If $f(a)$ is divisible by either $101$ or $107$ for each $a\in\Bbb{Z}$, then $f(a)$ is divisible by at least one of them for all $a$ I've been struggling with this problem for a while, I really don't know where to start:

Let $f(x) \in \mathbb{Z}[X]$ be a polynomial such that for every value of $a \in \mathbb{Z}$, $f(a)$ is always a multiple of $101$ or $107$. Prove that $f(a)$ is always divisible by $101$ for all values of $a$, or that $f(a)$ is divisible by 107 for all values of $a$.


Hint: Divisiblility of $f(x)$ by $101$ depends only on the residue class of $x \mod 101$.
