the numbers 1, 12, 23, 34, and so on, or of a series which continually
increases by 11, up to 100. Let us suppose that the first person, who
knows the game, makes choice of 1; it is evident that his adversary, as
he must count less than 11, can at most reach 11, by adding 10 to it. The
first will then take 1, which will make 12; and whatever number the second
may add the first will certainly win, provided he continually add the
number which forms the complement of that of his adversary to 11; that is
to say, if the latter take 8, he must take 3; if 9, he must take 2; and so
on. By following this method he will infallibly attain to 89, and it will
then be impossible for the second to prevent him from getting first to
100; for whatever number the second takes he can attain only to 99; after
which the first may say--"and 1 makes 100." If the second take 1 after 89,
it would make 90, and his adversary would finish by saying--"and 10 make
100." Between two persons who are equally acquainted with the game, he who
begins must necessarily win.
THE ASTONISHED FARMER.
A and B took each 30 pigs to market, A sold his at 3 for a dollar, B at 2
for a dollar, and together they received $25. A afterwards took 60 alone,
which he sold _as before_, at 5 for $2, and received, but $24; what became
of the other dollar?
This is rather a catch question, the insinuation that the first lot were
sold at the rate of five for $2, being only true in part. They commence
selling at that rate, but after making ten sales, A's pigs are exhausted,
and they have received $20: B still has 10 which he sells at "2 for a
dollar" and of course receives $5; whereas had he sold them at the rate of
5 for $2, he would have received but $4. Hence the difficulty is easily
settled.
MAGICAL CENTURY.
If the number 11 be multiplied by any one of the nine digits, the two
figures of the product will always be alike, as appears in the following
example:--
11 11 11 11 11 11 11 11 11
1 2 3 4 5 6 7 8 9
-- -- -- -- -- -- -- -- --
11 22 33 44 55 66 77 88 99
-- -- -- -- -- -- -- -- --
Now, if another person and yourself have fifty counters a-piece, and agree
never to stake more than ten at a time, you may tell him that if he permit
you to stake first, you always complete the even century before him.
In order to succeed, you must first stake 1, and remembering the order of
the above series, constantly add to what he stakes as many as will make
one more than the numbers 11, 22, 33, &c., of which it is composed, till
you come to 89, after which your opponent cannot possibly reach the even
century himself, or prevent you from reaching it.
If your opponent has no knowledge of numbers, you may stake any other
number first, under 10, provided you subsequently take care to secure one
of the last terms, 56, 67, 78, &c.; or you may even let him stake first,
if you take care afterward to secure one of these numbers.
This exercise may be performed with other numbers; but, in order to
succeed, you must divide the number to be attained by a number which is
a unit greater than what you can stake each time, and the remainder will
then be the number you must first stake. Suppose, for example, the number
to be attained be 52 (making use of a pack of cards instead of counters),
and that you are never to add more than 6; then, dividing 52 by 7, the
remainder, which is 3, will be the number which you must first stake; and
whatever your opponent stakes, you must add as much to it as will make it
equal to 7, the number by which you divided, and so in continuation.
THE UNLUCKY HATTER.
A blackleg passing through a town in Ohio, bought a hat for $8 and gave in
payment a $50 bill. The hatter called on a merchant near by, who changed
the note for him, and the blackleg having received his $42 change went his
way. The next day the merchant discovered the note to be a counterfeit,
and called upon the hatter, who was compelled forthwith to borrow $50
of another friend to redeem it with; but on turning to search for the
blackleg he had left town, so that the note was useless on the hatter's
hands. The question is, what did he lose--was it $50 besides the hat, or
was it $50 including the hat?
This question is generally given with names and circumstances as a real
transaction, and if the company knows such persons so much the better, as
it serves to withdraw attention from the question; and in almost every
case the first impression is, that the hatter lost $50 besides the hat,
though it is evident he was paid for the hat, and had he kept the $8 he
needed only to have borrowed $42 additional to redeem the note.
THE BASKET OF NUTS.
A person remarked that when he counted over his basket of nuts, two by
two, three by three, four by four, five by five, or six by six, there was
one remaining; but when he counted them by sevens, there was no remainder.
How many had he?
The least common multiple of 2, 3, 4, 5, and 6 being 60, it is evident,
that if 61 were divisible by 7, it would answer the conditions of the
question. This not being the case, however, let 60 × 2 + 1, 60 × 3 + 1,
60 × 4 + 1, &c., be tried successively, and it will be found that 301 =
60 × 5 + 1, is divisible by 7; and consequently this number answers the
conditions of the question. If to this we add 420, the least common
multiple of 2, 3, 4, 5, 6 and 7, the sum 721 will be another answer; and
by adding perpetually 420, we may find as many answers as we please.
