Take a small card and place it on its longer
edge upon the dotted line. Now set the picture
in a good light on the table, and let your head
drop gradually towards the card until you almost
touch it with your nose. You will see the bird
fly into the jaws of the snake!
1. A carpenter was called in to mend a hole
in a wooden floor. The gap was two feet wide,
and twelve feet long, while the only board at
hand was three feet wide, and eight feet long.
This had been put aside as useless, but, on
catching sight of it, the carpenter ran his rule
over it and said that he could make a perfect
fit, and cover all the hole by cutting the board
into two pieces. How did he do this?
Solution
This most interesting optical illusion was devised
by Professor Thompson some years ago:—
If the illustration is moved by hand in a small
circle on the level, with such motion as is given
in rinsing out a bowl, the circles of the larger
diagram will seem to revolve in the direction
in which the paper is moved, while the cogs of
the smaller diagram will apparently turn slowly
in the opposite direction.
Here is another combination of the clever
illusion of the whirling wheels.
If a rapid rotating motion is given to the
diagram, each circle will seem to revolve, and
the cog wheel in the centre will appear to move
slowly round in the opposite direction.
2. A man leaves an orchard of forty choice
apple trees to his ten sons. On the first tree is
one apple, on the second there are two, on the
third three, and so on to the fortieth, on which
there are forty.
Each son is to have four of the trees, and on
them an equal number of the apples. How can
they thus apportion the trees, and how many
apples will each son have? Here is one way:—
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
| 20 |
19 |
18 |
17 |
16 |
15 |
14 |
13 |
12 |
11 |
| 21 |
22 |
23 |
24 |
25 |
26 |
27 |
28 |
29 |
30 |
| 40 |
39 |
38 |
37 |
36 |
35 |
34 |
33 |
32 |
31 |
| 82 |
82 |
82 |
82 |
82 |
82 |
82 |
82 |
82 |
82 |
Can you find another perfect solution?
Solution
We call very particular attention to this fascinating
illustration of the fact that the mind
and eye may receive and register false impressions
under quite simple conditions:—
Hold this at rather more than reading distance,
upright, and move it steadily up and down. The
dark line will soon seem to slide up and down upon
the perpendicular line. It will be better seen if
drawn to pattern on a card.
As the eye falls upon the principal lines of this
interesting diagram, an immediate impression is
formed that they are not parallel.
This, however, is a most curious illusion, created
in the mind entirely by the short sloping lines,
as is found at once by the simple test of measurement.
3. With no knowledge of the surrounding district,
I was making my way to a distant town
through country roads, guided by the successive
sign-posts that were provided.
Coming presently to four cross-roads I found
to my dismay that some one had in mischief
uprooted the sign-post and thrown it into the
ditch. In this perplexing fix how could I find
my way? A bright thought struck me. What
was it?
Solution
4. As I stood on the platform at a quiet country
station, an engine, coming along from my left
at thirty miles an hour, began to whistle when
still a mile away from me. The shrill sound
continued until the engine had passed a mile and
a half to my right. For how long was I hearing
its whistle?
Solution
In this curious optical illusion the lines are
exactly equal in length.
The eye is misled by the effect which the lines
drawn outward and inward at their ends produce
upon the mind and sight.
5.
In room marked A two men were placed,
A third he lodged in B;
The fourth to C was next assigned,
The fifth was sent to D.
In E the sixth was tucked away,
F held the seventh man;
For eighth and ninth were G and H,
Then back to A he ran.
Thence taking one, the tenth and last,
He lodged him safe in I;
Thus in nine rooms ten men found place,
Now can you tell me why?
Solution
6. Are there any particular days of the week
with which no new century can begin?
Solution
These straight lines, at right angles to each
other, are, though they do not seem to be, exactly
equal in length.
This and similar illusions are probably due to
the variation of the vague mental standard which
we unconsciously employ, and to the fact that
the mind cannot form and adhere to a definite
scale of measurement.
7. Why has a spliced cricket bat such good
driving power? and why is the “follow through”
of the head of a golf club so telling in a driving
stroke?
Solution
8. On my bookshelf in proper order stand two
volumes. Each is two inches thick over all, and
each cover is an eighth of an inch in thickness.
How far would a bookworm have to bore in
order to penetrate from the first page of Vol. I.
to the last page of Vol. II.?
Solution
Can you decide at a glance which of the two
lines below the thick band is a continuation of
the line above it?
Make up your mind quickly, and then test your
decision with a straight edge.
9. From what height must a man fall out of
an airship—screaming as he goes overboard—so
as to reach the earth before the sound of his cry?
N.B.—Resistance of the air, and the acoustical
fact that sound will not travel from a rare
to a dense atmosphere, are to be disregarded.
Solution
10. Imagine a man on a perfectly smooth table
surface of considerable size, in a vacuum, where
there is no outside force to move him, and there
is no friction. He may raise himself up and
down, slide his feet about, double himself up,
wave his arms, but his centre of gravity will be
always vertically above the same point of the
surface.
How could he escape from this predicament,
if it was a possible one?
Solution
It is difficult, even after measurement, to believe
that these figures are of the same size.
But they will stand the test of measurement.
11. A rope passes over a single fixed pulley. A
monkey clings to one end of the rope, and on
the other end hangs a weight exactly as heavy
as the monkey. The monkey presently starts
to climb up the rope. Will he succeed?
Solution
12. Seeing that the tension on a pair of traces
tends as much to pull the horse backward as it
does to pull the carriage forward, why do the
traces move on at all?
Solution
A very curious and interesting form of optical
illusion is well illustrated by what may be called
“the shifting brick.”
The central brick, drawn to show all its edges,
as though it were made of glass, will assume the
form indicated by one or other of the smaller
bricks at its right and left, according to the way
in which the eyes accommodate themselves for
the moment to one pattern or to the other. If
you do not see this at first, look steadily for
awhile at the pattern you desire.
13. Why does a rubber tyre leave a double
rut in dust, and a single one in mud?
Solution
14. If two cats, on opposite sides of a sharply
sloping roof, are on the point of slipping off,
which will hold on the longest?
Solution
If you place four coins in the positions shown
at the top of this diagram, and attempt, or challenge
some one to attempt, without any measuring,
to move the single coin down in a straight
line until the spaces from C to D on either side
exactly equal the distance from A to B—
It must drop as far as is shown here, which
seems to the unaided eye to be too far.
This excellent illusion can be shown as an after-dinner
trick with four napkin-rings.
Here is another excellent optical illusion.
Look attentively at the diagram below, and notice
in which direction you apparently look into it,
as though it were an open cask.
Now shake the paper, or move it slightly, and
you will find, more often than not, that you seem
to see into it in quite the opposite direction.
15. I hold a penny level between my finger
and thumb, and presently let it fall from the
thumb by withdrawing my finger. It makes
exactly a half-turn in falling through the first
foot. If it starts “heads,” how far must it fall
to bring it “heads” to the floor?
Solution
16. “They call these safety matches,” said
Funnyboy at his club one day, “and say that
they strike only on the box. Don’t believe it! I
can strike them quite easily on my boot.”
No sooner said than done. He took out a
match, struck it on his boot, and—phiz!—it was
instantly alight. The box was handed round,
and match after match was struck by the bystanders
on their boots, but not one of them
could succeed.
“You don’t give the magic touch,” said Funnyboy,
as he gaily struck another. How did he
do it?
Solution
How many cubes can you see as you look at
the large diagram? The two smaller ones should
be looked at first alternately, and they will assist
the eye to see at one time six, and at another time
seven, very distinct cubes.
This curious optical illusion is not easily followed
by eye to the finish of the several lines.
Each short line is, in fact, part of the circumference
of a circle, and the circles when completed
will be found to be accurately concentric. It
would seem at first sight that the lines are taking
courses which would eventually meet at some
point common to them all.
17. We commend this curious point to the
special attention of cyclists:—
A bicycle is stationary, with one pedal at its
lowest point. If this bicycle is lightly supported,
and the bottom pedal is pulled backward, what
will happen?
Solution
A most remarkable optical illusion is produced
by the blending of the dark and light converging
rays of this diagram. Stand with your back to
the light, hold the page, or better still, the diagram
copied on a card, by the lower right-hand
corner, give it a continuous revolving movement
in either direction, and the visible ghost
of a silver coin, sometimes as large as sixpence,
sometimes as large as a shilling, will appear!
Where can it come from?
18. A merchant has a large pair of scales,
but he has lost his weights, and cannot at the
moment replace them. A neighbour sends him
six rough stones, assuring him that with them
he can weigh any number of pounds, from 1 to
364. What did each stone weigh?
Solution
Here is a pretty form of our first illusion:—
Place the edge of a card on the dotted line,
look down upon it in a good light, and, as you
drop your face till it almost touches the card, you
will see the goose move towards the sugar in the
little maiden’s hand.
19. A wheel is running along a level road,
and a small clot of mud is thrown from the
hindermost part of the rim. What happens to
it? Does it ever renew its acquaintance with
the wheel that has thus rejected it?
Solution
Here is another method by which an optical
illusion of length is very plainly shown:—
Judged by appearances, the line A B in the
larger figure is considerably longer than the line
A B below it, but tested by measurement they
are exactly equal.
20. A village carpenter undertook to make
a cupboard door. When he began to put it in
its place it was too big, so he took it back to
his workshop to alter it. Unfortunately he now
cut it too little. What could he do? He determined
to cut it again, and it at once became a
good fit. How was this done?
Solution
Here is another excellent illustration that
seeing is not always believing.
No one could suppose at first sight that these
four lines are perfectly straight and parallel,
but they will stand the test of a straight edge.
The divergent rays distract the vision.
21. If from the North Pole you start sailing
in a south-westerly direction, and keep a straight
course for twenty miles, to what point of the
compass must you steer to get back as quickly
as possible to the Pole?
Solution
The optical illusion in the picture which we
reproduce is due to the defective drawing of the
two men on the platform. In actual size upon
the paper the further man looks much taller than
the other.
Measurement, however, shows the figures to
be exactly of a height. This illusion is due to
the fact that the head of the further man is quite
out of perspective. If he is about as tall as the
other, and on level ground, both heads should
be about on the same line. As drawn, he is, in
fact, a monster more than eight feet high.
22. If Dick, who is five feet in height, stands
bolt-upright in a swing, the ropes of which are
twenty feet long, how much further in round
numbers do his feet travel than his head in
describing a semi-circle?
Solution
Here is an excellent and very simple illustration
of a well-known optical curiosity:—
Hold this picture at arm’s length in the right
hand, hold the left hand over the left eye, and
draw the picture towards you gradually, looking
always at the black cross with the right eye.
The black disc will presently disappear, and then
come into sight again as you continue to advance
the paper.
23. Can you name nine countries in Europe
of which the initial letters are the same as the
finals?
Solution
Here is a delightfully simple way in which
market gardeners, or others who buy or sell
weighty produce, can check their invoices for
potatoes or what not.
Say, for example, that a consignment weighs
6 tons, 10 cwts., 1 qr. Then, since 20 cwts. are
to a ton as 20s. are to a pound, and each quarter
would answer on these lines to 3d., we can at once
write down £6 10s. 3d., as the price at £1 the
ton. On this sure basis any further calculation
is easily made.
Look steadily, in a good light, for thirty seconds
at the cross in the eye of the pictured skull;
then look up at the wall or ceiling, or look fixedly
at a sheet of paper for another thirty seconds,
when a ghost-like image of the skull will be
developed.
A gardener, when he had planted 100 trees on
a line at intervals of 10 yards, was able to walk
from the first of these to the last in a few
seconds, for they were set on the circumference
of a circle!
Here is another example of what is known as
the persistence of vision:—
Look fixedly for some little time at this grotesque
figure, then turn your eyes to the wall or
ceiling, and you will in a few seconds see it appear
in dark form upon a light ground.
The sum of nine figures a number will make,
Of which just a third will remain
If fifty away from the whole you should take,
Thus turning a loss to a gain.
It needs something more than mere arithmetic
to discover that the solution to this puzzle is
XLV, the sum of the nine digits, for if the L is
removed, XV, the third of XLV, remains.
Here is another curious illusion:—
The four straight lines are perfectly parallel,
but the contradictory herring-bones disturb the
eye.
If our penny had been current coin in the first
year of the Christian era, and had been invested
at compound interest at five per cent., it would
have amounted in 1905 to more than
£132,010,000,000,000,000,000,000,000,000,000,000,000.
This gigantic sum would afford an income of
£101,890,000,000,000,000,000 every second to
every man, woman, and child in the world, if we
take its population to be 1,483,000,000 souls!
Absurdly small in contrast to these startling
figures is the modest eight shillings which the
same penny would have yielded in the same time
at simple interest.
Here in another form is shown the illusion of
length.
At first sight it seems that the two upright
lines are distinctly longer than the line that slopes,
but it is not so.
Here is a neat method of discovering the age
of a person older than yourself:—
Subtract your own age from 99. Ask your
friend to add this remainder to his age, and then
to remove the first figure and add it to the last,
telling you the result. This will always be the
difference of your ages. Thus, if you are 22,
and he is 35, 99 - 22 = 77. Then 35 + 77 = 112.
The next process turns this into 13, which, added
to your age, gives his age, 35.
In this diagram one hundred and twenty-one
circular spots are grouped in a diamond.
If we half close our eyes, and look at this through
our eye-lashes, we find that it takes on the appearance
of a section of honeycomb, with hexagonal
cells.
Here is a ready method for multiplying together
any two numbers between 12 and 20.
Take one of the two numbers and add it to
the unit digit of the other. Beneath the sum
thus obtained, but one place to the right, put
the product of the unit digits of the two original
numbers.
The sum of these new numbers is the product
of the numbers that were chosen. Thus:—
| 19 × 13. |
(19 + 3) |
= |
22 |
| (9 × 3) |
= |
27 |
| |
247 |
Black has made an illegal move. He must
replace this, and move his king as the penalty.
White then mates on the move.
Solution
Here is a method for determining the total
number of balls in a solid pyramid built up on
a square base:—
Multiply the number of shot on one side of
the base line by 2, add 3, multiply by the number
on the base line, add 1, multiply again by the
number on the base, and finally divide by 6.
Thus, if the base line is 12—
12 × 2 + 3 = 27; 27 × 12 + 1 = 325; 325 × 12 = 3900;
and 3900 ÷ 6 = 650, which is the required number.
White might have given mate on the last move.
White now to retract his move, and mate at once.
Show by analysis the mating position.
Solution
White to play and draw.
Solution
“Now, boys,” said Dr Bulbous Roots to
class, “you shall have a half-holiday if you
prove in a novel way that 10 is an even number.”
Next morning, when the doctor came into
school, he found this on the blackboard:—
| |
SIX |
= |
6 |
| SIX |
= |
9 |
| By subtraction |
S |
= |
-3 |
| |
SEVEN |
= |
7 |
| S |
= |
-3 |
| Therefore |
SEVEN |
= |
10 |
Q. E. D.
(Quite easily done!)
The half-holiday was won.
White to play, and draw.
Solution
White to play, and mate in two moves.
There are twelve variations in this beautiful
problem.
Solution
It will interest all who study short cuts and
contrivances to know that a novice at arithmetic
who has mastered simple addition, and can
multiply or divide by 2, but by no higher numbers,
can, by using all these methods, multiply any
two numbers together easily and accurately.
This is how it is done:—
Write down the numbers, say 53 and 21, divide
one of them by 2 as often as possible, omitting
remainders, and multiply the other by 2 the same
number of times; set these down side by side,
as in the instance given below, and wherever there
is an even number on the division side, strike out
the corresponding number on the multiplication
side. Add up what remains on that side, and
the sum is done. Thus:—
| 53 |
21) |
| 26 |
(42) |
| 13 |
84) |
| 6 |
(168) |
| 3 |
336) |
| 1 |
672) |
| |
1113) |
which is 53 multiplied by 21.
White to play, and mate in two moves.
Solution
White to play, and mate in two moves.
Solution
White to play, and mate in two moves.
Solution
The following particulars about a very rare
property of numbers will be new and interesting
to many of our readers:—
The number 6 can only be divided without
remainder by 1, 2, and 3, excluding 6 itself. The
sum of 1 + 2 + 3 is 6. The only exact divisors
of 28 are 1, 2, 4, 7, and 14, and the sum of these
is 28; 6 and 28 are therefore known as perfect
numbers.
The only other known numbers which fulfil
these conditions are 496; 8128; 33,550,336;
8,589,869,056; 137,438,691,328; and 2,305,843,008,139,952,128.
This most remarkable rarity of
perfect numbers is a symbol of their perfection.
White to play, and mate in two moves.
There are no less than twelve variations!
Solution
White to play, and mate in two moves.
Solution
Somewhat akin to perfect numbers are what
are known as amicable numbers, of which
there is a still smaller quantity in the realm of
numbers.
The number 220 can be divided without remainder
only by 1, 2, 4, 5, 10, 11, 22, 44, 55, and
110, and the sum of these divisors is 284. The
only divisors of 284 are 1, 2, 4, 71, and 142, and
the sum of these is 220.
The only other pairs of numbers which fulfil
this curious mutual condition, that the sum of
the divisors of each number exactly equals the
other number, are 17,296 with 18,416, and
9,363,584 with 9,437,056. No other numbers,
at least below ten millions, are in this way
“amicable.”
White to play, and mate in two moves.
Solution
White to play, and mate in three moves.
Solution
| 1 |
× |
9 |
+ |
2 |
= |
11 |
| 12 |
× |
9 |
+ |
3 |
= |
111 |
| 123 |
× |
9 |
+ |
4 |
= |
1111 |
| 1234 |
× |
9 |
+ |
5 |
= |
11111 |
| 12345 |
× |
9 |
+ |
6 |
= |
111111 |
| 123456 |
× |
9 |
+ |
7 |
= |
1111111 |
| 1234567 |
× |
9 |
+ |
8 |
= |
11111111 |
| 12345678 |
× |
9 |
+ |
9 |
= |
111111111 |
White to play, and mate in three moves.
Solution
There is a pleasant touch of mystery in the
following method of discovering a person’s age:—Ask
any such subjects of your curiosity to write
down the tens digit of the year of their birth, to
multiply this by 5, to add 2 to the product, to
multiply this result by 2, and finally to add the
units digit of their birth year. Then, taking the
paper from them, subtract the sum from 100.
This will give you their age in 1896, from which
their present age is easily determined.
White to play, and mate in three moves.
Solution
As I came in after a day among the birds and
rabbits, the keeper asked me—“Well, sir, what
sport?” I replied, “36 heads and 100 feet.”
It took him some time to calculate that I had
accounted for 22 birds and 14 rabbits.
White to play, and mate in three moves.
Solution
The nine digits can be arranged to form fractions
equivalent to
13
14
15
16
17
18
19
thus:—
582317469 =
13
795631824 =
14
297314865 =
15
294317658 =
16
527436918 =
17
932174568 =
18
836175249 =
19
White to play, and mate in three moves.
Solution
White to play, and mate in three moves.
Solution
This motor problem will be new and amusing
to many readers:—
Let m be the driver of a motor-car, working
with velocity v. If a sufficiently high value is
given to v, it will ultimately reach pc. In most
cases v will then = o. For low values of v, pc
may be neglected; but if v be large it will generally
be necessary to square pc, after which v will
again assume a positive value.
By a well-known elementary theorem, pc + lsd
= (pc)2, but the squaring may sometimes be
effected by substituting x3 (or × × ×) for lsd.
This is preferable, if lsd is small with regard to m.
If lsd be made sufficiently large, pc will vanish.
Now if jp be substituted for pc (which may
happen if the difference between m and pc be
large) the solution of the problem is more difficult.
No value of lsd can be found to effect the squaring
of jp, for, as is well-known, (jp)2 is an impossible
quantity.
White to play, and mate in three moves.
Solution
White to play, and mate in three moves.
Solution
To divide any sum easily by 99, cut off the
two right-hand figures of the dividend and add
them to all the others. Set down the result of
this in line below, and then repeat this process
until no figures remain on the left to be thus
dealt with.
Now draw a line down between the tens and
hundreds columns, and add all up on the left of
it, thus:—
| 8694 |
|
|
32 |
|
120 |
|
|
78 |
|
| 87 |
|
|
26 |
1 |
|
|
98 |
| 1 |
|
|
13 |
|
|
|
99 |
| |
|
|
14 |
121 |
and |
99 |
over. |
| 8782 |
and |
14 |
over. |
In other words, 122. |
The last number on the right of the lines shows
always the remainder. If this should appear as
99 (as in the second example above), add one to
the number on the left.
White to play, and mate in three moves.
Solution
White to play, and mate in three moves.
Solution
The question, “How many times can 19 be
subtracted from a million?” was set by an
examiner, who no doubt expected that the
answer would be obtained by dividing a million
by 19. One bright youth, however, filled a
neatly-written page with repetitions of
| 1,000,000 |
1,000,000 |
1,000,000 |
| 19 |
19 |
19 |
| 999,981 |
999,981 |
999,981 |
and added at the foot of the page, “N.B.—I can
do this as often as you like.”
There was a touch of unintended humour in
this, for, after all, the boy gave a correct answer
to a badly worded question.
White to play, and mate in three moves.
Solution
White to play, and mate in three moves.
Solution
Here is a very smart and very simple method
of dividing any multiple of 9 by 9, from the fertile
brain of Lewis Carroll:—Place a cypher over the
final figure, subtract the final figure from this,
place the result above in the tens place, subtract
the original tens figure from this, and so on to
the end. Then the top line, excluding the intruded
cypher, gives the result desired. Thus:—
36459 ÷ 9 = 4051,0
36459 = 4051.
White to play, and mate in three moves.
Solution
| 1 |
× |
8 |
+ |
1 |
= |
9 |
| 12 |
× |
8 |
+ |
2 |
= |
98 |
| 123 |
× |
8 |
+ |
3 |
= |
987 |
| 1234 |
× |
8 |
+ |
4 |
= |
9876 |
| 12345 |
× |
8 |
+ |
5 |
= |
98765 |
| 123456 |
× |
8 |
+ |
6 |
= |
987654 |
| 1234567 |
× |
8 |
+ |
7 |
= |
9876543 |
| 12345678 |
× |
8 |
+ |
8 |
= |
98765432 |
| 123456789 |
× |
8 |
+ |
9 |
= |
987654321 |
White to play, and mate in three moves.
Solution
Here is one of the methods by which we can
readily discover a number that is thought of.
The thought-reader gives these directions to his
subject: “Add 1 to three times the number you
have thought of; multiply the sum by 3; add
to this the number thought of; subtract 3, and
tell me the remainder.” This is always ten times
the number thought of. Thus, if 6 is thought of—6
× 3 + 1 = 19; 19 × 3 = 57; 57 + 6 - 3 = 60, and
60 ÷ 10 = 6.
White to play, and mate in three moves.
Solution
Here is a curious rough rule for remembering
distances and sizes:—
The diameter of the earth multiplied by 108
gives approximately the sun’s diameter. The
diameter of the sun multiplied by 108 gives the
mean distance of the earth from the sun. The
diameter of the moon multiplied by 108 gives
the mean distance of the moon from the earth.
White to play, and mate in three moves.
Solution
Says Giles, “My wife and I are two,
Yet faith I know not why, sir.”
Quoth Jack, “You’re ten, if I speak true,
She’s one, and you’re a cypher!”
White to play, and mate in three moves.
Solution
Here is a curious and quite uncommon method
of dividing any multiple of 11 by 11.
Set down the multiple of 11, place a cypher
under its last figure, draw a line, and subtract,
placing the first remainder under the tens place.
Subtract this from the next number in order,
and so on throughout, adding in always any
number that is carried. Thus:—
| 363 |
56408 |
375034 |
| 0 |
0 |
0 |
| 33 |
5128 |
34094 |
White to play, and mate in three moves.
Solution
Perhaps the old saying, “there is luck in odd
numbers,” may have some connection with the
curious fact that the sum of any quantity of
consecutive odd numbers, beginning always with
1, is the square of that number. Thus:—
| 1 + 3 + 5 |
= |
9 |
= |
3 |
× |
3. |
| 1 + 3 + 5, etc., up to 17 |
= |
81 |
= |
9 |
× |
9. |
| 1 + 3 + 5, etc., up to 99 |
= |
2500 |
= |
50 |
× |
50. |
White to play, and mate in three moves.
Solution
In the number 142857, if the digits which
belong to it are in succession transposed from
the first place to the end, the result is in each
case a multiple of the original number. Thus:—
| 285714 |
= |
142857 |
× |
2 |
| 428571 |
= |
142857 |
× |
3 |
| 571428 |
= |
142857 |
× |
4 |
| 714285 |
= |
142857 |
× |
5 |
| 857142 |
= |
142857 |
× |
6 |
White to play, and mate in three moves.
Solution
By the following simple method, a plausible
attempt is made to prove that 1 is equal to 2:—
Suppose that a = b, then
| ab |
= |
a2 |
| ∴ |
ab - b2 |
= |
a2 - b2 |
| ∴ |
b(a - b) |
= |
(a + b)(a - b) |
| ∴ b |
= |
a + b |
| ∴ b |
= |
2b |
| ∴ 1 |
= |
2 |
This process only proves in reality that 0 × 1 =
0 × 2, which is true.
