Any number, therefore, being used as a divisor, may be made a check upon the correctness of an operation. To provide a check which may be most fit for use, we must take a divisor the remainder to which is most easily found. The most convenient divisors are 3, 9, and 11, of which 9 is far the most useful. As to the numbers 3 and 9, the remainder is always the same as that of the sum of the digits. For instance, required the remainder of 246120377 divided by 9. The sum of the digits is 2 + 4 + 6 + 1 + 2 + 0 + 3 + 7 + 7, or 32, which gives the remainder 5. But the easiest way of proceeding is by throwing out nines as fast as they arise in the sum. Thus, repeat 2, 6 (2 + 4), 12 (6 + 6), say 3 (throwing out 9), 4, 6, 9 (throw this away), 7, 14, (or throwing out the 9) 5. This is the remainder required, as would appear by dividing 246120377 by 9. A proof may be given thus: It is obvious that each of the numbers, 1, 10, 100, 1000, &c. divided by 9, leaves a remainder 1, since they are 1, 9 + 1, 99 + 1, &c. Consequently, 2, 20, 200, &c. leave the remainder 2; 3, 30, 300, the remainder 3; and so on. If, then, we divide, say 1764 by 9 in parcels, 1000 will be one more than an exact number of nines, 700 will be seven more, and 60 will be six more. So, then, from 1, 7, 6, 4, put together, and the nines taken out, comes the only remainder which can come from 1764. To apply this process to a multiplication: It is asserted, in page 32, that 10004569 × 3163 = 31644451747. In casting out the nines from the first, all that is necessary to repeat is, one, five, ten, one, _seven_; in the second, three, four, ten, one, _four_; in the third, three, four, ten, one, five, nine, four, nine, eight, twelve, three, ten, _one_. The remainders then are, 7, 4, 1. Now, 7 × 4 is 28, which, casting out the nines, gives 1, the same as the product. Again, in page 43, it is asserted that 23796484 = 130000 × 183 + 6484. Cast out the nines from 13000, 183, 6484, and we have 4, 3, and 4. Now, 4 × 3 + 4, with the nines cast out, gives 7; and so does 23796484. To avoid having to remember the result of one side of the equation, or to write it down, in order to confront it with the result of the other side, proceed as follows: Having got the remainder of the more complicated side, into which two or more numbers enter, subtract it from 9, and carry the remainder into the simple side, in which there is only one number. Then the remainder of that side ought to be 0. Thus, having got 7 from the left-hand of the preceding, take 2, the rest of 9, forget 7, and carry in 2 as a beginning to the left-hand side, giving 2, 4, 7, 14, 5, 11, 2, 6, 14, 5, 9, 0. Practice will enable the student to cast out nines with great rapidity.