a property of numbers com­
posed of the digit 1 (see problem 129)?
 PROBLEMS · 77

138. The Trisection of Pappus

We have already encountered the trisection of the angle
in connection with the devices of the eighteenth and nine­
teenth centuries, which were invented to perform that
geometric task automatically. But the construction of one­
third of a given angle is far more ancient than that, since
Archimedes and Pappus found a way of achieving this
(not, of course, with compass and straightedge alone)
before our era.

In this construction, attributed to Pappus, an arbitrary

distance oa is marked on a side of the angle . The line ab
is perpendicular to the other side, and ad is perpendicular
to ab; od is such that cd = 2oa, which could not, of course,
be done using only a compass and straightedge. Why is
hoc one-third the angle oab?
78 . THE GARDEN OF THE SPHINX

139. Archimedes Cuts It Up

Twenty-three centuries before our time, Archimedes played
at Tangram (the cut-up square below), a puzzle suppos­
edly of Chinese origin . In Archimedes' version of the
game, the goal is to cut up a square into fourteen ele­
ments, each of which is in a rational relation to the whole
figure.
In Archimedes' solution (below), what is the relation
of each piece to the whole?

140. Tails I Win

Two contestants take turns tossing a coin. The first one
to toss tails wins. Of course, it's obvious that the one who
tosses first has a greater chance of winning than his ad­
versary. But what, in fact, are their respective chances
exactly?
 PROBLEMS · 79

141. The Barrier of Thirst

The desert march under the conditions imposed has been
made with increasing efficiency in the solutions provided
in problems I , 40, and 80. Each time progress has been
made by discarding another prej udice. This is no time to
abandon such a promising course. As the result of con­
quering an additional prej udice-more serious than the
earlier ones-the supply of water necessary is no longer
eleven and one-half days but rather nine and one-half
days.
How is that possible?

142. Regrouping the Pieces

Problem 139 also entails three additional puzzles that are
not as simple as they may seem at first glance . How can
you regroup the pieces, without changing their positions,
into different parts of the square representing:
1) three equal integers;
2) three consecutive integers;
3) the first eight numbers and 12?

143. A Square for a Compass

When you are without a ruler, a compass can by itself be
used to make a number of geometric constructions . In­
deed, one of the most famous achievements along that
line is to locate the center of a given circle-already drawn.
For the present, let us attempt something not quite so
difficult. Using only a compass, see if you can draw a
circle and then determine on its circumference the four
apexes of a square .
80 • THE GARDEN OF THE SPHINX

144. Words and Numbers

It is easy to express numbers in terms of letters. All you
have to do is spell them out: 1 becomes ONE, etc. But
how does one express words by numbers in such a way
that every word has one and only one numerical equiv­
alent and is easy to decipher? One solution is to use the
numbers of an even pair of digits in such a way that each
two-digit section of the number is a letter of the alphabet:
ART = 011820
POT = 161520
Would you be able to come up with other ways of doing
these and in particular some of the techniques logicians
use to express words by numbers and explore the limits
of language?

145 . Words, Numbers, and

Sentences
Following up on the last problem (144), can you find a
way of extending the procedure described there for as­
sociating a number with a word to an unambiguous way
of associating a number with a sentence (composed of
words)?
 PROBLEMS ' 81

146. A Triangle in a Square

The equilateral triangle is second only to the square as a
perfect figure in the planes. Does one give sufficient thought
to the possibility of inscribing the former in the latter?
Given a square, what is the length of a side of the smallest
equilateral triangle that can be inscribed in it?

147. Mysterious Powers

1 2890625
is a sequence of digits that imparts very special arithmetic
properties to numbers terminating in that way. If you are
told that these properties involve powers, can you dis­
cover what they are?

148. Superior Antimagic

Strengthened by the experience of solving problem 16, let
us proceed to the sixth degree of antimagic.
Can you construct a square of thirty-six places that con­
tain the first thirty-six integers placed in such a way that
the sums of the six rows and the six columns and the two
diagonals are consecutive numbers?
82 . THE GARDEN OF THE SPHINX

149. Coupled Couples

Let us explore a variation of the "river crossing," problem
119.
Once again we shall accept the constraints imposed by
the antiquated passion of j ealousy. Six couples are pre­
paring to cross a river in a boat that can take only five
persons at a time. Furthermore, no husband will allow
his wife to be unaccompanied by him either on the boat
or on the shore in the presence of one or two other men.
Given these conditions, how many crossings are reqUired?

150. No Holds Barred

The desert to travel in remains the same, and the con­
ditions of the trip are identical with those in problem 1.
But now you must perform the feat not with 20, 14, 11%,
or 9% days' worth of water; you must reduce the supply
by several more days' worth. You will find that this op­
timal solution was hidden by an inherent prejudice far
stronger than that which you had to deal with in the
previous solution.
 SOLUTIONS • 85

1. Water in the Desert

The mission can be accomplished by enlisting three companions
and setting out as a foursome with a twenty-day supply of
water all told.
At the end of the first day, there will remain only a sixteen­
day supply of water. One of your companions returns to the
starting point, carrying with him a one-day reserve of water
and leaving you a fifteen-day supply of water to carry on as a
threesome.
At the end of the second day, you will have a twelve-day
supply of water. A second companion returns to the starting
point with a two-day supply of water, leaving you with a ten­
day supply to make the trip as a twosome.
At the end of the third day, you will have an eight-day supply
of water, and your last companion makes the trip back to the
starting point with a three-day supply of water, leaving you a
five-day supply. That will allow you to make the one-day march
to your destination, plant the flag, and make the four-day trip
back to the starting point.
Did you discover this solution? That's fine, but it's far from
the best one. Problem 40 will ask you to reduce considerably
your consumption of water.

2. Pure Reason
In 7 seconds the train passes by a stationary observer, or, in
other words, travels its own length. To traverse the length of
the station means to travel the length of the station and its own
length. Thus, the train travels the length of the station (alone)
in 26 7= 19 seconds. In 1 second it travels 380 + 19 = 20
meters. In one hour it travels 72 kilometers. Since it travels its
-

own length in 7 seconds, that amounts to 7 x 20 = 140 meters.

86 . THE GARDEN OF THE SPHINX

3. The Age of His Digits

Although it is possible to solve this problem in a lengthy and
roundabout way, there is a simple and elegant solution based
on congruences.
Any number and the sum of its digits will both leave the
same remainder when divided by 9. Therefore, dividing both
the year of birth and the age in question by 9 must leave the
same remainder. The sum of the two numbers (1898) when
divided by 9 must leave a remainder twice as large (in this case
8) . Therefore, the age in question is 4 plus a multiple of 9; for
instance, 13, 22, 31, etc. The correct solution (22) can be found
immediately by trial and error.
One reader has pointed out an ambiguity arising from the
fact that one's age changes on one's birthday while the year
changes on January 1 . Taking that into account, the problem
admits of two solutions.
 S O L UTIO N S · 87

4. The Exchange of Knights

Let us change our way of thinking about the moves of the
knights on the twelve squares and imagine them carried out
only on the files, the numbers of the squares being assigned
to the files. In that way we can untangle the skein and give it
a simpler form. In this new model, equivalent to the other, a
knight's jump can be considered as a move of one square in
one direction or the other.
My proposed solution of twenty-six moves has been bettered
by numerous readers. The record minimum is sixteen moves.
Here is the solution:
10 9 4
2 - 9 - 10
- -

12 - 7 - 2
1 - 6 - 7 - 12
11 - 6 - 1 -
4 - 9 8"
,, "
3 - 4 - 11
" ,
9 4 3
- -

1 3
I I
! I
6 4
' "
I " " I
"
I 11 I
I I
I I
I I
I 2 I
" "
I ,, I
7" '
9,
I
I
1 2, ,,1 0
...
' "
'5"
88 . THE GARDEN OF THE SPHINX

5. Squaring the Maltese Cross

This solution achieves in eight pieces what up to now no one
could do in fewer than nine.

\ I
\I � V
...... I J
-� '" �:'
,, � il ""
tr�- J
-
1-0�
I �,
_ � �I
7 , "",r-
� I "
II .... 1\
J \

6. Reconstructing Octagons
The dimensions of the little octagons must be multiplied by
v'3 in order to obtain the big octagon with three times the
surface. It is therefore necessary to produce the hypotenuse of
a right-angle triangle with sides V2 and VI. The rest is inspired
trial and error. (See Greg Frederickson, "More Geometric Dis­
sections," Journal of Recreational Mathematics [summer 1974] . )
 SOLUTIONS • 89

7. Basic Multiplication
Replacing each odd number in the left-hand column by 1 and
each even number by 0 is equivalent to expressing the first
number in base 2 and reading it from bottom to top. The op­
erations on the right-hand side then represent a multiplication
in base 2.

8. A World of Fours
Let us avail ourselves of a little-used mathematical entity; namely,
logarithms to the base 4 instead of the more familiar base 10,
where x = lo� y when y = 4X . Then we can write:

- lo� { IO� V\lv. . . v'W4}

For a given n, the expression contains 2n nested square-root
signs: something that is possible to do for any given integer n .
This expression is thus equivalent to:

- lo� { lo� 4(1I2)2" }

= - IOg4 Gr = n
(See A. P. Domoryad, Mathematical Games and Pastimes [Perga­
mon Press, 1963] . )
90 • THE GARDEN O F THE SPHINX

9. Circular Uncertainties
The third method of calculation proposed by Joseph Bertrand
consisted in starting, for reasons of symmetry, with an inter­
section of the chord with the circle and the angle between the
chord and the tangent. The chord must lie within one of the
three angles of 60°. Therefore, the probability must be V3.
Actually, one knows now that such a problem is not defined
until one has specified precisely the stochastic variable, which
was not done in the statement of the problem. The three meth­
ods of solution refer to three types of experiment and therefore
to three different variables:
1) The distance of the chord from the center
2) The angle between the chord and the tangent
3) The position of the midpoint of the chord
Dr. R. J. Denichou has proposed a fourth method of calcu­
lation that is equally contradictory. Let us assume a basketful
of all the straight lines able to serve as chords, of lengths ranging
from 0 to d (the diameter). Those that are suitable have a length
V3
fallin g between 7 and d. The probability is therefore
2 - V3
. To explore this apparent paradox further, see problem
2
10.
n'

0 t--�-�

s
 SOLUTIONS · 91

10. The Random Chord

Let ABC be an equilateral triangle and ON a radius perpendic­
ular to OA. 5 is a point on OA above O. 5N intersects AC at
C' and the tangent to the circle at A at N ' . Any chord passing
through A corresponds to a point on 5N' . So the probability
that it will be greater than the side of the triangle is ��:. But
5 may vary, and this "probability" may take any value from 0
(when 5 = 0) to 1 when 5 rises.