White to play, and mate in four moves.
Solution
Few people know a very singular but simple
method of calculating rapidly how much any
given number of pence a day amounts to in a
year. The rule is this:—Set down the given
number of pence as pounds; under this place
its half, and under that the result of the number
of original pence multiplied always by five.
Take, for example, 7d a day:—
| £7 |
0 |
0 |
| 3 |
10 |
0 |
| |
2 |
11 |
| £10 |
12 |
11 |
The reason for this is evident as soon as we
remember that the 365 days of a year may be
split up into 240, 120, and 5, and that 240 happens
to be the number of pence in a pound.
A small wheel with ten teeth is geared into a
large fixed wheel which has forty teeth. This
small wheel, with an arrow mark on its highest
cog, is revolved completely round the large wheel.
How often during its course is the arrow pointing
directly upwards? Here is a diagram of the
starting position.
Solution
Here is a most curious and interesting question:—When
an engine is drawing a train at
full speed from York to London, what part of the
train at any given moment is moving towards
York?
At any time, when the engine is drawing a train
at full speed from York to London, that part of
the flange of each wheel which is for the moment
at its lowest is actually moving backwards towards
York.
For any point, such as A, on the circumference
of the tyre, describes in running along a series
of curves, as shown by full lines in the diagram;
and any point, B, on the outer edge of the flange,
follows a path shown by the dotted curves.
If these lines are followed round with a pencil
in the direction of the arrows, it will be found
that the point on the flange actually moves
backwards as it passes below the track, while the
point A, as it completes each curve, is at rest for
the instant on the track, just before it starts
afresh. The speed of the train does not affect
these very curious facts.
In the subjoined diagram A and B represent
two islands, round which a river runs as is indicated,
with fifteen connecting bridges, that lead
from the islands to the river’s banks.
Can you contrive to pass in turn over all these
bridges without ever passing over the same one
twice?
Solution
In a school where two boys were taught to
think out the bearings of their work, a sharp
pupil remarked that 100 is represented on paper
by the smallest digit and two cyphers, which are
in themselves symbols of nothing. The master,
quick to catch any signs of mental activity, took
the opportunity to propound to his class the
following ingenious puzzle:—How can the sum
of 100 be represented exactly in figures and signs
by making use of all the nine digits in their reverse
order? This is how it is done:—
9 × 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 100.
Another ingenious method of using the nine
digits, so that by simple addition they sum up
to exactly 100, and each is used once only, is
this:—
15 + 36 + 47 = 98 + 2 = 100
Here is another arrangement by which the
nine digits written in their inverse order can be
made to represent exactly 100:—
98 - 76 + 54 + 3 + 21 = 100.
Here is yet another way of arriving at 100 by
using each of the digits, this time with an 0:—
Here is quite a pretty scientific experiment,
which any one of a handy turn can construct and
arrange:—
The spiral track is formed of two wires bent,
and connected by curved cross-pieces. The upper
twist is turned so that the ball starts on a horizontal
course.
During the accelerated descent the ball acquires
momentum enough to keep it on the vertical
track, held outwardly against the wires by centrifugal
force.
Convenient proportions are: height of spiral
two feet, diameter six inches, and wire rails three-quarters
of an inch apart.
A close approach to an ideal flying machine
can be made with a little ingenuity. Two Y-shaped
standards, secured to the backbone rod,
support two wires which carry wings of thin silk,
provided with light stays, and connected at their
inner corners with the backbone by threads.
Rubber bands are attached to a loop on the inner
end of the crank shaft, and secured to a post at
the rear. These are twisted by turning the shaft
with the cross wire, and when the tension is released
the wings beat the air and carry the bird
forward. It is known as Penaud’s mechanical
bird, and has been sold as an attractive toy.
A simple apparatus constructed on the lines
of this illustration will give an interesting proof
of the laws which govern falling bodies on an
inclined plane or on a curved path.
In the case of the inclined plane the ball is
governed by the usual law which controls falling
bodies. In that of the concave circular curve,
as it is accelerated rapidly at the start, it makes
its longer journey in quicker time. In the case
of the cycloidal curve it acquires a high velocity.
This curve has therefore been called “the curve
of swiftest descent,” as a falling body passes over
it in less time than upon any path except the
vertical.
Here is another very simple and pretty illustration
of the natural forces which come into play
in “looping the loop.”
This scientific toy on a small scale may be
easily made, if care is taken that the height of the
higher end of the rails is to the height of the
circular part in a greater ratio than 5 to 4.
A ball started at the higher end follows the
track throughout, and at one point is held by
centrifugal force against the under side of the
rails, against the force of gravity.
If a ball is fired point blank from a perfectly
horizontal gun, and travels half a mile over a
level plain before it touches ground, and another
similar ball is at the same moment dropped from
the same height by some mechanical means, the
two balls will touch ground simultaneously.
The flight, however long, of one through the
air has no influence upon the force of gravity,
which draws it earthward at the same resistless
rate as it draws the other that is merely
dropped.
A quick method of multiplying any number of
figures by 5 is to divide them by 2, annexing a
cypher to the result when there is no remainder,
and if there is any remainder annexing a 5.
Thus:—
464 × 5 = 2320; 464 ÷ 2 = 232, annex 0, = 2320.
753 × 5 = 3765; 753 ÷ 2 = 376, annex 5, = 3765.
A duck begins to swim round the edge of a
circular pond, and at the same moment a water
spaniel starts from the middle of the pond in
pursuit of it.
If both swim at the same pace, how must the
dog steer his course so that he is sure in any case
to overtake the duck speedily?
Solution
Louis Napoleon, Emperor 1852
| |
1852 |
|
1852 |
|
1852 |
| date |
1 |
date |
1 |
date |
1 |
| of |
8 |
of |
8 |
of |
8 |
| his |
0 |
Empress’s |
2 |
their |
5 |
| birth |
8 |
birth |
6 |
marriage |
3 |
| |
1869 |
|
1869 |
|
1869 |
Thus, by a most remarkable series of coincidences,
the principal dates of the Emperor and
Empress of the French added, as is shown above,
to the year of the Emperor’s accession, express
in each instance the year before his fall.
In this domino diagram we have a pretty and
practical proof that the squares of the sides containing
the right angle in any right-angled triangle
are together equal to the square of the side opposite
to the right angle.
Each stone forms two squares, and it is easily
seen that the number of squares which make up
the whole square on the line opposite the right
angle are equal to the number of those which
make up the two whole squares on the lines which
contain that angle.
A second point to be noticed is that the number
of pips on the large square are equal to the number
on the other two squares combined, an arrangement
of the stones which forms quite a game of
patience to reproduce, if this pattern is not at
hand.
Four colours at most are needed to distinguish
the surfaces of separate districts on any plane
map, so that no two with a common boundary
are tinted alike.
On this diagram A, B, and C, are adjoining
districts, on a plane surface, and X borders, in
one way or another, upon each.
It is clearly impossible to introduce a fifth area
which shall so adjoin these four districts as to
need another tint.
Here is another freak of figures:—
| 9 |
× |
1 |
- |
1 |
= |
8 |
| 9 |
× |
21 |
- |
1 |
= |
188 |
| 9 |
× |
321 |
- |
1 |
= |
2888 |
| 9 |
× |
4321 |
- |
1 |
= |
38888 |
| 9 |
× |
54321 |
- |
1 |
= |
488888 |
| 9 |
× |
654321 |
- |
1 |
= |
5888888 |
| 9 |
× |
7654321 |
- |
1 |
= |
68888888 |
| 9 |
× |
87654321 |
- |
1 |
= |
788888888 |
| 9 |
× |
987654321 |
- |
1 |
= |
8888888888 |
A bird made fast to a pole six inches in diameter
by a cord fifty feet long, in its flight first uncoils
the cord, keeping it always taut, and then recoils
it in the reverse direction, rewinding the coils
close together. If it starts with the cord fully
coiled, and continues its flight until it brings up
against the pole, how far does it fly in its double
course?
Solution
Ask a person to write down in a line any number
of figures, then to add them all together as
units, and to subtract the result from the sum
set down. Let him then strike out any one
figure, and add the others together as units,
telling you the result.
If this has been correctly done, the figure struck
out can always be determined by deducting the
final total from the multiple of 9 next above it.
If the total happens to be a multiple of 9, then
a 9 was struck out.
A fly, starting from the point A, just outside
a revolving disc, and always making straight for
its mate at the point B, crosses the disc in four
minutes, while the disc is revolving twice. What
effect has the revolution of the disc on the path
of the fly?
Solution
This Magic Square is so arranged that the
product of the continued multiplication of the
numbers in each row, column, or diagonal is 4096,
which is the cube of the central 16.
The railway, D E F, has two sidings, D B A
and F C A, connected at A. The rails at A,
common to both, are long enough to hold a single
wagon such as P or Q, but too short to admit
the whole of the engine R, which, if it runs up
either siding, must return the same way.
How can the engine R be used to interchange
the wagons P and Q without allowing any flying
shunts?—From Ball’s Mathematical Recreations.
Solution
It is a little known and very interesting fact
that an equilateral triangle can easily be drawn
by rule of thumb in the following way:—Take
a triangle of any shape or size, and on each of its
sides erect an equilateral triangle. Find and
join the centres of these, and a fourth equilateral
triangle is always thus formed, as shown by the
dotted lines.
These centres are centres of gravity, and they
are symmetrically distributed around the centre
of gravity of the original triangle.
The figure formed by joining them must therefore
be symmetrical, and, as in this case, it is a
triangle, it must be always equilateral.
There can be no better instance of how the eye
may be deceived than is so strikingly afforded in
these very curious diagrams:—
The square which obviously contains sixty-four
small squares, is to be cut into four parts,
as is shown by the thicker lines. When these
four pieces are quite simply put together, as
shown in the second figure, there seem to be
sixty-five squares instead of sixty-four.
This phenomena is due to the fact that the
edges of the four pieces, which lie along the
diagonal A B, do not exactly coincide in direction.
In reality they include a very narrow diamond,
not easily detected, whose area is just equal
to that of one of the sixty-four small squares.
Very curious are the results when the nine
digits in reverse order are multiplied by 9 and
its multiples up to 81. Thus:—
| 9 |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
× |
9 |
= |
|
8 |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
9 |
| |
× |
18 |
= |
1 |
7 |
7 |
7 |
7 |
7 |
7 |
7 |
7 |
7 |
8 |
| |
× |
27 |
= |
2 |
6 |
6 |
6 |
6 |
6 |
6 |
6 |
9 |
9 |
7 |
| |
× |
36 |
= |
3 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
6 |
| |
× |
45 |
= |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
5 |
| |
× |
54 |
= |
5 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
4 |
| |
× |
63 |
= |
6 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
3 |
| |
× |
72 |
= |
7 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
2 |
| |
× |
81 |
= |
8 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
It will be seen that the figures by which the
reversed digits are multiplied reappear at the
beginning and end of each result except the first,
and that the figures repeated between them are
to be found by dividing the divisors by 9 and subtracting
the result from 9. Thus, 54 ÷ 9 = 6, and
9 - 6 = 3.
In this nest of triangles there are no less than
six hundred and fifty-three distinct triangles of
various shapes and sizes.
In a chamber of the Great Pyramid an ancient
Egyptian jar was found, marked with the device
now known as Pharaoh’s seal.
Can you count the number of triangles or
pyramids, of many sizes, but all of similar shape
that are expressed on it? Solvers should draw
the figure on a larger scale.
Solution
It is interesting to note that the repeated
addition of odd numbers to one another can be
so arranged as to produce cube numbers in due
sequence. Thus:—
| 1 |
= |
1 × 1 × 1 |
| 3 + 5 |
= |
2 × 2 × 2 |
| 7 + 9 + 11 |
= |
3 × 3 × 3 |
| 13 + 15 + 17 + 19 |
= |
4 × 4 × 4 |
| 21 + 23 + 25 + 27 + 29 |
= |
5 × 5 × 5 |
and so on, to any extent.
In a large old-fashioned garden walks were
arranged round a central fountain in the shape
of a Maltese cross.
If four persons started at noon from the fountain,
walking round the four paths at two, three, four
and five miles an hour respectively, at what time
would they meet for the third time at their starting-point,
if the distance on each track was one-third
of a mile?
Solution
When the tens of two numbers are the same,
and their units added together make ten, multiply
the units together, increase one of the tens by
unity, and multiply it by the other ten. The
result is the product of the two original numbers,
if the first result follows the other. Thus:—
43 × 47 = 2021.
Can you cut Fig. A into two parts, and so rearrange
these that they form either Fig. B or
Fig. C?
The two parts of A must not be turned round
to form B or C, but must retain their original
direction.
Solution
Coal may fail us, but we can never run short
of material for “words that burn.” It has been
calculated that if a man could read 100,000 words
in an hour, and there were 4,650,000 men available,
they could not pronounce the possible variations
which could be formed from the alphabet
in 70,000 years!
It is possible, in a sense, by the following neat
method, to take 45 from 45, and find that 45
remains:—
| 987654321 |
= |
45. |
| 123456789 |
= |
45. |
| 864197532 |
= |
45. |
Cut out in stiff cardboard four pieces shaped
as Fig. 1, four as Fig. 2, and
four as Fig. 3, taking care
that they are all exactly
true to pattern in shape and
proportion to one another.
Now see whether you can
put the twelve pieces together so as to form a
perfect octagon.
Solution
Here is a proof that 7, if it cannot rival the
mystic 9, has quaint properties of its own:—
| |
1 |
5 |
8 |
7 |
3 |
× |
7 |
= |
1 |
1 |
1 |
1 |
1 |
1 |
| |
3 |
1 |
7 |
4 |
6 |
× |
7 |
= |
2 |
2 |
2 |
2 |
2 |
2 |
| |
4 |
7 |
6 |
1 |
9 |
× |
7 |
= |
3 |
3 |
3 |
3 |
3 |
3 |
| |
6 |
3 |
4 |
9 |
2 |
× |
7 |
= |
4 |
4 |
4 |
4 |
4 |
4 |
| |
7 |
9 |
3 |
6 |
5 |
× |
7 |
= |
5 |
5 |
5 |
5 |
5 |
5 |
| |
9 |
5 |
2 |
3 |
8 |
× |
7 |
= |
6 |
6 |
6 |
6 |
6 |
6 |
| 1 |
1 |
1 |
1 |
1 |
1 |
× |
7 |
= |
7 |
7 |
7 |
7 |
7 |
7 |
| 1 |
2 |
6 |
9 |
8 |
4 |
× |
7 |
= |
8 |
8 |
8 |
8 |
8 |
8 |
| 1 |
4 |
2 |
8 |
5 |
7 |
× |
7 |
= |
9 |
9 |
9 |
9 |
9 |
9 |
Here is an arithmetical curiosity:—
| 9 |
× |
9 |
|
+ |
7 |
= |
8 |
8 |
|
| 9 |
× |
9 |
8 |
|
+ |
6 |
= |
8 |
8 |
8 |
|
| 9 |
× |
9 |
8 |
7 |
|
+ |
5 |
= |
8 |
8 |
8 |
8 |
|
| 9 |
× |
9 |
8 |
7 |
6 |
|
+ |
4 |
= |
8 |
8 |
8 |
8 |
8 |
|
| 9 |
× |
9 |
8 |
7 |
6 |
5 |
|
+ |
3 |
= |
8 |
8 |
8 |
8 |
8 |
8 |
|
| 9 |
× |
9 |
8 |
7 |
6 |
5 |
4 |
|
+ |
2 |
= |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
|
| 9 |
× |
9 |
8 |
7 |
6 |
5 |
4 |
3 |
|
+ |
1 |
= |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
|
| 9 |
× |
9 |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
+ |
0 |
= |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
8 |
A farmer’s wife kept a pure strain of Aylesbury
ducks for market on a square pond, with a duck-house
at each corner. As trade grew brisk she
found that she must enlarge her pond. An ingenious
neighbour undertook to arrange this
without altering the shape of the pond, and without
disturbing the duck-houses. What was his
plan?
Solution
It would seem impossible to subtract 69 from
55, but it can be arranged thus, with six as a
remainder:—
Cut out in cardboard
twenty triangular pieces exactly
the size and shape of
this one, and try to place
them together so that they
form a perfect square.
Solution
The decimal equivalent of 1⁄13 is .076923. This
(omitting the point), multiplied by 1, 3, 4, 9, 10,
or 12, yields results in which the same figures
appear in varied order, but similar sequence, and
multiplied by 2, 5, 6, 7, 8, or 11, it yields a different
series, with similar characteristics. Thus:—
| 76923 |
× |
1 |
= |
76923 |
76923 |
× |
2 |
= |
153846 |
| × |
3 |
= |
230769 |
× |
5 |
= |
384615 |
| × |
4 |
= |
307692 |
× |
6 |
= |
461538 |
| × |
9 |
= |
692307 |
× |
7 |
= |
538461 |
| × |
10 |
= |
769230 |
× |
8 |
= |
615384 |
| × |
12 |
= |
923076 |
× |
11 |
= |
846153 |
A kaleidoscope cylinder contains twenty small
pieces of coloured glass. As we turn it round,
or shake it, so as to make ten changes of pattern
every minute, it will take the inconceivable space
of time of 462,880,899,576 years and 360 days
to exhaust all the possible symmetrical variations.
(The 360 days is good!)
Here is an amusing little exercise for the ingenuity
of our solvers.
Take six sharp pins, and puzzle out how to
stick them into six of the black dots, so that no
two pins, are on the same line, in any direction,
vertical, horizontal, or diagonal.
Solution
The middle of a large playground was paved
with sixty-four square flagstones of equal size,
which are numbered on this diagram from one
to sixty-four.
| 1 |
9 |
17 |
25 |
33 |
41 |
49 |
57 |
| 2 |
10 |
18 |
26 |
34 |
42 |
50 |
58 |
| 3 |
11 |
19 |
27 |
35 |
43 |
51 |
59 |
| 4 |
12 |
20 |
28 |
36 |
44 |
52 |
60 |
| 5 |
13 |
21 |
29 |
37 |
45 |
53 |
61 |
| 6 |
14 |
22 |
30 |
38 |
46 |
54 |
62 |
| 7 |
15 |
23 |
31 |
39 |
47 |
55 |
63 |
| 8 |
16 |
24 |
32 |
40 |
48 |
56 |
64 |
One of the schoolmasters, who had a head for
puzzles, took his stand upon the square here
numbered 19, and offered a prize to any boy who,
starting from the square numbered 46, could
make his way to him, passing through every
square once, and only once. It was after many
vain attempts that the course was at last discovered.
Can you work it out?
Solution
Place twelve draughtsmen, or buttons, in a
square, so that you count four along each side
of it, thus:—
Now take the same men or buttons, and arrange
them so that they form another square, and you
can count five along each side of it.
Solution
Here is a good specimen of the eccentricities
and powers of numbers:—
| 153846 |
× |
13 |
= |
1999998 |
| 230769 |
× |
13 |
= |
2999997 |
| 307692 |
× |
13 |
= |
3999996 |
| 384615 |
× |
13 |
= |
4999995 |
| 461538 |
× |
13 |
= |
5999994 |
| 538461 |
× |
13 |
= |
6999993 |
| 615384 |
× |
13 |
= |
7999992 |
| 692307 |
× |
13 |
= |
8999991 |
She was quite an old maid, and her age was
a most absolute secret. Determined to discover
it, her scapegrace nephew, on Christmas Eve,
produced these tables, and asked her with well
simulated innocence on which of them she could
see the number of her age.
| 1 |
23 |
45 |
|
2 |
23 |
46 |
|
16 |
27 |
54 |
| 3 |
25 |
47 |
3 |
26 |
47 |
17 |
28 |
55 |
| 5 |
27 |
49 |
6 |
27 |
50 |
18 |
29 |
56 |
| 7 |
29 |
51 |
7 |
30 |
51 |
19 |
30 |
57 |
| 9 |
31 |
53 |
10 |
31 |
54 |
20 |
31 |
58 |
| 11 |
33 |
55 |
11 |
34 |
55 |
21 |
48 |
59 |
| 13 |
35 |
57 |
14 |
35 |
58 |
22 |
49 |
60 |
| 15 |
37 |
59 |
15 |
38 |
59 |
23 |
50 |
61 |
| 17 |
39 |
61 |
18 |
39 |
62 |
24 |
51 |
62 |
| 19 |
41 |
A |
19 |
42 |
B |
25 |
52 |
C |
| 21 |
43 |
22 |
43 |
26 |
53 |
| 8 |
27 |
46 |
|
4 |
23 |
46 |
|
32 |
43 |
54 |
| 9 |
28 |
47 |
5 |
28 |
47 |
33 |
44 |
55 |
| 10 |
29 |
56 |
6 |
29 |
52 |
34 |
45 |
56 |
| 11 |
30 |
57 |
7 |
30 |
53 |
35 |
46 |
57 |
| 12 |
31 |
58 |
12 |
31 |
54 |
36 |
47 |
58 |
| 13 |
40 |
59 |
13 |
36 |
55 |
37 |
48 |
59 |
| 14 |
41 |
60 |
14 |
37 |
60 |
38 |
49 |
60 |
| 15 |
42 |
61 |
15 |
38 |
61 |
39 |
50 |
61 |
| 24 |
43 |
62 |
20 |
39 |
62 |
40 |
51 |
62 |
| 25 |
44 |
D |
21 |
44 |
E |
41 |
52 |
F |
| 26 |
45 |
22 |
45 |
42 |
53 |
From her answer he was able to calculate that
the old lady was fifty-five.
The tell-tale tables disclosed her age thus:—As
it appeared in tables A, B, C, E, and F, he
added together the numbers at the top left-hand
corners, and found the total to be fifty-five.
This rule applies in all cases.
Of the many paper-cutting tricks which appeal
to us none is more simple and attractive than
this:—
Take a piece of paper, say 5 inches by 3
inches, but any oblong shape and size will do,
and after folding it four times cut it lengthways
up the centre. Unfold the pieces, and to your
surprise you will find a perfect cross and other
pieces in pairs of
the shapes shown
above. The
puzzle is how to
fold the paper.
The paper must
be folded first so
that B comes
upon C, then so
that A comes
upon D, then from D to C, and lastly from
E to C. If it is now cut lengthways exactly
along the centre the figures shown on the original
diagram will be formed, which resemble a cross
and lighted candles on an altar.
Take a thin board, about eight inches square,
and mark it out
into thirty-six
equal parts; bore
a hole in the centre
of each part, and
then fit in a small
wooden peg, leaving
about a quarter
inch above the
surface, as is
shown in Fig. 1,
the section below
the diagram.
Prepare thirty-six
pieces of white
or coloured cardboard
of the
length A to B,
and place them
over the pegs in any direction in which they will
fit so as to form
some such symmetrical
pattern
as is given on
the second diagram,
putting
two holes only
on each peg.
Chess-players
will see that
this is the regular
knight’s
move.
Quite a number
of beautiful
designs can be thus formed, and those who have
not the means at hand for making a complete set
can enjoy the puzzle by merely marking out
thirty-six squares, and drawing lines from centre
to centre of the exact length from A to B, with
black or coloured pencils.
Divide this figure into four similar and equal
parts.
Solution
The solution of the pretty little problem: place
three twos in three different groups, so that twice
the first group, or half the third group equals
the second group, is this:—
22 + 2 =
12
2 - 22 = 1
2 + 22 = 2
The following figure, which represents part of
a brick wall, cannot be marked out along all the
edges of the bricks in less than six continuous
lines without going more than once over the same
line:—
Here, in strong contrast to the simple figure
given above, which could not be traced without
lifting the pen six times from the paper, is an
intricate design, the lines of which, on the upper
or on the lower half, can be traced without any
break at all.
The general rule that governs such cases is,
that where an uneven number of lines meet a
fresh start has to be made. In the diagram now
given the only such points are at the extremities
of the upper and lower halves of the figure at
A and X. At all other points two, or four, or six
lines converge, and there is no break of continuity
in a tracing of the figure.
Can you suggest quite a simple and practical
way to fix the points on the sides of a square
which will be at the angles of an octagon formed
by cutting off equal corners of the square, as
shown below?
