Where one or more of the numbers are 10, or more than 10, and where an
even number of numbers has been thought of.
Suppose he fixes on six numbers, viz: 2, 6, 7, 15, 16, 18. He must add
together the numbers as follows, and tell you the sum in each case:
1. The sum of the 1st and 2d 8
2. The sum of the 2d and 3d 13
3. The sum of the 3d and 4th 22
4. The sum of the 4th and 5th 31
5. The sum of the 5th and 6th 34
6. The sum of the 2d and last 24
You must then add together the 2d, 4th, and 6th sums, 13 + 31 + 24 =
68, and the 3d and 5th sums, 22 + 34 = 56. Subtract one from the other,
leaving 12; the 2d number will be 6, the half of this; take the 2d from
the sum of the 1st and 2d, and you will get the 1st; take the 2d from
the sum of the 2d and 3d, and you will have the 3d, and so on.
How Many Counters Have I in My Hands?
A person having an equal number of counters in each hand, it is
required to find how many he has altogether.
Suppose he has 16 counters, or 8 in each hand. Desire him to transfer
from one hand to the other a certain number of them, and to tell you
the number so transferred. Suppose it be 4, the hands now contain 4
and 12. Ask him how many times the smaller number is contained in the
larger; in this case it is three times. You must then multiply the
number transferred, 4, by the 3, making 12, and add the 4, making 16;
then divide 16 by the 3 minus 1; this will bring 8, the number in each
hand.
In most cases fractions will occur in the process; when 10 counters are
in each hand and if four be transferred, the hands will contain 6 and
14.
He will divide 14 by 6 and inform you that the quotient is 2-1/3.
You multiply 4 by 2-1/3, which is 9-1/3.
Add four to this, making 13-1/3 equal to 40/3.
Subtract 1 from 2-1/3, leaving 1-1/3 or 4/3.
Divide 40/3 by 4/3, giving 10, the number in each hand.
The Three Travelers.
Three men met at a caravansary or inn, in Persia; and two of them
brought their provisions along with them, according to the custom of
the country; but the third, not having provided any, proposed to the
others that they should eat together, and he would pay the value of his
proportion. This being agreed to, A produced 5 loaves, and B 3 loaves,
all of which the travelers ate together, and C paid 8 pieces of money
as the value of his share, with which the others were satisfied, but
quarreled about the division of it. Upon this the matter was referred
to the judge, who decided impartially. What was his decision?
At first sight it would seem that the money should be divided according
to the bread furnished; but we must consider that as the 3 ate 8
loaves, each one ate 2-2/3 loaves of the bread he furnished. This from
5 would leave 2-1/3 loaves furnished the stranger by A; and 3 - 2-2/3
= 1/3 furnished by B, hence 2-1/3 to 1/3 = 7 to 1, is the ratio in
which the money is to be divided. If you imagine A and B to furnish,
and C to consume all, then the division will be according to amounts
furnished.
The Money Game.
A person having in one hand a piece of gold, and in the other a piece
of silver, you may tell in which hand he has the gold, and in which
the silver, by the following method: Some value, represented by an
even number, such as 8, must be assigned to the gold; and a value
represented by an odd number, such as 3, must be assigned to the
silver; after which, desire the person to multiply the number in the
right hand by any even number whatever, such as 2, and that in the left
by an odd number, as 3; then bid him add together the two products,
and if the whole sum be odd, the gold will be in the right hand, and
the silver in the left; if the sum be even, the contrary will be the
case.
To conceal the artifice better, it will be sufficient to ask whether
the sum of the two products can be halved without a remainder, for in
that case the total will be even, and in the contrary case odd.
It may be readily seen that the pieces, instead of being in the two
hands of the same person, may be supposed to be in the hands of two
persons, one of whom has the even number, or piece of gold, and the
other the odd number, or piece of silver. The same operations may then
be performed in regard to these two persons, as are performed in regard
to the two hands of the same person, calling the one privately the
right, and the other the left.
The Philosopher’s Pupils.
