Part 8

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.