Solution
Very interesting and curious are the properties
of the figures 142857, used in varied order but
always in similar sequence, in connection with
7 and 9:—
| 1 |
4 |
2 |
8 |
5 |
7 |
× |
7 |
= |
|
9 |
9 |
9 |
9 |
9 |
9 |
÷ |
9 |
= |
1 |
1 |
1 |
1 |
1 |
1 |
| 2 |
8 |
5 |
7 |
1 |
4 |
× |
7 |
= |
1 |
9 |
9 |
9 |
9 |
9 |
8 |
÷ |
9 |
= |
2 |
2 |
2 |
2 |
2 |
2 |
| 4 |
2 |
8 |
5 |
7 |
1 |
× |
7 |
= |
2 |
9 |
9 |
9 |
9 |
9 |
7 |
÷ |
9 |
= |
3 |
3 |
3 |
3 |
3 |
3 |
| 5 |
7 |
1 |
4 |
2 |
8 |
× |
7 |
= |
3 |
9 |
9 |
9 |
9 |
9 |
6 |
÷ |
9 |
= |
4 |
4 |
4 |
4 |
4 |
4 |
| 7 |
1 |
4 |
2 |
8 |
5 |
× |
7 |
= |
4 |
9 |
9 |
9 |
9 |
9 |
5 |
÷ |
9 |
= |
5 |
5 |
5 |
5 |
5 |
5 |
| 8 |
5 |
7 |
1 |
4 |
2 |
× |
7 |
= |
5 |
9 |
9 |
9 |
9 |
9 |
4 |
÷ |
9 |
= |
6 |
6 |
6 |
6 |
6 |
6 |
The subjoined diagram shows how a square
with sides that measure each 12 yards can
be divided into five triangles, no two of which
are of equal area, and of which the sides and
areas can be expressed in yards by whole numbers:—
The areas of these triangles are 6, 12, 24, 48,
and 54 square yards respectively, and the sum
of these, 144 square yards, is the area of the
square.
Our readers may remember the remarkable fact
that the figures of the sum, £12, 12s. 8d., when
written thus, 12,128, exactly represent the number
of farthings it contains. Now this, so far as
we know, is the only instance of the peculiarity,
but there are at least five other cases which come
curiously near to it. They are these:—
| £ |
s. |
d. |
|
| 9 |
9 |
6 |
|
= |
9096 |
farthings |
| 6 |
6 |
4 |
= |
6064 |
„ |
| 3 |
3 |
2 |
= |
3032 |
„ |
| 10 |
10 |
6 |
1⁄2 |
= |
10106 |
„ |
| 13 |
13 |
8 |
1⁄2 |
= |
13138 |
„ |
If a ladder, with rungs 1 foot apart, rests
against a wall, and its thirteenth rung is 12 feet
above the ground, the foot of the ladder is 25 feet
from the wall.
Proof.—Drop a perpendicular from A to B.
Then, as A B C is a right angle, and the squares
on A C, A B, are 169 feet and 144 feet, the square
on C B must be 25 feet, and the length of C B
is 5 feet. We thus move 5 feet towards the wall
in going 13 feet up the ladder, and in mounting
65 feet (five times as far) we must cover 25 feet.
A prettily ingenious method of dividing the
area of a circle into quarters, each of them
a perfect curve, with perimeter (or enclosing
line) equal to the circumference of the circle, and
with which four circles can be formed, is clearly
shown by the subjoined diagrams:—
The host of a large hotel at Cairo noticed that
his Visitors’ Book contained the names of an
Austrian, a Brazilian, a Chinaman, a Dane, an
Englishman, a Frenchman, a German, and a
Hungarian. Moved by this curious alphabetical
list, he offered them all free quarters and the
best of everything if they could arrange themselves
at a round dining-table so that not one of
them should have the same two neighbours on
any two occasions for 21 successive days.
The following is one of many ways in which
this arrangement can be made, and it seems to
be the simplest of them all.
Number the persons 1 to 8; and for our first
day set them down in numerical order except that
the two centre ones (4 and 5) change places:
| (1st day)— |
1 |
2 |
3 |
5 |
4 |
6 |
7 |
8 |
Keep the 1 and the 7 unaltered but double
each of the other numbers. When the product
is greater than 8, divide by 7, and only set down
the remainder. Thus we get:
| (8th day)— |
1 |
4 |
6 |
3 |
8 |
5 |
7 |
2 |
(Here the fourth figure 3 is 5 × 2 ÷ 7, giving
remainder 3, and so on.)
Repeat this operation once more:
| (15th day)— |
1 |
8 |
5 |
6 |
2 |
3 |
7 |
4 |
To fill in the intermediate days we have only
to keep 1 unchanged and let the remaining numbers
run downwards in simple numerical order,
following 8 with 2, 2 with 3, and so on. Thus:—
| 1st day— |
1 |
2 |
3 |
5 |
4 |
6 |
7 |
8 |
| 2nd day— |
1 |
3 |
4 |
6 |
5 |
7 |
8 |
2 |
| 3rd day— |
1 |
4 |
5 |
7 |
6 |
8 |
2 |
3 |
| 4th day— |
1 |
5 |
6 |
8 |
7 |
2 |
3 |
4 |
| 5th day— |
1 |
6 |
7 |
2 |
8 |
3 |
4 |
5 |
| 6th day— |
1 |
7 |
8 |
3 |
2 |
4 |
5 |
6 |
| 7th day— |
1 |
8 |
2 |
4 |
3 |
5 |
6 |
7 |
| 8th day— |
1 |
4 |
6 |
3 |
8 |
5 |
7 |
2 |
| 9th day— |
1 |
5 |
7 |
4 |
2 |
6 |
8 |
3 |
| 10th day— |
1 |
6 |
8 |
5 |
3 |
7 |
2 |
4 |
| 11th day— |
1 |
7 |
2 |
6 |
4 |
8 |
3 |
5 |
| 12th day— |
1 |
8 |
3 |
7 |
5 |
2 |
4 |
6 |
| 13th day— |
1 |
2 |
4 |
8 |
6 |
3 |
5 |
7 |
| 14th day— |
1 |
3 |
5 |
2 |
7 |
4 |
6 |
8 |
| 15th day— |
1 |
8 |
5 |
6 |
2 |
3 |
7 |
4 |
| 16th day— |
1 |
2 |
6 |
7 |
3 |
4 |
8 |
5 |
| 17th day— |
1 |
3 |
7 |
8 |
4 |
5 |
2 |
6 |
| 18th day— |
1 |
4 |
8 |
2 |
5 |
6 |
3 |
7 |
| 19th day— |
1 |
5 |
2 |
3 |
6 |
7 |
4 |
8 |
| 20th day— |
1 |
6 |
3 |
4 |
7 |
8 |
5 |
2 |
| 21st day— |
1 |
7 |
4 |
5 |
8 |
2 |
6 |
3 |
This completes the schedule. It will be found
on examination that every number is between
every pair of the other numbers once, and once
only.
In order to reduce our first-day ring to exact
numerical order we have only to interchange
the numbers 4 and 5 throughout. The first three
lines for example would then become:
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
| 1 |
3 |
5 |
6 |
4 |
7 |
8 |
2 |
| 1 |
5 |
4 |
7 |
6 |
8 |
2 |
3, |
etc. |
or, by putting letters for figures,
| A |
B |
C |
D |
E |
F |
G |
H |
|
| A |
C |
E |
F |
D |
G |
H |
B |
| A |
E |
D |
G |
F |
H |
B |
C, |
etc. |
An arrangement of the guests is thus arrived at
for twenty-one successive days, so that not one
of them has the same two neighbours on any
two occasions.
Can you apply the two oblongs drawn below
to the two concentric squares, so as to produce
thirty-one perfect squares?
Solution
Take a piece of cardboard in the form of a
Greek cross with arms, as shown here,
and divide it by two straight cuts, so that the
pieces when reunited form a perfect square.
Solution
The diagrams which we give below show how
a hollow square can be formed of the pieces of
three-quarters of another square from which a
corner has been cut away:—
The subjoined diagram shows how a square
of paper or cardboard may be cut into nine pieces
which, when suitably arranged, form five perfect
squares.
On how many days can fifteen schoolgirls go
out for a walk so arranged in rows of three, that
no two are together more than once?
Fifteen schoolgirls can go out for a walk on
seven days so arranged in rows of three that no
two are together more than once.
It is said, on high authority, that there are no
less than 15,567,522,000 different solutions to
this problem. Here is one of them, given in
Ball’s Mathematical Recreations, in which k stands
for one of the girls, and a, b, c, d, e, f, g, in their
modifications, for her companions on the seven
different days:—
| Sunday |
Monday |
Tuesday |
Wednesday |
Thursday |
Friday |
Saturday |
| ka1a2 |
kb1b2 |
kc1c2 |
kd1d2 |
ke1e2 |
kf1f2 |
kg1g2 |
| b1d1f1 |
a1d2e2 |
a1d1e1 |
a2b2c2 |
a2b1c1 |
a1b2c1 |
a1b1c2 |
| b2e1g1 |
a2f2g2 |
a2f1g1 |
a1f2g1 |
a1f1g2 |
a2d2e1 |
a2d1e2 |
| c1d2g2 |
c1d1g1 |
b1d2f2 |
b1e1g2 |
b2d1f2 |
b1e2g1 |
b2d2f1 |
| c2e2f2 |
c2e1f1 |
b2e2g2 |
c1e2f1 |
c2d2g1 |
c2d1g2 |
c1e1f2 |
It is an excellent game of patience, for those
who have time and inclination, to place the
figures 1 to 15 inclusive in seven such columns,
so as to fulfil the conditions.
It is possible from a Greek cross to cut off four
equal pieces which, when put together, will form
another Greek cross exactly half the size of the
original, and by this process to leave a third
Greek cross complete.
This is how to do it:—
Bisect C D at N, F G at O, K L at P, and B I
at Q.
Join N H, O M, P A, Q E, intersecting at R, S,
T, U.
Bisect A R at V, E S at W, T H at X, and
M U at Y.
Join V Q, N Y, W P, O X, N W, V O, Q X, Y P.
Carefully cut out from the original Greek cross
the newly-formed Greek cross, and the odd pieces
from around it can be arranged to form another
Greek cross.
The greatest number of plane figures that can
be formed by the union of ten straight lines is
thirty-six.
The two equal lines at right angles are first
drawn, and each is divided into eight equal parts.
The other eight straight lines are then drawn
from a to a, from b to b, and so on, until the
hammock-shaped network of thirty-six plane
figures is produced.
It will be seen on the diagram below that seven
straight vertical cuts with a table-knife will
divide a crumpet into twenty-eight parts.
The nine digits can be so adjusted as to form
an equation, or, if taken as weight, to balance
the scales. Thus:—
9, 61⁄2 = 3, 5, 74⁄8
How large is the sea? This is a bold big question,
and any possible answer involves a considerable
stretch of the imagination. Here is a
startling illustration of its vast volume:—
If the water of the sea could be gathered into
a round column, reaching the 93,000,000 miles
which separate us from the sun, the diameter of
this column would be nearly two miles and a
half!
It is perhaps even more difficult to realise that
this mighty mass of waters could be dissipated
in a few moments, if the column we have imagined
could become ice, and if the entire heat of the
sun could be concentrated upon it. All would
be melted in one second of time, and converted
into steam in eight seconds!
Many of our readers may be glad to know an
easy way to make an envelope of any shape or
size.
This diagram speaks for itself. When the lines
A B, A C, D B, D C have been drawn, the corners
of the rectangle, H E G F, are folded over, as
shown by the dotted lines, after the corners have
been rounded, and the margins touched with
gum.
The diagrams below will show how a piece of
paper, 15 inches long and 3 inches wide, can be
cut into five parts, and rearranged to form a
perfect square.
The following astounding calculation is answer
enough to a question put by one of the authors
of “Rejected Addresses:”—“Who filled the
butchers’ shops with big blue flies?”
A pair of blow-flies can produce ten thousand
eggs, which mature in a fortnight. If every egg
hatches out, and there are equal numbers of
either sex, which forthwith increase and multiply
at the same rapid rate, and if their descendants
do the like, so that all survive at the end of six
months, it has been calculated that, if thirty-two
would fill a cubic inch of space, the whole innumerable
swarm would cover the globe, land
and sea, half a mile deep everywhere.
Here is quite a neat way to make an equilateral
triangle without using compasses:—
Take a piece of paper exactly square, which
we will call A B C D, fold it across the middle,
so as to form the crease E F; unfold it, and fold
it again so that the corner D falls upon the crease
E F at G, and the angle at G is exactly divided.
Again unfold the square, and from G draw the
straight lines G C and G D. Then G C D is the
equilateral triangle required.
How can we be sure that the value of a Frenchwoman
is just 1 franc 8 centimes?
We can be sure that the exact value of a Frenchwoman
is 1 franc 8 centimes, for
| |
Two Frenchwomen |
= |
Deux Françaises. |
| |
Deux Françaises |
= |
deux francs seize. (2 francs 16). |
| Therefore, |
One Frenchwoman |
= |
1 franc 8! |
If a 52-feet ladder is set up so as just to clear
a garden wall 12 feet high and 15 feet from the
building, it will touch the house 48 feet from the
ground.
Our diagram shows this, and also, by a dotted
line, the only other possible position in which it
could fulfil the conditions, if it were then of any
practical use.
If one pin could be dropped into a vessel this
week, two the next, four the next, and so on,
doubling each time for a year, the accumulated
quantity would be 4,503,599,627,370,495, and their
weight, if we reckon 200 pins to the ounce, would
amount to 628,292,358 tons, a full load for 27,924
ships as large as the Great Eastern, whose capacity
was 22,500 tons.
When at the signpost which said “To A
4 miles, to B 9 miles” on one arm, and on the
other “To C 3 miles, to D —— miles,” and the boy
whom I met could only tell me that the farm he
worked at was equidistant from A, B, C, and D,
and nearer to them than to the signpost, and
that all the roads ran straight, I found, thanks
to memories of Euclid, that I was 12 miles
from D.
Since B A and D C intersect outside the circle
at the signpost E,
| therefore |
A E × E B = C E × E D. |
| but |
A E × E B = 4 × 9 = 36, |
| therefore |
C E × E D = 36, |
| and |
C E = 3, therefore E D = 12. |
Q.E.D.
This seems to be quite a poor attempt at a
Maltese cross, but there is method in the madness
of its make.
It is possible by two straight cuts to divide
this uneven cross into four pieces which can be
arranged together again so that they form a
perfect square. Where must the cuts be made,
and how are the four pieces rearranged?
Solution
We all remember that splendidly terse message
of success sent home to the authorities by Napier
when he had conquered the armies of Scinde—“Peccavi!”
(I have sinned).
History had an excellent opportunity for repeating
itself when Admiral Dewey defeated the
Spaniel fleet, for he might have conveyed the
news of his victory by the one burning word—“Cantharides”—“The
Spanish fly!”
In the diagram below a square is subdivided
into twenty-five cells.
Can you, keeping always on the straight lines,
cut this into four pieces, and arrange these as
two perfect squares, in which every semicircle
still occupies the upper half of its cell?
Solution
The following very simple recipe for a home-made
microscope has been suggested by a Fellow
of the Royal Microscopical Society:—
Take a piece of black card, make a small pinhole
in it, put it close to the eye, and look at
some small object closely, such as the type of a
newspaper. A very decided magnifying power
will be shown thus.
Where on this diagram must we place twenty-one
pins or dots so that they fall into symmetrical
design and form thirty rows, with three
in each row?
Solution
Those who are fond of figures will find it a
most interesting exercise to see how far they are
able to represent every number, from one up to
a hundred, by the use of four fours. Any of the
usual signs and symbols of arithmetic may be
brought into use. Here are a few instances of
what may thus be done:—
3 = 4 + 4 + 44;
9 = 4 + 4 + 44;
36 = 4(4 + 4) + 4;
45 = 44 + 44;
52 = 44 + 4 + 4; 60 = 4 × 4 × 4 - 4.
This figure, which now forms a square, and
the quarter of that square, can be so divided by
two straight lines that its parts, separated and
then reunited, form a perfect square. How is
this done?
Solution
Here we have arranged five rows of five cards
each, so that no two similar cards are in the same
lines. Counting the ace as eleven, each row,
column, and diagonal adds up to exactly twenty-six.
After you have looked at this Magic Square,
and set it out on the table, shuffle the cards,
and try to re-arrange them so as to give the same
results.
Cut a square of paper or cardboard into seven
such pieces as are marked in this diagram.
Can you rearrange them so that they form the
figure 8.
Solution
There are several ways in which strange juggling
with figures and numbers is to be done, but
none is more curious than this:—
Ask someone, whose age you do not know,
to write down secretly the date and month of
his birth in figures, to multiply this by 2, to add
5, to multiply by 50, to add his age last birthday
and 365. He then hands you this total only
from this you subtract 615. This reveals to you
at a glance his age and birthday.
Thus, if he was born April 7 and is 23, 74 (the
day and the month) × 2 = 148; 148 + 5 = 153;
153 × 50 = 7650; 7650 + 23 (his age) = 7673; and
7673 + 365 = 8038. If from this you subtract
615, you have 7423, which represents to you the
seventh day of the fourth month, 23 years age!
This rule works out correctly in all cases.
Cut out in stiff paper or cardboard two pieces
of the shape and size of the small triangle, and
four pieces of the shapes and sizes of the other
three patterns—fourteen pieces in all.
The puzzle is to fit these pieces together so
that they form a perfect oblong.
Solution
Here is a string of sentences, which may be
used as stimulating mental gymnastics when we
leave the “Land of Nod.”
A sleeper runs on sleepers, and in this sleeper
on sleepers sleepers sleep. As this sleeper carries
its sleepers over the sleepers that are under the
sleeper, a slack sleeper slips. This jars the
sleeper and its sleepers, so that they slip and no
longer sleep.
Clever calculation has established a fact which
we shall not be able to verify by personal experience.
Whatever else may happen, the first
day of a century can never fall on Sunday,
Wednesday, or Friday.
On this cross there are seventeen distinct and
perfect squares marked out at their corners by
asterisks.
How few, and which, of these can you remove,
so that not a single perfect square remains?
Solution
On a globe 2 feet in diameter the Dead Sea
appears but as a small coloured dot. If it were
frozen over there would be standing room on its
surface for the whole human race, allowing 6
square feet for each person; and if they were
all suddenly engulfed, it would merely raise the
level of the lake about 4 inches.
Here is quite a good exercise for ingenious
brains and fingers. Cut a piece of stiff paper
or cardboard into such a right-angled triangle
as is shown below.
Can you divide this into only three pieces,
which, when rearranged, will form the design
given as No. 2?
Solution
I sent an order for dwarf roses to a famous
nursery-garden, asking for a parcel of less than
100 plants, and stipulating that if I planted them
3 in a row there should be 1 over; if 4 in a
row 2 over; if 5 in a row 3 over, and if 6 in a
row 4 over, as a condition for payment.
The nurseryman was equal to the occasion,
charged me for 58 trees, and duly received his
cheque.
This puzzle is not so easy of solution as it may
seem at first sight.
Take a counter, or a coin, and place it on one
of the points; then push it across to the opposite
point, and leave it there. Do this with a second
counter or coin, starting on a vacant point, and
continue this process until every point is covered,
as we place the eighth counter or coin on the last
point.
Solution
“Write down,” said a schoolmaster, “the
nine digits in such order that the first three shall
be one third of the last three, and the central
three the result of subtracting the first three
from the last.”
The arrangement which satisfies these conditions
is, 219, 438, 657.
There is a sum of money of such sort that
its pounds, shillings, and pence, written down
as one continuous number, represent exactly
the number of farthings which it contains. What
is it?
Solution
2. If on a level track a train, running all the
time at 30 miles an hour, slips a carriage,
which is uniformly retarded by the brakes, and
which comes to rest in 200 yards, how far has
the train itself then travelled?
Solution
A traveller said to the landlord of an inn,
“Give me as much money as I have in my hand,
and I will spend sixpence with you.” This was
done, and the process was twice repeated, when
the traveller had no money left. How much
had he at first?
Solution
How can we obtain eleven by adding one-third
of twelve to four-fifths of seven?
Solution
5. Can you replace the missing figures in this
mutilated long division sum?
| 2 |
1 |
5 |
) |
* |
7 |
* |
9 |
* |
( |
1 |
* |
* |
| |
* |
* |
* |
|
|
| |
* |
5 |
* |
9 |
|
| |
* |
5 |
* |
5 |
|
|
| |
* |
4 |
* |
|
| |
|
* |
* |
* |
|
Solution
I buy as many heads of asparagus in the
market as can be contained by a string 1 foot
long. Next day I take a string 2 feet long,
and buy as many as it will gird, offering double
the price that I have given before. Was this a
reasonable offer?
Solution
7. As “one good turn deserves another,”
first reverse me, then reverse my square, then
my cube, then my fourth power. When all this
is done no change has been made. What am I?
Solution
8. How can a thousand pounds be so conveniently
stored in ten sealed bags that any sum
in pounds from £1 to £1000 can be paid without
breaking any of the seals?
Solution
This is at once a problem and a puzzle:—
Though you twist and turn me over,
Yet no change can you discover.
Take me thrice, and cut in twain,
You will find but one remain.
Solution
Three gamblers, when they sit down to
play, agree that the loser shall always double
the sum of money that the other two have before
them. After each of them has lost once, it is
found that each has eight sovereigns on the
table. How much money had each at starting?
Solution
11. When Tom’s back was turned, the boy
sitting next to him rubbed out almost all his
sum. Tom could not remember the multiplier,
and only this remained on his slate—
| |
3 |
4 |
5 |
| |
. |
. |
| |
. |
. |
. |
. |
| . |
. |
. |
. |
|
| . |
. |
7 |
6 |
. |
Can you reconstruct the sum?
Solution
12. The sum of the nine digits is 45. Can you
hit upon other arrangements of 1, 2, 3, 4, 5, 6, 7,
8, 9, writing each of them once only, which will
produce the same total. Of course fractions
may be used.
Solution
13. The combined ages of Mary and Ann are
forty-four years. Mary is twice as old as Ann
was when Mary was half as old as Ann will be
when Ann is three times as old as Mary was when
Mary was three times as old as Ann. How old,
then, is Mary?
Solution
Mr Oldboy was playing backgammon with
his wife on the eve of his golden wedding, and
could not make up his mind whether he should
leave a blot where it could be taken up by an
ace, or one which a tré would hit.
His grandson, at home from Cambridge for the
Christmas vacation, solved the question for him
easily. What was his decision?
Solution
15. I am aboard a steamer, anchored in a bay
where the needle points due north, and exactly
1200 miles from the North Pole. If the course
is perfectly clear, and I steam continuously at
the rate of 20 miles an hour, always steering
north by the compass needle, how long will it
take me to reach the North Pole?
Solution
Three persons, A, B, and C, share twenty-one
wine casks of equal capacity, of which seven
are full, seven are half full, and seven are empty.
How can these be so apportioned that each person
shall have an equal number of casks, and an
equal quantity of wine, without transferring any
of it from cask to cask?
Solution
A hungry mouse, in search of provender,
came upon a box containing ears of corn. He
could carry three ears home at a time, and only
had opportunity to make fourteen journeys to
and fro. How many ears could he add to his
store?
Solution
Take exactly equal quantities of lard and
butter; mix a small piece of the butter intimately
with all the lard. From this blend take a piece
just as large as the fragment removed from the
butter, and mix this thoroughly with the butter.
Is there now more lard in the butter or more
butter in the lard?
Solution
19. Here is an apparent proof that 2 = 3:—
4 - 10 = 9 - 15
4 - 10 + 25⁄4 = 9 - 15 + 25⁄4
and the square roots of these:—
2 - 5⁄2 = 3 - 5⁄2
therefore 2 = 3.
Solution
20. Our Parcel Post regulations limit the
length of a parcel to 3 feet 6 inches, and the
length and girth combined to 6 feet. What is
the largest parcel of any size that can be sent
through the post under these conditions?
Solution
How can we show, or seem to show, that
either four, five, or six nines amount to one
hundred?
Solution
22. Can you arrange nine numbers in the
nine cells of a square, the largest number 100,
and the least 1, so that the product by multiplication
of each row, column and diagonal is 1000?
Solution
23. Seven London boys were at the seaside
for a week’s holiday, and during the six week-days
they caught four fine crabs in pools under the
rocks, when the tide was out at Beachy Head.
Hearing of this, the twenty-one boys of a school
in the neighbourhood determined to explore the
pools; but with the same rate of success they
only caught one large crab. For how long were
they busy searching under the seaweed?
Solution
24. On how many nights could a watch be
set of a different trio from a company of fifteen
soldiers, and how often on these terms could one
of them, John Pipeclay, be included?
Solution
25. If Augustus Cæsar was born on September
23rd, B.C. 63, on what day and in what year did
he celebrate his sixty-third birthday, and by
what five-letter symbol can we express the date?
Solution
A was born in 1847, B in 1874. In what
years have the same two digits served to express
the ages of both, if they are still living?
Solution
27.
A hundred and one by fifty divide,
To this let a cypher be duly applied;
And when the result you can rightly divine,
You find that its value is just one in nine.
Solution
28. A man started one Monday morning with
money in his purse to buy goods in a neighbouring
town. He paid a penny to cross the ferry, spent
half of the money he then had, and paid another
penny at the ferry on his return home.
He did exactly the same for the next five days,
and on Saturday evening reached home with just
one penny in his pocket. How much had he in
his pocket on Monday before he reached the
ferry?
Solution
29. Three men gathered mangoes, and agreed
that next day they would give one to their
monkey and divide the rest equally. The first
who arrived gave one to the monkey, and then
took his proper share; the second came later
and did likewise, and the third later still, neither
knowing that any one had preceded him. Finally
they met, and, as there were still mangoes, gave
one to the monkey, and shared the rest equally.
How many mangoes at least must there have
been if all the divisions were accurate?
Solution
I look at my watch between four and five,
and again between seven and eight. The hands
have, I find, exactly changed places, so that the
hour-hand is where the minute-hand was, and
the minute-hand takes the place of the hour-hand.
