Who was the first man condemned to hard labor for life? Adam.
Why was the giant Goliath very much astonished when David hit him with a
stone? Because such a thing had never entered his head before.
Which are the two smallest things mentioned in the Bible? The widow’s
mite and the wicked flee.
How is it that Methusalah was the oldest man when he died before his
father? His father was translated.
How many soft boiled eggs could the giant Goliath eat on an empty
stomach? One, after which his stomach was not empty.
What was the difference between Joan of Arc and Noah’s ark? One was Maid
of Orleans, the other was made of wood.
Where did Noah strike the first nail in the ark? On the head.
Why was Eve not afraid of the measles? Because she’d Adam (had ’em).
What church did Eve belong to? Adam thought her Eve-angelical.
What two animals carried the least into the ark? The fox and cock,
because they carried only a brush and comb between them.
Who had the first entrance into a theatre? Joseph, when he was taken
from the family circle and put into the pit.
In what place did the cock crow so loud that all the world heard him? In
the ark.
What became of Lot when his wife was turned into a pillar of salt? He
took a fresh one.
Who first introduced salt meat into the navy? Noah, when he took Ham
into the ark.
What animal took most baggage into the ark? The elephant, who took his
trunk.
What confection did they have in the ark? Preserved pairs (pears).
What man mentioned in the Bible had no father? Joshua, the son of Nun.
What reason have we to think that Moses wore a wig? Because he was
sometimes seen with Aaron and sometimes without ’Air on (hair on).
Why was Noah like a hungry cat? Because he went forty days and forty
nights without finding Ararat.
If Solomon was the son of David and Joab was the son of Zeruiah what
relation was Zeruiah to Joab? His mother.
NOTE.—Most persons will answer “his father,” not remembering that
Zeruiah was a woman.
If “Moses was the son of Pharaoh’s daughter,” then he must have been the
daughter of Pharaoh’s son.
NOTE.—Most persons will say that it was impossible for Moses to have
been a daughter, etc. It will aid in understanding it to connect the
words thus: “daughter-of-Pharoah’s son.”
POETICAL CONUNDRUMS
Legs I have, but seldom walk;
I backbite all, yet never talk.
A flea.
I came to a field and couldn’t get through it;
So I went to a school and learned how to do it.
Fence.
My first I hope you are,
My second I see you are,
My whole I know you are,
Wel-come.
My first’s a dirty little brute,
My second’s at the end on’t;
My third like many an honest man,
Is on a fool dependent.
Pig-tail.
By equal division—I know I am right—
The half of thirteen you’ll find to be eight.
VIII
XIII ———— VIII.
ΛIII
My number, definite and known,
Is ten times ten told ten times o’er;
One-half of me is one alone,
The other exceeds all count and score.
Thou-sand.
There’s a word composed of three letters alone,
Which reads backwards and forwards the same;
It expresses the sentiments warm from the heart,
And to beauty lays principal claim.
Eye.
The cat did my first with a curl of her tail,
When the game she had made quite secure
By means of my second and not of my whole,
As she ought to have done, I am sure.
Pur-chase.
Pray tell me, ladies, if you can,
Who is that highly favored man,
Who though he has married many a wife,
May still live single all his life?
A clergyman.
Can you tell me why a hypocrite’s eye
Can better descry than you or I,
On how many toes a pussy cat goes?
A man of deceit can best count-er-feit;
And so, I suppose, can best count her toes.
Without my first you’d look very strange,
My second you much want to be;
My whole is what many a lady has worn,
At a ball, an assembly, or play.
Nose-gay.
Two letters often tempt mankind,
And those who yield will surely find
Two others ready to enforce
The punishment that comes of course.
X-S and D-K (excess and decay).
My first doth affliction denote,
Which my second is destined to feel.
My whole is a sweet antidote
That affliction to soothe and to heal.
Wo-man!
My first of anything is half,
My second is complete;
And so remains until once more
My first and second meet.
Semi-circle.
My first makes company,
My second shuns company,
My third assembles company,
My whole puzzles company.
Co-nun-drums.
