Prime numbers as factors

Fermat's Little Theorem provides the answer!

It states that for any prime $p$

$$a^{p-1}-1 \text{ mod(p)}\equiv 0$$

If a and p are relatively prime

Plugging in $a=10$ gives us

$$10^{p-1}-1 \text{ mod(p)} \equiv 0$$

Doing some algebra, we get

$$10^{p-1}-1 \text{ mod(p)} \equiv (10-1) * (\sum^{p-2}_{j=0}{10^j})\text{ mod(p)}\equiv0$$

And note that $$ \sum^{p-2}_{j=0}{10^j}\text{ mod(p)}\equiv0$$ is just the sum of a bunch of 1 digits, a repunit as @prets would call it.

Also, note that this is not true when $p=5$, since 10 and 5 are not relatively prime, but it is true for all prime examples greater than 5.