At what time did I first look at my
watch?
Solution
31. There is a number consisting of twenty-two
figures, of which the last is 7. If this is
moved to the first place, the number is increased
exactly sevenfold. Can you discover this lengthy
number?
Solution
A farmer borrowed from a miller a sack
of wheat, 4 feet long and 6 feet in circumference.
He sent in repayment two sacks, each
4 feet long and 3 feet in circumference.
Was the miller satisfied?
Solution
33. Five gamblers, whom we will call A, B,
C, D, and E, play together, on the condition that
after each hazard he who loses shall give to all
the others as much as they then have in hand.
Each loses in turn, beginning with A, and when
they leave the table each has the same sum in
hand, thirty-two pounds. How much had each
at first?
Solution
Knowing that the square of 87 is 7569,
how can we rapidly, without multiplication,
determine in succession the squares of 88, 89,
and 90?
Solution
35.
Two numbers seek which make eleven,
Divide the larger by the less,
The quotient is exactly seven,
As all who solve it will confess.
Solution
36. If there are twenty sorts of things from
which 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
different selections can be made, how many of
each sort are there?
Solution
Three women, with no money, went to
market. The first had thirty-three apples, the
second twenty-nine, the third twenty-seven.
Each woman sold the same number of apples for
a penny. They all sold out, and yet each received
an equal amount of money. How was
this?
Solution
38.
Take five from five, oh, that is mean!
Take five from seven, and this is seen.
Solution
If a bun and a half cost three halfpence,
how many do you buy for sixpence?
Solution
How many times in a day would the
hands of a watch meet each other, if the minute-hand
moved backward and the hour-hand forward?
Solution
How can half-a-crown be equally divided
between two fathers and two sons so that a penny
is the coin of smallest value given to them?
Solution
42. If the number of the revolutions of the
wheel of a bicycle in six seconds is equal to the
number of miles an hour at which it is running,
what is the circumference of the wheel?
Solution
Hearing a clock strike, and being uncertain
of the hour, I asked a policeman. He
had a turn for figures, and replied: “Take half,
a third, and a fourth of the hour that has just
struck, and the total will be one larger than that
hour.” What o’clock was it?
Solution
44.
If cash is lent at five per cent.
To those who choose to borrow,
How soon shall I be worth a pound
If I lend a crown to-morrow?
Solution
45. If I jump off a table with a 20-lb. dumb-bell
in my hand, what is the pressure upon me
of its weight while I am in the air?
Solution
46. In charitable mood I went recently to a
bazaar where there were four tents arranged to
tempt a purchaser. At the door of each tent
I paid a shilling, and in each tent I spent half of
the money remaining in my purse, giving the
door-keepers each another shilling as I came out.
It took my last shilling to pay the fourth door-keeper.
How much money had I at first, and
what did I spend in each tent?
Solution
47. Rain is falling vertically, at a speed of
5 miles an hour. I am walking through it at
4 miles an hour. At what angle to the vertical
must I hold my umbrella, so that the raindrops
strike its top at right angles?
Solution
48. The following interesting problem, translated
from the original Sanscrit, is given by Longfellow
in his “Kavanagh”:—
“A tree, 100 cubits high, is distant from a
well 200 cubits. From the top of this tree
one monkey descends, and goes to the well.
Another monkey leaps straight upwards from
the top, and then descends to the well by the
hypotenuse. Both pass over an equal space.
How high does the second monkey leap?”
Solution
A steamboat 105 miles east of Tynemouth
Lighthouse springs a leak. She puts back at
once, and in the first hour goes at the rate of
10 miles an hour.
More and more water-logged, she decreases
her speed each succeeding hour at the rate of
one-tenth of what it has been during the previous
hour. When will she reach the lighthouse?
Solution
If a hen and a half lays an egg and a half
in a day and a half, how many eggs will twenty-one
hens lay in a week?
Solution
If the population of Bristol exceeds by
two hundred and thirty-seven the number of
hairs on the head of any one of its inhabitants,
how many of them at least, if none are bald,
must have the same number of hairs on their
heads?
Solution
A benevolent uncle has in his pocket a
sovereign, a half-sovereign, a crown, a half-crown,
a florin, a shilling, and a threepenny piece.
In how many different ways can he tip his
nephew, using only these coins, and how is this
most easily determined?
Solution
Here is a prime problem, in more senses
than one, which will tax the ingenuity of our
solvers:—I am a prime number of three figures.
Increased by one-third, ignoring fractions, I
become a square number. Transpose my first
two figures and increase me by one-third, and
again I am a square number. Put my first
figure last, and increase me by one-third, and I
am another square number. Reverse my three
figures, and increase as before by one-third, and
for a fourth time I become a square number.
What are my original figures?
Solution
In how many different ways can six different
things be divided between two boys?
Solution
What is quite the highest number that
can be scored at six card cribbage by the dealer,
if he has the power to select all the cards, and to
determine the order in which every card shall be
played?
Solution
56. A fanciful collector, who bought pictures
with more regard to quantity than quality, gave
instructions that the area of each frame should
exactly equal that of the picture it contained,
and that the frames should be of the same width
all round.
At an auction he picked up a so-called “old
master,” unframed, which measured 18 inches
by 12 inches. What width of frame will fulfil
his conditions?
Solution
57. Our family consists of my mother, a
brother, a sister, and myself. Our average age
is thirty-nine. My mother was twenty when I
was born; my sister is two years my junior, and
my brother is four years younger than she is.
What are our respective ages?
Solution
A spider in a dockyard unwittingly attached
her web to a mechanical capstan 1 foot
in diameter, at the moment when it began to
revolve. To hold her ground she paid out
73 feet of thread, when the capstan stopped,
and she found herself drained of silk.
To make the best of a bad job she determined
to unwind her thread, walking round and round
the capstan at the end of the stretched thread.
When she had gone a mile in her spiral path she
stopped, tired and in despair. How far was she
then from the end of her task?
Solution
A mountebank at a fair had six dice, each
marked only on one face 1, 2, 3, 4, 5, or 6,
respectively. He offered to return a hundredfold
any stake to a player who should turn up
all the six marked faces once in twenty throws.
How far was this from being a fair offer?
Solution
60.
If ninety groats for twenty cats
Will furnish three weeks’ fare,
How many hounds for forty pounds,
Less one, may winter there?
Just ninety days and one assume
The winter’s space to be;
And note that what five cats consume
Will serve for dogs but three.
(A groat = 4d.)
Solution
Two wineglasses of equal size are respectively
half and one-third full of wine. They
are filled up with water, and their contents are
then mixed. Half of this mixture is finally
poured back into one of the wineglasses. What
part of this will be wine and what part will be
water?
Solution
A legend goes that on a stout ship on
which St Peter was carried with twenty-nine
others, of whom fourteen were Christians and
fifteen Jews, he so arranged their places, that
when a storm arose, and it was decided to throw
half of the passengers overboard, all the Christians
were saved. The order was that every ninth
man should be cast into the sea. How did he
place the Christians and the Jews?
Solution
63. Farmer Southdown was the proud possessor
of a prize cow, which had a fine calf every
year for sixteen years. Each of these calves
when two years old, and their calves also in their
turn, followed this excellent example. How
many head did they thus muster in sixteen years?
Solution
A shepherd was asked how many sheep
he had in his flock. He replied that he could not
say, but he knew that if he counted them by
twos, by threes, by fours, by fives, or by sixes,
there was always one over, but if he counted
them by sevens, there was no remainder. What
is the smallest number that will answer these
conditions?
Solution
65. If a number of round bullets of equal size
are arranged in rows one above another evenly
graduated till a single bullet crowns the flat
pyramid, how can their number be readily
reckoned, however long the base line may be?
Solution
66.
Old General Host
A battle lost,
And reckoned on a hissing,
When he saw plain
What men were slain,
And prisoners, and missing.
To his dismay
He learned next day
What havoc war had wrought;
He had, at most,
But half his host
Plus ten times three, six, ought.
One-eighth were lain
On beds of pain,
With hundreds six beside;
One-fifth were dead,
Captives, or fled,
Lost in grim warfare’s tide.
Now, if you can,
Tell me, my man,
What troops the general numbered,
When on that night
Before the fight
The deadly cannon slumbered?
Solution
A farmer sends five pieces of chain, of
three links each, to be made into one continuous
length. He agrees to pay a penny for each link
cut, and a penny for each link joined. What was
the blacksmith entitled to charge if he worked
in the best interest of the farmer?
Solution
In a parcel of old silver and copper coins
each silver piece is worth as many pence as there
are copper coins, and each copper coin is worth
as many pence as there are silver coins, and the
whole is worth eighteen shillings. How many
are there of each?
Solution
69. Take the natural numbers 1 to 11, inclusive,
and arrange them in five groups, not
using any of them more than once, so that these
groups are equal. Any necessary signs or indices
may be used.
Solution
70. John Bull passed one-sixth of his life in
childhood and one-twelfth as a youth. When
one-seventh of his life had elapsed he had a son
who died at half his father’s age, and John himself
lived on four years more. How old was he at
the last?
Solution
71. There are two numbers under two thousand,
such that if unity is added to each of them,
or to the half of each, the result is in every case
a square number. Can you find them?
Solution
A cheese in one scale of a balance with
arms of unequal length seems to weigh 16 lbs.
In the other scale it weighs but 9 lbs. What
is its true weight?
Solution
73. Can you divide 100 into two such parts
that if the larger is divided by the lesser the
quotient is also 100?
Solution
I have marbles in my two side pockets.
If I add one to those in the right-hand pocket,
and multiply its increased contents by the number
it held at first, and then deal in a similar way
with those in the other pocket, the difference
between the two results is 90. If, however,
I multiply the sum of the two original quantities
by the square of their difference, the result is
176. How many marbles had I at first in each
pocket?
Solution
75. A friendly circle of twenty-one persons
agreed to meet each week, five at a time, for an
afternoon of bridge, so long as they could do
so without forming exactly the same party on
any two occasions.
As a central room had to be hired, it was
important to have some idea as to the length
of time for which they would require it. How
long could they keep up their weekly meetings?
Solution
A herring and a half costs a penny and
a half; what is the price of a dozen?
Solution
What sum of money is in any sense seen
to be the double of itself?
Solution
78. At the close of his lecture upon unknown
quantities, Dr Bulbous Roots, in playful mood,
wrote this puzzle on his blackboard:—
Divide my fifth by my first and you have my
fourth; subtract my first from my fifth and you
have my second; multiply my first by my fourth
followed by my second, and you have my third;
place my second after my first and you have my
third multiplied by my fourth. What am I?
Solution
79. If we can imagine the earth at a standstill
for the purpose of our experiment, and if a
perfectly straight tunnel could be bored through
its centre from side to side, what would be the
course of a cannon ball dropped into it from one
end, under the action of gravity?
Solution
80.
A lady to her lover cried,
“How many notes have you of mine?”
“Six more I’ve sent,” the youth replied,
“Than I have had of thine.”
“But if from one pound ten you take
The pennies we on stamps have spent,
One eighth their cost you thus will make.”
How many had they had and sent?
Solution
A man, on the day of his marriage, made
his will, leaving his money thus:—If a son should
be born, two-thirds of the estate to that son
and one-third to the widow. If a daughter
should be born, two-thirds to the widow and one-third
to that daughter. In the course of time
twins were born, a boy and a girl. The man
fell sick and died without making a fresh will.
How ought his estate to be divided in justice to
the widow, son, and daughter?
Solution
My carpet is 22 feet across. My stride,
either backwards or forwards, is always 2 feet,
and I make a stride every second. If I take
three strides forwards and two backwards
continuously until I cross the carpet, how long
does it take me to reach the end of it?
Solution
83. A merchant at Lisbon has an urgent
business call to New York. Taking these places
to be, as they appear on a map of the world, on
the same parallel of latitude, and at a distance
measured along the parallel, of some 3600 miles,
if the captain of a vessel chartered to go there
sails along this parallel, will he be doing the best
that he can for the impatient merchant?
Solution
84. Two schoolboys, John and Harry, start
from the right angle of a triangular field, and
run along its sides. John’s speed is to Harry’s
as 13 is to 11.
They meet first in the middle of the opposite
side, and again 32 yards from their starting
point. How far was it round the field?
Solution
85. The following question is given and spelt
exactly as it was contributed to a puzzle column
by “John Hill, Gent.,” in 1760:—
“A vintner has 2 sorts of wine, viz. A and B,
which if mixed in equal parts a flagon of mixed
will cost 15 pence; but if they be mixed in a
sesqui-alter proportion, as you should take two
flagons of A as often as you take three of B, a
flagon will cost 14 pence. Required the price of
each wine singly.”
Solution
86. A man met a beggar and gave him half
the money he had in his pocket, and a shilling
besides. Meeting another he gave him half of
what was left and two shillings, and to a third,
he gave half of the remainder and three shillings.
This left a shilling in his pocket. How much
had he at first?
Solution
A young clerk wishes to start work at an
office in the City on January 1st. He has two
promising offers, one from A of £100 a year, with
a yearly rise of £20, the other from B of £100
a year with a half-yearly rise of £5. Which
should he accept, and why?
Solution
88. I have an abundance of florins and half-crowns,
but no other coins. In how many different
ways can I pay my tailor £11, 10s. without
receiving change?
Solution
A monkey climbing up a greased pole
ascends 3 feet and slips down 2 feet in
alternate seconds till he reaches the top. If the
pole is 60 feet high, how long does it take him
to arrive there?
Solution
90. Old Adze, the village carpenter, who kept
his tools in an open chest, found that his neighbours
sometimes borrowed and forgot to return
them.
To guard against this, he secured the lid of the
chest with a letter lock, which carried six revolving
rings, each engraved with twelve different
letters. What are the chances against any one
discovering the secret word formed by a letter
on each ring, which will open the lock, and be
the only key to the puzzle?
Solution
Five merry married couples happened to
meet at a Swiss hotel, and one of the husbands
laughingly proposed that they should dine together
at a round table, with the ladies always
in the same places so long as the men could seat
themselves each between two ladies, but never
next to his own wife. How long would their
nights at the round table be continued under
these conditions?
Solution
92. In calm water the tip of a stiff rush is 9
inches above the surface of a lake. As a steady
wind rises it is gradually blown aslant, until at
the distance of a yard it is submerged. Can you
decide from these data the depth of the water
in which the rush grows?
Solution
93.
If to Aminta’s age exact
Its square you add, and eighteen more,
And from her age its third subtract,
And to the difference add three score,
The latter to the former then
Will just the same proportion bear
As eighteen does to nine times ten.
Can you Aminta’s age declare?
Solution
94. A bag of nuts was to be divided thus
among four boys:—Dick took a quarter, and
finding that there was one over when he made
the division, gave it to the parrot. Tom dealt in
exactly the same way with the remainder, as did
Jack and Harry in their turns, each finding one
nut from the reduced shares to spare for the
parrot. The final remainder was equally divided
among the boys, and again there was one for the
bird. How many nuts, at the lowest estimate,
did the bag contain?
Solution
Here is an easy one:—
If five times four are thirty-three,
What will the fourth of twenty be?
Solution
What fraction of a pound, added to the
same fraction of a shilling, and the same fraction
of a penny, will make up exactly one pound?
Solution
97. “Now, boys!” said Dr Tripos, “I think
of a number, add 3, divide by 2, add 8, multiply
by 2, subtract 2, and thus arrive at twice the
number I thought of.” What was it?
Solution
Two club friends, A and B, deposit similar
stakes with C, and agree that whoever first wins
three games at billiards shall take the whole of
them. A wins two games and B wins one. Upon
this they determine to divide the stakes in proper
shares. How must this division be arranged?
Solution
Not so simple as it sounds is the following
compact little problem:—If I run by motor from
London to Brighton at 10 miles an hour, and
return over the same course at 15 miles an hour,
what is my average speed?
Solution
“I can divide my sheep,” says Farmer
Hodge, who from his schooldays had a turn for
figures, “into two unequal parts, so that the
larger part added to the square of the smaller
part shall be equal to the smaller part added to
the square of the larger part.”
How many sheep had the farmer?
Solution
101. The following question was proposed in
an old book of Mathematical Curiosities published
more than a hundred years ago:—
“It often happens that if we take two horses,
in every respect alike, yet, if both are put to the
draught, that horse which is most loaded shall
be capable of performing most work; so that
the horse which carries the heavier weight can
draw the larger load. How is this?”
Solution
In the king’s treasury were six chests.
Two held sovereigns, two shillings, and two pence,
in equal numbers of these coins. “Pay my
guard,” said the king, “giving an equal share to
each man, and three shares to the captain; give
change if necessary.” “It may not be possible,”
replied the treasurer, “and the captain may
claim four shares.” “Tut, tut,” said the king,
“it can be done whatever the amount of the
treasure, and whether the captain has three
shares or four.”
Was the king right? If so, how many men
were there in the guard?
Solution
103. I bought a parcel of nuts at forty-nine
for twopence. I divided it into two equal parts,
one of which I sold at twenty-four, the other at
twenty-five for a penny. I spent and received
an integral number of pence, but bought the
least possible number of nuts. How many did
I buy? What did they cost? What did I gain?
Solution
104. My purse contained sovereigns and shillings.
After I had spent half of its contents
there were as many pounds left as I had shillings
at first. With what sum did I start?
Solution
105. A lady was asked her age in a letter, and
she replied by postcard thus:—
If first my age is multiplied by three,
And then of that two-sevenths tripled be,
The square root of two-ninths of this is four,
Now tell my age, or never see me more!
What was her age?
Solution
If cars run at uniform speed on the
twopenny tube, from Shepherd’s Bush to the
Bank at intervals of two minutes, how many shall
I meet in half an hour if I am travelling from
the Bank to Shepherd’s Bush?
Solution
107. What would it cost me to keep my word
if I were to offer my greengrocer a farthing for
every different group of ten apples he could
select from a basket of a hundred apples?
Solution
If the minute-hand of a clock moves
round in the opposite direction to the hour-hand,
what will be the real time between three and
four, when the hands are exactly together?
Solution
Two monkeys have stolen some filberts
and some walnuts. As they begin their feast
they see the owner of the garden coming with a
stick. It will take him two and a half minutes
to reach them. There are twice as many filberts
as walnuts, and one monkey finishes the walnuts
at the rate of fifteen a minute in four-fifths of
the time and bolts. The other manages to finish
the filberts just in time.
If the walnut monkey had stopped to help
him till all was finished, when would they have
got away if they ate filberts at equal rates?
Solution
A cashier, in payment for a cheque, gives
by mistake pounds for shillings and shillings for
pounds. The receiver spends half-a-crown, and
then finds that he has twice as much as the
cheque was worth. What was its value?
Solution
What five uneven figures can be added
together so as to make up 14?
Solution
Three posts which vary in value are
vacant in an office. In how many ways can the
manager fill these up from seven clerks who apply
for the appointments?
Solution
“It is now 5⁄11 of the time to midnight,”
said the fasting man, who began his task at noon.
What time was it?
Solution
114. If a clock takes six seconds to strike six
how long will it take to strike eleven?
Solution
115. How would you arrange twenty horses
in three stalls so as to have an odd number of
horses in each stall?
Solution
Here is a pretty little problem, which has
at any rate an Algebraic form, and is exceedingly
ingenious:—
Given a, b, c, to find q.
Solution
117. Tom Evergreen was asked his age by
some men at his club on his birthday in 1875.
“The number of months,” he said, “that I have
lived are exactly half as many as the number
which denotes the year in which I was born.”
How old was he?
Solution
Draw three circles of any size, and in any
position, so long as they do not intersect, or lie
one within another. How many different circles
can be drawn touching all the three?
Solution
We have seen that the nine digits can
be so dealt with, using each once, as to add up
to 100. How can 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 be
arranged so that they form a sum which is equal
to 1?
Solution
How is it possible, by quite a simple
method, to find the sum of the first fifty numbers
without actually adding them together?
Solution
Two tram-cars, A and B, start at the
same time. A runs into a “lie by” in four
minutes, and waits there five minutes, when B
meets and passes it. Both complete the whole
course at the same moment. In what time can
A complete it without a rest?
Solution
122. Take 10, double it, deduct 10, and tell me
what remains.
Solution
The average weight of the Oxford crew
is increased by 2 lbs., when one of them, who
weighs 12 stone, is replaced by a fresh man.
What is the fresh man’s weight?
Solution
124. A motor car is twice as old as its tyres
were when it was as old as its tyres are. When
these tyres are as old as the car is now, the united
ages of car and tyres will be two years and a
quarter. What are their respective ages now?
Solution
A and B on the edge of a desert can each
carry provisions for himself for twelve days.
How far into the desert can an advance be made,
so that neither of them misses a day’s food?
Solution
A bottle of medicine and its cork cost
half-a-crown, but the bottle and the medicine cost
two shillings and a penny more than the cork.
What did the cork cost?
Solution
127. A boat’s crew are afloat far from land
with no sail or oars. How can they, without
making any use of wind or stream, and without
any outside help, regain the shore by means of
a coil of rope which happens to be at hand.
Solution
What is the largest sum in silver that I
can have in my pockets without being able
to give change for a half-sovereign.
Solution
I have apples in a basket. Without
cutting an apple I give half of the number and
half an apple to one person; half of what then
remains and half an apple to another, and half
of what are still left and half an apple to a third.
One apple now remains in the basket. How
many were there at first?
Solution
130.
A third of twelve divide
By just a fifth of seven;
And you will soon decide
That this must give eleven.
Solution
A motor goes 9 miles an hour uphill,
18 miles an hour downhill, and 12 miles an hour
on the level. How long will it take to run 50
miles and return at once over the same course?
Solution
In firing at a mark A hits it in two out
of three shots, B in three out of four, and C in
four out of five. The mark was hit 931 times.
If each fired the same number of shots, how many
hits did each make, and how many shots were
fired?
Solution
If a cat and a dog, evenly matched in
speed, run a race out and back over a course
75 yards in all, and the dog always takes
5 feet at a bound and the cat 3 feet, which
will win?
Solution
134. In a fog a man caught up a wagon going
in the same direction at 3 miles an hour. If
the wagon was just visible to him at a distance
of 55 yards, and he could see it for five
minutes, at how many miles an hour was he
walking?
Solution
Three horses start in a race. In how
many different ways can they be placed by the
judge?
Solution
136. The New Zealanders, winning a match
against Oxbridge, scored 34 points, from tries
and tries converted into goals.
If every try had been converted they would
have made four-fifths of the maximum which a
score of 34 points from tries and goals can yield.
What is this maximum, and what was their
actual score?
Solution
What is the smallest number, of which
the alternate figures are cyphers, that is divisible
by 9 and by 11?
Solution
Here is an interesting little problem:—A,
with 8d. in his hand, meets B and C, who have
five and three loaves respectively. In hungry
mood they all agree to share the loaves equally,
and to divide the money fairly between B and
C. How much does each receive?
Solution
139. When four money-boxes, containing pennies
only, were opened and counted, it was found
that the number in the first with half those
in all the others, in the second with a third
of all the others, in the third with a fourth of all
the others, and in the fourth with a fifth of all
the others, amounted in each case to 740. How
much money did the boxes contain, and how was
it divided?
Solution
Two steamers start together to make a
trip to a far-off buoy and back. Steamer A runs
all the time at 10 miles an hour. Steamer B
does the passage out at 8 miles an hour, and
the return at 12 miles. Will they regain port
together?
Solution
A golf player has two clubs mended in
London. One has a new head, the other a new
leather face. The head costs four times as much
as the face. At St Andrews it costs five times
as much, and the leather face at St Andrews is
half the London price. Including a shilling for
a ball he pays twice the St Andrews charges for
these repairs. What is the London charge for
each?
Solution
142. Two children were asked to give the total
number of animals in a pasture, where sheep
and cattle were grazing. They were told the
numbers of sheep and of cattle, but one subtracted,
and gave 100 as the answer, and the
other arrived at 11,900 by multiplication. What
was the correct total?
Solution
143. Fifty-two stones are placed at intervals
along a straight road. The distance between
the first and the second is 1 yard, between the
second and the third it is 3 yards, between
the next two 5 yards, and so on, the intervals
increasing each time by 2 yards.
How far would a tramp have to travel to earn
five shillings promised to him when he had brought
them one by one to a basket placed at the first stone?
Solution
On a division in the House of Commons,
if the Ayes had been increased by fifty from the
Noes, the motion would have been carried by
five to three. If the Noes had received sixty
votes from the Ayes, it would have been lost by
four to three. Did the motion succeed? How
many voted?
Solution
145. How many positions are there on the face
of a watch in which the places of the hour and
minute-hands can be interchanged so as still to
indicate a possible time?
Solution
146. In a cricket match the scores in each successive
innings are a quarter less than in the
preceding innings. The match was played out,
and the side that went in first won by fifty runs.
What was the complete score?
Solution
147. A boy throws a cricket ball vertically
upwards, and catches it as it falls just five seconds
later. How high from his hands does the ball
go?
Solution
148. It may be said of a section of the gigantic
carpet at Olympia that had it been 5 feet broader
and 4 feet longer it would have contained 116
more feet; but if 4 feet broader and 5 longer
the increase would have been but 113 feet. What
were its actual breadth and length?