Nt

11. What Coincidences?

Every twelve hours the hands of a clock make 1 revolution, 12
revolutions, and 720 revolutions, respectively. The fIrst and the
second coincide 1 1 times and the first and the third coincide
719 times, at regular intervals in each case. But since 719 and
11 are relatively prime, the three cannot align themselves exactly
during the twelve hours.
92 • THE GARDEN O F THE SPHINX

12. Meetings on the Dot

Th e solution is in the vicinity o f an instant when the two slowest
hands are superimposed and the third just misses a perfect
alignment as it approaches them. The hour and minute hands
meet eleven times in twelve hours. At which of these meetings
is the second hand closest to them?
The two slower hands coincide at every eleventh part of the
clock dial. At the third of these (and its symmetrical counter­
part), the second hand is only 1/11 of the dial away from them.
The time is 16 411 1 minutes after 3 o'clock.
One obtains the exact solution by moving back the second
hand until it coincides with the hour hand, the minute hand
being moved back slightly. The hour is that multiple of 11719
(the coincidence points of hours and seconds) that comes closest
to 3/11 : namely, 1961719; that is to say, 16 minutes, 16 2561719
seconds after 3 o'clock. The symmetrical solution is 43 minutes,
43 463/719 seconds after 8 o'clock.

13. Steer Metabolism

One can reasonably argue that steers are similar in shape. Now
when two solid figures are similar, their areas are proportional
to the squares of their linear dimensions, and their volumes,
like their weights, are proportional to the cubes. Thus, their
areas are determined by the square of the cube root of the
V3
weight. The number of calories required is
[420J
13,5 00x
630
= 00 ��/49 =
13,5 x approximately 10,300.
 S O L U T I O N S · 93

14. Lightning Calculators

Actually, he does not have to know any of the digits. The only
possible answer is 2.
Let us make the calculation with the help of logarithms. The
logarithm of a twenty-digit number is 19 . . . . The logarithm of
its sixty-fourth root is
19
� .
. It lies between 0.29 and 0.32.
The only integer represented in that interval is 2, or 0.30103.
To take a more direct approach, 1064 is a number of 65 digits.
(v'fO)64 = 1032 is a number of 33 digits. (V'IO)64 lies between
1021 and 1022 and is therefore a number of 22 digits.
({lTIij64 = 1016 is a number of 17 digits. The result is therefore
between V'IO and V'lO. That is equivalent to saying that its
cube is less than 10 and its fourth power is greater than 10.
The only whole number meeting that condition is 2.

15. Heterogeneous Squares

A square of the second order contains the numbers I, 2, 3, 4.
The smallest sum possible is 3 and the largest is 7. Thus, it is
impossible to have six different sums. For the third order, one
contributor has reported the existence of 3, 120 heterogeneous
squares, obtained by selection out of a total of 45,360 squares
of order 3 of all possible kinds, generously produced by the
computer. Here are some examples:

1 2 3 9 8 7 6 5 4
4 5 9 2 I 6 7 2 3
6 8 7 3 4 5 8 1 9
94 . THE GARDEN OF THE SPHINX

16. Antimagic
Here is an antimagic square of order 4 in which the sums of
the aligned numbers run from 29 to 38:

6 8 9 7
3 12 5 11
10 1 14 13
16 15 4 2

In this square of order 5, the sums run from 59 to 70:

21 18 6 17 4
7 3 13 16 24
5 20 23 11 1
15 8 19 2 25
14 12 9 22 10

1 7. Strategic Dates
The first to play (A) is certain to win if he commits to memory
the following sequence of dates: January 20, February 21, March
22, April 23, May 24, June 25, July 26, August 27, September
28, October 29, November 30. In the actual contest A must pick
one of these dates as soon as possible.
For obviously, if A says November 30, B can reply only with
December 30, which lets A win by saying December 31 .
If A says October 29, B has only four replies: October 31 and
December 29, which allows A to win immediately, or October
30 and November 29, which allows A to win by saying Novem­
ber 30, which wins in two steps as shown above.
By starting with January 20, A is able to keep the upper hand
and win .
 SOLUTIONS · 95

18. Manual Tactics

The key number is 23: it is the first prime number to be more
than five places from the next prime (29) . The objective, then,
is to be the one to arrive at that cumulative total. In order to
do that, the winner must force his opponent to say 13, which
can be done by saying 11 and forcing him to say 7 by first
saying 5. A wins by starting with 5.

19. Two-Handed Confrontations

In this game, the first jump greater than 10 lies between 113
and 127.
The winning strategy can be deduced as follows: 113 succeeds
103, 107, or l09-therefore, say 101; 101 succeeds 97-therefore,
say 89; 89 succeeds 79 or 83-therefore, say 73; 73 succeeds 67
or 71-therefore, say 61; 61 succeeds 53-therefore, say 47; 47
succeeds 37 or 41-therefore, say 31; 31 succeeds 23 or 29-
therefore, say 19; 19 succeeds 11, 13, or 17-therefore, say 7.
The first player wins by starting his attack with 7.
96 • THE GARDEN OF THE SPHINX

20. The Best Bridge

A problem involving the shortest distance between one point
and another can be thought of as an optical problem involving
the path of a ray of light. To travel from a to b on the specified
road is obviously the same as going in from a to a point b'
symmetrical to b with respect to the bank. One solution would
consist of tracing the line ab' , cutting the bank at c, and then
choosing the route acb. The solution obtained in this intuitive
way can be confirmed in terms of inequalities. Any point (c')
other than c makes ac' + c'b' longer and thus ac'b longer.

c
,
,
,
,
,
,

b' , , ' b
•
 SOLUTIONS • 97

21. The Best Place for the Bridge

Taking the solution to the preceding problem as a guide, let us
think of this problem in terms of an appropriate optical analogy.
If the canal were a pane of glass and the road a ray of light,
the rays to and from the canal would be parallel, but the ray
would obviously not cross the bridge perpendicularly to the
banks.
To find the shortest path, we establish the auxiliary point a '
lying the distance cd from a in the direction of the canal on the
perpendicular drawn from a to the canal. From a ' , draw the
straight line connecting a ' and b, which crosses b's side of the
canal at the point d' . From d' draw the line d' c' crossing the
canal perpendicularly back to a's side . The line d'c' determines
the best place to put the bridge since it permits the route
ac' + c'd' + d'b, which is the shortest route from a to b, being
equal to the distance aa ' + a ' b "as the crow flies. "
T o prove the last equality, w e need merely point out that
c'd' = aa ' (by construction); ac' = a'd' (opposite sides of par­
allelogram) . Therefore, ac' + c'd' + d'b = aa' + a ' b.
98 . THE GARDEN OF THE SPHINX

22. Thrift in Mi"ors

Oddly enough, the answer depends neither on the height of
my eyes nor on my distance from the mirror. Whatever the
height of my gaze in front of an infinite vertical mirror, (1) the
ray of light passing from my eyes to the top of the image of
my head strikes the mirror at the midpoint of its (the ray's)
trajectory and therefore midway between my eyes and the top
of my head; (2) the ray proceeding from my eyes to my feet
strikes the mirror midway between my eyes and my feet. All
told, the part of the mirror I am actually using is only one-half
of my height: say, two feet nine inches.

23. Multiple Images

Let a be the angle between the mirrors. The object A has in
each mirror an original image Al and A ' 2' These two produce
images in the opposing mirrors: the images A2 (of A ' l) and A' 2
(of AI) . . .
Thus, one mirror reflects a sequence of images AI' A2, A2
. . . A2, located in the planes that form with the bisecting plane
the angles a, 2a, 3a . . na, where na is the greatest term less
.

than or equal to 1T. If na is exactly equal to 1T, the two images

An and A ' n will not overlap and will produce a total of 2n - 1
1T
images. If a is between -
1 + and �, one will obtain a total of
n n
2n images.
 SOLUTIONS · 99

24. A False Proof

p

A straight line is not perpendicular to a plane unless it is per­

pendicular to two distinct and nonparallel lines belonging to
the plane. Accordingly, one has to prove that in the two pairs
ca, cb and da, db, the straight lines are distinct. But that can be
the case for only one of these two at the same time .
For clearly, if p is projected onto the point c of the plane, the
center 0 of the sphere havi!lg the diameter ap is projected onto
the plane ac in q. But q is the center of the circle of intersection,
and thus ac is a diameter. Likewise, be is a diameter of the other
circle. Each of these two diameters is tangent to the other circle
and thus are perpendicular. Thus, either acb or adb are not
distinct, and either pc or pd is not perpendicular to the plane.
100 . THE GARDEN OF THE SPHINX

25. Visualizing Tetrahedrons

The triangles abd, bee, eaf, and def all have, taken in pairs, a
common side. A point i, located above, determines four tetra­
hedrons fulfilling the condition by means of the surfaces passing
through i.
Similarly, with a point j (not shown) located below and the
triangles kIp, Imq, mkn, and npq, one obtains four more suitable
tetrahedrons . The common portions of the triangles in the dia­
gram ensure that the group of eight tetrahedrons taken in pairs­
but only in pairs-shares a portion of a common surface.

"J""1::::;::oo.....::'--+-�� c
k w;r;;t;-..t:-----",....::::;;;...-� m

26. Truth and Brotherhood

Since each brother does not lie more than once a year, he must
tell the truth at least once in the time period of this problem.
Thus, the first brother was born either on December 30 or 31 .
If he was born on the thirtieth he cannot lie on January 1 by
saying that he was born on the thirty-first. But if he was born
on the thirty-first, he could lie on that day and tell the truth
on the next.
The second brother was born either on January 1 or 2. If he
was born on the second, he does not have the right to lie on
the thirty-first by saying that he was born on January 1 . Being
born on the first, he tells the truth the first time and lies the
second. Thus, the first brother was born on December 31 and
the second on January 1 .
 SOLUTIONS ' 101

27. Doubts About Euclid

The most disconcerting thing about this proof is that it makes
no use at all of the fact that the two lines are parallel. In reality,
when two straight lines are not parallel, the sum of the two
angles is greater than two right angles on one side and less
than two right angles on the other. But this possibility was
missing from the set of hypotheses used in the proof. This
possibility also arises with parallel lines without postulating or
proving the theorem.

28. The Sum of the Angles

In calling x the sum of the angles of a triangle, we have made
the implicit assumption that it is a constant for all triangles.
But to begin with, that remains to be proved. It is open to
doubt, all the more so because it is obviously false in spherical
geometry, for example .

