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+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+7=NINE.
And the quotient of 362,880, divided by 9, will be, 40,320; that is 4+0+3+2+0=NINE.
To add a figure to any given number, which shall render it divisible by Nine: Add the figures together in your mind, which compose the number 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 those 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 make 100.” Between two persons who are equally acquainted with the game, he who begins must necessarily win.