Solution
In estimating the cost of a hundred
similar articles, the mistake was made of reading
pounds for shillings and shillings for pence in
each case, and under these conditions the estimated
cost was £212 18s. 4d. in excess of the real
cost. What was the true cost of the articles?
Solution
150. The square of the number of my house
is equal to the difference of the squares of the
numbers of my next door neighbour’s houses on
either side.
My brother, who lives in the next street, can
say the same of the number of his house, though
his number is not the same as mine. How are
our houses numbered?
Solution
Two men of unequal strength are set
to move a block of marble which weighs 270 lbs.,
using for the purpose a light plank 6 feet long.
The stronger man can carry 180 lbs. How must
the block be placed so as to allow him just that
share of the weight?
Solution
A man has twenty coins, of which some
are shillings and the rest half-crowns. If he were
to change the half-crowns for sixpences and the
shillings for pence he would have 156 coins. How
many shillings has he?
Solution
153. Some coins are placed at equal distances
apart on a table, so that they form the sides of
an equilateral triangle.
From the middle of each side as many are then
taken as equal the square root of the number on
that side, and placed on the opposite corner
coin. The number of coins on each side is then
to the original number as 5 is to 4. How many
coins are there in all?
Solution
154. A gardener, wishing to fence round a
piece of ground with some light posts, found that
if he set them a foot apart there would be 150
too few, but if placed a yard apart there would
be 70 to spare. How many posts had he?
Solution
A gives B £100 to buy 100 animals,
which must be cows at £5 each, sheep at £1, and
geese at 1s. How many of each sort can he buy?
Solution
John is twice as old as Mary was when
he was as old as Mary is. John is now twenty-one.
How old is Mary?
Solution
In a cricket match A made 35 runs;
C and D made respectively half and one-third
as many as B, and B’s score was as much below
A’s as C’s was above D’s. What did B, C, and
D each score?
Solution
158. What is the least number which, divided
by 2, 3, 4, 5, 6, 7, 8, 9, or 10, leaves respectively
as remainders 1, 2, 3, 4, 5, 6, 7, 8, and 9?
Solution
159. A square table stands on four legs, which
are set at the middle point of its sides. What
is the greatest weight that this table can uphold
upon one of its corners?
Solution
160. Well pleased with the inspector’s report,
the rector of a country parish came into his school
with 99 new pennies in his pocket, and said that
he would give them to the five boys in Standard
VII. if they could, within an hour, show him how
to divide them so that the first share should
exceed the second by 3, be less than the third by
10, be greater than the fourth by 9, and less than
the fifth by 16. What was the answer which
would satisfy these conditions?
Solution
161. Some Indian raiders carried off a third
of a flock of sheep, and a third of a sheep. Another
party took a fourth of what remained, and a
fourth of a sheep. Others took a fifth of the rest
and three-fifths of a sheep. What was the
number of the full flock, if there were then 409
left?
Solution
Three boys begin to fill a cistern. A
brings a pint at the end of every three minutes,
B a quart every five minutes, and C a gallon
every seven minutes. If the cistern holds fifty-three
gallons, in what time will it be filled, and
who will pour in the last contribution?
Solution
163. A man late in the last century said that
his age was the square root of the year in which
he was born. In what year did he say this?
Solution
164. A dealer in Eastern curios sold a Satzuma
vase for £119, and on calculation found that the
number which expressed his profit per cent. expressed
also the cost price in pounds of the vase.
What was this number?
Solution
What is the chance of throwing at least
one ace in a single throw with a pair of dice?
Solution
A thief starts running from a country
house as fast as he can. Four minutes later a
policeman starts in pursuit. If both run straight
along the road, and the policeman gets over the
ground one-third faster than the thief, how soon
will he catch him?
Solution
Twenty-seven articles are exposed for
sale on one of the stalls of a bazaar. What choice
has a purchaser?
Solution
168. “How old are you, dad?” said Nellie on
her birthday, as her father gave her as many
shillings as she was years old. His answer was
quite a puzzle for a time, but with the help of
her schoolfellows Nellie worked it out.
This is what he said:—
“I was twice as old as you are
The day that you were born.
You will be just what I was then
When fourteen years are gone.”
How old was Nellie, and how old was her dad?
Solution
This Magic Cocoon is so cleverly spun that the
word can be traced and read in many ways.
| |
N |
|
| |
N |
O |
N |
|
| |
N |
O |
O |
O |
N |
|
| |
N |
O |
O |
C |
O |
O |
N |
|
| |
N |
O |
O |
C |
O |
C |
O |
O |
N |
|
| N |
O |
O |
C |
O |
C |
O |
C |
O |
O |
N |
| |
N |
O |
O |
C |
O |
C |
O |
O |
N |
|
| |
N |
O |
O |
C |
O |
O |
N |
|
| |
N |
O |
O |
O |
N |
|
| |
N |
O |
N |
|
| |
N |
|
How many readings can you discover starting
from one or other of the Cs, and passing up and
down or sideways, but not diagonally, and never
over the same letter twice in a reading? There
are 756!
The Chronogram, severely classed by Addison
as “a species of false wit” is a sentence in which
the salient letters represent in Roman numerals
some particular year. A good English specimen
is this: “My Day Closed Is In Immortality.”
The capital letters in these words give MDCIII.,
or 1603, the year in which Queen Elizabeth died.
The battle-cry at Montlhéry in 1465 was:—“à
CheVaL, à CheVaL, gendarMes, à CheVaL!”
Taking the letters printed in capitals—
| M |
= |
1000 |
| CCC |
= |
300 |
| LLL |
= |
150 |
| VVV |
= |
15 |
| we have the date of the battle |
1465 |
QVI CHRISTI IAVDES CANTANT
SANCTÆ PASSIONIS SVÆ VIRTVTE
IN IPSO ET PATRE VNVM SINT.
This curious inscription is placed over the
organ at Ober Ammergau. Add together its
Roman numerals and they give the date at which
the organ was dedicated. The English of it is:—
May those who sing the praises of Christ be
by virtue of His Sacred Passion one in The
Father and in Himself.
On a damaged inscription to Bishop Berkeley
in Winchester Cathedral are the words—VIXI,
LVXI—I have lived, I have shone. Added
together in their values as Roman numerals the
letters of these two words give his age exactly
at his death—eighty-three.
On the return of the C.I.V. from the Boer War
a prize was offered by Truth for the best motto
appropriate to them. This was to consist of
three words of which the first must begin with
C the second with I and the third with V.
The prize was taken by the following Latin
motto which is singularly happy both in construction
and in meaning:—
| CIVI |
IVI |
VICI |
| I roused |
I went |
I won. |
The sequence of events is perfect; no letters
but C.I.V. are used and the motto is a palindrome
if read by syllables.
As in olden days some of the Psalms and other
writings were constructed in acrostic form, so
in the Middle Ages even serious writers would
juggle with letters, as though they felt that such
tricky methods were an aid to memory.
It was in this spirit that Guido Aretino, a
Benedictine monk of Tuscany in 1204, gave
names to the notes used in the musical scale
from the first syllables of the lines of a Latin
hymn. “Ut” is still used in France, though
we and the Italians have substituted “do.”
| UT queant laxis |
REsonare fibris, |
| MIra gestorum |
FAmuli tuorum, |
| SOLve polutis |
LAbii reatis |
| O Pater alme! |
Many of us know that there is a long verse in
the Book of Ezra in which all the letters of the
alphabet are used, taking “j” as “i” (Ezra vii.,
v. 21).
This very curious coincidence also occurs in a
comparatively short sentence in “The Beth Book,”
by Sarah Grand:—“It was an exquisitely deep
blue just then, with filmy white clouds drawn up
over it like gauze;” and here “j” is itself in
evidence.
Schopenhauer, the famous German philosopher,
who was a confirmed bachelor and misogynist,
was compelled while living at Frankfort to support
an old lady who had been crippled by his
violence. When her death came as a welcome
relief to him, he composed the following clever
epitaph:—
which by the interchange of two letters pictured
the position. It may be freely rendered:—
Old lady dies,
My burden flies.
The following clever composition, which appeared
in the pages of Truth, contains a double
sequence of words, which increase a letter at a
time, the same letters appearing in varied order
until at last “o” culminates in thornless, and
“a” in restrainest. It is quite a remarkable
tour-de-force.
O lack-a-day! at eve we sat,
One star had lit its lamps on high.
We did not note the circling bat,
Start from the stone when flitting nigh.
For the strait gate of honest doubt
Shut off the thrones of Love and Gain;
We dreamed not, as we mourned without,
That Time’s swift transit shortens pain.
O Thou, Who trainest souls to shine,
Though once we craved a thornless lot,
This gracious truth we now divine:
The bruised reed Thou strainest not;
But by restraints, that gently tame;
Restrainest Passion’s kindling flame.
During the Reign of Terror, France and her
people and position were thus alphabetically
described:—
| Le peuple Français |
A B C. |
(abaissé). |
| La gloire nationale |
F A C. |
(effacée). |
| Les places fortes |
O Q P. |
(occupées). |
| Quarante trois députés |
C D. |
(cédés). |
| L’armée |
D P C. |
(dépaysée). |
| Les ministres |
A J. |
(agés). |
| La liberté |
O T. |
(ôtée). |
| La charte |
L U D. |
(éludée). |
This elaborate method of piling up no less
than seven consecutive “thats,” so that they
make tolerable sense, was told to his boys during
school-time by Dr Moberly, then headmaster of
Winchester, and afterwards Bishop of Salisbury,
just fifty years ago:—
I saw that C saw.
C saw that that I saw.
I saw that that that C saw was so.
C saw that, that that that I saw was so.
I saw that, that that that that C saw was so.
C saw that that, that that that that I saw was so.
I saw that that, that that that that that C saw was so.
If the Roman ladies and children, at their
equivalent for Christmas, amused themselves by
acting verbal charades, an excellent word was
at their disposal, “sustineamus”—“let us endure,”
which can be broken up exactly into sus,
tinea, mus—a sow, a moth, a mouse.
1. Can you complete this word square, so that
its four words read alike from top to bottom and
from left to right?
| * |
E |
* |
* |
| E |
* |
* |
* |
| * |
* |
* |
E |
| * |
* |
E |
* |
Solution
2. Can you fill in this word square?
| C |
* |
* |
C |
* |
E |
| * |
N |
U |
* |
E |
S |
| * |
U |
* |
E |
S |
* |
| C |
* |
E |
* |
S |
* |
| * |
E |
S |
* |
* |
E |
| E |
* |
T |
E |
* |
* |
Solution
3. In this sentence, when complete,
So * * * * AG * * * * LATI * * * * X * * * * ITH
each group of four missing letters contains two
pairs of letters which are alike. Can you on
these lines complete the sentence?
Here is a similar sentence by way of illustration:
T * * * * M * * * * TERTAIN * * * * MUND,
which becomes when filled in—
Two women entertained Edmund.
Solution
4. Can you complete this word square by substituting
letters for the dots?
| W |
* |
* |
* |
E |
| * |
* |
T |
* |
* |
| * |
T |
O |
N |
* |
| * |
* |
N |
* |
* |
| E |
* |
* |
* |
T |
Solution
5. What word can be made with these?
L S D U D O D U D.
Solution
A lovelorn youth consulted a married lady
on his condition, and was asked by her on a slip
of paper:—
“Loruve?”
When he had deciphered this, and had answered
in the affirmative, she handed to him another
slip, on which this advice was written:—
| |
L |
|
| “Prove |
A |
|
F |
and ensure success.” |
| |
D |
|
What did it all mean?
Solution
7.
Saint of Spain, whose daily word
Twenty years hath London heard!
Sweet days, elastic metal, motion sharp.
My halls once echoed to an Irish harp.
The “son of sorrow,” honourable of yore.
Dread goddess, loosing the loud dogs of war.
Time’s atom—total of eternities.
This name an insect bears, a patriot bore.
“So do, yet hear me,” said Themistocles.
Solution
8. Can you complete this word square?
| * |
M |
* |
N |
* |
S |
| M |
* |
N |
* |
O |
* |
| * |
N |
* |
B |
* |
E |
| N |
* |
B |
* |
L |
* |
| * |
O |
* |
L |
* |
R |
| S |
* |
E |
* |
R |
* |
Solution
9. A spruce young Frenchman at a fête
consulted a modern oracle as to how he could
best please the ladies. This was the mystic
response:—
MEC DO BIC.
Can you interpret it?
Solution
10. In our young days we have often wrestled
with vulgar fractions, but apart from Algebra
we have had no serious concern with any in which
letters take the place of figures. A specimen of
this sort, not known to science, is the following
curiosity:—
m
ot
y = mo.
Solution
The puzzle in Truth was recently founded
upon “ourang-outang,” which had been cleverly
buried. We will give a few of the best results.
This is one:—
Poor wretch! a moisture filled his eye,
“Do not rebuff a lonely boy,”
Said he, “If ere I sink and die
Your smile—O! pardon will be joy!”
Another is:—
Though I jump, and row, and run,
Cap or cup I never won.
What animals are buried in these lines?
Solution
12.
Fourteen letters here we fix,
Vowels only two are spoken;
All together these we mix
Into what can not be broken.
Solution
13. There is an English word of thirteen
letters in which the same vowel occurs four times,
the same consonant six times, another consonant
twice, and another once. Can you hit upon it?
Solution
14. Though brevity is said to be the soul of
wit, we are too often flooded nowadays with a
superabundance of words.
Here is an attempt at modest condensation.
A familiar English proverb is quite clearly expressed
to the solver’s seeing eyes in this brief
phrase:—
WE IS DO
What is the proverb?
Solution
15. Can you complete this word square?
| W |
* |
* |
* |
* |
S |
| * |
R |
* |
* |
R |
* |
| * |
* |
O |
R |
* |
* |
| * |
* |
R |
M |
* |
* |
| * |
R |
* |
* |
N |
* |
| S |
* |
* |
* |
* |
M |
Solution
16. The following puzzle lines are attributed
to Dr Whewell:—
| |
O |
O |
N |
O |
O. |
|
| U |
O |
A |
O |
O |
I |
O |
U |
|
| O |
N |
O |
O |
O |
O |
M |
E |
T |
O |
O. |
| U |
O |
A |
O |
I |
D |
O |
S |
O |
|
| I |
O |
N |
O |
O |
I |
O |
U |
T |
O |
O! |
Solution
17. Can you fill in the vacancies in this
diamond?
| |
P |
|
| |
F |
O |
* |
|
| |
C |
* |
R |
* |
* |
|
| |
F |
* |
* |
C |
* |
* |
* |
|
| P |
O |
R |
C |
E |
L |
A |
I |
N |
| |
R |
* |
* |
L |
* |
* |
* |
|
| |
S |
* |
A |
* |
* |
|
| |
S |
I |
* |
|
| |
N |
|
Its words must read alike from left to right and
from top to bottom.
Solution
18.
Tan HE Edsa VEN in
It N Gja SmeTs AsgN
aD Az Rett De.
Solution
19. I can travel first-class on the Great Eastern
Railway from 2 2 2 2 2 2 2 2 4 4 4 4 4 5 0 0. What
is the cost of my journey, and its length in time?
Solution
20.
Seogeh sreve ereh wcisume vahl
Lah sehs se otreh nos llebdnas
Regni freh nos gnires rohyer
Ganoed iryd ale nifae esots sorcy
Rub nabot es rohk co caed ir.
Can you decipher it?
Solution
21. I wrote the following note recently:—
Dear Jack,—Meet me at Simpson’s to-morrow
at 1.30. We will sample their excellent Irish
stew. Here are some catchwords that will
remind you of the invitation:—
Join me at and
Join me at ai
Join me at as
Why should they remind him of it?
Solution
22. This was the exact text of a letter sent
to the master of an English village school by a
labourer as an excuse for his boy’s absence:—
“Cepatomtogoatatrin”
Can you decipher it?
Solution
23. This double Acrostic will afford an easy
exercise in mental gymnastics for those to whom
such pastime appeals:—
Now we are fain
To rack your brain.
1. More fit for babes and sucklings than for you.
2. Robbed of externals this is very true.
3. Diminutive in measure and in weight.
4. Pen-name of one a true pen potentate.
5. A palindrome quite plain is here in sight.
6. Sans head and tail it also yields this light.
7. Here is in short what anyone may write.
Solution
| 24. |
c |
e |
f |
h |
i |
m |
n |
o |
r |
s |
t |
v |
y |
| 3 |
2 |
2 |
2 |
7 |
1 |
1 |
2 |
6 |
5 |
8 |
3 |
9 |
| |
4 |
|
4 |
1 |
|
2 |
1 |
|
6 |
1 |
|
| 3 |
|
1 |
5 |
8 |
2 |
3 |
| 5 |
7 |
9 |
|
|
6 |
| 4 |
|
|
|
The letters with ones under them are the first
letters of words, those with twos under them are
second letters of words, and so on.
Solution
25. Having occasion to make a few slight
additions to my will, I called in my lawyer to
arrange the matter. How far forward did the
instructions contained in the following lines carry
him in his work?
Set down a hundred in my will,
Add nothing to the text;
Five hundred now a space may fill,
And one be added next.
Another hundred write as well,
And yet another one;
Then fifty more, and try to tell
The deed that now is done.
Solution
26. Can you complete this word square?
| * |
D |
* |
* |
O |
* |
| D |
* |
* |
I |
* |
E |
| * |
S |
* |
A |
* |
D |
| T |
* |
A |
* |
A |
* |
| * |
R |
* |
A |
* |
E |
| R |
* |
D |
* |
E |
* |
Solution
27. Over the grave of a French musician, who
was choked by a fish bone, the following epitaph
was inscribed in notes of music:—A. G. A. E. A.
Over the porch of the house of Gustave Doré
these musical notes were placed on a tablet:—C.
E. B. A. C. D.
What do these inscriptions signify?
Solution
28. “Where can we meet to-morrow?” said
Jack Spooner to his best girl.
“We will go,” she replied, “at 222222222222
LEY STREET.”
When and where did they meet?
Solution
29. What does this spell?
CT T T T T T T T T T
Solution
30. What English word is it which may be
so treated as to affirm or disallow the use of its
own initial or final letter?
Solution
31. Can you arrange these letters
E I O O O U
B C N N R R S S
so that they form the title of a book well-known
to boys?
Solution
32. Keeping these letters in their present order
make a sensible sentence by inserting among
them as often as is necessary another letter, which
must be in every case the same.
A DEN I I CAN DOCK.
Solution
33. Here is a well-known quotation from
Shakespeare, which seems to need some straightening
out:—
OXXU8 MAAULGIHCTE
NOR
Solution
34. Here, as an awful warning to those who
are ready to accept the definition of English
spelling given by a former headmaster of Winchester—“Consonants
are interchangeable, and
vowels do not count”—is a common English
word of twelve letters, in “linked sweetness long
drawn out.”
Iewkngheaurrhphthewempeighghtips.
Can you decipher it?
Solution
35. Can you, by filling in letters, complete
this word square so that it shall read alike across
and from top to bottom?
| * |
A |
* |
* |
| A |
* |
* |
A |
| * |
E |
* |
* |
| * |
A |
* |
E |
Solution
36. The following curious inscription may be
seen on a card hanging up in the bar of an old
riverside inn in Norfolk:—
THEM * ILL * ERSLEA * VET * HEMI
LLT * HEW * HER * RYMEN * LOW
ERTH * EIRS * AILTH * EMA
LTS * TER * SLE * AVET * HE * KI
LN * FORAD * ROPO * FTH
EWHI * TESW * AN * SALE.
Can you decipher it?
Solution
37.
TONDEBNIOTOCHUMFOARYHUR
OTDIRECTTHAWHOTERSOFKLSYA;
TIKATESTUBALIGHTSTILLETRUFLYR
OTBOWLALLNEFESLEAVARFWYAA.
In this printer’s pie the words are in their
proper sequence, but the letters are tangled.
Solution
38. In the following short sentences five names
of places are buried—that is to say, the letters
which spell them in proper order form parts of
more words than one. Thus, for example
“Paris” might be buried in the words “go up a rise:”
“The men could ride all on donkeys, the
skipper, though, came to a bad end.”
When you have discovered these places, try
to find out what very unexpected word of more
than four letters is buried in the sentence, “On
Christmas Eve you rang out angel peals.”
Solution
39. According to an old poet, Sir John Harrington
(1561-1612):—
“Treason doth never flourish; what’s the reason?
For if it prosper none dare call it treason!”
The classic lines may possibly have been the
germ of the flippant modern riddle, “Why is it
no offence to conspire in the evening?”
Solution
40.
Inscribe an m above a line
And write an e below,
This woodland flower is hung so fine
It bends when zephyrs flow.
Solution
41. The following letters, if they are properly
rearranged, will fall into the words which form
a popular proverb:—
A A E E G G H I L L M N N N O O O O R R S S S S T T
Can you place them in position?
Solution
42. Can you fill in the gaps of this proverb?
E**t* *e*s**s *a*e *h* *o** **i*e.
Solution
43. There is an English word of five syllables
which has only eight letters, five of them vowels—an
a, an e, twice i, and y. What are its consonants?
Solution
44. Can you turn TORMENT to RAPTURE,
using four links, changing only one letter each
time, and varying the order of the letters?
Solution
45. Can you arrange these letters so that they
form a sentence of five words?
a a c e e e f f h h i i i i i m n n o o o p r r s s t t t t t.
The result is a well-known English proverb.
Solution
46. Add one letter, and make this into a
sensible English sentence:—
G D L D P R T F R R T H D X X F R D D N S
Solution
47.
One vowel in an English word is found,
Which by eight consonants is hedged around.
Solution
Can you form an English word with these
letters?
A A A A A B B N N I I R S S T T.
Solution
What is this? It is found in Shakespeare:—
K I N I.
Solution
50. There are two English words which contain
each of them ten letters, and six of these are
a, b, c, d, e, f, the first six letters of the alphabet.
Can you build up either or both of them without
looking at the solution?
Solution
51.
Of a band of true kinsmen I stand at the head,
Who, to keep themselves warm, cluster three in a bed.
Put four into gaol and their number has risen,
So that six can be counted together in prison.
Take the six and recount them, they dwindle to three;
Count again, and a change into five you will see.
With no number from one to one hundred I mix,
Yet with five of my mates I am seen to make six.
Solution
52. The prodigal son of a wealthy colonial
farmer received a letter from his father, to suggest
that a considerable part of his inheritance should
be safeguarded before he squandered it. His
reply ran thus:—“Dear dad, keep 1000050.”
As such a sum, even in dollars, was out of the
question, the father was completely puzzled.
What did the prodigal mean?
Solution
53. The names of eight famous British poets
are buried in these lines, that is to say, the letters
that spell the names form in their proper order
parts of different words:—
The sun is darting rays of gold
Upon the moor, enchanting spot,
Whose purpled heights, by Ronald loved,
Up open to his Shepherd cot.
And sundry denizens of air
Are flying, aye, each to his nest;
And eager make at such an hour
All haste to reach the mansions-blest.
Can you dig them up?
Solution
54. When A. B. gave up the reins of government,
and C. B. took office in his place, it was
found that their political positions could be
exactly described by two quite common English
verbs, which differ only in this, that the one is
longer by one letter than the other, while the
rest of the letters are the same, and in the same
order. What are these two verbs?
Solution
55. The following is a prize Rebus:—
done
mutt
and
i
you make me
Solution
56.
First a c and a t, last a c and a t,
With a couple of letters between,
Form a sight that our eyes are delighted to see,
Unless in their sight it is seen.
Solution
57.
Cut off my tail and set it at my head,
What was an island is a little bear instead.
Solution
T S.
58. What English word do these two letters
indicate? There are two possible solutions of
equal merit.
Solution
59. How can we read this?
I N X I N X I N.
Solution
60.
IFS
Solution
No. XXVI
Black has made the false move Kt from Q sq
to Kt 3. When this is replaced, and the king
is moved as the proper penalty, White mates
at once with one or other of the Knights.
Back to Cameo
No. XXVII
Replace the White Kt at B 7, and a Black
Pawn at K 4; then P takes P en pas. Mate.
ANALYSIS AND PROOF
It can be proved that Black’s last move must
have been P from K 2 to K 4, so that White may
take the P en pas.
The Black King cannot have moved from any
occupied square.
(The White Kt now occupies B 7.)
Nor from Kt 3 or 4, as both are now doubly
guarded, so that he cannot have moved out of
a check.
(The White Kt now helps to guard Kt 5.)
Nor can he have moved from K 2, as the White
P on Q 6 cannot have moved to give check.
No other P can have moved.
The K P cannot have moved from K 3, became
of the position of the White King.
Therefore Black’s last move must have been
P from K 2 to K 4, which White can take en pas
giving Mate.
Back to Cameo
No. XXVIII
B—Q 2P—R 7
B—R 5P—R 7 becomes Q
P—Kt 4any
White hasno move.
The way in which the B runs to earth and is
shut in is most ingenious. Black with the new
Q cannot anyhow give White a move.
