I'm trying to figure out if given a rational number $\frac{a}{p^n}$ with $0 < a \leq p^n$ if you will always be able to find an integer multiple of the rational $\frac{b}{p^m}$ with $0 < b \leq p^m$ give that $n <<< m$ . I know it would be number theory but am unsure exactly what to search. I feel like knowing more about $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/p^n\mathbb{Z}$, with $n \in \mathbb{Z}^{+}$ will help but I'm not 100% sure. What should I be reading through and searching for?
For clarity the exact problem is as follows.
Let $p$ be a prime. Given $x = exp(\frac{a2i\pi}{p^n})$ and $y = exp(\frac{b2i\pi}{p^m})$ with $n, m, a, b \in \mathbb{Z}^{+}$, ($n < m$), and $ 0 < a < p^{n}$ and $ 0 < b < p^{m}$. Will some integer power of $y$ generate $x$ if $m$ is large enough (but we can't know exactly what $m$ is)?
For $p$ prime and $p \nmid a, p \nmid b$ then $$\exists k,\quad \exp(2i \pi a/p^n)^k=\exp(2i \pi b/p^m)$$ iff $m \le n$.
Proof : take $ac\equiv 1 \bmod p^n$ (for example $c = a^{(p-1)p^{n-1}-1}$) and $k =cb p^{n-m} $.