Safe on a fair one’s arm my first may rest,
And raise no tumult in a husband’s breast;
To those who neither creep, nor run, nor fly,
The want of legs my second will supply.
My whole’s a rival of the fairest toast,
And when I’m liked the best I suffer most.
Muf-fin.
What is that which is
The beginning of eternity,
The end of time and space,
The beginning of every end,
The end of every race?
Letter E.
Your initials begin with an A,
You’ve an A at the end of your name,
The whole of your name is an A,
And its backward and forward the same.
Anna!
We are airy little creatures,
Each have different forms and features;
One of us in glass is set,
Another you will find in jet;
A third, less bright, is set in tin,
A fourth a shining box within;
And the fifth, if you pursue,
It will never fly from you.
Vowels.
My first a baby does when you pinch it;
My second a lady says when she doesn’t mean it;
My third exists and no one e’er has seen it;
And my whole contains the world’s best half within it.
Cri-no-line.
Formed long ago, yet made to-day,
I’m most employed while others sleep;
What none would like to give away,
Yet no one likes to keep.
Bed.
What’s that? What’s that? Oh! I shall faint,
Call, call the priest to lay it!
Transpose it, and to king and saint,
And great and good you pay it.
Spectre; respect.
How shall the following be read?
Yy u r yy u b
I c u r yy 4 me.
ANSWER.—
Too wise you are,
Too wise you be;
I see you are
Too wise for me.
How shall the following be read?
U o a o, but I o thee;
O o no o, but O o me;
Then let my o thy o be
And give o o I o thee.
ANSWER.—
You sigh for a cipher, but I sigh for thee;
Oh! sigh for no cipher, but oh! sigh for me;
They let my cipher thy cipher be;
And give sigh for sigh, for I sigh for thee.
How shall the following stanza be read that it may be true?
There is a lady in the land
With twenty nails on each hand,
Five-and-twenty on hands and feet,
This is true without deceit.
ANSWER.—
There is a lady in the land
With twenty nails; on each hand
Five, and twenty on hands and feet,
This is true without deceit.
FRENCH CONUNDRUMS
Je suis le capitaine de vingt-quatre Soldats, et sans moi Paris serait
pris.
The letter A.
Pourquoi les amateurs du beau sexe vont-ils souvent à l’hippodrome?
Afin d’être en cirque assis.
Quelle différence y-a-t-il entre le souverain de la Perse et le Viceroi
d’Egypte.
C’est que l’un est chat (Shah) et l’autre pas chat (Pacha).
Quel est le peuple le moins gai de l’univers?
Le peuple Persan, parce qu’il est gouverné par un Schah (chat), et que
le schah fait fuir les souris.
A French friend wishes to know: Vai ze Keeng ov Eatalee ees laike von
seengair at ze opera oo ees loozeng ess voice?
Parce qu’il a perdu sa voix (Savoie).
Pourquoi les Carthaginois portaient-ils toujours des gants?
Parce qu’ils n’aimaient pas l’air aux mains (les Romains)!
Quand un gant ressemble-t-il au numéro vingt-cinq.
Quand il est neuf et très étroit (et treize et trois)!
Un félon peut-il prendre pour devise,—“Honneur à Dieu”?
Non, car il faut qu’il dise,—“Adieu, honneur!”
Je ne suis pas ce que je suis; car si j’étais ce que suis, je ne serais
pas ce que je suis, cependant je suis ce que je suis; devinez ce que je
suis?
Un domestique qui suit sa maîtresse.
ARITHMETICAL PUZZLES
Write eleven thousand eleven hundred and eleven.
METHOD.— 11,000 + 1,100 + 11 = 12,111.
What four United States coins will amount to fifty-one cents?
ANSWER.—Two twenty five cent pieces and two half cents.
Place three 6’s together so as to amount to 7.
METHOD.— 6-6/6 = 7.
Place three 2’s together so as to make 24.
METHOD.— 22 + 2 = 24.
Place three 3’s together so as to make 24.
METHOD.— 3^3 - 3 = 24.
Take one from nine and make it ten.
METHOD.—Write nine thus, IX; take away the I we have X.