THE UNITED DIGITS.
Arrange the figures 1 to 9 in such order that, by adding them together,
they amount to 100.
15
36
47
--
98
2
---
100
DECEMBER AND MAY.
An old man married a young woman; their united ages amounted to C. The
man's age multiplied by 4 and divided by 9, gives the woman's age. What
were their respective ages?
ANSWER.--The man's age, 60 years 12 weeks; the woman's age, 30 years 40
weeks.
THE TWO DROVERS.
Two drovers, A and B, meeting on the road, began discoursing about the
number of sheep they each had. Says B to A, "Pray give me one of your
sheep and I will have as many as you." "Nay," replied A, "but give me one
of your sheep and I will have as many again as you." Required to know the
number of sheep they each had?
A had seven and B had five sheep.
THE BASKET AND STONES.
If a hundred stones be placed in a straight line, at the distance of a
yard from each other, the first being at the same distance from a basket,
how many yards must the person walk who engages to pick them up, one
by one, and put them into the basket? It is evident that, to pick up
the first stone, and put it into the basket, the person must walk two
yards; for the second, he must walk four; for the third, six: and so on
increasing by two, to the hundredth.
The number of yards, therefore, which the person must walk will be equal
to the sum of the progression, 2, 4, 6, &c., the last term of which is 200
(22). But the sum of the progression is equal to 202, the sum of the two
extremes, multiplied by 50, or half the number of terms: that is to say,
10,100 yards, which makes more than 5½ miles.
THE FAMOUS FORTY-FIVE.
How can number 45 be divided into four such parts that, if to the
first part you add 2, from the second part you subtract 2, the third
part you multiply by 2, and the fourth part you divide by 2, the sum
of the addition, the remainder of the subtraction, the product of the
multiplication, and the quotient of the division be all equal?
The 1st is 8; to which add 2, the sum is 10
The 2nd is 12; subtract 2, the remainder is 10
The 3rd is 5; multiplied by 2, the product is 10
The 4th is 20; divided by 2, the quotient is 10
--
45
Required to subtract 45 from 45, and leave 45 as a remainder?
Solution.--9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
--------------------------------------
8 + 6 + 4 + 1 + 9 + 7 + 5 + 3 + 2 = 45
SUBTRACTION.
From 1 mile subtract 7 furlongs, 39 rods, 5 yards, 1 foot, 5 inches.
miles, furlongs, rods, yards, feet, inches.
From 1 0 0 0 0 0
Take 0 7 39 5 1 5
--------------------------------------
0 0 0 0 0 1
In this problem, instead of borrowing 1 foot, we borrow ½ a foot = 6
inches, from which we take 5 inches, and 1 remains; we then carry ½ to
1, and borrowing ½ a yard = 1½ feet, we have 1½ from 1½ = 0,
and afterwards proceed as usual.
THE EXPUNGED FIGURE.
In the first place desire a person to write down secretly, in a line,
any number of figures he may choose, and add them together as units;
having done this, tell him to subtract that sum from the line of figures
originally set down; then desire him to strike out any figure he pleases,
and add the remaining figures in the line together as units, (as in the
first instance,) and inform you of the result, when you will tell him the
figure he has struck out.
76542-24
24
------
76518
Suppose, for example, the figures put down are 76542; these, added
together, as units, make a total of 24: deduct 24 from the first line, and
76518 remain; if 5, the center figure be struck out, the total will be 22.
If 8, the first figure be struck out, 19 will be the total.
In order to ascertain which figure has been struck out, you make a mental
sum one multiple of 9 higher than the total given. If 22 be given as the
total, then 3 times 9 are 27, and 22 from 27 show that 5 was struck out.
If 19 be given, that sum deducted from 27 shows 8.
Should the total be equal multiplies of 9, as 18, 27, 36, then 9 has been
expunged.
With very little practice any person may perform this with rapidity, it is
therefore needless to give any further examples. The only way in which a
person can fail in solving this riddle is, when either the number 9 or a
cipher is struck out, as it then becomes impossible to tell which of the
two it is, the sum of the figure in the line being an even number of nines
in both cases.
THE MYSTERIOUS ADDITION.
It is required to name the quotient of five or three lines of
figures--each line consisting of five or more figures--only seeing the
first line before the other lines are even put down. Any person may write
down the first line of figures for you. How do you find the quotient?