29. Jet Psychology

Vertically, the fall of each drop of water from each hole is gov­
erned by the law of distance = Y2gt2 . The time periods of the
fall of each are tv t2, and t3, such that Y2g(tl)2 = 1, Y2g(t ) 2 = 2,
and Y2.[(t3)2 = 3. Thus, the times are proportional to VI, '\12,
2
and V3.
Horizon tally, each drop of water is expelled with a uniform
motion with velocity v corresponding to the kinetic energy Y2mv2
equal to the energy lost by the column of water above the hole.
Thus, the velocity is proportional to the square root of the height
of the water above the hole, and Vv V2' and V3 are proportional
to 0, '\12, and VI. It follows that the horizontal paths are
proportional to 0, '\12, and 0. The first and third jets will
fall in the same place, closer to the container than the second
jet.
102 . T H E GARD E N O F T H E S P H I N X

30. The Parson and His Sexton

The product of the ages of the parishioners, 2,450, has as its
prime factors 1, 2, 5, 5, 7, and 7. Among the nineteen possible
triplets (1, 2, 1225), (1, 5, 490) . . . (7, 10, 35), (7, 14, 25), only
two have the same sum:
5 + 10 + 49 = 64
7 + 7 + 50 = 64
Thus, there are two possible ways of arriving at one age for
the sexton. In this case, the oldest parishioner is either 49 or
50. The ambiguity can be resolved only if the parson is 49, so
the parishioners are 7, 7, and 50.
 SOLUTIONS ' 103

31. Revealing Ambiguities

The sum of the two numbers must be an ambiguous one arising
out of several different pairs:
5 = a + a ' = b + b' = n + n '
Therefore, Simon knows that Paul is confronted with one of
the products aa ' , bb' , . . . nn ' in such circumstances that the
ambiguity of one of them, pp ' , reveals the solution to Simon.
Reciprocally, pp ' must correspond to several pairs of which at
least two present Simon with ambiguous sums in order for his
ignorance not to be revealing to Paul and one of which permits
the preceding logical operation.
Let us explain some ambiguous sums beginning with the
smallest:
6 = 2 + 4 = 3 + 3
Neither of the two products, 8 and 9, is ambiguous.
7 = 2 + 5 = 3 + 4
The products are 10 and 12. Only the second is ambiguous,
furnishing Simon with the solution 3 and 4. Let us check this
from Paul's point of view. He knows that the product is 12 and
can deduce that Simon has the sum 7 or 8 (if the pair is 2 and
6). He reasons that 7 would give Simon sufficient information
but 8 would not, for 8 = 2 + 6 = 3 + 5 = 4 + 4, correspond­
ing to three products, 12, 15, and 16, of which two are ambig­
uous instead of only one. Therefore, Paul in tum is able to
arrive at the answer 3 and 4. The problem calls for only one
pair, but others exist-84 and 84, 69 and 96, 98 and 78---that
readers have brought to my attention.
104 • THE GARDEN OF THE SPHINX

32. An Illegitimate Equilateral

Triangle
To be proved: the triangle formed by the trisectors of an arbi­
trary triangle is equilateral.
Draw the trisectors of angles B and C, which measure 3b and
3c, A measuring 3a.
a + b + c = 600
Let F and H be such that DFH = FDH = a + c and E and
K such that KDE = DEK = a + b. One can determine that
FDE = 60° and that BFD = a + 60°. The point D on the bisec­
tors of FBC and ECB is eqUidistant from BF and CEo As one
calculates that DFB and DEC are equal angles, DE = DF and
DEF is equilateral. But HF and KE are of course the trisectors
of angle A. For note that FK is the bisector of DKE, BF the
bisector of DEK, and F therefore the point of intersection of the
bisectors o f the triangle BK, KE, B X . Since
BFH = BFD HFD = a + b, the angle between BX and HF must
2a - 2b), or 2a. HF is the bisector of
-

be 180° 2d (180°
BX and KE, and likewise, KE is the bisector of HF and CY.
- - -

x y
 SOLUTIONS • 1D5

33. Division by Geometry

It is not possible to cover the compartments of a rectangle
missing two comers with triple dominoes.
Let us fill in the squares of the rectangle by using three
symbols (squares, diamonds, and circles) in such a way that no
two neighboring compartments will have the same symbol.
Whenever a domino is placed, three different symbols will be
covered. To cover the whole figure with dominoes, there would
have to be the same number of each symbol. But there are 19
squares, 18 circles, and 17 diamonds.

• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • •

34. Dismembering One Thousand

The two ways are equally numerous since the odd and even
partitions can be placed in one-to-one correspondence. Each
even partition-1,DDD = a + b + c + d-can be paired with an
odd partition-1,OOO = (a - 1) + (b - 1) + (c - 1) + (d + 3)­
and each odd partition-1,DDD = a ' + b' + c' + d'-with
an even partition-1 ,DDD = (a ' + 1) + (b' + 1) + (c' + 1) +
(d' - 3) . With this one-to-one correspondence, the two sets are
equal.
106 . THE GARDEN OF THE SPHINX

35. How Many Elevators?

There are fIfteen floor-to-floor connections to be provided within
the six intermediate floors, and each elevator provides at most
three of these, so a priori there are at least fIve elevators.
This situation is eqUivalent to incorporating in 15 different
triangles the 15 line segments joining the apexes of a hexagon.
If the "triangle" 135 is chosen, then the line 4-5 is either bound
to 2, which leaves 5-6 unusable, or to 6, which links 3-4 to 2
and leaves 1-4 unusable.
Each triangle, then, must have at least one external side, and
only one triangle can have two external sides, for example, 1-
2-3. But in that case, 3-4 is bound to 6 and 1-6 cannot belong
to any triangle. The partitioning of segments is therefore im­
possible, and one is forced to use a less economical six-elevator
system. For example:
G 1 247
G2357
G3467
G 1 457
G2567
G 1 367
 S OLUT I O N S ' 107

36. The Queen Takes a Walk

The chess queen traverses the board in fourteen moves, as
follows:

- - - -
.... ... � - ,1
I '
, ..,
I , " , I
I'
I ,
,
-t I
, ,
, ,
I , ,, ,
,
,
, , /1 I
I
I ,
\. I I
I <
" ,
I
"
, ' , , ,
I " , ,
I " I
I
; , ,
I
,
;<.
, .(
, ,
I , ,
, , , I
I , ,
v
, ,
I
"
I
,
/
, ,
, ,
t , ,
I
,
� - - - - -,, - _I
t , ,
I � 7 - ,-
I" ,

,
L �.
,
- - - - - - - -
7 - - "'" .
,
,
;

�- - - - - - - - - - - - -
,
108 . THE GARDEN OF THE SPHINX

37. An Attractive Number

If in the three-digit number a,b,c, a is greater than c,
then, allowing for the necessary carryovers:
a,b,c - c,b,a = a - I - c, 9, c + 10 - a.
In this expression, it is dear by inspection that the expressions
on either side of the constant middle digit 9 add up to 9.
Thus, the three-digit numbers produced by the subtraction
a,b,c - c,b,a must have one of the following values: 099, 198,
297, 396, or 495, each one following the other in the sequence
of operations of rearrangement and subtraction described in the
original statement of the problem. In other words, the numbers
on either side of 9 will increase and decrease by 1, respectively.
That is because c + 10 - 9 will decrease by 1 for successive
subtractions and of course (9 - c + 10 - 9) on the other side
of 9 will increase by 1 correspondingly.
Thus, the numbers on the left of 9 will increase by 1 at the
end of each operation step, and the numbers on the right of 9
will decrease by 1 until the number 495 is reached, at which
point that number will repeat indefinitely (since 594 and 495
have the same digits) . This theorem can be extended to four
digits as well as to numbers in bases other than 10. For four­
digit numbers in the decimal system, convergence begins with
the number 6174. This algorithm was discovered by the math­
ematician Kaprekar.
 SOLUTIONS ' 109

38. If and Only If

Two affirmative statements are linked by "if and only if . . .

then": (A) I raise Tom or Dick, or both, and do not raise Harry;
(B) I raise Tom, and if I raise Dick, I raise Harry. A and B are
both true or both false. A is true if both affirmations joined by
"and" are true and false if one of them is false. Since the plan
is to raise two of the three employees, A is true only if Tom
and Dick are raised. That verifies the fIrst part of B: Tom is
raised. But then the second part becomes false: Dick is raised,
but Harry is not. A and B cannot both be true. Can they both
be false? If Tom and Harry are raised, A is false with respect
to the second part. But the fIrst part of B is true, and the second
part is equally true, for a false premise is compatible with either
a true or a false conclusion.
Finally, if Dick and Harry are raised, A remains false, and B
becomes false by virtue of its fIrst part.
Therefore, it is Dick and Harry who get the raises.
110 • THE GARDEN OF THE SPHINX

39. Curvilinear Mystery

fact that the area of a circle of radius 1 is '11'.

In tackling this problem, you must remember the elementary
Draw all the
additional lines shown in the diagram. The points G and H
indicate points of intersection of the quarter circles. The point
o is the center of the square ABCD. The two circular segments
0
CGC and DHD are equal to % circle minus V2 of the uilateral
triangles BGC and EDH, respectively, that is, � - -;. By in­
spection, the area of the two circular segments defined above
and the square HOFC equals % circle plus % the circular square
('11'- -) - -'11' -
( 5) or 2 6 - 8
\13
+
1
4
= + S' which
4 4
.
leads to 5 =
3
-'11' -
\13 + 1, or approximately 0.31 .
 SOLUTIONS ' 111

40. Less Water in the Desert

It is possible to accomplish the feat by yourself with a fourteen­
day supply of water if you are willing to make a number of
trips back and forth to place reserve supplies of water along
the way.
1) Start with five days of water and leave three of these a
day's march from the base.
2) Set out again with five days of water. After one day's
walk, take a one-day supply from your reserve, proceed
one day further, leave a two-day supply there, and return
to the base.
3) Leave with water for four days, take another day's worth
at each station on the way. That will enable you to reach
your destination and return.
This solution is better than the first, since six fewer days of
water are consumed. But in spite of that, it is not the best
solution. See problem 80.
112 . THE GARDEN OF THE S PHINX

41 . Superfluous Digits
The three-digit divisor that produces a three-digit number when
it is multiplied by 8 has to be a number between 100 and 124,
inclusive; 9 is the only number that, when multiplied by such
a divisor, produces a four-digit number. Therefore, 9 is the last
digit of the quotient.
Since the first subtractions result only in a two-digit number,
the first digit of the quotient must be greater than 7. Since it
cannot be 9, it is 8. The remaining two digits of the quotients
are zeros. Again, because the first subtractions have only two
digits, 8 times the divisor must produce a number of at least
990 and that in turn requires that the divisor must be 124. It
follows that the division in question is:

1 00203 1 6 I 1 24
992 80809
1 003
992
1 1 16
1 1 16
 SOLUTIONS · 113

42. A Difficult Juncture

At the intersection, let us inscribe a sphere that is contained
within both cylinders . Now cut through the planes parallel to
the two axes. A calculation of the volume will require the sum­
mation of the intersection. Since they are both squares, the
planes cutting the cylinders along the generatrices, the inscribed
circles are the intersections with the sphere . Since the calcula­
tion of the volume of the sphere will be a summation on the
circles, the two volumes are in proportion to the areas of the
square and the circle:
V 4
=
4 1T y'3 ;
V
16
=
3