Add one to nine and make it twenty.
METHOD.—Nine is IX; cross the I we have XX.
Prove that one taken from nineteen leaves twenty.
METHOD.—Take the I from XIX, and we have XX.
Make four straight lines and then add five straight lines and make ten.
METHOD.— | | | | ; TEN.
Prove that the half of eleven is six.
METHOD.—In XI draw a line thus,
VI
——,
ΛI
the upper half is VI.
Prove that one added to twenty is nineteen.
METHOD.—Add I to XX we have XIX.
What number of three figures multiplied by 8 will make exactly 10?
ANSWER.— 1¼ or 1.25.
Does the top of a carriage-wheel move faster than the bottom? If so,
explain the reason.
ANSWER.—The top always moves faster than the bottom.
Which is greater, and how much, six dozen dozen or a half a dozen dozen;
or is there no difference between them?
The former.
Which is heavier, a pound of gold or a pound of feathers?
ANSWER.—The pound of feathers, because it is weighed by Avoirdupois
weight, while gold is weighed by Troy weight.
Take fifty, add a cipher, add five, add the fifth of eight, and the
total is the sum of human happiness.
ANSWER.—LOVE.
Six ears of corn are in a hollow stump; how long will it take a squirrel
to carry them all out if he takes out three ears a day?
REMARK.—The “catch” is on the word ears. He takes out two ears on his
head and one ear of corn each day; hence, it takes six days.
How to prove, by mathematical principles, that two unequal numbers are
equal, as 4 = 2.
METHOD.—All will admit that 8 - 8 = 4 - 4. Divide both of these by 2 -
2, and the quotient will be equal. Thus,
(8 - 8) (4 - 4)
—————— = ——————, or 4 = 2
(2 - 2) (2 - 2)
Supposing there are more persons in the world than any one has hairs on
his head; then there must be at least two persons who have the same
number of hairs on their head to a hair. Show how this is.
Place 17 little sticks—matches for instance—making 6 equal squares, as
in the margin. Then remove 5 sticks, and leave three perfect squares of
the same size.
+---+---+---+
| | | |
+---+---+---+
| | | |
+---+---+---+
ANSWER.—The method of doing this is indicated below this answer.
+---+
| |
+---+---+---+
| | | |
+---+ +---+
A and B have an 8 gallon cask full of wine, which they wish to divide
into two equal parts, and the only measures they have are a 5-gallon
cask and a 3-gallon cask. How shall they make the division with these
two vessels?
METHOD.—First fill the 3-gallon cask from the 8-gallon cask; then pour
these 3 gallons into the 5-gallon cask; then fill the 3-gallon cask
again, and fill the 5-gallon cask from the 3-gallon cask; this will
leave 1 gallon in the 3-gallon cask; then empty the 5-gallon cask into
the 8-gallon cask, pour the 1 gallon from the 3-gallon cask into the
5-gallon cask, and fill the 3-gallon cask from the 8-gallon cask. There
will then be 4 gallons in the 8-gallon cask.
Two men in an oyster saloon laid a wager as to which could eat the most
oysters. One ate ninety-nine and the other ate a hundred, and won. How
many did both eat?
REMARK.—The catch is on a hundred and _won_. When spoken it sounds as if
it meant “one ate ninety-nine and the other ate a hundred and one;”
hence, the result usually given is two hundred. The correct result is
one hundred and ninety-nine.
If a room with 8 corners had a cat in each corner, seven cats before
each cat, and a cat on each cat’s tail, what would be the whole number
of cats?
ANSWER.—Eight cats.
Tell a person to think of a number, multiply by 3, multiply the product
by 2, divide the result by 6, add 20, subtract the number thought of,
divide by 4, and then tell him what his result is.
METHOD.—The result will be five. The reason is clear. By multiplying by
3 and 2 and dividing by 6 he has obtained the number thought of. Add 20,
he has the number thought of, plus 20; then subtract the number thought
of, and he has twenty. Now I know he has twenty; hence, I can tell him
what he has if he divides by 4.