EXAMPLE.--When the first line of figures is set down, subtract 2 from the
last right-hand figure, and place it before the first figure of the line,
and that is the quotient for five lines. For example, suppose the figures
given are 86,214, the quotient will be 286,212. You may allow any person
to put down the two first and the fourth lines, but you must always set
down the third and fifth lines, and in doing so, always make up 9 with the
line above, as in the following example:
Therefore in the annexed diagram you will see that you have made 9 in the
third and fifth lines with the lines above them. If the person desire to
put down the figures should set down a 1 or 0 for the last figure, you
must say we will have another figure, and another, and so on until he sets
down something above 1 or 2.
86,214
42,680
57,319
62,854
37,145
------
Qt. 268,212
67,856
47,218
52,781
------
Qt. 167,855
In solving the puzzle with three lines, you subtract 1 from the last
figure, and place it before the first figure, and make up the third line
yourself to 9. For example: 67,856 is given, and the quotient will be
167,855, as shown in the above diagram.
TO TELL AT WHAT HOUR A PERSON INTENDS TO RISE.
Let the person set the hand of the dial of a watch at any hour he pleases,
and tell you what hour that is; and to the number of that hour you add in
your mind 12; then tell him to count privately the number of that amount
upon the dial, beginning with the next hour to that on which he proposes
to rise, and counting backwards, first reckoning the number of the hour at
which he has placed the hand. For example:
Suppose the hour at which he intends to rise be 8, and that he has placed
the hand at 5; you will add 12 to 5, and tell him to count 17 on the
dial, first reckoning 5, the hour at which the index stands, and counting
backwards from the hour at which he intends to rise; and the number 17
will necessarily end at 8, which shows that to be the hour he chose.
TO FIND THE DIFFERENCE BETWEEN TWO NUMBERS, THE GREATEST OF WHICH IS
UNKNOWN.
Take as many nines as there are figures in the smallest number, and
subtract that sum from the number of nines. Let another person add the
difference to the largest number, and talking away the first figure of the
amount add it to the last figure, and that sum will be the difference of
the two numbers.
For example: John, who is 22, tells Thomas, who is older, that he can
discover the difference of their ages; he therefore privately deducts 22
from 99 (his age consisting of two figures, he of course takes two nines);
the difference, which is 77, he tells Thomas to add to his age, and to
take away the first figure from the amount, and add it to the last figure
and that will be the difference of their ages; thus,
The difference between John's age and 99 is 77
To which Thomas adding his age 35
---
The sum is 112
Then by taking away the first figure 1, and adding
it to the figure 2, the sum is 13
Which add to John's age 22
---
Gives the age of Thomas 35
THE REMAINDER.
A very pleasing way to arrive at an arithmetical sum, without the use of
either slate or pencil, is to ask a person to think of a figure, then to
double it, then add a certain figure to it, now halve the whole sum, and
finally to subtract from that the figure first thought of. You are then to
tell the thinker what is the remainder.
The key to this lock of figures is, that HALF of whatever sum you request
to be added during the working of the sum is THE REMAINDER. In the example
given, five is the half of ten, the number requested to be added. Any
amount may be added, but the operation is simplified by giving only even
numbers, as they will divide without fractions.
_Example._
Think of 7
Double it 14
Add 10 to it 10
---
Halve it 2)24
---
Which will leave 12
Subtract the number thought of 7
---
THE REMAINDER will be 5
A PERSON HAVING AN EQUAL NUMBER OF COUNTERS, OR PIECES OF MONEY, IN EACH
HAND, TO FIND HOW MANY HE HAS ALTOGETHER.
Request the person to convey any number, as 4, for example, from the
one hand to the other, and then ask how many times the less number
is contained in the greater. Let us suppose that he says the one is
the triple of the other; and, in this case, multiply 4, the number of
the counters conveyed, by 3, and add to the product the same number,
which will make 16. Lastly, take 1 from 3, and if 16 be divided by the
remainder 2, the quotient will be the number contained in each hand, and
consequently the whole number is 16.
This curious problem deserves another example. Let us again suppose that
4 counters are passed from one hand to the other, and the less number is
contained in the greater 2⅓ times. In this case, we must, as before,
multiply 4 by 2⅓, which will give 9⅓; to which, if 4 be added, we
shall have 13⅓, or 40/3; if 1, then, be taken from 2⅓, the remainder
will be 1⅓, or 4/3, by which, if 40/3 be divided, the quotient 10 will
be the number of counters in each hand.