43. Using Your Imagination

The value of the required final digit does not depend on the
three-digit number from which it was derived and which I hope
you did not bother to calculate . For it is a fact that every integer
and its fifth power have the same last digit (i. e . , in the units
place) . Strange as it is, the original three-digit number would
have led you to this result one step at a time. It can be easily
verified.
114 . THE GARDEN OF THE SPHINX

44. Father, Son, and Horse

Unless one is careful in considering this problem, it is possible
to fall into the trap of believing that the overall time of the trip
can be shortened by traveling any part of it at the speed of the
horse. If the father and the son are to arrive at their destination
at the same time, the person riding cannot average better time
than the person walking. If the rider goes faster than the walker,
he must wait for him to catch up after he has fallen behind.
In fact, there cannot be a single solution to this problem
without supplying what part of the trip each person is allowed
to ride . To arrive together in the shortest time (7% hours), the
son will have to do all the walking. Doubtless, the father will
prefer this solution for other reasons, as well.
 SOLUTIONS ' 115

45. Archimedes Trisects the Triangle

Given an arbitrary angie x, draw a semicircle at 0 with a radius
equal to the (arbitrary) distance between two points on a
straightedge. A and B represent the points of intersection of
the circle with the initial and terminal sides of L x, respectively.
Extend the base of L x along the diameter to the left with the
straightedge. Now, keeping the edge of the straightedge ex­
tending through B, manipulate the straightedge in such a way
that one of the marked points lies on the semicircle at some
point 0 and so the other marked point lies on some point C
of the diameter. The angle BCA is V3 the given angle AOB.
Proof: b,. COO is isosceles and L DOC = L OCO. Then, since
the external angle BOO = L OCO + L D O C = 2BCO,
L BOO = 2 x L OC O . But b,. BOD is also isosceles and
L BOO = L OBO. External angle AOB = L BOO + L OCO
or L AOB = 2 x L OCO + L DCa = 3 L OCO. Therefore,
L OCO = L BCA = V3 L AOB.
116 . THE GARDEN OF THE SPHINX

46. Predictable Pain

Instead of focusing on the intervals between the irritations, let
us study the abscissas. In this way one obtains 2, 4, 9, 12, 13,
16, 20, 23, 26, 2S, 30, 36, 37, 39. If he had but one eyelash
behaving normally and growing regularly, we would have an
arithmetic sequence. As it is not that, the phenomenon must
be a more complex one . Is it possible that there are several
eyelashes growing at different rates?
One notices an arithmetic series beginning at the origin, or
zero: 0, 13, 26, 39. Thus, the eyelash that was plucked out was
able to grow back in 13 days. The other observations, as well,
can be rationalized only by separating these into two arithmetic
series: one beginning at 2 with constant increments of 7 and
one beginning at 4 with constant increments of S.
The irritation can be produced by three eyelashes growing
back in 7, S, and 13 days.

47. Watch Out for the Train

In the same period of time, the man can escape by running
forward the last third of the bridge or backward a third. At that
moment, the train has just reached the bridge. If the man is
able to escape by running backward, it means that the train
covers the whole length of the bridge in the time the man covers
the remaining third. Thus, the man runs at one-third the speed
of the train, or 15 miles per hour.
 SOLUTIONS ' 117

48. Knight Moves

A knight's move is either from one file to a contiguous one or
from one rank to a contiguous one. From each of the p - 1
first files, it can make q - 2 jumps to the next column, either
up or down. Including its jumps backward, one has 4 (p - 1)
(q - 2 ) jumps.
Similarly, from each of the q - 1 first ranks it can make
(p - 2) jumps to the following rank, making a total of 4 (q - 1)
(p - 2) j u m p s . Taken together, we have 4 (q - 1 )
(p - 2) + (p - 1) (q - 2) . On the 7 x 9 rectangle, 328 jumps
are possible.

49. Playing with Blocks

This problem is equivalent to solving in terms of positive in­
tegers the equation x2 + y2 = (x - y? Let us begin by creating
the equivalent equation (x - y)3 - (x - y)2 = 2xy. That means
that for any cube n 3 that represents the right number of blocks,
there are two numbers x and y such that x - y = n and
(n 3 - n2)
=
xy 2 .
These conditions for the values of x and y for a given n can
be expressed by the quadratic equation
(n 3 - n 2)
x2 - nx - = o.
2
Since the discriminant of this quadratic equation is 2n 3 - n2 for
any number of blocks meeting the conditions of the problem,
2n 3 - n 2 must be a perfect square. Using this as a test for n = 1,
2, 3, . . one finds that n = 5 has the "acceptable" discriminant
250 - 25 = (15)2. Solving the quadratic gives x = 10. Since y
.

must then be 5, the two squares are 102 and 52, and the set has
125 blocks.
11 8 • THE GARDEN OF THE SPHINX

50. IQ Logic
The fractions can be written in the form
1 1 1 1 1
1 x 5 5 x 9 9 x 13 13 x 17 17 x 21
Therefore, the next term could be
1 1
x
= -

21 25 525
and the general term is
[1 + 4(n _

In order to arrive at the general case for the sum of n terms,

\)] [1 + 4n]
we observe that
1 2 2 3 3
1 + 4 2 9
51 5 5
1 + (2 x 4) 3 1 + (3 x 4)
= = = = =

13
In order to confirm our intuition as to this pattern, note that if
5n n
1 + 4n
=

n
+ +
1
5n
(1 + 4n) [1 + 4(n + 1) ]
1
1 + 4n
=
------

n + 1
5n + 1
1 + 4(n + 1 )
=

Thus, the sum of thirteen terms is !�.

 S OLUT I O N S ' 119

51 . Magic Cubes
Top part: 10 26 6
24 1 17
8 15 19
Middle part: 23 3 16
7 14 21
12 25 5
Bottom part: 9 13 20
11 27 4
22 2 18
It will be noted that this cube has been filled by a series of
identical jumps interrupted by regular deviations. One sup­
poses that the cube reproduces itself identically in all directions
and:

a general jump: a step downward to the level below.

after each multiple of 9, a step below.
after each multiple of 3, a step down and to the right toward
the bottom.
120 . THE GARDEN OF THE SPHINX

52. Three Strokes of the Compass

On a straight line, ab and ac are lengths a and b, respectively.
One draws the circle of radius ac and determines the point such
that bd = ac. The circle with center d has the same radius as
the first and cuts the first circle at e. The two triangles eab and
cae are both similar and isosceles. Their base angles are equal.
Therefore,
=
ab ea
ea
- -

ac
This construction supposedly dates back to the seventeenth
century (Thomas Strode).

d 4----f�----"'="'"-- c
a
 SOLUTIONS · 121

53. To Be Continued
Here are some continuations of the sequence 1, 2, 3, 4, 5:
1) 1, 2, 3, 4, 5, 3, 7, 4, 6, 8, 11 . . .
Un is the smallest number the factorial of which is divisible
by n .
2) 1, 2 , 3 , 4, 5 , 5 , 7, 6 , 6 , 7, 1 1 , 7, 13 . . .
Un is the sum of the prime factors.
3) 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2 . . .
Un is the smallest number of cubes required to represent
n.
4) 1, 2, 3, 4, 5, 6 , 7, 8, 9 , 1 0 , 1 1 , 12, 13, 14, 15, 1 . . .
Un is the smallest number of fourth powers necessary to
represent n .
5) 1 , 2 , 3 , 4, 5 , 6 , 7, 8 , 9 , 10, 11, 12, 15, 20, 22 . . .
Un is a sequence of numbers divisible by each of their
digits.
6) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14 . . .
Un = numbers that do not have a prime factor greater
than 7.
7) 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33 . . .
Un = palindrome numbers symmetrical with respect to
their center.
8) 1, 2, 3, 4, 5, 7, 8, 9, 11, 13 . . .
Un = powers of a single prime.
(See N.J.A. Sloane, A Handbook of Integer Sequences [New York:
Academic Press, 1973] . )
122 . THE GARDEN OF THE SPHINX

54. Five Folds

1) ef is determined by folding be to ad.
2) g by making a fold from d such that e comes down along
ef·
3) hi by making a fold from g such that a comes down along
ad and b along be.
4) j by folding ie to ih .
5) jk by making a fold from j such that d comes down along
de and a along ab.
ijkc is a square.
And in the triangle hgd, dg = 1 and hg = 1h . Thus,
hd2 = 1 - % = % ;= jk2.

a e b
,-------,---,

h �r__--+"-----t

"--"--
- --'----..... c
k
 SOLUTIONS • 123

55. Nine Lines Make Twenty-One

Triangles
Here is how to make twenty-one triangles with nine straight
lines.

56. A Cube of Bricks

Twenty-seven bricks measuring 2 x 1 x % cannot be assem­
bled into a cube 3 x 3 x 3. Let us construct a cube as an
assemblage of 27 little cubes 1 x 1 x 1 composed alternately
of black-and-white material. Cubes of the same color come into
contact only al()ng straight lines.
Now if the bricks are assembled to form a cube, none of them
placed at a slant, each brick must be parallel to the edges and
occupy as much of the black as white volume.
Therefore, all told, the 27 bricks must produce 13% white
cubes and 13V2 black cubes, which is obviously impossible .
124 . THE GARDEN OF THE SPHINX

57. Kicking a Goal

Let us assume that the player is at X , an arbitrary point on the
sideline. We proceed to draw the circle that passes through X
and the two goalposts Bl and B2 • If the circle retraverses the
sideline at a second point X ' , it makes no difference whether
one shoots from X or X ' : at both points, the angle is the same,
since it is contained in equal arcs of the circle. But it is preferable
to shoot from any point other than X or X' in the circular
segment XX' . (At all these other points, the angle is greater. )
X is the optimal point if and only if the circle is tangent to the
sideline and does not retraverse it. In that particular case, we
can write the square of the distance from the comer of the field
C to X in relation to the circle:
CX2 = CB1 X CB2 = (35 - 3.66) (35 + 3.66) and CX = 34.81
meters.
 SOLUTIONS · 125

58. Other People's Husbands

Five crossings suffice to transport n couples on a boat capable
of carrying (n - 1) persons.
1) (n - 1) wives get in the boat and cross over to the other
side.
2) (n - 2) wives stay where they are, and the remaining one
brings back the boat.
3) The wife who has done the rowing remains on the original
shore to rest up in the company of her husband and one
other couple, while (n - 2) husbands take the boat across
again to rejoin their wives.
4) One couple returns to the original shore .
5) The remaining three couples make the fifth trip .