A farmer having an ox-chain consisting of 15 links, broke it into five
equal parts, and took it to a blacksmith to be welded together. The
blacksmith agreed to repair it for 50 cents for each welding; but when
he presented his bill he charged for four weldings, making the bill
$2.00. The farmer objected to the bill, saying that it should have been
repaired with only three weldings. How was it to be done?
METHOD.—Each piece consisted of three links; cut open the three links of
one piece and use these to connect the other four pieces of the chain.
Think of a number, multiply by six, divide by three, add forty, divide
by two; name the result, and I will name the number thought of.
METHOD.—Multiplying by six and dividing by three gives twice the number;
add forty we have twice the number, plus forty, divide by two we have
once the number, plus twenty; hence, if I subtract twenty from the
result he gives me I have the number thought of.
Let a person select a number greater than 1 and not exceeding 10. I will
add to it a number not exceeding 10, alternately with himself; and,
although he has the advantage in selecting the number to start with, I
will reach the even hundred first.
METHOD.—I make my additions so that the sums are 12, 23, 34, 45, etc.,
to 89, when it is evident I can reach the hundred first. With one who
does not know the method, I need not run through the entire series, but
merely aim for 89, and when the secret of this is seen aim at 78, then
67, etc.
Think of a number of 3 or more figures, divide by nine, and name the
remainder; erase one figure of the number, divide by 9, and tell me the
remainder and I will tell you what figure you erased.
METHOD.—If the second remainder is less than the first, the figure
erased is the difference between the remainders; but if the second
remainder is greater than the first, the figure erased equals 9, minus
the difference of the remainders.
Let a person think of any number on the dial face of a watch. I will
then point to various numbers, and at each he will silently add one to
the number selected, until he arrives at twenty, which he will announce
aloud; and my pointer will then be on the number he selected.
METHOD.—I point promiscuously about the face of the watch until the
eighth point, which should be on the “12.” I then pass regularly around
toward the “1” pointing at “11,” “10,” “9,” etc., until “twenty” is
called, when my pointer will be over the number selected.
Take _nine_ from _six_ and _ten_ from _nine_ and _fifty_ from _forty_,
and _six_ will remain.
SIX IX XL
IX X L
—— —— ——
S I X
Two-thirds of six is nine, one-half of twelve is seven, the half of five
is four, and six is half of eleven.
METHOD.—Two-thirds of SIX is IX, the upper half of XII is VII, the half
of FIVE is IV, and the upper half of XI is VI.
Two men have 24 ounces of fluid which they wish to divide between them
equally. How shall they effect the division, provided they have only
three vessels; one containing 5 oz., the other 11 oz., and the third 13
oz.?
METHOD.—The method is similar to the division of 8 gallons in the
question on page 78.
Three persons own 51 quarts of rice, and have only two measures; one a
4-quart, the other a 7-quart measure. How shall they divide it into
three equal parts?
METHOD.—One-third of 51 is 17; so each must have 17 quarts. To measure
17 quarts fill the 7-quart measure twice and pour into some large
vessel, making 14 quarts; then fill the 7-quart measure, draw off 4
quarts in the 4-quart measure, and then pour the remaining 3 quarts in
the vessel containing the 14 quarts.
Think of a number composed of two unequal digits, invert the digits,
take the difference between this and the original number, name one of
the digits and I will name the other.
METHOD.—The sum of the digits in the difference is always nine; hence,
when one is named the other equals 9, minus the one named.
Take any number, consisting of three consecutive digits and permutate
them, making 6 numbers, and take the sum of these numbers, divide by 6,
and tell me the result and I will tell you the digits of the number
taken.
METHOD.—The quotient consists of three equal digits; the digits of the
number taken are: 1st. one of these equal digits; 2d. this digit
increased by a unit; 3d. this digit diminished by a unit. The same
principle holds when the digits of the number taken differ by 2, 3, or
4. It is a very pretty problem to prove that the sum is always divisible
by 9 and 18.
Take any number, divide it by 9, and name the remainder. Multiply the
number by some number which I name, and divide the product by 9, and I
will name the remainder.