THE THREE JEALOUS HUSBANDS.
Three jealous husbands, A, B, and C, with their wives, being ready to
pass by night over a river, find at the water side a boat which can carry
but two at a time, and for want of a waterman they are compelled to row
themselves over the river at several times. The question is how those six
persons shall pass, two at a time, so that none of the three wives may be
found in the company of one or two men, unless her husband be present?
This may be effected in two or three ways; the following may be as good
as any: Let A and wife go over--let A return--let B's and C's wives go
over--A's wife returns--B and C go over--B and wife return, A and B go
over--C's wife returns, and A's and B's wives go over--then C comes back
for his wife. Simple as this question may appear, it is found in the works
of Alcuin, who flourished a thousand years ago, hundreds of years before
the art of printing was invented.
THE FALSE SCALES.
A cheese being put into one of the scales of a false balance, was found to
weigh 16 lbs., and when put into the other only 9 lbs. What is the true
weight?
The true weight is a mean proportional between the two false ones, and is
found by extracting the square root of their product. Thus 16 × 9 = 144;
and square root 144 = 12 lbs., the weight required.
THE APPLE WOMAN.
A poor woman, carrying a basket of apples, was met by three boys, the
first of whom bought half of what she had, and then gave her back 10; the
second boy bought a third of what remained, and gave her back 2; and the
third bought half of what she had now left, and returned her 1; after
which she found she had 12 apples remaining. What number had she at first?
From the 12 remaining, deduct 1, and 11 is the number she sold the last
boy, which was half she had; her number at that time, therefore, was 22.
From 22 deduct two, and the remaining 20 was 2/3 of her prior stock, which
was therefore 30. From 30 deduct 10, and the remainder 20 is half her
original stock; consequently she had at first 40 apples.
THE GRACES AND MUSES.
The three Graces, carrying each an equal number of oranges, were met by
the nine Muses, who asked for some of them; and each Grace having given
to each Muse the same number, it was then found that they had all equal
shares. How many had the Graces at first?
The least number that will answer this question is twelve; for if we
suppose that each Grace gave one to each Muse, the latter would each have
three, and there would remain three for each Grace. (Any multiple of 12
will answer the conditions of the question.)
THE JESUITICAL TEACHER.
A teacher, having fifteen young ladies under her care, wished them to take
a walk each day of the week. They were to walk in five divisions of three
ladies each, but no two ladies were to be allowed to walk together twice
during the week. How could they be arranged to suit the above conditions?
SUN. MON. TUES. WEDN. THURS. FRID. SAT.
a b c|a d g|a k n|a e l|a h o|a f p|a i m
d e f|b e h|b l o|b f m|b i p|b d n|b g k
g h i|c m p|c f i|c g n|c d k|c h l|c e o
k l m|f k o|d h m|d i o|e m n|e i k|d l p
n o p|i l n|e g p|h k p|f g l|g m o|h f n
QUAINT QUESTIONS.
What is the difference between twenty four quart bottles, and four and
twenty quart bottles?
_Ans._--56 quarts difference.
What three figures, multiplied by 4, will make precisely 5?
_Ans._--1¼, or 1·25.
What is the difference between six dozen dozen, and half-a-dozen dozen?
_Ans._--792: Six dozen dozen being 864, and half-a-dozen dozen, 72.
Place three sixes together, so as to make seven.
_Ans._--6⁶/₆.
Add one to nine and make it twenty.
_Ans._ IX--cross the _1_, it makes XX.
Place four fives so as to make six and a half. _Ans._--5⁵/₅ ·5.
A room with eight corners had a cat in each corner, seven cats before each
cat, and a cat on every cat's tail. What was the total number of cats?
_Ans._--Eight cats.
Prove that seven is the half of twelve. _Ans._--Place the Roman figures on
a piece of paper, and draw a line through the middle of it, the upper will
be VII.
THE FOX, GOOSE AND CORN.
A countryman having a Fox, a Goose, and a peck of Corn, came to a river,
where it so happened that he could carry but one over at a time. Now as no
two were to be left together that might destroy each other, he was at his
wit's end, for says he "Though the corn can't eat the goose, nor the goose
eat the fox; yet the fox can eat the goose, and the goose eat the corn."
How shall he carry them over, that they shall not destroy each other?
Let him first take over the Goose, leaving the Fox and Corn; then let him
take over the Fox and bring the Goose back; then take over the Corn; and
lastly take over the Goose again.
MULTIPLYING MONEY BY MONEY.
