Divisibility

Let us consider an arbitrary base ten number in expanded form,

where each of the coefficients is a digit from 0 to 9.

For divisibility by 2, 5, or 10 we simply split off the last term and factor out 10 getting

For each of the numbers 2, 5, or 10 the term on the left [ ] 10 is a multiple. So, divisibility depends only on the last digit.

For divisibility by 4 we operate similarly pulling off the last two terms and factoring out 100 getting

Since 100 is a multiple of 4 the original number is a multiple iff the number made by the last two digits is a multiple.

Divisibility by 8 works similarly except that here we need 1000 to get a multiple of 8 and so we must split off the last three digits.

Divisibility by 3 & 9 proceeds somewhat differently. Since a power of ten minus 1 is a string of 9’s (e.g. 1000-1 is 999) we have the following expansion

If we factor out each of the terms on the right of the square brackets we get the sum of two expansions

For either 3 or 9 each of the terms in the first expression is a multiple so divisibility by these numbers depends only on the second expression. Thus divisibility depends on the sum of the digits. Since we are only interested in divisibility we can discard any sums which add up to the number in question and subtract a 9 whenever the sum gets to ten or more.

Divisibility by 11 depends on an interesting point. Even powers of 10 are (a multiple of 99) + 1, while odd powers of 10 are (a multiple of 11) - 1. Thus when you group the multiples of 11 on the left you are left with where you alternately add and subtract the digits, starting by adding the units digit. For divisibility it will not matter if you start at the left but you will have problems later with clock arithmetic.

Divisibility by 6 requires divisibility by both 2 & 3.

Divisibility by 12 requires divisibility by both 3 & 4.

There is no easy rule for divisibility by 7.