AN EASY PROOF FOR SUMS IN MULTIPLICATION.
As boys are always interested in short cuts in arithmetical processes, it may be well to insert for the benefit of those who are studying multiplication, a method of proving their examples which I learned a short time ago from an old banker of New York. This rule is simply to add the digits of both multiplicand and multiplier, divide both answers by 9, and multiply the remainders; divide this product by 9 and the remainder will be, if the example is correct, the same as that obtained by adding the digits of the product and dividing that answer by 9. For instance, suppose after multiplying 4359 by 2786 we have 12144174 for the answer; now instead of performing this operation over a second time to make sure our answer is correct, we simply add the digits in 4359 and divide the sum 21 by 9, we find we have 3 left. As it is the only remainder we have to deal with, we need not keep the other figures. By adding the digits in the multiplier we obtain 23, which divided by 9 gives 2 and 5 remainder. Now, multiplying the first remainder by the second we have 15: this product divided by 9 gives 1 and 6 remainder. If the product 12144174 is correct, the sum of its digits divided by 9 will leave 6 for a remainder. Performing the operation, we find the sum of its digits is 24, divided by 9 equals 2 and 6 remainder. As both the remainders correspond, the answer was correct. After a little practice you will find you can prove your examples very quickly by this method, and where a number of sums are given without the answers it will be of invaluable assistance, besides saving you a great amount of labor.
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