APPENDIX II.
ON VERIFICATION BY CASTING OUT
NINES AND ELEVENS.
The process of casting out the nines, as it is called, is one which the young computer should learn and practise, as a check upon his computations. It is not a complete check, since if one figure were made too small, and another as much too great, it would not detect this double error; but as it is very unlikely that such a double error should take place, the check furnishes a strong presumption of accuracy.
The proposition upon which this method depends is the following: If a, b, c, d be four numbers, such that
a = bc + d,
and if m be any other number whatsoever, and if a, b, c, d, severally divided by m, give the remainders p, q, r, s, then
p and qr + s
give the same remainder when divided by m (and perhaps are themselves equal).
For instance, 334 = 17 × 19 + 11;
divide these four numbers by 7, the remainders are 5, 3, 5, and 4. And 5 and 5 × 3 + 4, or 5 and 19, both leave the remainder 5 when divided by 7.
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.
This process of casting out the nines does not detect any errors in which the remainder to 9 happens to be correct. If a process be tedious, and some additional check be desirable, the method of casting out elevens may be followed after that of casting out the nines. Observe that 10 + 1, 100-1, 1000 + 1, 10000-1, &c. are all divisible by eleven. From this the following rule for the remainder of division by 11 may be deduced, and readily used by those who know the algebraical process of subtraction. For those who have not got so far, it may be doubted whether the rule can be made easier than the actual division by 11.
Subtract the first figure from the second, the result from the third, the result from the fourth, and so on. The final result, or the rest of 11 if the figure be negative, is the remainder required. Thus, to divide 1642915 by 11, and find the remainder, we have 1 from 6, 5; 5 from 4, -1; -1 from 2, 3; 3 from 9, 6; 6 from 1, -5; -5 from 5, 10; and 10 is the remainder. But 164 gives-1, and 10 is the remainder; 164291 gives-5, and 6 is the remainder. With very little practice these remainders may be read as rapidly as the number itself. Thus, for 127619833424 need only be repeated, 1, 6, 0, 1, 8, 0, 3, 0, 4, -2, 6, and 6 is the remainder.
When a question has been tried both by nines and elevens, there can be no error unless it be one which makes the result wrong by a number of times 99 exactly.