59. Sequences and Gaps

The sequence of the divisors of 27720 (i. e . , 2, 32, 5, 7, 1 1 ) is 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 1 5 , 18 . . . Excluded is the
number 13. In the same way, definitions of sequences in the
form of groups of numbers that share a common divisor with
a given number allow one to exclude any given prime number,
at the same time including all the integers that precede it.
One would not suppose it possible similarly to exclude a
composite number in the same way. To do that one makes use
of the function E (the whole part of a rational number) . Thus,
for example, 14 is excluded from this sequence, which contains
the numbers that come before and after it:
126 . THE GARDEN OF THE SPHINX

60. Trick Squares

Let us examine four squares about the same point:
A B
C D
If A contains a and B contains ab, a multiple of b, it is possible
to fill 0 with b (provided that b and a are relatively prime) . But
now one cannot fill C, which cannot have a common factor
with a or b without having one with abo Therefore, two squares
meeting along a line can never contain a number and one of
its multiples, but must contain two numbers of the form ab and
ac, where b and c are relatively prime. The first line can therefore
take the form:
ab ac ad ae
and the flrst column the form:
ab fb gb hb etc.
One is led to create a kind of multiplication table of the flrst
eight primes. In order to keep their products as low as possible,
the four biggest primes are paired with the four smallest, which
yields the solution:
2 3 5 7
11 22 33 55 77
13 26 39 65 91
17 34 51 85 119
19 38 57 95 113
 S O L UT I O N S ' 127

61. Circles and Square Roots of Seven

Let pq be the unit length. We can trace these circles:
1) Having center p passing through q.
2) Having center q passing through p and intersecting the
first at r and m .
3) Having center r passing through p and q and intersecting
the first two at t and s .
4) Having center t passing through p and r and intersecting
the first circle at u .
The sought-after length i s SU, for
s u 2 = m s 2 + mu
2
= (1 + 1) 2 + ( 0) 2
= 4 + 3
= 7
128 • THE GARDEN OF THE SPHINX

62. An Adequate Supply of Digits

Write down all the numbers in a column, one above the other,
and line them up on the right-hand side so that the unit digits
will be in the same column. Add a line of m zeros along the
top containing enough zeros so that each number will have m
digits. Accordingly, this table will contain m(N + 1) digits. How
many zeros does one have to eliminate?
The numbers from 10m 1 to N have none, but there are
10m 1 in the left-hand column, 10m 2 in the second column,
-

- -

10 in the next-to-Iast column on the right, and 1 in the last

column.
10m - 1
The total must be (a number composed of 1s) .
9
10m - 1
That leaves m (N + 1) - digits.
9

63. Pawn Parity

There is invariably an even number of pawns on the white
squares.
Let us assign coordinates to each square running from bottom
to top vertically and from left to right horizontally. In the 3 x 3
board, the coordinates of the black squares (1, 1), (1,3), (3,1)
. . . have both coordinates of the same parity and of different
parity from the white squares.
An arrangement of n pawns on different rows and columns
can be expressed (1, Yl)' (2, Y2) • • (n, Yn) where the y's are all
•

different. Therefore, the sum (1 + Yl) + (2 + yJ + . . .

+ (n + Yn) is even since it is double the sum of the frrst n
numbers. If one excludes all the parentheses with even sums
(L e . , the black squares), the sum of the remaining odd numbers
is even, that is, there is an even number of white squares.
 SOLUTIONS ' 129

64. Trick and Countertrick

The present record is:
A contributor has raised a subtle question: can a square be
simultaneously trick and magic? (Can its lines, columns, and
two diagonals have the same sum?)

5 22 3 7

2 15 14 9

27 26 33 4

13 21 10 11

65. Squaring a Star

Here is a simple way to cut up the star in order to form a
square:
130 . THE GARDEN OF THE SPHINX

66. Matchmaking
The figure shows the 12 ways of connecting five matches that
touch each other in at least one extremity; 28 ways have been
identified for six matches, 74 ways for seven, and 207 ways for
eight.
 SOLUTIONS ' 131

67. Prime Magic

A little experimenting shows that 111 is the smallest sum of
three primes summed in the eight required ways.

7 73 31

61 37 13

43 1 67

68. Constructive Folds

The square must be folded in two twice in order to produce
the medians EF and GH with the center O.
Next, one refolds AD and Be on EF in order to determine IJ
and KL, equidistant parallel lines. The part to the right of EF
is folded back, as is the lower part of GH in order to set off
angle EOG. That makes it possible to place 0 on IJ and to fold
GM in such a way that GM = GO. One obtains GQ = HP = HN
in the same way.
Since angle MGO is 60° and GO = MN = PQ, GMNHPQ is
indeed a regular hexagon.

A'-�---TE---r--� B

D��J --�F---L�� C
132 • THE GARDEN OF THE SPHINX

69. Three and Five Make Four

Let us construct a parallelogram on the two axes Ox and Oy at
an angle of 60° and divided into equilateral triangles. Ox rep­
resents the 5-pint container, empty at 0 and full at A. Oy
represents the 3-pint container B. Let us call C the 8-pint con­
tainer.
The three possible kinds of decantations are:
1 ) y remains constant; that is, pouring from C into A or from
A into C.
2) x remains constant; that is, pouring from C into B or from
B into C.
3) x + y remains constant; that is, pouring from A to B or
from B to A.
The aim is to start from 0 and by means of a broken line to
end up at Z. Each segment will traverse the parallelogram. Also,
it will always be more efficient to bounce off a boundary line
with a symmetrical ricochet (as in billiards) rather than at any
other angle.
One can start along either axis. For example, OA-Ae-ef-fg-gh­
hi; or OB-Bj-jk-kl-lm-mn-no-oZ.
The flrst solution requires only six operations.
 S O L UTI O N S • 133

70. The Price Is Right

The prize can be won as follows. Let us suppose that the correct
price P happens to be within the $51 1 range. The price can then
be found by making corrections in intervals equal to decreasing
powers of 2. Suppose the first estimate is Nt . If that is incorrect,
the response N2 should be:
N2 = Nt + 256 if N t < P
N2 = N t - 256 if Nt > P
N3 = N2 + 128 if N2 < P
N3 = N2 - 128 if N2 > P
etc.
NlO = N9 + 1 if N9 < P
NlO = N9 - 1 if N9 > P
At the worst, Nt - NlO = 256 + 128 + 64 + 32 + 16 + 8
+ 4 + 2 + 1 = 511 . Thus, one sweeps through the whole range
between Nt - 511 and Nt + 511, and the right number is found.

71. Newton 's Steers

The key to solving this problem easily is to calculate the amount
of grass growing on one acre in one day, as expressed in terms
of the surface area of the same amount of grass at the original
height. The value for the appropriate constant factor h can be
obtained from the two equal expressions for the amount of grass
cropped by a steer in a day, as calculated from the preliminary
data of the problem.
60 + (12 x 6Oh) 72 + (15 x 72h )
- l tll3 , then h
_

Thus, if -
75 x 12 81 x 15
must equal V12.
If one allows for the growth of the grass, cropping a field of
96 acres in 18 days is the equivalent of cropping 96 +
(18 x 96h) = 240 acres. At the rate of 1 V3 acres per steer per
day, 100 steers would be needed.
134 . THE GARDEN OF THE SPHINX

72. Sixteen Equals Four Times Fifteen

73. Truncated Cubes

If a cube is truncated at one apex, only one axis will survive:
the axis of symmetry of order 3 perpendicular to the equilateral
triangle .
If it is truncated at two apexes, there will remain one axis of
order 3 and three axes of order 2.
If it is truncated at four nonadjacent axes, there will remain
three axes of order 2 (the original axes of order 4) and four axes
of order 3.
 SOLUTIONS • 135

74. Leaping Counters

The situation is greatly simplified by rearranging the positions
in such a way that they are next to each other if they are linked
by a simple jump. In that case, the displacements will consist
of moving to an adjacent numbered position. The counters are
all on their original positions. One sees that they cannot leap
over one another and can only achieve the circular permutations
on 1 4 2 3:
a d b c, d b c a, b c a d, and c a d b. This gives for 1 2 3 4: a b c
d, d c a b, b a d c, c d b a .

a
I
8 6
c 3
11

10 4 d

5 9
12 2
7 b
136 . THE GARDEN OF THE SPHINX

75. Trickery and Magic

This square is both magic insofar as its rows, columns, and
diagonals add up to 90 and antitrick insofar as the numbers
that are adjacent vertically or hOrizontally are relatively prime
and have a common divisor when they touch each other at an
angle.
Are there larger magic and antitrick squares? Are there magic
and trick squares in which the numbers that touch diagonally
are relatively prime and have a common divisor if they touch
at an angle?

3 32 39 16

40 9 14 27

21 34 33 2

26 15 4 45
 S O L UT I O N S ' 137

76. Black and White Dots

The numbers 21 and 14 are multiples of 7, suggesting an in­
tersecting heptagon for constructing the following solution of
21 dots in 14 alignments of two black dots and two white dots.
138 • THE GARD E N OF THE SPHINX

77. Christmas Dinner

By a point on a circle, let us represent each brother, and let us
join each to the position of his wife by a chord. If we system­
atically explore all the ways of arranging the six chords, we can
count nine different figures, each of which can be permutated
by rotating the arrangement of the chords.
The first figure, the figure itself and its mirror image, gives
two arrangements for the wives. Equally, the second figure
gives two arrangements. The following figures give, respec­
tively, 4, 6, 6, 12, 12, 12, and 24 arrangements, yielding a total
of 80. Thus, the possibilities will be exhausted in 40 years.
 SOLUTIONS · 139

78. Flawless Cubes

Here is an assemblage of tricubes in which no pair of tricubes
overlap the same vertical or horizontal layer .

• •
-

• • •

79. What Times Are They?

There does not appear to b e an algebraic solution that eliminates
trial and error. It is simpler, therefore, to imagine (or actually
use) two strips of paper divided by vertical lines into intervals
of five units and four units, respectively. In testing the seven
positions from second to second that do not allow the chimes
of the two clocks to differ more than three seconds from second
to second one finds thirteen strokes from five positions and in
each instance for nine o'clock. One can assume that it is nine
'
0 clock and overlook any divergence in the rates of the clocks.
There is another solution. If one accepts that the two clocks
do not have a difference in synchrony of an integral number
of seconds, one can consider the possibility of its being eight
o'clock for an asynchrony of between two and three seconds.
140 . THE GARDEN OF THE SPHINX

80. Beyond Thirst

Here is a way o f planting the flag by a system requiring only
eleven and a half days worth of water. It requires the slight
innovation of accepting a solution involving fractions of days.
Let us designate A as the starting point, B a point a quarter
of a day's march away from A, C a day and a half away from
A, and D the point of arrival. The man carrying the flag leaves
with a five-day supply of water of which he leaves a four-and­
a-half-day supply at B and then returns to the starting point.
He sets out again with a five-day supply; sets aside a one­
quarter day's supply at B, and a two-and-a-half-day supply at
C before returning. On the return trip he uses one quarter of
a day's water at B. The final time he leaves with one and one­
half days' water and picks up three and three-quarters at B and
one and one-quarter at C. He marches the remaining distance
and plants the flag. On the way back, he takes one and one­
quarter from C and two and one-quarter from B.
Do you think that this is the best pOSSible solution? Tum to
problem 141 .
 PROBLEMS ' 141

81. Kings, Queens, and Knaves

Let us follow the situation of the first player and suppose that
he holds c cards of which f are face cards. The probability of
c - f
his failing to draw a face card on his first try is , and the
c
(c - 1) - f
probability of his failing on his second try is . If one
c - 1
equates the product of those twoJ:robabilities to %, one can
calculate that c = % (4f + 1 ± V8F + 1), but f can take only
the values between 1 and 12. Since c and f must be whole
numbers, the fIrst player can only have 21 cards, of which 6
are face cards, or 4 cards, of which one is a face card. And the
second player must be subject to the same calculations. In order
for the third player to have the smallest hand, it is necessary
that both of the fIrst two have 21 cards, of which 6 are face
cards. And if that is so, the third player can hold no face cards.
(See Jacoby and Benson, Mathematics for Pleasure [New York:
Fawcett, 1973] .)
142 . THE GARDEN OF THE SPHINX

82. Folding, Folding

It's easy to determine an octagon by another method of creating
folds in a square, if we change dimensions. An octagon of half
the size appears when we use folds to join each midpoint of a
side to its two opposite angles.

