If no elements of a sequence $a_n$ are divisible by $\pi$, does $\forall n, a_n \mod \pi \in (0;\pi)$ hold?

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Given a sequence like $a_n = n$ or $a_n = 50n$, (or any arbitrary constant), and that no element of the sequence is divisbile by $\pi$, would $b_n = a_n \mod \pi$ eventually take on all values in the interval $(0; \pi)$?

In more general sense, it doesn't have to be $\pi$, since if the above would work, the same would apply to any arbitrary number $p$, where no element of $a_n$ is divisible by $p$ and the interval would be $(0;p)$.

If such property is true, how would one prove it?