The divisibility test for 3 and 9 is fairly well known. To find out of any given integer is divisible by 9, add up the digits of the number; if the result is divisible by 9 then the number is divisible by 9.

What’s a bit less well known is that in any base *b*, the same trick applies for *b-1* and all of its divisors.

For example, if we are doing arithmetic in hexadecimal (base 16), the divisibility rule we use for base 10 applies not for 3 and 9, but instead for 3, 5, and 15.

Suppose we were to confirm that is divisible by . Adding up the digits:

And since is divisible by , so is the original number.

### Proof

Let *n *be an integer, *b *be the base, and *m *be the number of digits in *n*. Then we can represent *n*:

We can rearrange this:

Each of the expressions , , up to are divisible by . I’ve explored the proof for this in an earlier blog post.

Thus the entire first group of the above expression is divisible by *b-1*. What remains is the second group, which happens to be the sum of digits in *n*.

The sum of digits in *n* is divisible by *b-1* if and only if n is divisible by *b-1*, which is what we wanted to prove.

For a factor of *b-1*, this also works as the entire first expression is divisible by any factor of *b-1*.