/
"
/
,
/
' /
"
/
"
/
,
 SOLUTIONS • 143

83. Whodunit?
The two last statements of A and C are the most important
because they respond to each other and risk contradiction. If
A is right when he says that C will tell the truth, C should be
believed when he says that A has lied, which is impossible. A
is mistaken, therefore, and the last statement of C is not true.
For this to be contrary to truth, at least one statement of A
must be true. This can only be the first, so B is guilty.

84. A False Unknown

The digit on the left of a number of two or more digits cannot
be equal to O. Since the digits in the thousands column are all
different, their sum must be at least 1 + 2 + 3 + 4 = 10. But
since the result has only four digits, one of the following is
wrong : A, B, C, or D . But their s u m m u s t be at l e a s t
1 + 1 + 2 + 3 = 7.
Now, according to the units column, x must be an even
number. Therefore, x = 8. In the same column one can deduce
that 0 is 1 or 6. But there must be a carryover of 2 to the second
column in order to maintain its parity. Therefore, 0 is 6 and C
is 4. Therefore, 0 is incorrect in the thousands column, since
C is 4 and A and B are 1 or 2 or 2 and 1 and 0 is 1 . The
hundreds column cannot contribute a carryover but receives a
carryover of 1 . Therefore, E must be 3, and B must be 1 .
The error arose in writing a 0 instead o f a B in the thousands
column.
(See E. R. Emmet, A Diversity of Puzzles [Totowa, N.J . : Barnes
& Noble, 1977] . )
144 . THE GARDEN OF THE SPHINX

85. Practical Trisection

Here is the articulated device invented by Laisant in 1885 (?)
for trisecting angles. The circled parts represent joints, and the
parts within squares are sliding points. Two deformable rhom­
buses have a common angle that guarantees that the three
angles originating at the apex will all be equal.
 S OLUTIO N S ' 145

86. The Bandbox

There are eleven different ways of assembling three bandboxes
illustrated below. But they are not limited to these few unless
one assumes implicitly that the constructions may not place
smaller lengths beside the square of the base. If this prohibition
is waived, then there are an infinite number of constructions.
In only one of these constructions does a part of a bandbox
protrude over its base.
146 • T H E G A R D E N OF T H E S P H I N X

87. A Multicolored Chessboard

It will require at least four colors: on a board 2 x 2 any two
squares can be linked by one move of the king.
Four colors are enough, for if one juxtaposes the 2 x 2 squares,
each one filled in the same manner with four colors, one arrives
at the solution:
A B A B . . .

C O C O
A B A B
C O C O

88. Breaking the Chain

To open one link is clearly insufficient. That leaves 3 segments
of chain of lengths 1, a, and b. Recombining those will produce
a maximum of seven different lengths: 1, a, b, a + 1, b + 1,
a + b, and a + b + 1, which cannot cover the twenty-three
lengths. Opening 2 links, which leaves 5 segments of chain,
permits many more combinations. In fact, one can solve the
problem by the following division: 3-1-6-1-12.
Opening two links suffices.
 SOLUTIONS ' 147

89. What's It Good For?

Let p be a point on the vertical tangent to the circle and t the
point of contact of another tangent from p. If the angle apt is
given, it is possible to make the apparatus coincide so that pt
is tangent to the circle. In that case b and e determine the two
trisectors of the angle, since the angles pab, pbe, and pet are
equal.

a c
148 . THE GARDEN OF THE SPHINX

90. Reversible Magic

If the five rows of the square have the same sum, the sum of
the twenty-five palindromes has to be divisible by five. Let us
see whether that sum ends in a 5 or a O. The twenty-five
numerical palindromes beginning with 1 1 are 11, 22 . . , 99,
.

101, 1 1 1 , 121 .
. , 191, 202 . . . , 252. Since the sum of their
.

units in the digits place is 67, the sum of the twenty-five num­
bers cannot be divisible by 5. Therefore, no magic square can
be made with them.

91 . A Cross and a Square

92. The Finishing Touch

The second player has an assured win. He is certain to fill in
the last square if he leaves his opponent either two empty
squares that are not aligned or two different alignments of two
empty squares.
The second player can always reintroduce that position re­
gardless of his opponent's tactics. If the offensive is on two or
three squares, he completes a T, an L, or a cross of five squares.
If the offensive is on one square, he makes a right angle of
three squares. Regardless of the other's countermoves, he main­
tains the winning position.
(See Silverman, Your Move [New York: McGraw-Hill, 1974] .)
 SOLUTIONS ' 149

93. Back to Trisection

Here is the apparatus that will trisect any angle. All the fixed
arms are of equal length. The square indicates a sliding jOint,
and the rhombuses provide the fundamental property of the
structure.
150 • THE GARDEN OF THE SPHINX

94. The Symmetrical Cross

This solution requires seven scissor cuts, but it produces a
perfectly symmetrical combination of pieces.

'" y
"' IA'/'
"' 'V
" 1/
"
L�
�I\
" /�
I\.

11
11 '�
11 ""
11 ""
l"-
/[ r....
LV r....
� r....
 SOLUTIONS · 151

95. Century Sundays

In determining what centuries start on Sunday, the first pitfall
to avoid is, of course, to start a century on a year that is a
multiple of 100. In other words, the next century begins on
January 1, 200l .
What one must do is to make certain calculations, modulo 7;
that is, to consider only remainders to that modulus. Every
normal year containing 365 days shifts the day by one notch,
and leap years by two. Thus, in calculating from January 1,
1978, to January 1, 2001, there are 23 years of which 6 are leap
years, which leaves a remainder of 1 and places January 1, 2001,
on a Monday.
Only years divisible by 400 have the extra leap-year day.
Therefore, it follows, that a century requires an adjustment of
2 for the 100 years plus 3 for the 24 leap years, giving a total
of 5. The result is the following first days for the centuries:
2001 : Monday; 2101 : Saturday; 2201: Thursday; 2301 : Tuesday;
2401: Monday.
The adjustment at the end of 4 centuries being 0, these are
the only days that come up; none of these centuries starts on
a Sunday.

96. To Have or Not to Have

The distributions have the same probability, since it is obvious
that if you and your partner do not hold any of the cards in
question, your opponents will hold them all .
152 . THE GARDEN OF THE SPHINX

97. The Area of the Slice

In this problem, abc is equilateral, and its area if -rr/6. To it must
be added the areas x and y, which belong to the rectangle dfge,
V3
along with the rest of the section, or 1 - 2' The total of these
V3
areas is -rr/6 + 1 - 2' which comes to approximately 0.658.

a b
-------..

d f
e .....:;;.__ ...;=--.
;:: ... 9
 SOLUTIONS · 153

98. Integral Triangles

Let us designate the three sides x, y, z and apply Heiron's
formula for the area:
x + y + z =
� v'(x + y + z) (x + y - z) (x - y + z) ( - x + y + z)
And then:
16 (x + Y + z) = (x + y - z) (x - y + z) ( - x + y + z)
Now those factors are either all odd or all even, since they differ
from one another by twice two of the sides . They are all even,
therefore, since their product is an even number. One can there­
fore conclude:
x + y - z = 2k
(/(2 + 4) (x- k)
Y kx - (/(2 + 4)
=

z = x + y - 2k
One obtains for k = 1 and k = 2:
5, 12, 13
6, 8, 10
6, 25, 29
7, 15, 20
9, 10, 17
The other values repeat these solutions.
(See Stephen Ainley, Mathematical Puzzles, 1977.)
154 . THE GARDEN OF THE SPHINX

99. Vases
This is a standard problem when the capacity of the third vase
is given. But I have seen it solved in the following way, without
knowing in advance the capacity of the third vase. Two equa­
tions can be set up . Let 5x - 3y = 4, with the solution that
x = 2 and y = 2, which suggests 3 pints in the flrst vase, 5 in
the second, and none in the third. We start with this combi­
nation (3, 5, 0) and then 3, 0, 5; 0, 3, 5; 3, 3, 2; 1, 5, 2; 1, 0, 7;
0, 1, 7; 3, 1, 4.
Let 3x - 5y = 4, which gives the solution x = 3 and y = 1,
suggesting 3, 5, 0; 0, 5, 3; 3, 2, 3; 0, 2, 6; 2, 0, 6; 2, 5, 1; 3, 4,
1; 0, 4, 4. With the last method, 6 pints suffice for the third
vase.

100. Writer's Cramp

Since he produces 1150 page a day when he has only 1 page
left to write, daily writing speed is 1/50 the balance as a general
rule. As he starts with a speed of 1110, that means that the
work contains 5 pages. The first page took him 10 days; the
second page, at 4150 a day, 12% days; the third page, at 3/50 a
day, 167"3 days; the fourth day, at 2/50 a day, 25 days; the flfth
page, 50 days, making 114% days all told.
 S OLUTIONS ' 155

101. Lining Up Blacks and Whites

156 . THE GARDEN OF THE SPHINX

102. More Alignments

 S O L U T l O N S · 157

103. Trisection in the Enlightenment

For trisecting angles, the apparatus is made up only of two
external arms with rods of equal lengths:
CD = BD = DF = DE = AB = AC
The little circles indicate fixed joints, and the little rectangles
indicate sliding ones. The angle EDF is three times the angle
BAC. For the angle DBE is the exterior angle of the isosceles
triangle DAB, and accordingly equal to BAC, and also to DEB
since BDE is isosceles. The angle GDE is exterior to ADE and
therefore equal to one and one-half times the angle BAC. Its
double FDE is three times BAC. To make a division by five
(quintisection?), one must attach additional arms and of equal
length: GF, GE, GH, and GI. The angle HGI is the quintuple
of angle ABC. It is apparent that the device described in problem
89 can be adapted to dividing an angle by five, with the ad­
vantage of also dividing it by four; but with the disadvantage
of having to add five supplementary rods instead of four.
158 . THE GARDEN OF THE SPHINX

105. Pi in an Unfamiliar Context

The first decoupage gives a regular hexagon, of which the sides
are the six linear parts. The second decoupage gives a rectangle.