To find a number of which the half, fourth, and seventh, added to
three, shall be equal to itself.
This was a favorite problem among the ancient Grecian arithmeticians,
who stated the question in the following manner: “Tell us, illustrious
Pythagoras, how many pupils frequent thy school?” “One-half,” replied
the philosopher, “study mathematics, one-fourth natural philosophy,
one-seventh preserve silence, and there are three females besides.”
The answer is 28: 14 + 7 + 4 + 3 = 28.
The Certain Game.
Two persons agree to take, alternately, numbers less than a given
figure, for example, 11, and to add them together till one of them
has reached a certain sum, such as 100. By what means can one of them
infallibly attain to that number before the other?
The whole artifice in this consists in immediately making choice of 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 Dice Guessed Unseen.
A pair of dice being thrown, to find the number of points on each die
without seeing them. Tell the person who cast the dice to double the
number of points upon one of them, and add 5 to it; then to multiply
the sum produced by 5, and to add to the product the number of points
upon the other die. This being done, desire him to tell you the amount,
and having thrown out 25, the remainder will be a number consisting of
two figures, the first of which, to the left, is the number of points
on the first die, and the second figure, to the right, the number of
the other. Thus:
Suppose the number of points of the first die which comes up to be 2,
and that of the other 3; then, if to 4, the double of the points of
the first, there be added 5, and the sum produced, 9, be multiplied
by 5, the product will be 45; to which, if 3, the number of points on
the other die, be added, 48 will be produced, from which, if 25 be
subtracted, 23 will remain; the first figure of which is 2, the number
of points on the first die, and the second figure 3, the number on the
other.
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 first is 8; to which add 2, the sum is 10
The second is 12; subtract 2, the remainder is 10
The third is 5; multiplied by 2, the product is 10
The fourth 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.
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 5 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 “two 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.
The Expunged Figure.
In the first place we 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 twenty-four 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. If 19 be given, that sum deducted from 27 shows 8.
Should the total be equal multiples 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 a number
9 or a 0 is struck out, as it then becomes impossible to tell which of
the two it is, the sum of the figures in the line being an even number
of nines in both cases.
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?
86,214
42,680
57,319
62,854
37,145
------
286,212
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 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.
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 you
request 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.
67,856
47,218
52,781
------
167,855
In solving the puzzle with 3 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.
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 abstract 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, 5 is the half of 10, 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.
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
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 Arithmetical Mouse-Trap.
One of the best and most simple mouse-traps in use may be constructed
as follows: Get a slip of smooth pine, about the eighth of an inch
thick, a quarter of an inch broad, and of sufficient length to cut out
the following parts of a trap: First, an upright piece, three or four
inches high, which must be square at the bottom, and a small piece to
be cut off at the top to fit a notch in No. 2.
The second piece must be of the same length as the first, with the
notch cut across nearly at the top of it, to fit the top of No. 1, and
the other end of it trimmed to catch the notch in No. 3. The third
piece should be twice as long as either of the others; a notch, similar
to that in No. 2, must be cut in one end of it to catch the lower end
of No. 2. Having proceeded thus far, you must put the pieces together,
in order to finish it, by adding another notch in No. 3, the exact
situation of which you will discover as follows: Place No. 1 upright,
then put the notch of No. 2 in the thinned top of No. 1; then get a
flat piece of wood, or a slate, one end of which must rest on the
ground, and the center of the edge of the other on the top of No. 2.
You will now find the thinned end of No. 2 elevated by the weight of
the flat piece of wood or slate; then put the thinned end of it in the
notch of No. 3, and draw No. 2 down by it, until the whole forms a
resemblance of a figure 4; at the exact place where No. 3 touches the
upright, cut a notch, which, by catching the end of No. 1, will keep
the trap together. You may now bait the end of No. 3 with pieces of
cheese; a mouse, by nibbling the bait, will pull down No. 3, the other
pieces immediately separate, and the slate or board falls upon the
mouse. We have seen numbers of mice, rats and birds caught by this.