When students learn about decimals, they’re told that every fraction either has a terminating decimal expansion or a repeating decimal expansion. The previous post gives a constructive proof of the converse: given a repeating fraction (in any base) it shows how to find what rational number it corresponds to.

Maybe you learned this so long ago that it seems obvious. But here’s a corollary that I don’t believe is so obvious.

For any positive integer *n* and for any integer *b* > 1, with *n* and *b* relatively prime, there is some number of the form *b*^{k} − 1 which is divisible by *n*.

So if I take some number, say 37, I can find a hexadecimal number consisting entirely of Fs which is divisible by 37. In fact FFFFFFFFF will do the trick. This is just the statement above with *b* = 16.

How does this follow from every rational number having a repeating decimal expansion?

Write 1/*n* as a repeating decimal in base *b*. (The assumption that *n* and *b* are relatively prime implies that the decimal repeats.) If the period of this decimal is *k*, then the previous post shows that 1/*n* is a multiple of 1/(*b*^{k} − 1). It follows that *n* is a divisor of *b*^{k} − 1.

**Related post**: A bevy of ones

I think you need the word ‘coprime’ in here somewhere. The number 999…9 never divides by 10. I guess this must have something to do with decimals that end 000… .

Thanks. I updated the post to say that n and b must be relatively prime.