106. Toward Unity

The sum of the two numbers will obviously be 1 without any
need to invoke the concept of limits if one can fmd two fractions
a b-a
b and - b
- such that the two numbers are those of the decimal
expressions.
The first fraction must be a little less than 2110. What is the
denominator? % = 0 . 17 . .is too large; 1h = 0.14 . . . is too
.

small. Let us try 2111, 2112, 2113 . . . . This last fraction works,
and so the two fractions are the decimal expressions of 2113
and 11113, respectively.
 SOLUTIONS ' 159

104. The Odds on the White Card

Be careful about calculating the probabilities in this experiment.
There are not two equally probable events, but actually three.
In fact, the white side showing can be either one side of the
white card, the other side of the white card, or the white side
of the red-and-white card. Thus, there are two chances in three,
and not one in two, that the other side is white. So I should
give you two to one odds on the white card or six dollars against
three. If I offer you less than that, the odds are on my side.
 160 . THE GARDEN OF THE SPHINX

107. An Accidental Ellipse

An ellipse is described by each point (except the top) of a pair
of folding household steps, in the shape of an old-fashioned
stile . This happens when the hinge at the top breaks and the
two sets of steps tumble to the ground, one of the feet of one
set being propped against a wall. The diagram below shows
the cross-section of the steps in boldface lines, the one half
blocked by the wall at point O . If the two steps are of length a,
the prolongation of the second as far as the wall is the same
length. An arbitrary point is the distance r from the top and
has as its coordinates x and y. The lengths x' and y' complement
them on the sides of the large right-angle triangle. The similarity
a - 2
of the two triangles gives us 1J.. and . By the Pythagorean
y a + r
theorem, x2 + y' 2 = (a + rf. From this one can deduce that
x2 y2
--- + = 1.
(a + 2) 2 (a - r) 2
Thus, the point must describe a quarter Ellipse with axes
a - r and a + r.

,.
.
,,
,

,,
\

,
, a
,
,
'
,
y ,
,
\
, p

O L-���--�Q�----+ x
 SOLUTIONS ' 161

108. Your Basic Boat

Although i t would appear t o violate the laws o f physics, such
a boat is possible and has been entered in special races in
England . The "motor" consists of a rope attached to the stem.
To move the boat forward, the crew gives the rope a series of
short, sharp jerks. The system depends on a certain friction
effect of the water and would not work in the absence of such
a medium (say, in outer space) .
It is thus possible to displace the center of gravity of such a
craft by a process involving three steps:
1) A displacement of the center of gravity toward the stem;
as one leans forward this creates a slight forward motion
of the boat checked by the resistance of the water.
2) A displacement toward the bow similarly retarded by fric­
tion.
3) The sudden tension of the rope, which transmits kinetic
energy to the boat, providing an obviously more powerful
impulse than the effect of the two preceding steps.
It is possible to achieve a speed of 3 knots by this means.
(See Rouse-Ball, Mathematical Recreations, rev. ed. [New York:
Macmillan, 1960] .)
 162 . THE GARDEN OF THE SPHINX

109. Points and Lines

110. Squaring the Square

If you have tried to juxtapose two geom e tric squares to form a
third, you have bumped up against what seems-so far-an
impossibility .

On the other hand, it is possible to juxtapose the squares of

two numbers. For example, 4 and 9, the squares of 2 and 3,
can be put together to make 49, the square of 7. The exploration
of such numbers appears to require trial and error. Here are
others up to 1 00, 000 : 169, 361, 1225, 1444 , 1681, 3249, 4225,
15625, 16900, 36100, 42025, 49729, 64009, 81225, 93025.
(Problem 134 is a further extension of this problem. )
 S OLUTI ONS ' 163

111. The Missing Token

The sum total of the tokens being 1 + 2 + 3 + 4 +
5 + 6 + 7 + 8 + 9 = 45, the sum of the remaining tokens is
not divisible by 3 unless the missing one is also so divisible.
Thus, it has to be 0, 3, 6, or 9, which reveals that the sum of
the remaining tokens is 45, 42, 39, or 36. Only one of these
totals is divisible by 4, which proves that it must have been the
9 token that was lost. But these conditions are not sufficient.
One must first prove that the groups can, in fact, be formed.
There are (8,3, 1), (0, 7,5), (6,4,2), and (8, 1), (7,2), (6,3) , and
(5,4,0) .

112. Quadrature and Dissection

Two cuts suffice to tum two juxtaposed squares into a third.
(The arrows indicate two equal lengths. )
164 " THE GARDEN OF THE SPHINX

113. An Uneasy Cryptogram

In his bathtub Jack saw an indigo cockroach trying to climb out.
It had a jaunty air and a wisp of grass in its mouth. This was
Jack's first warning of drug addiction.
This message has the unusual feature of containing no E's,
in spite of the frequency of that letter in English spelling. The
cipher itself involves a method of alphabetical substitution as­
cribed to Julius Caesar.
For each letter of the plaintext the letter 15 places subsequent
in the alphabet has been substituted. Because the text uses no
Es, that letter has been omitted altogether, leaving a 25-letter
alphabet.

114. Roots
Assume that you wish to find the square root of n for which a
Vn
is a rough approximation greater than " This approximation
T
a2
a can be improved systematically by the formula m =
n
�.
-

If m is positive, a is too large by that amount, if m is negative,

a must be increased accordingly. The corrected value of a is
used to calculate a second value for m and so on iteratively
until one arrives at the desired result a 2 - n = O. The advan­
tage of performing these steps on a pocket calculator is obvious.
If it has a memory function, this can be performed without
pencil and paper.
 SOLUTIONS ' 165

115. The Triangulation of the Triangle

The lines joining the midpoints of the sides yield an immediate
solution, if we remember that they are parallel to the respective
corresponding sides.
From this we can draw a general principle: if two triangles
have two isometric sides and angles corresponding to these
supplementary sides, then there is a way of cutting that trans­
forms one into the other.
166 . THE GARDEN OF THE SPHINX

116. Travels on the Hexagon

11 7. A Bailout Fee
One can assume that the workman's charge is based on the
price for raising his pulley one unit of length. Let us call x the
price after emptying the first pailful, which he has raised a
mean height of Vz yard. The second pailful raised a mean height
of 1 Vz yards costs three times as much. The third costs
five times as much, and so on. Thus the first 10 yards
cost 100x = x(1 + 3 + . . . + 19) while 20 yards cost
x(1 + 3 + . . . + 39) = 400x. Accordingly, the workman is
owed $100. Calculus buffs can verify that this example comes
out. If the worker had used a pump and worked continuously,
the result would be the same.
 S O L UT I O N S ' 167

118. A Prize Hamster

Naturally, there is a logical pattern governing the succession.
Each price was arrived at by adding to the previous price the
sum of the square of its digits:
425 + 16 + 4 + 25 470, etc.
=

802 + 64 + 0 + 4 = 870.

119. The Women Man the Oars

There are eight crossings of the boat; the letters designate the
men and the numbers the wives:
A1B2C3D4

A1BCD 234
A1B2CD 34
ABCD 1234
A1BCD 234
A1 B2C3D4
A1B2 C3D4
A1 B2C3D4

A1B2C3D4

120. Breaking Up a Year

As the first operation shows, it is not helpful to deal in 1s. It's
not any more helpful to introduce three 2s as factors, since
3 x 3 > 2 x 2 x 2. 4s are not helpful as 4 = 2 + 2 = 2 x 2.
One should avoid the digits 5 and larger: 3 x 2 > 5, 3 x 3 > 6,
and more generally, (n 2) x 2 > n. There remain 3s and a
pair of 2s since 1980 = 660 x 3. The number in question is 3660•
-
168 • THE GARDEN OF THE SPHINX

121. Protagoras at the Bar

If the court decides that Protagoras is in the right and wins his
case, the problem is resolved. But actually the court can only
make its decisions relative to events that have already taken
place . It cannot deliver a judgment bearing on an event that is
in the process of taking place. Consequently, since up to the
most recent past the student has not won a case, the court must
sustain his right not to pay.
But afterward, at the very moment the trial is over, the status
of the student has changed: he has, in fact, won a case. Pro­
tagoras can now, through a second suit, make a valid claim for
payment of his fee. He will win that one with certainty and get
paid-if the student has the money. But the situation called for
two lawsuits instead of one.
 S O L UT I O N S · 169

122. Squaring Polygons

Here is one solution for transforming the irre gular but sym­
metrical decagon into a square by cutting it into eight pieces.
Note that it has been necessary to tum over one of the pieces.
Some puzzle fans have done even better, coming up with dif­
ferent ways of creating a square out of seven pieces. At least
one person believes it is possible to do even better than that!

rr -t-
I l
- I

I
,-
t
-;-'
-
l-

I i-'
I

I �
-

+, . " , \
" oJ
170 THE G A R DEN OF THE SPH I N X
.

123. Drawing the Line

The solution is presented in two steps, the start and the finish.
They would be illegible if one tried to combine them.
 S O L UT I O N S ' 171

124. A Band of Five Squares Creates

Another Square
It is obviously necessary to introduce V5 in order to guide the
process of trial and error.

125. The Fifth Power

This can be proved by induction by showing that if it is true
for n = k, it will be true for n = k + 1 .
A Simpler "empirical" approach is to note that all numbers
end in 0, 1, . . or 9, and the fifth power of 0, 1, . . . 9 end,
.

respectively, in 0, 1, . . . 9, as you can verify by a table or

laborious calculation. Therefore, the last digit of the fifth power
of a number is the last digit of the number itself. Thus, nS n
ends with a ° and is divisible by 5.
-
172 . THE GARDEN OF THE SPHINX

126. A Line to Follow

 SOLUTIONS · 1 73

127. A Tricky Division

Here is a way to cut up the rest of a square into four congruent
parts after a quarter of it has been removed in the form of a
triangle. One has to be bold enough to imagine the unfamiliar
shapes Ail, Bb, Cc, and Dd.
(See Royal Magaz ine, November 1930)

128. Unlisted Numbers

In order to eliminate the six-digit numbers containing 12, we
shall distinguish five classes of numbers according to these
forms:
A 1 2
B 1 2
C 1 2
D 1 2
E . 1 2
Each class contains 104 numbers. There is no number common
to AB, BC, CD, DE. On the other hand, AC, AD, AE, BD, BE,
CE each have 102 numbers in common. Additionally, there is
a number common to AEC: 121212.
Therefore, the number o f numbers to be excluded is
5 x 104 + 1 - 6 x 1()2 = 49,401 .
174 . T H E G A R D E N OF T H E S P H I N X

129. No Way to Make a Square

Except for the number 1 itself, no number composed of Is
exclusively can be the square of an integer.
This is made clear by an examination of the composition of
such numbers, each of which can be expressed in the form
1 1 + lOOn 4(25n + 2} + 3.
It is obvious that such numbers, if divided by 4, will always
=

leave a remainder 3.
As for the squares of integers, they are either even numbers
o f the form 4p 2 or odd numbers of the form
(2n + 1} 2 = 4n 2 + 4n + 1. Therefore, the remainder of any
square of a whole number upon division by 4 is either 0 or 1 .
For that reason, n o number made up o f 1 (other than I) can be
the square of an integer.