METHOD.—To tell the remainder, I multiply the first remainder by the
number by which I told them to multiply the given number, and divide
this product by 9. The remainder is the second number obtained.
Think of a number greater than 3, multiply it by 3; if even, divide it
by 2; if odd, add 1, and then divide by 2. Multiply the quotient by 3;
if even, divide by 2; if odd, add 1, and then divide by 2. Now divide by
9 and tell the quotient without the remainder, and I will tell you the
number thought of.
METHOD.—If _even_ both times, multiply the quotient by 4; if even 2d and
odd 1st, multiply by 4 and add 1; if even 1st and odd 2d, multiply by 4
and add 2; if odd both times, multiply by 4 and add 3.
Suppose it were possible for a man in Cincinnati to start on Sunday
noon, when the sun is in the meridian, and travel westward with the sun
so that it might be in his meridian all the time. Now it was Sunday noon
when he started, it has been noon with him all the way round, and is
Monday noon when he returns. The question is, at what point did it
change from Sunday noon to Monday noon?
Take any number, subtract the sum of the digits, strike out any digit
from the remainder, tell me the sum of the remaining digits, and I will
tell you the digit struck out.
METHOD.—Subtract the “sum of the remaining digits” from the smallest
multiple of nine greater than “the sum.” The remainder will be the digit
struck out.
In the bottom of a well, 45 feet deep, there was a frog which commenced
traveling toward the top. In his journey he ascended 3 feet every day,
but fell back 2 feet every night. In how many days did he get out?
METHOD.—He gains 1 foot a day, and in 42 days he is 3 feet from the top;
and on the 43d day he reaches the top.
Think of any three numbers less than 10. Multiply the first by 2 and add
5 to the product. Multiply this sum by 5 and add the second number to
the product. Multiply the last result by 10 and add the third number to
the product; then subtract 250. Name the remainder and I will name the
numbers thought of and in the order in which they were thought of.
METHOD.—The three digits composing this remainder will be the numbers
thought of: and the order in which they were thought of will be the
order of hundreds, tens, and units.
If a man had a triangular lot of land, the largest side being 136 rods,
and each of the other sides 68 rods; what would be the value of the
grass on it at the rate of $10 an acre?
REMARK.—The “catch” in this is that the sides given will form no
triangle.
Says A to B: “Give me four weights and I can weigh any number of pounds
not exceeding 40.” Required the weights and the method of weighing.
ANSWER.—The weights are 1, 3, 9, and 27 pounds. In weighing we must put
one or more in both scales, or some in one scale and some in another:
thus, 7 lbs. = 9 lbs. + 1 lb. - 3 lbs.
Three men traveling with their wives came to a river which they wished
to cross. There was but one boat and but two could cross at one time;
and since the husbands were jealous no woman could be with a man unless
her own husband was present. In what manner did they get across the
river?
METHOD.—Let the persons be denoted A, B, and C, and Mrs. A, Mrs. B, and
Mrs. C. First Mr. A and Mrs. A go over; then A comes back and Mrs. B and
Mrs. C go over; then Mrs. A comes back and Mr. B and Mr. C go over; then
Mr. B and Mrs. B return and Mr. A and Mr. B go over; then Mrs. C returns
and Mrs. A and Mrs. B go over; then Mr. C returns and takes his wife,
Mrs. C, over.
A man having a fox, a goose, and some corn came to a river which it was
necessary to cross. He could, however, take only one across at a time,
and if he left the goose and corn while he took the fox over, the goose
would eat the corn; but if he left the fox and goose, the fox would kill
the goose. How shall he get them all safely over?
METHOD.—First he takes the goose over, then returns and takes the fox
over, then brings the goose back and takes the corn over, and then
returns and takes the goose over again.
How may the 9 digits be arranged in a rectangular form so that the sum
of any row, whether horizontal, vertical, or diagonal, shall equal 15?
ANSWER.—As below.
+---+---+---+
| 4 | 9 | 2 |
+---+---+---+
| 3 | 5 | 7 |
+---+---+---+
| 8 | 1 | 6 |
+---+---+---+
How may the first 16 digits be arranged so that the sum of the vertical,
the horizontal, and the two oblique rows may equal 34?
