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. |