Divisibility Tests

Article by Tim Rowland

Published January 1997,October 2010,February 2011.

Multiples of 6 and 12

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

This is not at all obvious: it is true because 2×3=6 and because 2 and 3 are 'coprime' - i.e. they have no common factor (apart from 1).

Example: 1638 is even and its digital root is 9. Therefore it is a multiple of 6.

Similarly, a number is divisible by 12 if and only if it is divisible by both 3 and 4 - because 3×4=12, and 3 and 4 are coprime.

Multiples of 11

The test for 11 is a modified version of that for 3 and 9.

Whereas every power of 10 is 1 more than a multiple of 3 (or 9), an alternating pattern emerges for multiples of 11. That is to say, 10 is 1 less than 11, 100 is 1 more than 9×11, 1000 is 1 less than 91×11, 10000 is 1 more than 909×11, and so on. If we write `m11' as shorthand for 'a multiple of 11', we see that odd powers of 10 are m11−1, and even powers of 10 are m11+1.

Example:

Is 54637 divisible by 11?

Solution: Start with the units digit and work 'left':

54637=7+3×(m11−1)+6×(m11+1)+4×(m11−1)+5×(m11+1),

which equals m11+(7−3+6−4+5) or m11+11. Therefore 54637 must be a multiple of 11.

This is only slightly more complicated than finding the digital root of a number, because we alternately add and subtract the digits, starting from the right. (We could call the answer the 'alternating digital root').

Example:

What is the remainder when 710−7 is divided by 11?

Solution:

710−7=282475242 whose alternating digital root is 2−4+2−5+7−4+2−8+2=−6. Our number is 6 less than a multiple of 11, so if we divide it by 11, the remainder will be 5.

A number is divisible by 11 if its alternating digital root is 0 or 11 (or any other multiple of 11).