Suppose $K$ is a number field (you may suppose $K$ is imaginary quadratic if necessary, but I doubt that matters) with ring of integers $A$, and suppose $\mathfrak{p}$ is a prime of $A$. Choose $t, u\in \mathfrak{p}^{n-1}\backslash \mathfrak{p}^n$. Is it always the case that you can find $\alpha\in A$ such that $\alpha u - t\in\mathfrak{p}^n$?
Certainly any such $\alpha$ must not lie in $\mathfrak{p}$, so that $\mathfrak{p} + (\alpha) = A$ and then $\mathfrak{p}^n + \alpha \mathfrak{p}^{n-1} = \mathfrak{p}^{n-1}$; this at least makes it plausible, but I can't find either a construction or a proof. Experimentation with a few small rings shows it to be the case.
(This is part of a larger problem which I can post if necessary.)
The classes $\bar t$ and $\bar u$ are non-zero in the quotient $\mathfrak{p}^{n-1}/\mathfrak{p}^n$.
But $\mathfrak{p}^{n-1}/\mathfrak{p}^n$ is an $A/\mathfrak{p}$-vector space of dimension $1$, thus there exists $\bar a\in A/\mathfrak{p}$ such that $$ \bar t=\bar a\cdot\bar u $$ (in fact $\bar a\neq0$).
Lifting to $A$, i.e, choosing a representant $a\in A$ the displayed formula reads $$ t-au\in\mathfrak{p}^n. $$