ANSWER.—As below.
+----+----+----+----+
| 1 | 16 | 11 | 6 |
+----+----+----+----+
| 13 | 4 | 7 | 10 |
+----+----+----+----+
| 8 | 9 | 14 | 3 |
+----+----+----+----+
| 12 | 5 | 2 | 15 |
+----+----+----+----+
In what manner may the first 25 digits be arranged so that the sum of
each row of five figures may equal 65?
ANSWER.—As below.
+----+----+----+----+----+
| 1 | 10 | 12 | 18 | 24 |
+----+----+----+----+----+
| 9 | 11 | 20 | 22 | 3 |
+----+----+----+----+----+
| 13 | 19 | 21 | 5 | 7 |
+----+----+----+----+----+
| 17 | 23 | 4 | 6 | 15 |
+----+----+----+----+----+
| 25 | 2 | 8 | 14 | 16 |
+----+----+----+----+----+
An old Jew took a diamond cross to a jeweler to have the diamonds reset,
and fearing the jeweler might be dishonest, he counted the diamonds and
found that they numbered 7 in three different ways. Now the jeweler
stole two diamonds, but arranged the remainder so that they counted 7
each way as before. How was it done?
Fig. 1. Fig. 2.
7 7
6 7 6 7
7 6 5 6 7 5
4 4
3 3
2 2
1 1
METHOD.—The form of the cross when left is represented by Fig. 1, and
when returned by Fig. 2. It will be seen by the figures how the diamonds
were counted by the old Jew, and how they were arranged by the jeweler,
who “jewed” the Jew.
Take 10 pieces of money, lay them in a row, and require some one to put
them together into heaps of two in each heap by passing each piece over
two others.
METHOD.—Let the pieces be denoted by the numbers 1, 2, 3, 4, 5, 6, 7, 8,
9, 10. Then place 7 on 10, 5 on 2, 3 on 8, 1 on 4, and 9 on 6.
A man goes to a store and purchases a pair of boots worth $5 and hands
out a $50 bill to pay for them. The merchant, not being able to make the
change, goes over the street to a broker and gets the bill changed and
then returns and gives the man who bought the boots his change. After
the purchaser of the boots had been gone a few hours the broker, finding
the bill to be a counterfeit, comes and demands $50 of good money from
the merchant. How much does the merchant lose?
REMARK.—At first glance some say $45 and the boots, some $50 and the
boots, some $95 and the boots, and others $100 and the boots. Which is
correct?
A vessel with a crew of 30 men, half of whom were black, became short of
provisions and fearing that unless half the crew were thrown overboard
all would perish, the captain proposed to the sailors to stand upon deck
in a row and every ninth man be thrown overboard until half the crew
were destroyed. It so happened that the whites were saved. Required, the
order of arrangement.
ANSWER.—W W W W B B B B B W W B W W W B W B B W W B B B W B B W W B.
This can easily be proved by trial, using letters or figures to
represent men.
Suppose a hare is 10 rods before a hound, and that the hound runs 10
rods while the hare runs 1 rod. Now, when the hound has run 10 rods the
hare has run 1 rod; hence they are now 1 rod apart, and when the hound
has run that one rod the hare has run 1/10 of a rod; hence they are now
1/10 of a rod apart, and when the hound has run the 1/10 of a rod they
are 1/100 of a rod apart; and in the same way it may be shown the hare
is always 1/10 of the previous distance ahead of the hound; hence the
hound can never catch the hare. How is the contrary shown
mathematically? How far will the hound run to catch the hare.
ANSWER.—The distance the hound runs will be represented by the series
1 1 1
10 + 1 + ——— + ——— + ————,
10 100 1000
to infinity. The sum of this series can be found by the algebraic
formula
_a_
S = ———————,
(1 - _r_)
in which _a_ = 10 and _r_ = 1/10. Substituting the value of _a_ and _r_
we have
10 10
——————— ————— 100
S = 1 = 9 = ——— = 11-1/9.
1 - ——— ——— 9
10 10