The amount of the first product (viz. 9), when added to the other product, whose respective component figures make 9, is 81; which is the square of Nine.
The said number 81, when added to the above-mentioned amount of the several products, or multiples, of 9 (viz. 405), makes 486; which, if divided by 9, gives, for a quotient, 54; that is 5 plus 4 = Nine.
It is also observable, that the number of changes that may be rung on nine bells, is 362,880; which figures added together, make 27; that is, 2 plus 7 = Nine.
And the quotient of 362,880, divided by 9, will be 40,320; that is, 4 plus 0 plus 3 plus 2 plus 0 = Nine.
To add a figure to any given number, which shall render it divisible by Nine: Add the figures named; and the figure which must be added to the sum produced, in order to render it divisible by 9, is the one required. Thus
Suppose the given number to be 7521: Add these together, and 15 will be produced; now 15 requires 3 to render it divisible by 9; and that number 3, being added to 7521, causes the same divisibility; 7521 plus 3 gives 7524, and divided by 9, gives 836. This exercise may be diversified by your specifying, before the sum is named, the particular place where the figure shall be inserted, to make the number divisible by 9; for it is exactly the same thing whether the figure be put at the head of the number, or between any two of its digits.
THE MAGIC HUNDRED.
Two persons agree to take, alternately, numbers less than a given number, 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 secret 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 makes 100." Between two persons who are equally acquainted with the game, he who begins must necessarily win.
TO GUESS THE MISSING FIGURE
To tell the figure a person has struck out of the sum of two given numbers: Arbitrarily command those numbers only, that are divisible by 9; such, for instance, as 36, 63, 81, 117, 126, 162, 261, 360, 315, and 432. Then let a person choose any two of these numbers; and, after adding them together in his mind, strike out from the sum any one of the figures he pleases. After he has so done, desire him to tell you the sum of the remaining figures; and it follows, that the number which you are obliged to add to this amount, in order to make it 9 or 18, is the one he struck out. Thus:—Suppose he chooses the numbers 162 and 261, making altogether 423, and that he strike out the center figure; the two other figures will, added together, make 7, which, to make nine, requires 2, the number struck out.