Back to Cameo
No. XXIX
R—Kt 7, ch.K moves
R—Kt 5P becomes Q
R—B 5, ch.Q × R, stalemate.
Back to Cameo
No. XXX
B—Kt 8.
Back to Cameo
No. XXXI
B—R 4.
Back to Cameo
No. XXXII
B—R sq.
Back to Cameo
No. XXXIII
B—Q 4.
Back to Cameo
No. XXXIV
B—B sq.
Back to Cameo
No. XXXV
K—R 4.
Back to Cameo
No. XXXVI
Kt—Kt’s 6.
Back to Cameo
No. XXXVII
R—Kt 5Kt × R
Kt—B 6Kt—B 6 ch.
Kt × Kt mate
Back to Cameo
No. XXXVIII
Kt—R 7B moves
Q—KB 8B returns
Q—R 8 mate.
Any other move of the Kt would impede the
movements of the Q.
Back to Cameo
No. XXXIX
R—R sq.B moves.
Q—Kt sq.B returns.
Q to Kt sq. mate.
Back to Cameo
No. XL
This beautiful problem is solved by:—
Q—Kt 6P × Q
K—B 2any
mates accordingly.
or
Kt—K 3
R—B 3any
mates accordingly.
Back to Cameo
No. XLI
R—R 2.B × Kt
Q—R sq.any
Q mates.
if
B—Q sq.
Kt—QB 8,any
Q or Kt mates.
if
B elsewhere
Kt—QB 6,any
Q or Kt mates.
Back to Cameo
No. XLII
Kt—Kt 4, dis. ch.K—R 8.
Q—KR 2, ch.P × Q.
Kt—B 2, mate.
There are other variations.
Back to Cameo
No. XLIII
B—Kt sq.P × Kt,
Q—QR7,K × Kt.
Q mates.
If K × Kt, Q × P, and mates next move.
Back to Cameo
No. XLIV
K—Q 7,K moves
R—Q 5,K × R
Q mates.
Back to Cameo
No. XLV
R—Q 8K moves
Q × P, ch.K × Q
B mates.
if
B moves
Q—K 7any
Q mates.
Back to Cameo
No. XLVI
B—B 6K × R
Q—Q 7, ch.K × Q
R mates.
if
B × R
R—B 6, ch.K × P
Q mates.
There are other variations.
Back to Cameo
No. XLVII
Q—R 8Kt × Q
Kt—B 6any
mates accordingly.
Back to Cameo
No. XLVIII
B—Kt 8B—K Kt 2
Q × B, ch.Kt—K 4
Q—QR 7, mate.
if
B—B 3
Q—Q 2, ch.K moves
P mates.
Back to Cameo
No. XLIX
B—Kt 2K—K 4
Q—K 3 ch.any
mates.
There are other variations.
Back to Cameo
No. L
Q—KR 2K—B 3 or B4
Q—Q 6, ch.Q × Q
Kt—K 5. double ch. mate.
There are other variations.
Back to Cameo
No. LI
R—QR sq.P moves
R—R 2P × R
P mates.
Back to Cameo
No. LII
Q—B sq.P—K 7
Q × BP ch.K—Q 3
Q mates.
if
K—Q 2
Q × BP ch.K—B 3
Q mates.
There are other variations.
Back to Cameo
No. LIII
B—R 8K—R 2
Q—QR sq.K moves
Q—Q Kt. 7, mate.
Back to Cameo
No. LIV
Q—R 5B × Q or B—B 2
Kt—Kt 5any
Kt mates.
Back to Cameo
No. LV
K—Kt 7Kt—B 3
Q—Q 5, ch.Kt × Q
Kt. mates.
if
P—K 3
Q—B 8, ch.K moves
Q mates.
Back to Cameo
No. LVI
R—Kt 6P moves
B—Kt 4P × B
R × P mate.
Back to Cameo
No. LVII
Kt from R 3—Kt 5P × Kt
B—B 4P × B
P—Q 4 ch.P × P en pas
R—B 5 mate.
if
R × B
R—Kt 5 ch.any
mates accordingly.
and if
Q—K 6
Kt—Q 6Q—Kt 3
R—B 5 ch.Q × R
Kt—B 7 mate.
Back to Cameo
No. LVIII.—THE GEARED WHEELS
The arrow head at the top of a small wheel
with ten teeth, which is geared into and revolved
round a large fixed wheel with forty teeth, will
point directly upwards five times in its course
round the large wheel. Four of these turnings
are due to the rotation of the small wheel on its
own axis, and one of them results from its revolution
round the large wheel.
Back to Description
No. LX.—THE FIFTEEN BRIDGES
It is possible to pass over all the bridges which
connect the islands A and B and the banks of
the surrounding river without going over any of
them twice.
The course can be shown thus, using capital
letters for the different regions of land, and italics
for the bridges:—Ea Fb Bc Fd Ae Ff Cg Ah Ci Dk
Am En Ap Bq ElD.
This order of the bridges can, of course, be
reversed.
Back to Description
No. LXVI.—A DUCK HUNT
In order that a spaniel starting from the middle
of a circular pond, and going at the same pace
as a duck that is swimming round its edge, shall
be sure to catch it speedily, the dog must always
keep in the straight line between the duck and
the centre of the pond.
The duck can never gain an advantage by
turning back, and if it swims on continuously
in a circle it will be overtaken when it has passed
through a quarter of the circumference, for the
dog will in the same time have described a semi-circle
whose diameter is the radius of the pond,
ending at the point where the duck is caught.
Back to Description
No. LXIX.—THE TETHERED BIRD
When a bird tethered by a cord 50 feet long
to a post 6 inches in diameter uncoils the full
length of the cord, and recoils it in the opposite
direction, keeping it always taut, it flies 10,157
feet, or very nearly 2 miles, in its double course.
To avoid possible misunderstanding, we point
out that, in order to pass from the uncoiling
to the recoiling position, the bird must fly through
a semicircle at the end of the fully extended cord.
Back to Description
No. LXX.—THE MOVING DISC AND THE
FLY
When a fly, starting from the point A, just
outside the revolving disc, and always making
straight for its mate at the point B, crosses the
disc in four minutes, during which time the disc
is turning twice, the revolution of the disc has
a most curious and interesting effect on the path
of the fly.
The fly is a quarter of a minute in passing from
the outside circle to the next, during which the
disc has made an eighth of a revolution, and the
fly has reached the point marked 1. The succeeding
points up to 16 show the position of the fly
at each quarter of a minute, until, by a prettily
repeated curve, B is reached.
Back to Description
No. LXXI.—A SHUNTING PUZZLE
The following method enables the engine R
to interchange the positions of the wagons, P
and Q, for either of which there is room on the
straight rails at A, while there is not room there
for the engine, which, if it runs up either siding,
must return the same way:—
1. R pushes P into A. 2. R returns, pushes
Q up to P in A, couples Q to P, draws them both
out to F, and then pushes them to E. 3. P is
now uncoupled, R takes Q back to A, and leaves
it there. 4. R returns to P, pulls P back to C,
and leaves it there. 5. R, running successively
through F, D, B, comes to A, draws Q out, and
leaves it at B.
Back to Description
No. LXXV.—PHARAOH’S SEAL
It is quite puzzling to decide how many similar
triangles or pyramids are expressed on the seal
of Pharaoh. There are in fact 96.
Back to Description
No. LXXVI.—ROUND THE GARDEN
The four persons who started at noon from
the central fountain, and walked round the four
paths at the rates of two, three, four, and five
miles an hour would meet for the third time at
their starting point at one o’clock, if the distance
on each track was one-third of a mile.
Back to Description
No. LXXVII.—A JOINER’S PUZZLE
This diagram shows how to divide Fig. A into
two parts, and so rearrange these that they form
either Fig. B or Fig. C, without turning either
of the pieces.
Cut the five steps, and shift the two pieces as
is shown.
Back to Description
No. LXXVIII.—THE BROKEN OCTAGON
The Broken Octagon is repaired and made
perfect if its pieces are put together thus:—
Back to Description
No. LXXIX.—AT A DUCK POND
The pond was doubled in size without disturbing
the duck-houses, thus:—
Back to Description
No. LXXX.—ALL ON THE SQUARE
This is a perfect arrangement:—
Back to Description
No. LXXXI.—PINS AND DOTS
The pins may be placed thus:—
On the third dot in the top line; on the sixth
dot in the second line; on the second dot in the
third line; on the fifth dot in the fourth line;
on the first dot in the fifth line; on the fourth
dot in the sixth line.
Back to Description
No. LXXXII.—A TRICKY COURSE
To trace this course draw lines upon the diagram
from square 46 to squares 38, 52, 55, 23,
58, 64, 8, 57, 1, 7, 42, 10, 13, 27, and 19. This
gives fifteen lines which pass through every
square only once.
Back to Description
No. LXXXIII.—FOR THE CHILDREN
Make a square with three on every side, and
place the remaining four one on each of the
corner men or buttons.
Back to Description
No. LXXXVII.—LOYD’S MITRE PROBLEM
The figure given is thus divided into four
equal and similar parts:—
Back to Description
No. LXXXIX.—CUT OFF THE CORNERS
A very simple rule of thumb method for striking
the points in the sides of a square, which will
be at the angles of an octagon formed by cutting
off equal corners of the square, is to place another
square of equal size upon the original one,
so that the centre is common to both, and the
diagonal of the new square lies upon a diameter
of the other parallel to its side.
Back to Description
No. XCIII.—MAKING MANY SQUARES
The subjoined diagram shows how the two
oblongs, applied to the two concentric squares,
produce 31 perfect squares, namely, 17 small
ones, one equal to 25 of these, 5 equal to 9, and
8 equal to 4.
Back to Description
No. XCIV.—CUT ACROSS
The Greek Cross can be divided by two straight
cuts, so that the resulting pieces will form a
perfect square when re-set, as is shown in these
figures:—
Back to Description
No. CV.—A TRANSFORMATION
The diagram which is given below shows how
the irregular Maltese Cross can be divided by
two straight cuts into four pieces, which form
when properly rearranged, a perfect square.
Back to Description
No. CVI.—SHIFTING THE CELLS
The following diagram shows by its dark lines
how the whole square can be cut into four pieces,
and these arranged as two perfect squares in
which every semicircle still occupies the upper
half of its cell.
One piece forms a square of nine cells, and it
is easy to arrange the other three pieces in a
square of sixteen cells by lifting the three cells
and dropping the two.
Back to Description
No. CVII.—IN A TANGLE
It will be seen, on the subjoined diagram, how
twenty-one counters or coins can be placed on
the figure so that they fall into symmetrical
design, and form thirty rows, with three in each
row.
Back to Description
No. CVIII.—STILL A SQUARE
In order that a square and an additional
quarter may be divided by two straight lines
so that their parts, separated and then reunited,
form a perfect square, lines must be drawn from
the point A to the corners B and C. Draw the
figure on paper, cut through these lines, and
you will find that the pieces can be so reunited
that they form a perfect square.
Back to Description
No. CIX.—A TRANSFORMATION
The diagram below shows how the seven parts
of the square can be rearranged so that they
form the figure 8.
Back to Description
No. CX.—TO MAKE AN OBLONG
Here is an oblong formed by piecing together
two of the smaller triangles, and four of each
of the other patterns—
Here is another:—
Back to Description
No. CXI.—SQUARES ON THE CROSS
This diagram shows how every indication of
the seventeen squares is broken up by the removal
of seven of the asterisks which mark
their corners.
Those surrounded by circles are to be removed.
Back to Description
No. CXII.—A CHINESE PUZZLE
The dotted lines on the triangular figure show
how a piece of cardboard cut to the shape of
Fig. 1 can be divided into three pieces, and rearranged
so that these form a star shaped as in
Fig. 2.
Back to Description
No. CXIII.—FIRESIDE FUN
To solve this puzzle slip the first coin or counter
from A to D, then the others in turn from F to A,
from C to F, from H to C, from E to H, from
B to E, from G to B, and place the last on G.
It can only be done by a sequence of this sort, in
which each starting point is the finish of the next
move.
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1. THE CARPENTER’S PUZZLE
The carpenter cleverly contrived to mend a
hole 2 feet wide and 12 feet long, by cutting the
board which was 3 feet wide and 8 feet long, as
is shown in Fig. 1, and putting the two pieces
together as is shown in Fig. 2.
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2. GOLDEN PIPPINS
Here is another solution:—
| 1 |
3 |
5 |
7 |
9 |
11 |
13 |
15 |
17 |
19 |
| 39 |
37 |
35 |
33 |
31 |
29 |
27 |
25 |
23 |
21 |
| 2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
18 |
20 |
| 40 |
38 |
36 |
34 |
32 |
30 |
28 |
26 |
24 |
22 |
| 82 |
82 |
82 |
82 |
82 |
82 |
82 |
82 |
82 |
82 |
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3. AN AWKWARD FIX
I was able to find my way in a strange district,
when the sign-post lay uprooted in the ditch,
without any difficulty. I simply replaced the
post in its hole, so that the proper arm, with its
lettering, pointed the way that I had come, and
then, of necessity, the directions of the other
arms were correct.
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4. LINKED SWEETNESS LONG DRAWN
OUT
The train was whistling for 5 minutes. Sound
travels about a mile in 5 seconds, so the first I
heard of it was 5 seconds after it began. Its
last sound reached me 71⁄2 seconds after it ceased,
so I heard the whistle for 5 minutes, 21⁄2 seconds.
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5.
These quarters were not so elastic as they
are made to appear. In good truth, considering
that the second man who was placed in A was
afterwards removed to I, no real second man was
provided for at all.
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6.
The first day of a new century can never be
Sunday, Wednesday, or Friday. The cycle of
the Gregorian calendar is completed in 400 years,
after which all dates repeat themselves.
As in this cycle there are only four first days
of a century, it is clear that three of the seven
days of the week must be excluded. Any perpetual
calendar shows that the four which do
occur are Monday, Tuesday, Thursday, and
Saturday, so that Sunday, Wednesday, and Friday
are shut out.
A neat corollary to this proof is that Monday
is the only day which may be the first, or which
may be the last, day of a century.
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7.
A cricket bat with spliced handle has such
good driving power, because the elasticity of the
handle allows the ball to be in contact with the
blade of the bat for a longer time than would
otherwise be possible.
With similar effect the “follow through” of
the club head at golf maintains contact with the
ball, when it is already travelling fast.
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8.
When two volumes stand in proper order
on my bookshelf, each 2 inches thick over all,
with covers 1⁄8 of an inch in thickness, a bookworm
would only have to bore 1⁄4 of an inch, to penetrate
from the first page of Vol. I, to the last page of
Vol. II, for these pages would be in actual contact
if there was no binding. This very pretty
and puzzling question combines in its solution
all the best qualities of a clever catch with solid
and simple facts.
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9.
A man would have to fall from a height of
nearly 15 miles to reach earth before the sound
of his cry as he started. The velocity of sound
is constant, while that of a falling body is continually
accelerated. At first the cry far outstrips
the falling man, but he overtakes and passes
through his own scream in about 141⁄2 miles, for
his body falls through the 15 miles in 70 seconds,
and sound travels as far in 72 seconds. Air
resistance, and the fact that sound cannot pass
from a rare to a dense atmosphere, are disregarded
in this curious calculation.
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10.
A man on a perfectly smooth table in a
vacuum, and where there was no friction, though
no contortions of his body would avail to get
away from this position, could escape from the
predicament by throwing from him something
which he could detach from his person, such as
his watch or coat. He would himself instantly
slide off in the opposite direction!
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11.
The monkey clinging to one end of a rope
that passes over a single fixed pulley, while an
equal weight hangs on the other end, cannot
climb up the rope, or rise any higher from the
ground.
If he continues to try to climb up, he will
gradually pull the balancing weight on the other
end of the rope upwards, and the slack of the
rope will drop below him, while he remains in the
same place.
If, after some efforts, he rests, he will sink
lower and lower, until the weight reaches the
pulley, because of the extra weight of rope on
his side, if friction is disregarded.
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12.
Though the tension on a pair of traces
tends as much to pull the horse backward as
it does to pull the carriage forward, it is the
initial pull from slack to taut which sets the
traces in motion; and this, once started, must
continue indefinitely until checked by a counter
pull.
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13.
Some say that a rubber tyre leaves a double
rut in dust and a single one in mud, because the
air, rushing from each side into the wake of the
wheel, piles up the loose dust. Others hold that
the central ridge is caused by the continuous
contraction of the tyre as it passes its point of
contact with the road.
A correspondent, writing some years ago to
“Knowledge,” said:—“It is our old friend the
sucker. The tyre being round, the weight on
centre of track only is great enough to enable
the tyre to draw up a ridge of dust after it.”
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14.
If two cats on a sloping roof are on the
point of slipping off, one might think that whichever
had the longest paws (pause) would hold
on best. Todhunter, in playful mood, saw
deeper into it than that, and pronounced for the
cat that had the highest mew, for to his mathematical
mind the Greek letter mu was the coefficient
of friction!
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15.
If a penny held between finger and thumb,
and released by withdrawing the finger, starts
“heads” and makes half a turn in falling through
the first foot, it will be “heads” again on reaching
the floor, if it is held four feet above it at first.
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16. HE DID IT!
Funnyboy had secretly prepared himself for
the occasion by rubbing the chemical coating
from the side of the box on to his boot.
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17. THE CYCLE SURPRISE
If a bicycle is stationary, with one pedal at its
lowest point, and that pedal is pulled backwards,
while the bicycle is lightly supported, the bicycle
will move backwards, and the pedal relatively
to the bicycle, will move forwards. This would
be quite unexpected by most people, and it is
well worth trying.
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18.
The rough stones, by which any number
of pounds, from 1 to 364, can be weighed, are
respectively 1 ℔., 3 ℔s., 9 ℔s., 27 ℔s., 81 ℔s., and
243 ℔s. in weight.
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19.
If we disregard the resistance of the air,
a small clot of mud thrown from the hindermost
part of a wheel would describe a parabola, which
would, in its descending limb, bring it back into
kissing contact with the wheel which had rejected
it.
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20. THE CARELESS CARPENTER
When the carpenter cut the door too little,
he did not in fact cut it enough, and he had to cut
it again, so that it might fit.
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21.
If from the North Pole you start sailing
in a south-westerly direction, and keep a straight
course for twenty miles, you must steer due
north to get back as quickly as possible to the
Pole, if, indeed, it has been possible to start from
it in any direction other than due south.
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22. DICK IN A SWING
Dick’s feet will travel in round numbers nearly
16 feet further than his head, or to be exact,
15·707,960 feet.
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23. A POSER
The initial letters of Turkey, Holland, England,
France, Italy, Norway, Austria, Lapland, and
Spain spell, and in this sense are the same as,
“the finals.”
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1.
The only sum of money which satisfies the
condition that its pounds, shillings, and pence
written down as a continuous number, exactly
give the number of farthings which it represents,
is £12, 12s., 8d., for this sum contains 12,128
farthings.
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2.
If, when a train, on a level track, and running
all the time at 30 miles an hour, slips a
carriage which is uniformly retarded by brakes,
and this comes to rest in 200 yards, the train
itself will then have travelled 400 yards.
The slip carriage, uniformly retarded from
30 miles an hour to no miles an hour, has an
average speed of 15 miles an hour, while the
train itself, running on at 30 miles an hour all
the time, has just double that speed, and so
covers just twice the distance.
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3.
The traveller had fivepence farthing when
he said to the landlord, “Give me as much as I
have in my hand, and I will spend sixpence with
you.” After repeating this process twice he had
no money left.
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4.
This is the way to obtain eleven by adding
one-third of twelve to four-fifths of seven—
TW(EL)VE + S(EVEN) = ELEVEN
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5.
Here is the completed sum:—
| 2 |
1 |
5 |
) |
* |
7 |
* |
9 |
* |
( |
1 |
* |
* |
|
2 |
1 |
5 |
) |
3 |
7 |
1 |
9 |
5 |
( |
1 |
7 |
3 |
| |
* |
* |
* |
|
|
|
2 |
1 |
5 |
|
|
| |
* |
5 |
* |
9 |
|
|
1 |
5 |
6 |
9 |
|
| |
* |
5 |
* |
5 |
|
|
|
1 |
5 |
0 |
5 |
|
|
| |
* |
4 |
* |
|
|
6 |
4 |
5 |
|
| |
|
* |
* |
* |
|
|
|
6 |
4 |
5 |
|
The clue is that no figure but 3, when multiplied
into 215, produces 4 in the tens place.
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6.
If I attempt to buy as many heads of asparagus
as can be encircled by a string 2 feet
long for double the price paid for as many as half
that length will encompass, I shall not succeed.
A circle double of another in circumference is
also double in diameter, and its area is four times
that of the other.
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7.
If, when you reverse me, and my square,
and my cube, and my fourth power, you find
that no changes have been made, I am 11, my
square is 121, my cube 1331, and my fourth
power 14641.
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8.
A thousand pounds can be stored in ten
sealed bags, so that any sum in pounds up to
£1,000 can be paid without breaking any of the
seals, by placing in the bags 1, 2, 4, 8, 16, 32, 64,
128, 256, and 489 sovereigns.
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9.
It is the fraction 6⁄9 which is unchanged when
turned over, and which, when taken thrice, and
then divided by two becomes 1.
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10.
When the three gamblers agreed that the
loser should always double the sum of money
that the other two had before them, and they
each lost once, and fulfilled the conditions,
remaining each with eight sovereigns in hand,
they had started with £13, £7, and £4 as the
following table shows:—
| |
A |
B |
C |
| £ |
£ |
£ |
| At starts |
13 |
7 |
4 |
| When A loses |
2 |
14 |
8 |
| When B loses |
4 |
4 |
16 |
| When C loses |
8 |
8 |
8 |
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11.
Tom’s sum, which his mischievous neighbour
rubbed almost out, is reconstructed thus:—
| |
3 |
4 |
5 |
|
|
3 |
4 |
5 |
| |
* |
* |
|
3 |
7 |
| |
* |
* |
* |
* |
|
2 |
4 |
1 |
5 |
| * |
* |
* |
* |
|
1 |
0 |
3 |
5 |
|
| * |
* |
7 |
6 |
* |
1 |
2 |
7 |
6 |
5 |
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12.
Here are two other arrangements of the
nine digits which produce 45, their sum; each
is used once only:—
5 × 8 × 9 × (7 + 2)1 × 3 × 4 × 6 = 45
72 - 5 × 8 × 93 × 4 × 6 + 1 = 45
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13.
If, when the combined ages of Mary and
Ann are 44, Mary is twice as old as Ann was
when Mary was half as old as Ann will be when
Ann is three times as old as Mary was when
Mary was three times as old as Ann, Mary is 271⁄2
years old, and Ann is 161⁄2.
For, tracing the question backwards, when
Ann was 51⁄2 Mary was 161⁄2. When Ann is three
times that age she will be 491⁄2. The half of this
is 243⁄4, and when Mary was at that age Ann was
133⁄4. Mary’s age, by the question, was twice
this, or 271⁄2.
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14.
It is safer at backgammon to leave a blot
in the tables which can be taken by an ace than
one which a three would hit. In either the case
of an actual ace or a three the chance is one in
eleven; but there are two chances of throwing
deuce-ace, the equivalent of three.
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15.
If I start from a bay, where the needle
points due north, 1200 miles from the North Pole,
and the course is perfectly clear, I can never
reach it if I steam continuously 20 miles an hour,
steering always north by the compass needle.
After about 200 miles I come upon the Magnetic
Pole, which so affects the needle that it no longer
leads me northward, and I may have to steer
south by it to reach the geographical Pole.
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16.
The 21 casks, 7 full, 7 half full, and 7 empty,
were shared equally by A, B, and C, as follows:—
| |
Full cask. |
Half full. |
Empty. |
| A |
2 |
3 |
2 |
| B |
2 |
3 |
2 |
| C |
3 |
1 |
3 |
| or— |
|
| A |
3 |
1 |
3 |
| B |
3 |
1 |
3 |
| C |
1 |
5 |
1 |
Thus each had 7 casks, and the equivalent of
31⁄2 caskfuls of wine.
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17.
The foraging mouse, able to carry home
three ears at a time from a box full of ears of corn,
could not add more than fourteen ears of corn
to its store in fourteen journeys, for it had each
time to carry along two ears of its own.
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18.
If, with equal quantities of butter and
lard, a small piece of butter is taken and mixed
into all the lard, and if then a piece of this blend
of similar size is put back into the butter, there
will be in the end exactly as much lard in the
butter as there is butter in the lard.
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19.
The fallacy of the equation—
4 - 10 = 9 - 15
4 - 10 + 25⁄4 = 9 - 15 + 25⁄4
and the square roots of these—
2 - 5⁄2 = 3 - 5⁄2
therefore 2 = 3
is explained thus:—The fallacy lies in ignoring
the fact that the square roots are plus or minus.
In the working we have taken both roots as
plus. If we take one root plus, and the other
minus, and add 5⁄2, we have either 2 = 2, or 3 = 3.
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20.
The largest possible parcel which can be
sent through the post under the official limits
of 3 feet 6 inches in length, and 6 feet in length
and girth combined, is a cylinder 2 feet long and
4 feet in circumference, the cubic contents of
which are 26⁄11 cubic feet.