Amongst the various questions that are given for the purpose of puzzling
the unwary arithmetician, the multiplication of money by money is one of
the most curious: take for instance the following problems:
Multiply £99 19_s._ 11-3/4_d._ by £99 19_s._ 11-3/4_d._
Multiply £11 11_s._ 11_d._ by £11 11_s._ 11_d._
To the uninitiated they usually appear easy of solution but the various
modes of working them out, and the different results obtained, prove that
there is something absurd and wrong in the questions themselves. Some
reduce all to farthings, and after multiplying one term by the other,
return the product into pounds, shillings, and pence. Others convert them
into decimals; whilst some work the problem in the style of duodecimals.
Having sufficiently puzzled the tyros, the querist remarks: "The problem
itself is absurd, it is incapable of solution; for what is the nature
of the product of pounds, shillings, and pence multiplied by pounds,
shillings and pence? We know that a yard multiplied by a yard is a square
yard, but who can tell what is a penny multiplied by a penny, or a penny
by a pound?"
Now all this is quite correct, provided the question is limited, as
above to the product of pounds, shillings, and pence, into pounds,
shillings, and pence; but suppose the problem were put in this form--If
a capital of £1 produces by compound interest, in a certain time, £99
19_s._ 11¾_d._, how much would be produced by a capital of £99 19_s._
11¾_d.?_ It is evident that, to answer this, we must multiply £99
19_s._ 11¾_d._ by £99 19_s._ 11¾_d._: these are in fact the second
and, third terms of an ordinary "rule of three;" and though one of the
terms is a "concrete" quantity of pounds, shillings, and pence, the other
must be regarded as an "abstract" mathematical quantity, being 99 and a
fraction, of which the number of farthings in a pound is the denominator
960, and the number of farthings in the third term is the numerator, 959;
or, instead of this, the shillings and pence might be converted into
decimals of a pound, or into aliquot parts. The product of multiplying
£99 19_s._ 11¾_d._ by 99-959/960 is £9,999 15_s._ 10 1/3840_d._; the
quickest way of doing this, is to multiply by 100, and to subtract from
the product the 960th part of the multiplicand.
In the other question proposed, the product of £11 11_s._ 11_d._ into £11
11_s._ 11_d._, or 11 143/240, is £134 9_s._ 3 49/240_d._
Number and value are distinct abstract ideas, and cannot, without
committing a logical absurdity, be confused. To multiply is to repeat
a certain number of _times_, and it is obviously impossible to bring
_value_ into the question. Value is arbitrary; number is fixed. Put it
in this way, and the absurdity is evident: One pound is equivalent to 20
shillings, or 240 pence, or 960 farthings. In value there is no difference
whatever; but what an enormous difference between multiplying by 1, 20,
240, or 960!
THE UNFAIR DIVISION.
A gentleman rented a farm, and contracted to give to his landlord 2/5
of the produce; but prior to the time of dividing the corn, the tenant
used 45 bushels. When the general division was made, it was proposed to
give to the landlord 18 bushels from the heap, in lieu of his share of
the 45 bushels which the tenant had used, and then to begin and divide
the remainder as though none had been used. Would this method have been
correct?
The landlord would lose 7⅕ bushels by such an arrangement, as the rent
would entitle him to ⅖ of the 18. The tenant should give him 18 bushels
from his own share after the division is completed, otherwise the landlord
would receive but 2/7 of the first 63 bushels.
A POPULAR FALLACY.
It is often suggested from the pulpit and elsewhere, that enough persons
have lived and died in the world to cover its whole surface with bodies;
and even two or three strata deep. Is this probable?
Say the earth has existed 6000 years, the population always having been
800,000,000, and the average life of man 30 years; this being the utmost
that could be claimed. Allow then the State of Virginia to contain 70,000
square miles, and each grave to occupy a space of 6 feet by 2; the
territory of the State would contain 162,624,000,000; while the mighty
army of the dead would number only 160,000,000,000; leaving 2,624,000,000
graves yet unoccupied. How wide of truth then is the position often set
forth so positively!
FOOTNOTES:
[Footnote 10: Ancestor of the fighting and writing Napiers of our day.]
[Footnote 11: When an _exact half_ cannot be taken without a fraction, he
must take the _larger half_--you must tell him this before he commences.
Here it is the _larger half_.]
[Footnote 12: From Parkes' Philosophy of Arithmetic, a capital work
published by Moss & Bro. Philadelphia.]
TRICKS IN GEOMETRY.
"Let young beginners come and try
Their hands at our geometry."