In an attempt to present some of the most interesting and suggestive phases of mathematics, professor William F. White authored the expository book, which is entitled, ‘A Scrap Book of Elementary Mathematics’.

In this article, we bring you the various tests and procedures that are in use to determine divisibility of a number. The methods are illustrated elegantly by William F. White, who writes:

Let M represent any integer containing no prime factor that is not a factor of 10 (that is, no primes but 5 and 2). Then 1/M expressed decimally is terminate. Call the number of places in the decimal m. Let N represent any prime except 5, 2, 1. Then the reciprocal of N expressed decimally is a circulate. Call the number of places in the repetend n.

  1. The remainder obtained by dividing any integer, I, by M is the same as that obtained by dividing the number represented by the last (right-hand) m digits of I by M. If the number represented by those m digits is divisible by M, I is divisible by M, and not otherwise.
  2. The remainder obtained by dividing I by N is the same as that obtained by dividing the sum of the numbers expressed by the successive periods of n digits of I by N. If that sum is divisible by N, I is divisible by N, and not otherwise.

Additional reading *this depends on Fermat’s theorem, that Pp–1 – 1 is divisible by p when p and P are prime to each other*.

  1. If a number is composite and contains a prime factor other than 5 and 2, the divisibility of I by it may be tested by testing with the factors separately by (1) and (2).

It needs to be noted here in this section, that when we talk of ‘divisibility’ or say ‘divisible’, we mean to say ‘divisible without remainder’.

Thus it is possible to test the divisibility of any integer by any other integer. This is usually of only theoretic interest, as actual division is preferable. But in the case of 2, 3, 4, 5, 6, 8, 9, and 10 the test is easy and practical. A simple statement of it for each of these particular cases is found in almost any arithmetic.

For 11 a test slightly easier than the special application of the general test is usually given. That is, subtract the sum of the even numbered digits from the sum of the odd-numbered digits (counting from the right) and add 11 to the minuend if smaller than the subtrahend. The result gives the same remainder when divided by 11 as the original number gives. The original number is divisible by 11 if that result is, and not otherwise. These remainders may be used in the same manner as the remainders used in casting out the nines, but are not so conveniently obtained.

Test of divisibility by 7. No known form of the general test in this case is as easy as actually dividing by 7. From the point of view of theory it may be worth noticing that, as 7’s reciprocal gives a complementary repented, the general test admits of variety of form.* Let us consider, however, the direct application.

Since the repented has 6 places, the test for divisibility by 7 is as follows: A number is divisible by 7 if the sum of the numbers represented by the successive periods of 6 figures each is divisible by 7, and not otherwise; e. g.,

Given the number: 26, 436,080,216,581.

Dividing this large number by 7 to check for divisibility is a hard task. We use the divisibility method (the one delineated above), to check its divisibility by 7.

216581

436080

26

No remainder; therefore the given number is divisible by 7.

Test of divisibility by 7, 11, and 13 at the same time. Since 7x11x13 = 1001, divide the given number by 1001. If the remainder is divisible by 7, 11, or 13, the given number is also, and not otherwise.

To divide by 1001, subtract each digit from the third digit following. An inspection of a division by 1001 will show why this simple rule holds. The method may be made clear by an example, 4,728,350,169 ÷1001.

4728350169

4724626543

3

Quotient, 4723626; remainder, 543.

The third digit before the 4 being 0 (understood), write the difference, 4, beneath the 4. Similarly for 7 and 2. 8 – 4 = 4 (which for illustration is here written beneath the 8). We should next have 3 – 7. This changes the 4 just found to 3, and puts 6 under the original 3 (that is, 83 – 47 = 36). 5 – 2, 0 – 3 (always subtracting from a digit of the original number the third digit to the left in the difference, or lower, number), 1 – 6, etc. Making the corrections mentally we have the number as written. The number represented by the last three digits, 543, is the remainder after dividing the given number by 1001, and the number represented by the other digits, 4723626, is the quotient. With a little practice, this method can be applied rapidly and without making erasures. The remainder, 543, which alone is needed for the test, may also be obtained by subtracting the sum of the even-numbered periods of three figures each in the original number from the sum of the odd-numbered periods. A rapid method of obtaining the remainder thus is easily acquired; but the way illustrated above is more convenient.

However obtained, the remainder is divisible or not by 7, 11, or 13, according as the given number is divisible or not. (Here 543 are not divisible by 7, 11, or 13; therefore 4728350169 are not divisible by either of them.) The original number is thus replaced, for the purpose of investigation, by a number of three places at most. As these tests for three common primes at once, it is convenient for one factoring large numbers without a factor table.

*The content in italics is the added text*

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