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21.
We can show, or seem to show, that
either four, five, or six nines amount to 100,
thus:—
| 999⁄9 = 100 |
IX
IX
IX
IX
IX
100 |
9 × 9 + 9 + 99⁄9 = 100 |
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22.
This is the magic square arrangement, so
contrived that the products of the rows, columns,
and diagonals are all 1,000.
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23.
If seven boys caught four crabs in the rock-pools
at Beachy Head in six days, the twenty-one
boys who searched under the seaweed and
only caught one crab with the same rate of
success were only at work for half a day.
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24.
A watch could be set of a different trio
from a company of fifteen soldiers for 455 nights,
and one of them, John Pipeclay, could be included
ninety-one times.
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25.
If Augustus Cæsar was born September 23,
B.C. 63, he celebrated his sixty-third birthday
on September 23, B.C. 0; or, writing it otherwise,
September 23, A.D. 0; or again, if we wish to
include both symbols, B.C. 0 A.D. It is clear
that his sixty-second birthday fell on September
23, B.C. 1, and his sixty-fourth on September 23,
A.D. 1, so that the intervening year may be
written as above.
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26.
The difference of the ages of A and B who
were born in 1847 and 1874, is 27, or 30 - 03.
Hence, when A was 30 B was 03. And A was
30 in 1877. Eleven years later A was 41 and B
14, and eleven years after that A was 52 and B 25.
Thus the same two digits served to express the
ages of both in 1877, 1888, and 1899. This can
only happen in the cases of those whose ages
differ by some multiple of nine.
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27
A hundred and one by fifty divide,
To this let a cypher be duly applied;
And when the result you can rightly divine,
You find that its value is just one in nine—
is solved by CLIO, one of the nine Muses.
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28.
The man who paid a penny on Monday morning
to cross the ferry, spent half of what money
he then had left in the town, and paid another
penny to recross the ferry, and who repeated this
course on each succeeding day, reaching home
on Saturday evening with one penny in his pocket,
started on Monday with £1 1s. 1d. in hand.
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29.
When the three men agreed to share their
mangoes equally after giving one to the monkey,
and when each helped himself to a third after
giving one to the monkey, without knowing that
anyone had been before him, and they finally
met together, gave one to the monkey, and
divided what still remained, there must have
been at least seventy-nine mangoes for division
at the first.
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30.
If, after having looked at my watch
between 4 and 5, I look again between 7 and 8,
and find that the hour and minute-hands have
then exactly changed places, it was 3612⁄13
minutes past 4 when I first looked. At that
time the hour-hand would be pointing to 231⁄13
minutes on the dial, and at 231⁄13 minutes past
7 the hour hand would be pointing to 3612⁄13
minutes.
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31.
The number consisting of 22 figures, of
which the last is 7, which is increased exactly
sevenfold if this 7 is moved to the first place, is
1,014,492,753,623,188,405,797.
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32.
The two sacks of wheat, each 4 feet long
and 3 feet in circumference, which the farmer
sent to the miller in repayment for one sack 4 feet
long and 6 feet in circumference, far from being
a satisfactory equivalent, contained but half the
quantity of the larger sack, for the area of a circle
the diameter of which is double that of another
is equal to four times the area of that other.
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33.
The five gamblers, who made the condition
that each on losing should pay to the others as
much as they then had in hand, and who each
lost in turn, and had each £32 in hand at the
finish, started with £81, £41, £21, £11, and £6
respectively.
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34.
If we know the square of any number, we
can rapidly determine the square of the next
number, without multiplication, by adding the
two numbers to the known square. Thus if
we know that the square of 87 is 7569,
then the square of 88 = 7569 + 87 + 88 = 7744;
so too the square of 89 = 7744 + 88 + 89 = 7921;
and the square of 90 = 7921 + 89 + 90 = 8100.
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35.
The two numbers which solve the problem—
Two numbers seek which make eleven,
Divide the larger by the less,
The quotient is exactly seven,
As all who find them will confess—
are 13⁄8 and 95⁄8, for 13⁄8 +
95⁄8 = 11, and 77⁄8 ÷ 11⁄8 = 7.
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36.
There must be nine things of each sort, in
order that 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
different selections may be made from twenty sorts
of things.
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37.
The women who had respectively 33, 29,
and 27 apples, and sold the same number for a
penny, receiving an equal amount of money,
began by selling at the rate of three a penny.
The first sold ten pennyworth, the second eight
pennyworth, and the third seven pennyworth.
The first had then left three apples, the second
five, and the third six. These they sold at one
penny each, so that they received on the whole—
| The first |
10d. |
+ |
3d. |
= |
13d. |
| The second |
8d. |
+ |
5d. |
= |
13d. |
| The third |
7d. |
+ |
6d. |
= |
13d. |
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38.
The puzzle—
Take five from five, oh, that is mean!
Take five from seven, and this is seen—
is solved by fie, seen.
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39.
If a bun and a half cost three halfpence,
it is plain that each bun costs a penny, but, by
general custom, you buy seven for sixpence.
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40.
The hands of a watch would meet each
other twenty-five times in a day, if the minute-hand
moved backwards and the hour-hand
forwards. They are, of course, together at
starting.
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41.
The only way in which half-a-crown can
be equally divided between two fathers and two
sons, so that a penny is the smallest coin made
use of, is to give tenpence each to a grandfather,
his son, and his grandson.
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42.
If the number of the revolutions of a
bicycle wheel in six seconds is equal to the
number of miles an hour at which it is running,
the circumference of the wheel is 84⁄5 feet.
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43.
The hour that struck was twelve o’clock.
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44.
Sixty years.
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45.
If I jump off a table with a 20lb dumb-bell
in my hand there is no pressure upon me from
its weight while I am in the air.
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46.
If at a bazaar I paid a shilling on entering
each of four tents, and another shilling on
leaving it, and spent in each tent half of what
was in my pocket, and if my fourth payment
on leaving took my last shilling, I started with
45s., spending 22s. in tent 1, 10s. in tent 2, 4s.
in tent 3, and 1s. in tent 4, having also paid to
the doorkeepers 8s.
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47.
When rain is falling vertically at 5 miles
an hour, and I am walking through it at
4 miles an hour, the rain drops will strike the
top of my umbrella at right angles if I hold it at
an angle of nearly 39 degrees.
As I walk along, meeting the rain, the effect
is the same as it would be if I was standing still,
and the wind was blowing the rain towards me
at the rate of 4 miles an hour.
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48.
When one monkey descends from the top
of a tree 100 cubits high, and makes its way to
a well 200 yards distant, while another monkey,
leaping upwards from the top, descends by the
hypotenuse to the well, both passing over an
equal space, the second monkey springs 50 cubits
into the air.
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49.
The steamboat which springs a leak 105
miles east of Tynemouth Lighthouse, and, putting
back, goes at the rate of 10 miles an hour the
first hour, but loses ground to the extent in each
succeeding hour of one-tenth of her speed in
the previous hour, never reaches the lighthouse,
but goes down 5 miles short of it.
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50.
Twenty-one hens will lay ninety-eight
eggs in a week, if a hen and a-half lays an egg
and a-half in a day and a-half. Evidently one
egg is laid in a day by a hen and a-half, that is
to say three hens lay two eggs in a day. Therefore,
twenty-one hens lay fourteen eggs, in a
day, or ninety-eight in a week.
Q. E. D. (Quite easily done!)
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51.
If the population of Bristol exceeds by
237 the number of hairs on the head of anyone
of its inhabitants that are not bald, at least 474
of them must have the same number of hairs on
their heads.
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52.
In tipping his nephew from seven different
coins, the uncle may give or retain each, thus
disposing of it in two ways, or of all in 2 × 2 × 2 ×
2 × 2 × 2 × 2 ways. But as one of these ways
would be to retain them all, there are not 128,
but only 127 possible variations of the tip.
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53.
The prime number which fulfils the various
conditions of the question is 127. Increased
by one-third, excluding fractions, it becomes
169, the square of 13. If its first two figures
are transposed, and it is increased by one-third,
it becomes 289, the square of 17. If its first
figure is put last, and it is increased by one-third,
it becomes 361, the square of 19. If, finally,
its three figures are transposed, and then increased
by one-third, it becomes 961, the square of 31.
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54.
Six things can be divided between two
boys in 62 ways. They could be carried by two
boys in 64 ways (2 × 2 × 2 × 2 × 2 × 2), but they
are not divided between two boys if all are given
to one, so that two of the 64 ways must be rejected.
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55.
The highest possible score that the dealer
can make at six cribbage, if he is allowed to select
the cards, and to determine the order of play, is
78. The dealer and his opponent must each
hold 3, 3, 4, 4, the turn-up must be a 5, and
crib must have the knave of the suit turned up,
and 5, 5, 5. It will amuse many of our readers
to test this with the cards.
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56.
The picture frame must be 3 inches in
width all round, if it is exactly to equal in area
the picture it contains, which measures 18 inches
by 12 inches.
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57.
If my mother was 20 when I was born,
my sister is two years my junior, and my brother
is four years younger still, our ages are 56, 36,
34, and 30.
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58.
The spider in the dockyard, whose thread
was drawn from her by a revolving capstan 1 foot
in diameter, until 73 feet of it were paid out,
after walking for a mile round and round the
capstan at the end of the stretched thread in
an effort to unwind it all, had, when she stopped
in her spiral course, 49 more feet to walk to
complete her task.
Back to Description
59.
The mountebank at a fair, who offered to
return any stake a hundredfold to anyone who
could turn up all the sequence in twenty throws
of dice marked each on one face only with 1, 2, 3,
4, 5, or 6, should in fairness have engaged to
return 2332 times the money; for of the 46,656
possible combinations of the faces of the dice,
only one can give the six marked faces uppermost.
Thus the chance of throwing them all at
one throw is expressed by 1⁄46656, and in twenty
throws by about 1⁄2332.
Back to Description
60.
If 90 groats (each = 4d.) feed twenty cats
for three weeks, and five cats consume as much
as three dogs, seventy-two hounds can be fed
for £39 in a period of ninety-one days.
Back to Description
61.
When equal wine-glasses, a half and a third
full of wine, are filled up with water, and their
contents are mixed, and one wine-glass is filled
with the mixture, it contains 5⁄12 wine and 7⁄12 water.
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62.
The arrangement by which St Peter is
said to have secured safety for the fifteen Christians,
when half of the vessel’s passengers were
thrown overboard in a storm, is as follows:—
XXXXIIIIIXXIXXXIXIIXXIIIXIIXXI
Each Christian is represented by an X, and
if every ninth man is taken until fifteen have
been selected, no X becomes a victim.
Back to Description
63.
If Farmer Southdown’s cow had a fine
calf every year, and each of these, and their
calves in their turn, at two years old followed
this example, the result would be no less than
2584 head in sixteen years.
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64.
The number of the flock was 301. This
is found by first taking the least common multiple
of 2, 3, 4, 5, 6, which is 60, and then finding the
lowest multiple of this, which with 1 added is
divisible by 7. This 301 is exactly divisible by
7, but by the smaller numbers there is 1 as remainder.
Back to Description
65.
The rule for determining easily the number
of round bullets in a flat pyramid, with a base
line of any length, is this:—
Add a half to half the number on the base
line, and multiply the result by the number on
that line. Thus, if there are twelve bullets as
a foundation—
12 + 1⁄2 = 13⁄2; and 13⁄2 ×
12⁄1 = 78.
The same result is reached by multiplying the
number on the base line by a number larger by
one, and then halving the result. Thus—
12 × 13 = 156, 156 ÷ 2 = 78.
Back to Description
66.
We can gather from the lines—
Old General Host
A battle lost,
And reckoned on a hissing,
When he saw plain
What men were slain,
And prisoners, and missing.
To his dismay
He learned next day
What havoc war had wrought;
He had, at most,
But half his host
Plus ten times three, six, ought.
One-eighth were lain
On beds of pain,
With hundreds six beside;
One-fifth were dead,
Captives, or fled,
Lost in grim warfare’s tide.
Now, if you can,
Tell me, my man,
What troops the general numbered,
When on that night
Before the fight
The deadly cannon slumbered?
that old General Host had an army 24,000 strong.
Back to Description
67.
When the farmer sent five pieces of chain
of 3 links each, to be made into one continuous
length, agreeing to pay a penny for each link cut,
and a penny for each link joined, the blacksmith,
if he worked in the best interest of the farmer,
could only charge sixpence: for he could cut
asunder one set of 3 links, and use these three
single links between the other four sets.
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68.
If, in a parcel of old silver and copper coins,
each silver piece is worth as many pence as there
are copper coins, and each copper coin is worth
as many pence as there are silver coins, there
are eighteen silver and six copper coins, when the
whole parcel is worth eighteen shillings.
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69.
These are five groups that can be arranged
with the numbers 1 to 11 inclusive, so
that they are all equal:—
(82 - 52 + 1) = (112 - 92) = (72 -
32) = (62 + 22) = 4(10).
Back to Description
70.
John Bull, under the conditions given,
lived to the age of eighty-four years.
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71.
The two numbers to each of which, or to
the halves of which, unity is added, forming in
every case a square number, are 48 and 1680.
Back to Description
72.
The true weight of a cheese that seemed
to weigh 16 ℔s. in one scale of a balance with
arms of unequal length, and only 9℔s. in the
other, is 12℔. This is found by multiplying
the 16 by the 9, and finding the square root of
the result.
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73.
The two parts into which 100 can be
divided, so that if one of them is divided by the
other the quotient is again exactly 100 are 991⁄101
and 100⁄101.
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74.
If, with marbles in two pockets, I add one
to those in that on the right, and then multiply
its contents by the number it held at first, and
after dealing in a similar way with those on the
left, find the difference between the two results
to be 90; while if I multiply the sum of the two
original quantities by the square of their difference
the result is 176, I started with twenty-three in
the right-hand pocket and twenty-one in the
other.
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75.
The circle of twenty-one friends who arranged
to meet each week five at a time for Bridge
so long as exactly the same party did not meet
more than once, and who wished to hire a central
room for this purpose, would need it for no less
than 20,349 weeks, or more than 390 years, to
carry out their plan.
Back to Description
76.
If a herring and a half costs (not cost) a
penny and a half, the price of a dozen such quantities
is eighteenpence.
Back to Description
77.
The sum of money which in a sense appears
to be the double of itself is 1s. 10d., for we may
write it one and ten pence or two and twenty pence.
Back to Description
78.
The “comic arithmetic” question set by
Dr Bulbous Roots—
Divide my fifth by my first, and you have my
fourth; subtract my first from my fifth, and you
have my second; multiply my first by my fourth
followed by my second, and you have my third;
place my second after my first, and you have
my third multiplied by my fourth—is
solved by COMIC.
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79.
If the earth could stand still, and a straight
tunnel could be bored through it, a cannon ball
dropped into it, if there is no air or other source
of friction, would oscillate continually from end
to end.
Taking air into account, the ball would fall
short of the opposite end at its first lap, and in
succeeding laps its path would become shorter
and shorter, until its initial energy was exhausted,
when it would come to rest at the centre.
Back to Description
80.
He sent 163. She sent 157.
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81.
When twins were born the estate was
properly divided thus:—
| Taking the daughter’s share as |
1 |
|
| The widow’s share would be |
2 |
| And the son’s share |
4 |
| Total |
7 |
shares. |
So the son takes four-sevenths, the widow two-sevenths,
and the daughter one-seventh of the
estate.
Back to Description
82.
If each of my strides forwards or backwards
across a 22 feet carpet is 2 feet, and I make
a stride every second; and if I take three strides
forwards and two backwards until I cross the
carpet, I reach the end of it in forty-three seconds.
In three steps I advance 6 feet. Then in two
steps I retrace 4 feet, thus gaining only 2 feet
in five steps, i.e., in five seconds. I therefore
advance 16 feet in forty seconds, and three more
strides cover the remaining 6 feet.
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83.
If the captain of a vessel chartered to sail
from Lisbon to New York, which appear on a
map of the world to be on the same parallel
of latitude, and which are, along the parallel,
about 3600 miles apart, takes his ship along this
parallel, he will not be doing his best for the
impatient merchant who has had an urgent
business call to New York.
The shortest course between the two points
is traced by a segment of a “great circle,” having
its centre at the centre of the earth, and touching
the two points. This segment lies wholly north
of the parallel, and is the shortest possible course.
Back to Description
84.
When John and Harry, starting from the
right angle of a triangular field, run along its
sides, and meet first in the middle of the opposite
side, and again 32 yards from their starting point,
if John’s speed is to Harry’s as 13 to 11, the
sides of the field measure 384 yards.
Back to Description
85.
If two sorts of wine when mixed in a flagon
in equal parts cost 15d., but when mixed so that
there are two parts of A to three of B cost 14d.,
a flagon of A would cost 20d., and a flagon of B
10d.
Back to Description
86.
If, when a man met a beggar, he gave him
half of his loose cash and a shilling, and meeting
another gave him half what was left and two
shillings, and to a third half the remainder and
three shillings, he had two guineas at first.
Back to Description
87.
The clerk who has two offers of work from
January 1, one from A of £100 a year, with an
annual rise of £20, and the other from B of £100
a year, with a half-yearly rise of £5, should
accept B’s offer.
The half-yearly payments from A (allowing for
the rise), would be 50, 50, 60, 60, 70, 70, etc., etc.;
and from B they would be 50, 55, 60, 65, 70, 75,
etc., etc., so that B’s offer is worth £5 a year more
than A’s always.
Back to Description
88.
If I have a number of florins and half-crowns,
but no other coins, I can pay my tailor
£11, 10s. in 224 different ways.
This can be found thus by rule of thumb:
Start with 0 half-crowns and 115 florins. Then
4 half-crowns and 110 florins. Add 4 half-crowns
and deduct 5 florins each time till 92
half-crowns and 0 florins is reached.
Back to Description
89.
The monkey climbing a greased pole,
60 feet high, who ascended 3 feet, and slipped
back 2 feet in alternate seconds, reached the
top in 1 minute, 55 seconds, for he did not slip
back from the top.
Back to Description
90.
When Adze, the carpenter, secured his
tool-chest with a puzzle lock of six revolving
rings, each engraved with twelve different letters,
the chances against any one discovering the
secret word formed by a letter on each ring was
2,985,983 to 1; for the seventy-two letters may
be placed in 2,985,984 different arrangements,
only one of which is the key.
Back to Description
91.
The five married couples who arranged
to dine together in Switzerland at a round table,
with the ladies always in the same places, so
long as the men could seat themselves each
between two ladies, but never next to his own
wife, were able under these conditions to enjoy
thirteen of these nights at the round table.
Back to Description
92.
If in a calm the tip of a rush is 9 inches
above the surface of a lake, and as the wind
rises it is gradually blown aslant, until at the
distance of a yard it is submerged, it is growing
in water that is 5 feet 71⁄2 inches deep.
Back to Description
93.
Aminta was eighteen.
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94.
When Dick took a quarter of the bag of
nuts, and gave the one over to the parrot, and
Tom and Jack and Harry dealt in the same way
with the remainders in their turns, each finding
a nut over from the reduced shares for the bird,
and one was again over when they divided the final
remainder equally, there were, at the lowest
estimate, 1021 nuts in the bag.
Back to Description
95.
Eight and a quarter is the answer to the
nonsense question—
If five times four are thirty-three,
What will the fourth of twenty be?
Back to Description
96.
The similar fraction of a pound, a shilling,
and a penny which make up exactly a pound
are as follows:—
| |
s. |
d. |
| 240⁄253 of £1 = |
|
18 |
11 |
169⁄253 |
| 240⁄253 of 1s. = |
|
11 |
97⁄253 |
| 240⁄253 of 1d. = |
|
240⁄253 |
| |
£1 |
0 |
0 |
|
Back to Description
97.
When Dr Tripos thought of a number,
added 3, divided by 2, added 8, multiplied by 2,
subtracted 2, and thus arrived at double the
number, he started with 17.
Back to Description
98.
When A and B deposited equal stakes with
C, and agreed that the one who should first win
three games of billiards should take all, but
consented to a division in proper shares when
A had won two games and B one, it was evident
that if A won the next game all would go to him,
while if he lost he would be entitled to one half.
One case was as probable as the other, therefore
he was entitled to half of these sums taken together;
that is, to three quarters of the stakes,
and B to a quarter only.
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99.
The average speed of a motor which runs
over any course at 10 miles an hour, and returns
over the same course at 15 miles an hour, is
12 miles an hour, and not 121⁄2, as might be
imagined. Thus a run of 60 miles out takes,
under the conditions, six hours, and the return
takes four hours; so that the double journey
of 120 miles is done in ten hours, at an average
speed of 12 miles an hour.
Back to Description
100.
Farmer Hodge, who proposed to divide
his sheep into two unequal parts, so that the
larger part added to the square of the smaller
part should equal the smaller part added to the
square of the larger part, had but one sheep.
Faithful to his word, he divided this sheep into
two unequal parts, 2⁄3 and 1⁄3, and was able to
show that 2⁄3 + 1⁄9 = 7⁄9,
and that 1⁄3 + 4⁄9 = 7⁄9. He was
heard to declare further, and he was absolutely
right, that no number larger than 1 can be so
divided as to satisfy the conditions which he had
laid down.
The fact that sheep is both singular and plural,
adds much to the perplexing points of this attractive
problem.
Here is a very simple proof that the number
must be 1:—
| Let |
a + b |
= |
no. of sheep |
| then |
a2 + b |
= |
b2 + a |
| |
a2 - b2 |
= |
a - b |
| or |
(a + b)(a - b) |
= |
a - b |
| therefore |
a + b |
= |
1. |
Back to Description
101.
A horse that carries a load can draw a
greater weight up the shaft of a mine than a horse
that bears no burden. The load holds him
more firmly to the ground, and thus gives him
greater power over the weight he is raising from
below.
Back to Description
102.
In the six chests, of which two contained
pence, two shillings, and two pounds, there must
have been at least the value of 506 pence. This
can be divided into 22 (or 19 + 3) shares of 23d.
each, or 23 (19 + 4) shares of 22d. each. Evidently
then the treasure can be divided so that
19 men have equal shares, while their captain
has either 3 shares or 4 shares.
Back to Description
103.
If I bought a parcel of nuts at 49 for 2d.,
and divided it into two equal parts, one of which
I sold at 24, the other at 25 a penny; and if I
spent and received an integral number of pence,
but bought the least possible number of nuts,
I bought 58,800 nuts, at a cost of £10, and I
gained a penny.
Back to Description
104.
When, with a purse containing sovereigns
and shillings, after spending half of its contents,
I found as many pounds left as I had shillings
at first, I started with £13, 6s.
Back to Description
105.
When the lady replied to a question as
to her age—
If first my age is multiplied by three,
And then of that two-sevenths tripled be,
The square root of two-ninths of this is four;
Now tell my age, or never see me more—
she was 28 years old.
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106.
If cars run, at uniform speed, from
Shepherd’s Bush to the Bank, at intervals of
two minutes, and I am travelling at the same
rate in the opposite direction, I shall meet 30
in half-an-hour, for there are already 15 on the
track approaching me, and 15 are started from
the other end during my half hour’s course.
Back to Description
107.
If it was possible to carry out my offer
of a farthing for every different group of apples
which my greengrocer could select from a basket
of 100 apples, he would be entitled to the stupendous
sum of £18,031,572,350 19s. 2d.
Back to Description
108.
If the minute-hand of a clock moves
round between 3 and 4 in the opposite direction
to the hour-hand, the hands will be exactly together
when it is really 417⁄13 minutes past 3.
Back to Description
109.
If the walnut monkey had stopped to help
the other, and they had eaten filberts at equal
rates, they would have escaped in 21⁄4 minutes.
Back to Description
110.
The value of the cheque, for which the
cashier paid by mistake pounds for shillings,
was £5, 11s. 6d. The receiver to whom £11, 5s. 6d.
was handed, spent half-a-crown, and then found
that he had left £11, 3s., just twice the amount
of the original cheque.
Back to Description
111.
The number 14 can be made up by adding
together five uneven figures thus:—11 + 1 + 1 + 1.
It will be seen that although only four numbers
are used, 11 is made up of two figures.
Here is another, and quite a curious solution,
1 + 1 + 1 + 1 = 4, and with another 1 we can make
up 14!
Back to Description
112.
A business manager can fill up three
vacant posts of varying value from seven applicants
in 210 different ways. For the first
post there would be a choice among 7, for the
second among 6, and for the third among 5, so
that the possible variations would amount to
7 × 6 × 5 = 210.
Back to Description
113.
If the fasting man, who began his task
at noon, said it is now 5⁄11 of the time to midnight,
he spoke at 3.45 p.m., meaning that 5⁄11
of the remaining time till midnight had elapsed
since noon.
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114.
If a clock takes six seconds to strike 6,
it will take 12 seconds to strike 11, for there must
be ten intervals of 11⁄5 seconds each.
Back to Description
115.
Twenty horses can be arranged in three
stalls, so that there is an odd number in each,
by placing one in the first stall, three in the second,
and sixteen (an odd number to put into any stall!)
in the third.
Back to Description
116.