130. A Primary Law

f - 1 = (p + 1}(P - I)
Since p is a prime and therefore an odd number, and therefore
not divisible by 3, one of the two numbers (p I) and (p + I)
is divisible by 2 and the other by 4.
-

Also, one of the two numbers (p - I) and (p + I) is divisible

by 3. Therefore f - 1 = (p + 1}(P - I) = 2 x 3 x 4m = 24m.
This proof makes it abundantly clear that p need not be a
prime but need only be odd and not divisible by 3 (e.g., 25,
35, 49, etc.). Thus, the underlying equation is satisfied by one
out of every three integers.
 S O L UT I O N S ' 175

131. Ten Digits for One

Here are six fractions expressing 9 in which all ten digits appear
once and only once:
9 = = =
57429 58239 75249
06381 06471 08361
-- -- --
= = =
95742 95823 97524
10638 10647 10836
The search for this is helped by observing that the last digits
of the denominator and the numerator must be divisible by 9.
It is also helpful to try to start with an approximative division
by 9-approximately 54, giving 6, etc. Of course, sticklers will
object to the appearance of the Os in the first three expressions.
There seems to be no other way of tackling this problem.
(See Moscow Puzzles: Three Hundred Fifty-Nine Mathematical
Recreations, [New York: Charles Scribner's Sons, 1972] . )

132. Inside Out

An individual triangle remains on the inside or the outside . In
order to prove that, it is sufficient to trace a line starting on the
outside of the figure and ending up on the inside of the triangle.
Each time that that line follows the necessary course, it will
pass alternatively from the outside to the inside and from the
inside to the outside. Therefore, the position of the triangle
depends on the parity (odd or even) of the number of traversals.
But that number remains the same for each triangle regardless
of the nature of the solution.
176 . THE GARDEN OF THE SPHINX

133. Raindrops Are Falling . . .

Let us suppose that a man is h feet high, e feet thick, and r feet
wide and that he traverses a distance 1 in a time t. We shall
also assume that the amount of water in the air (i. e . , the number
of raindrops) is proportional to the volume m3• We also assume
that the rain is falling perpendicularly to the ground with the
velocity v (there being virtually no wind) . During the time t in
which he runs the distance 1, he will receive the amount of rain
contained in two parallelepipeds: (1) the horizontal surface he
exposes to the rain striking him from above multiplied by the
height of the column of rain that will fall in time t (this is the
same as the amount of rain that would fall on him if he stood
still for t seconds), that is, r x e x v x t; (2) the second par­
allelepiped is the vertical surface he exposes to the rain as he
runs through it, that is, r x h x I. Combining the volumes of
the two parallelepipeds, one obtains volume r(lh + evt), and
the amount of rain that strikes him = Q = dr(lh + evt) . It is
immediately apparent that the function Q increases with in­
creasing t. To stay as dry as possible, he must run as fast as
possible . Of course, perfectionists can carry this calculation fur­
ther by conSidering the advantage of leaning over, what to do
if there's a wind, etc.

134. Remarkable Quadratures

This wonderful number is 4,950,625 = 22252 . It can be decom­
p o s e d a s 4 = 2 2 and 950, 625 = 9 75 2 o r 49 = 72 and
50,625 = 2252 .
 SOLUTIONS ' 177

135. Mobius and His Better Half

This is how to cut the Mobius strip in a straight line without
touching more than one edge . The figure represents the strip
before it is given the half tum and fastened at the ends. Thus
the problem is reduced to cutting in the straight line a b c d e.
One can verify that the areas are equal by adding up the com­
ponents .

�t - - - - -
d - - _ _
=
e
- - - -

136. Quadruple Quadrature

One solution is 19,0252 = 361 ,950,625, where 361 = 192 and
950, 625 = 9752 and where 36 = 6 2 , 1 = P, 9 = 3 2 , and
50, 625 = 2252• Another solution is 90, 1252 = 8, 122,515, 625,
where one recognizes 81,225, 1 and 5,625 as squares, as well
as 81,225 = 2852 and 15,625 = 1252 • The reader might like to
explore the possibility of a systematic production of these quad­
ruple quadratics rather than discovering them through trial and
error.
178 . THE GARDEN OF THE SPHINX

137. Something New About Nines

We must prove that every integral square of two digits or more
contains at least one even digit.
All the two-digit squares have that property: 16, 25, 36, and
81 . Every other square is 100 or greater. Now, the square of a
number greater than 10 is greater than 100 and can be expressed
in the form (lOd + U) 2 = 100d2 + 20du + u 2, where u is not
less than 1 . Since the units digit is the same as the units digit
of u 2 , the property is established for u = 0, 2, 4, 6, 8. If u = I,
in 100d2 + 20d + I, the digit in the tens column is given by
2d, which is even. The reader can verify for himself that for
u = 3, 5, 7, 9 there will be an even number in the tens column.
 SOLUTIONS · 179

138. The Trisection of Papp us

To prove that the construction attributed to Pappus is in fact
one-third of the given angle, let us first produce a circle with
its center at e, the midpoint of cd, and passing through a . Since
aeo is isosceles, the angles numbered 2 are equal. One of them
is the exterior angle of ead and therefore equal to the sum of
the two angles numbered 1 .
(See Asger Aaboe, Episodes from the Early History of Mathematics
[New Mathematical Library, No. 13, 1975] . )

�--���b----
180 . THE GARDEN O F THE SPHINX

139. Archimedes Cuts It Up

Here is the square as cut up by Archimedes in order to obtain
fourteen pieces, each in a rational relation to the whole, the
fractions expressing forty-eighths. The computation of each is
elementary.

140. Tails I Win

The simplest way to approach this problem is to note that at
the very start of the game there is a probability of V2 that tails
will come up and the first player will win. For the second player
to win on his first toss, the probability is V2 x V2 = Y4 (the
probability of heads on the first player's toss times the proba­
bility of tails on the second player's toss) . Since the first player's
chances will always remain twice the second player's in every
pair of tosses and since the sum of their probabilities has to be
I, since one has to win, the probability of winning is ¥3 and V3
for the first and second players, respectively.
 SOLUTIONS · 181

141 . The Barrier of Thirst

Should the author of this solution be proud of it? Leave with
four and a half days' supply of water and a companion who
carries a five-day supply. At the end of a day and a half, your
companion decides not to go on living. You now have a six­
and-a-half-day supply of water. Leaving a one-and-a-half-day
supply where you are, you complete the trip to your goal and
return to your starting point, picking up the extra supply on
the way back. Can one do better? Yes, indeed. Take a look at
problem 150.

142. Regroup ing the Pieces

182 . THE GARDEN OF THE SPHINX

143. A Square for a Compass

Draw a circle and retain the opening of the legs of the compass.
Starting with a point A, chosen arbitrarily on the circumference,
and without changing the compass, find the three pOints B, C,
and D. Thus, A will be diametrically opposed to D. Using again
your center A, trace a circle passing through C and then at D
for an arc passing through B and meeting the preceding one at
E . OE is the side of the square. In fact,
OE = \lAE2 - R2
= y'7\C2 R2
_

= v'(R\73)2 - R2
= RV2

144. Words and Numbers

Here is one method inspired by the work of the mathematical
lOgician Kurt Godel that permits one to associate with each
word composed of letters or other symbols a number composed
of digits. One begins by assigning to each letter the number
representing its position in the alphabet from 1 to 26. Each
word is then represented by a product in which the first prime
numbers in succession are raised to the powers of the successive
numbers (L e . , places in the alphabet) of the letters spelling the
word. Thus, ART = 21 318 52°. The conversion of this number
back to the original word is accomplished by decomposing the
product into its prime factors and assigning to the power of
each prime factor the corresponding letter of the word.
 SOLUTIONS ' 183

145. Words, Numbers, and Sentences

To be able to associate a number with a sentence in an un­
ambiguous way, one can use again the method described in
problem 144, but one must take certain precautions in order to
avoid ambiguity. One must now assign to each letter an odd
number: 3 for A, 5 for B, etc., 53 for Z.
Let us assign to each word the product of the fIrst prime
numbers raised to the powers corresponding to the numbers
of the individual letters. Thus, ARTS = 2 3 3 3 754173 9 and
FINE = 21 3319529711 . We can now assign to an entire sentence
the product of the fIrst successive prime numbers raised to the
numbers of its words. Thus, FINE ARTS = 2FINE3ARTS. Ambi­
guity is eliminated since the power of 2 is odd for a word and
even for a sentence.

146. A Triangle in a Square

There are two possible ways of inscribing a triangle in a square:
1) Let the point of origin of one angle be at an apex of the
square with the two other angles touching the sides that
do not meet at the apex.
2) Let each of its three angles touch a side.
The fIrst possibility leads to an equilateral triangle with a side
that is greater than the side of the square.
The second possibility leads to an equilateral triangle of which
one side can at its minimum be equal to a side of the square if
there is parallelism. This is then the answer one is seeking.
One apex touches the middle of a side of the square, and the
opposite side of the triangle is parallel to the opposite side of
the square. The triangle and the square have equal sides.
184 . THE GARDEN OF THE SPHINX

147. Mysterious Powers

Have you defined the properties of this mysterious square?
When a number terminates in 12890625, all its integral powers
end in the same sequence:
if n = 12890625 = 58 X 33
n 1 = 12890624 = 28 X 5083
1) = 108 X q
-

n(n
n + 108 X q
-

n2 =

Thus, the numbers n and n 2 end in the same digits.

148. Sup erior Antimagic

One puzzle fan produces numerous magic squares of order 6.
Each original solution gave him twenty-three varieties having
at least twenty digits out of place. "It suffices to permute the
lines and columns in the same order. If the original is numbered
1, 2, 3, 4, 5, 6, one considers three couples 16, 25, 34, each
remaining symmetrically distributed with respect to the medi­
ans of the square . One permutes successively one couple, then
two, then three . " In that way you can achieve your own var­
iation of the square .

5 35 13 34 12 7
6 8 30 23 11 27
32 28 17 16 20 2
33 9 22 21 18 15
10 26 4 14 29 25
31 24 3 19 36
 PROBLEMS · 185

149. Coup led Coup les

Here is how they can all get across the river in seven crossings:
1) Five wives cross .
2) One wife returns .
3) Her husband stays with her, and four other men cross .
4) One couple returns.
5) Two couples cross.
6) One couple returns.
7) The last two couples cross.

150. No Holds Barred

Consider the simple possibility that the planting of the flag after
a four-day journey across the desert is the ultimate purpose of
your life. In that case you need only bring four days' worth of
water and, your goal once achieved, depart from this world.
This might be called the kamikaze solution!