The little problem, “Given a, b, c, to
find q,” is solved, without recourse to algebra,
thus: a, b, c, = c, a, b; take a cab and go
over Kew Bridge, and you find a phonetic Q!
Back to Description
117.
Tom Evergreen was 75 years old when he
was asked his age by some men at his club in 1875,
and said—“The number of months that I have
lived are exactly half as many as the number
which denotes the year in which I was born.”
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118.
Eight different circles can be drawn. A
circle can have one of the three inside and two
outside in three ways, or one outside and three
inside in three ways (each of the three being
inside or outside in turn), or all three may be
inside, or all three may be outside, the touching
circle.
Back to Description
119.
The way to arrange 1, 2, 3, 4, 5, 6, 7, 8, 9, 0,
so that used once each they form a sum which
is equal to 1 is this:—
3570 +
148296 = 1
Back to Description
120.
The sum of the first fifty numbers may be
found without any addition thus:—The first
fifty numbers form twenty-five pairs of fifty-one
each (1 + 50, 2 + 49, etc., etc.), and 51 × 25 is
practically 51 × 100 ÷ 4 = 1275.
Back to Description
121.
The tramcar A, which started at the same
time as B, but ran into a “lie by” in four minutes,
and waited there five minutes till B came along,
when they completed their courses at the same
moment in opposite directions, could have run
the whole distance in ten minutes.
Back to Description
122.
What remains will be 8 if we take 10 and
double it by writing one 10 over another so as
to form 18, and then deduct 10.
Back to Description
123.
If the average weight of the Oxford
crew is increased by 2℔s., when one of them
who weighs 12 stone, is replaced by a fresh man,
the weight of that substitute is 13 stone 2℔s.
Back to Description
124.
If a motor-car is twice as old as its tyres
were when it was old as its tyres are, and if, when
these tyres are as old as the car itself is now,
their united ages will be 21⁄4 years, the car is now
12 months old, and the tyres have had 9 months’
wear.
Back to Description
125.
A and B, who could each carry provisions
for himself for twelve days, started to penetrate
as far as possible into a desert, on the understanding
that neither of them should miss a day’s food.
After an advance of four days, each had provisions
still for eight days. One gave four
portions of his store to his companion, which
did not overload him, and returned with the
other four. His comrade was then able to
advance another four days’ journey, and still
have rations for the eight days’ return. Thus
the furthest possible penetration into the desert
under the conditions was an eight days’ march.
Back to Description
126.
If, when a bottle of medicine and its cork
cost half-a-crown, the bottle and the medicine
cost two and a penny more than the cork, the
cork cost twopence half-penny.
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127.
A boat’s crew far from land, with no sail
or oars, and with no assistance from wind or
stream, or outside help of any kind, can regain
the shore by means of a coil of rope. Motion
is given to the boat by tying one end of the rope
to the after thwart, and giving the other end a
series of violent jerks in a direction parallel to
the keel. This curious illustration of mechanical
principles is from “Ball’s Mechanical Recreations.”
(Macmillan.)
Back to Description
128.
It will be found that after a crown and as
many four-shilling pieces as possible have been
crammed into our pockets, there would still be
room for one sixpence and one threepenny-piece
in some corner or cranny. We can, therefore,
have one crown, one sixpence, one threepenny-piece,
and as many four-shilling pieces as our
pockets will hold, and yet be unable to give
change for a half-sovereign.
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129.
There were fifteen apples in the basket.
Half of these and half an apple, i.e., eight were
first given, then half the remainder and half an
apple, i.e., four, then on similar lines two, leaving
one in the basket.
Back to Description
130.
The Queer Division—
A third of twelve divide
By just a fifth of seven;
And you will soon decide
That this must give eleven—
is solved by LV ÷ V, or 55 ÷ 5 = 11.
Back to Description
131.
A motor that goes 9 miles an hour uphill,
18 miles an hour downhill, and 12 miles an hour
on the level, will take 8 hours and 20 minutes to
run 50 miles out and return at once over the
same course.
Back to Description
132.
The number of shots fired at a mark was
420 each by A, B, and C. A made 280 hits, B 315,
and C 336.
Back to Description
133.
If a dog and a cat, evenly matched in
speed, run a race out and back over a course of
75 yards in all, and the dog always takes 5 feet
at a bound, and the cat 3 feet, the cat will win,
because at the turning point the dog overleaps
the half distance more than the cat does, and so
has a longer run in.
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134.
When a man caught up a wagon going
at 3 miles an hour, which was just visible to him
in a fog at a distance of 55 yards, and which he
saw for five minutes before reaching it, he was
walking at the rate of 33⁄8 miles an hour.
Back to Description
135.
Three horses, A B C, can be placed after
a race in thirteen different ways, thus:—A B C,
A C B, B A C, B C A, C A B, C B A, or A B C as
a dead heat; or A B, A C, or B C equal for the
first place; or A first with B C equal seconds;
or B first with A C equal seconds; or C first with
A B equal seconds.
Back to Description
136.
The 34 points scored against Oxbridge
by the New Zealanders can be made up in two
ways, either by 8 tries and 2 converted tries, or
by 3 tries and 5 converted tries.
The highest possible score on these lines is 10
tries converted, equalling 50 points, and as the
New Zealanders’ score, if all tries are converted,
becomes four-fifths of this, their actual score
was 3 tries and 5 converted into goals.
Back to Description
137.
The smallest number, of which the
alternate figures are cyphers, which is divisible
by 9 and by 11 is 909090909090909090909!
Back to Description
138.
Our problem in which it is stated that A
with 8d. met B and C with five and three loaves,
and asked how the cash should be divided between
B and C, if all agreed to share the loaves.
Now each eats two loaves and two-thirds of a
loaf, and B gives seven-thirds of a loaf to A,
while C gives him one-third of a loaf. So B
receives 7d. and C 1d.
Back to Description
139.
When, on opening four money-boxes containing
pennies only, it was found that those in the
first with half of all the rest, those in the second
with a third of the others, those in the third with a
fourth, and those in the fourth with a fifth of all
the rest, amounted in each case to 740, the four
boxes held £6. 1s. 8d., and the numbers of pennies
were 20, 380, 500, and 560.
Back to Description
140.
If two steamers, A and B, start together
for a trip to a distant buoy and back, and A
steams all the time at ten knots an hour, while
B goes outward at eight knots and returns at
twelve knots an hour, B will regain port later
than A, because its loss on the outward course
will not have been recovered on the run home.
Back to Description
141.
If in London a new head to a golf club
costs four times as much as a new leather face,
while at St Andrews it costs five times as much,
and if the leather face costs twice as much in
London as in St Andrews, and if, including a
shilling paid for a ball, the charges in London
were twice as much as they would have been
at St Andrews, the London cost of a new head
is four shillings, and of a leather face a shilling.
Back to Description
142.
When two children were asked to give
the total number of sheep and cattle in a pasture,
from the number of each sort, and one by subtraction
answered 10, while the other arrived at
11,900 by multiplication, the true numbers were
170 sheep, 70 cattle, 240 in all.
Back to Description
143.
If a man picks up one by one fifty-two
stones, placed at such intervals on a straight road
that the second is a yard from the first, the third
3 yards from the second, and so on with intervals
increasing each time by 2 yards, and bring them
all to a basket placed at the first stone, he has to
travel about 52 miles, or, to be quite exact, 51
miles, 1292 yards.
Back to Description
144.
When, in the House of Commons, if the
Ayes had been increased by 50 from the Noes,
the motion would have been carried by 5 to 3;
and if the Noes had taken 60 votes from the Ayes
it would have been lost by 4 to 3, the motion
succeeded; 300 voted “Aye,” and 260 “No.”
Back to Description
145.
There are 143 positions on the face of a
watch in which the places of the hour and minute-hands
can be interchanged, and still indicate
a possible time. There would be 144 such
positions but for the fact that at twelve o’clock
the hands occupy the same place.
Back to Description
146.
If in a cricket match the scores in each
successive innings are a quarter less than in the
preceding innings, and the side which goes in
first wins by 50 runs, the complete scores of the
winners are 128 and 72, and of the losers 96 and
54.
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147.
When a ball is thrown vertically upwards,
and caught five seconds later, it has risen 100 feet.
It takes the same time to rise as to fall, and when
a body falls from rest, it travels a number of feet
represented by sixteen times the square of the
time in seconds.
Hence comes the rule that the height in feet
of a vertical throw is found by squaring the time
in seconds of its flight, and multiplying by four.
Back to Description
148.
The carpet which, had it been 5 feet
broader and 4 feet longer, would have contained
116 more feet, and if 4 feet broader and 5 longer
113 more, was 12 feet long and 9 feet broad.
Back to Description
149.
When, in estimating the cost of a hundred
similar articles, shillings were read as pounds,
and pence as shillings, and the estimated cost
was in consequence £212, 18s. 4d. in excess of
the real cost, the true cost of each article was
2s. 5d.
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150.
If the square of the number of my house
is equal to the difference of the squares of the
numbers of my next door neighbours’ houses,
and if my brother in the next street can say
the same of his house, though its number is not
the same as that of mine, our houses are numbered
8 and 4. In my street the even numbers
are all on one side, in my brother’s street they,
run odd and even consecutively, and so 82 = 102
- 62, and 42 = 52 - 32.
Back to Description
151.
When two men of unequal strength have
to move a block which weighs 270 ℔s., on a light
plank 6 feet long, if the stronger man can carry
180 ℔s., the block must be placed 2 feet from
him, so that he may have that share of the load.
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152.
If a man who had twenty coins, some
shillings, and the rest half-crowns, were to
change the half-crowns for sixpences, and the
shillings for pence, and then found that he had
156 coins, he must have had eight shillings at
first.
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153.
If, when coins are placed on a table at
equal distances apart, so as to form sides of an
equilateral triangle, and when as many are taken
from the middle of each side as equal the square
root of the number on that side, and placed on
the opposite corner, the number on each side is
then to the original number as five is to four,
there are forty-five coins in all.
Back to Description
154.
When the gardener found that he would
have 150 too few if he set his posts a foot apart,
and seventy to spare if he set them at every
yard, he had 180 posts.
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155.
In order to buy with £100 a hundred
animals, cows at £5, sheep at £1, and geese at
1s. each, the purchaser must secure nineteen
cows, one sheep, and eighty geese.
Back to Description
156.
If John, who is 21, is twice as old as
Mary was when he was as old as Mary is, Mary’s
age now is 153⁄4 years.
Back to Description
157.
If in a cricket match A makes 35 runs,
and C and D make respectively half and a third
of B’s score, and if B scores as many less than
A as C scores more than D, B made 30, C 15,
and D 10 runs.
Back to Description
158.
The least number which, divided by
2, 3, 4, 5, 6, 7, 8, 9, or 10, leaves remainders
1, 2, 3, 4, 5, 6, 7, 8, 9, is 2519, their least common
multiple less 1.
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159.
A square table standing on four legs,
which are set at the middle points of its sides,
can at most uphold its own weight upon one
of its corners.
Back to Description
160.
The division of ninety-nine pennies, so
that share 1 exceeds share 2 by 3, is less than
share 3 by 10, exceeds share 4 by 9, and is less
than share 5 by 16, is 17, 14, 27, 8, and 33.
Back to Description
161.
If Indians carried off a third of a flock
and a third of a sheep, and others took a fourth
of the remainder and a fourth of a sheep, and
others a fifth of the rest and three-fifths of a
sheep, and there were then 409 left, the full flock
was 1025 sheep.
Back to Description
162.
When a cistern which held fifty-three
gallons was filled by three boys, A bringing a pint
every three minutes, B a quart every five minutes,
and C a gallon every seven minutes, it took
230 minutes to fill it, and B poured in the final
quart, A and C coming up one minute too late to
contribute at the last.
Back to Description
163.
A man who said, late in the last century,
that his age then was the square root of the
year in which he was born, was speaking in the
year 1892.
Back to Description
164.
If when a dealer in curios sold a vase
for £119, his profit per cent., and the cost price
of the vase, were expressed by the same number,
it had cost him £70.
Back to Description
165.
The chance of throwing at least one ace
in a single throw with a pair of dice is 11⁄36, for there
are five ways in which each dice can be thrown
so as not to give an ace, so that twenty-five
possible throws exclude aces. Hence the chance
of not throwing an ace is 25⁄36, which leaves 11⁄36 in
favour of it.
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166.
The policeman who ran after a thief
starting four minutes later, and running one-third
faster, if they both ran straight along the
road, caught him in twelve minutes.
Back to Description
167.
At a bazaar stall, where twenty-seven
articles are exposed for sale, a purchaser may
buy one thing or more, and the number of choices
open to him is one less than the continued product
of twenty-seven twos, or 134217727.
Back to Description
VERY PERSONAL
168. When Nellie’s father said:—
I was twice as old as you are
The day that you were born.
You will be just what I was then
When fourteen years are gone—
he was 42, and she was 14.
Back to Description
1.
The word square is completed thus:—
| M |
E |
A |
D |
| E |
D |
G |
E |
| A |
G |
U |
E |
| D |
E |
E |
D |
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2.
The word square filled in is:—
| C |
I |
R |
C |
L |
E |
| I |
N |
U |
R |
E |
S |
| R |
U |
D |
E |
S |
T |
| C |
R |
E |
A |
S |
E |
| L |
E |
S |
S |
E |
E |
| E |
S |
T |
E |
E |
M |
Notice the curious diagonal of E’s.
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3.
In the incomplete sentence,
SO * * * * AG * * * * LATI * * * * X * * * * ITH
the duplicate letters are filled in thus:—
SOME MEAGRE RELATIVE VEXED EDITH
The two last letters of each word are repeated
as the two first of the word that follows.
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4.
The word square is completed thus:—
| W |
A |
S |
T |
E |
| A |
C |
T |
O |
R |
| S |
T |
O |
N |
E |
| T |
O |
N |
I |
C |
| E |
R |
E |
C |
T |
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5.
By twice building up two Ds into a B we
make BULBOUS.
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6.
The question put on paper to the love-lorn
youth, “Loruve?” is, when interpreted, “Are
you in love?” and the advice given
to him on another slip, “Prove
L
A FD and
ensure success,”
reads into, “Prove a fond lover, and
ensure success” (a f on d l over).
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7. DOUBLE ACROSTIC
ST. JAMES’ GAZETTE
| S |
prin |
G |
| T |
ar |
A |
| J |
abe |
Z |
| A |
t |
E |
| M |
omen |
T |
| E |
mme |
T |
| S |
trik |
E |
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8.
The completed word square is—
| A |
M |
E |
N |
D |
S |
| M |
I |
N |
I |
O |
N |
| E |
N |
A |
B |
L |
E |
| N |
I |
B |
B |
L |
E |
| D |
O |
L |
L |
A |
R |
| S |
N |
E |
E |
R |
S |
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9.
The oracular response to a young Frenchman
at a fête, who inquired how he could best
please the ladies—
MEC DO BIC
conceals this sage advice—
Aimez, cédez, obéissez!
Love, yield, obey!
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10.
The solution to our Letter Fraction
Problem is of a verbal character. The original statement
m
ot
y = mo
is dealt with thus:—
m on ot on y = mo not on y, and so the word
monotony solves the equation.
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11.
The buried beasts are chamois, buffalo,
heifer, and leopard; and when the Oxford athlete
cries—
“Though I jump and row and run,
Cap or cup I never won”
he introduces us to a porcupine.
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12.
The lines—
Fourteen letters here we fix,
Vowels only two are spoken;
All together these we mix
Into what can not be broken—
is solved by indivisibility, which has many an i,
like a peacock’s tail.
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13.
The English word of thirteen letters in
which the same vowel occurs four times, the same
consonant six times, another twice, and another
once, is Senselessness.
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14.
The condensed proverb “WE IS DO”
reads at its full length as “Well begun is half
done.”
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15.
This is the completed word square:—
| W |
A |
S |
H |
E |
S |
| A |
R |
T |
E |
R |
Y |
| S |
T |
O |
R |
M |
S |
| H |
E |
R |
M |
I |
T |
| E |
R |
M |
I |
N |
E |
| S |
Y |
S |
T |
E |
M |
Back to Puzzle
16.
Dr Whewell’s puzzle lines—
| |
O |
O |
N |
O |
O. |
|
| U |
O |
A |
O |
O |
I |
O |
U |
|
| O |
N |
O |
O |
O |
O |
M |
E |
T |
O |
O. |
| U |
O |
A |
O |
I |
D |
O |
S |
O |
|
| I |
O |
N |
O |
O |
I |
O |
U |
T |
O |
O! |
read thus:—
OH SIGH FOR NO CIPHER
You sigh for a cipher, O, I sigh for you,
Sigh for no cipher, O sigh for me too.
You sigh for a cipher, I decipher so,
I sigh for no cipher, I sigh for you too!
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17.
This is the completed diamond:—
| |
P |
|
| |
P |
O |
R |
|
| |
C |
O |
R |
E |
S |
|
| |
F |
O |
R |
C |
E |
P |
S |
|
| P |
O |
R |
C |
E |
L |
A |
I |
N |
| |
R |
E |
F |
L |
E |
C |
T |
|
| |
S |
P |
A |
C |
E |
|
| |
S |
I |
T |
|
| |
N |
|
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18.
The medley—
Tan HE Edsa VEN in
It N Gja SmeTs AsgN
aD Az Rett De
is read by taking first the capitals in their order,
and then the small type. It comes put out as “The
Evening Standard and Saint James’s Gazette.”
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19.
The statement, I can travel first-class on
the G.E.R. from 2222222244444500, reads into—from
22 to 2 to 22 to 4 for 44 4d; or, in plain
terms, from 1.38 to 3.38 for 14s. 8d. This works
out at about 3d. a mile, the usual allowance for
first-class, for two hours, at about 29 miles an
hour.
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20. A CURIOUS OLD INSCRIPTION
Read the inscription backwards, and it resolves
itself into the lines familiar to us in our
childhood:—
Ride a cock horse to Banbury Cross
To see a fine lady ride on a grey horse.
Rings on her fingers and bells on her toes,
She shall have music wherever she goes!
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21.
The reason why the invitation to Jack to
sample the Irish stew at Simpson’s was to be
kept in mind by the catch words—
Join me at and
Join me at ai
Join me at as
is because if you join me and at, and note that
and is on i, which is on s, you arrive at the suggestive
sentence—meat and onions.
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22.
The labourer’s quaint letter, which ran
“Cepatomtogoatatrin,” was, in plainer English,
“Kept at home to go a tatering.”
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23.
Our double acrostic comes out thus:—
Problem—Puzzles
| P |
a |
P |
| (T) R |
|
U (E) |
| O |
|
Z |
| B |
o |
Z |
| L |
eve |
L |
| E |
v |
E |
| M |
|
S |
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24.
The Hidden Proverb is—
“Necessity is the mother of invention.”
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25.
The “deed done” in our Will puzzle
is the making in Roman numerals of “Codicil.”
The lawyer was to set down, a hundred, to add
nothing, to set down five hundred, then one,
then another hundred, and then one more, and,
finally, fifty, and accordingly he wrote upon the
parchment the one word CODICIL.
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26.
The word square is—
| E |
D |
I |
T |
O |
R |
| D |
E |
S |
I |
R |
E |
| I |
S |
L |
A |
N |
D |
| T |
I |
A |
R |
A |
S |
| O |
R |
N |
A |
T |
E |
| R |
E |
D |
S |
E |
A |
Back to Puzzle
27.
The notes of music A. G. A. E. A. over
the grave of a French musician, who was choked
by a fish bone, are in the French notation, “La
sol la mi la,” which reads into “La sole l’a mis 1à.”
Similarly the inscription over the porch of
Gustave Doré’s house C. E. B. A. C. D. is equivalent
to “Do, mi, si, la, do, re,” which may be
taken to represent “Domicile à Doré.”
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28.
When his best girl said to Jack Spooner,
“We can go to-morrow at 222222222222 LEY
STREET,” he understood her to mean, “We
can go to-morrow, at two minutes to two, to
two twenty-two, to 222 Tooley Street.”
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29
CT T T T T T T T T T
spells contents (c on ten ts!).
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30.
We can treat the word disused so as to
affirm or to disallow the use of its initial or final
d, for we can write it d is used, or disuse d!
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31.
The title of the book shaken up into
E I O O O U
B C N N R R S S
is “Robinson Crusoe.”
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32.
If the letter M only is inserted in the proper
places in the line—
A DEN I I CAN DOCK
it will read: Madmen mimic and mock.
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33.
The quotation from Shakespeare—
OXXU8 MAAULGIHCTE
NOR
is by interpretation:—“Nothing extenuate, nor
set down aught in malice.”
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34.
The phonetic nightmare—
Ieukngheaurrhphthewempeighghteaps—
is merely the word unfortunates. It can be
justified thus by English spelling of similar sound
taken letter by letter:—
u—iew in view; n—kn in know; f—gh in
tough; o—eau in beau; r—rrh in myrrh; t—phth
in phthisis; u—ewe; n—mp in comptroller;
a—eigh in neigh; t—ght in light; e—ea in tea;
and s—ps in psalm.
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35.
The word square is completed thus:—
| F |
A |
R |
M |
| A |
R |
E |
A |
| R |
E |
N |
T |
| M |
A |
T |
E |
Back to Puzzle
36. A QUAINT INSCRIPTION
The millers leave the mill,
The wherrymen lower their sail;
The maltsters leave the kiln,
For a drop of the White Swan’s ale.
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37. A POET’S PI
TONDEBNIOTOCHUMFOARYHUR
OTDIRECTTHAWHOTERSOFKLSYA;
TIKATESTUBALIGHTSTILLETRUFLYR
OTBOWLALLNFESLEAVARFWYAA
is disentangled thus:—
Don’t be in too much of a hurry
To credit what other folks say;
It takes but a slight little flurry
To blow fallen leaves far away.
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38. BURIED PLACES
The five buried places are Deal, London, Esk,
Perth, and Baden. The word is Ourangoutang.
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39.
It is no offence to conspire in the evening,
because what is treasonable is reasonable after t!
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40.
The bit of botany—
Inscribe an m above a line,
And write an e below,
This woodland flower is hung so fine
It bends when zephyrs blow—
is solved by,
me,
an em on e, the delicately hung
wind-flower.
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41. A PIED PROVERB
The pied proverb, is “A rolling stone gathers
no moss.”
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42.
The Drop Letter Proverb—
E..t. .e.s..s .a.e .h. .o..
..i.e, is—Empty vessels make the most
noise.
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43.
The English word of five syllables, which
has eight letters, five of them vowels—namely
an a, an e, twice i, and y—is Ideality.
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44.
TORMENT may be turned into RAPTURE,
using four links, changing only one letter
each time, and varying the order of the letters,
thus: TORMENT, portent, protest, pratest,
praters, RAPTURE.
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45.
The pied sentence—
a a c e e e f f h h i i i i i m n n o o o p r r s s t t t t t
can be cast into the proverb—
“Procrastination is the thief of time.”
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46.
The English sentence, when the letter o is
added, reads:—
“Good old port for orthodox Oxford dons.”
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47
One vowel in an English word is found,
Which by eight consonants is hedged around—
is solved by Strengths.
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48.
The letters AAAAABBNNIIRSSTT form
the word Antisabbatarians.
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49.
The quotation from Shakespeare,
KINI
stands for “A little more than kin, and less than
kind.”
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50.
The two English words which have the
first six letters of the alphabet among their ten
letters are fabricated and bifurcated.
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51. SHIFTING NUMBERS
The letter A stands at the head of the letters
of the alphabet. For bed 3 of these are used;
for goal, 4; for prison, 6; for six, 3; for three, 5.
The letter A is not used in the spelling of the
name of any number from 1 to 100, but it makes
up, with the other vowels, the number 6.
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52.
The prodigal’s letter to his father, “Dear
Dad, keep 1000050,” in reply to a suggestion for
safeguarding some of his prospects, was written
in playful impudence; and its interpretation is,
“Dear Dad, keep cool!” for the figures in
Roman numerals are COOL.
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53. BURIED POETS
The poets’ names buried in the lines—
The sun is darting rays of gold
Upon the moor, enchanting spot;
Whose purpled heights, by Ronald loved,
Up open to his Shepherd cot.
And sundry denizens of air
Are flying, aye, each to his nest;
And eager make at such an hour
All haste to reach the mansions blest.
are Gray, Moore, Byron, Pope, Dryden, Gay,
Keats, and Hemans.
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54.
When A. B. gave up the reins of government,
and C.B. took office in his place, the two
verbs, similar in all respects, except that the one
is longer by one letter than the other, which
expressed the change, were resigns—reigns.
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55
Underdone mutton and onion make between you and me,
A glutton a little seedy after a capital tea.
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56
First a c and a t, last a c and a t,
With a couple of letters between,
Form a sight that our eyes are delighted to see,
Unless in their sight it is seen—
is solved by Cataract. The first line reads, First
a c and a t, last a c and a t, that is cat and act.
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57.
Cuba.
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58.
The Rebus T S is solved by the words
tones and tans, t before one s, or t before an s.
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59.
The phonetic phrase—
INXINXIN—
is, Ink sinks in!
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A GOOD END
60. Finis (F IN IS).
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