Problem: Prove that for every positive integer $n$
there exists an $n$-digit number divisible by $5^n$
all of whose digits are odd.

Solution: This is immediate by induction on $n$.
For $n = 1$ we take $5$;
moving forward if $M$ is a working $n$-digit number then exactly one of
\begin{align*}
  N_1 &= 10^n + M \\
  N_3 &= 3 \cdot 10^n + M \\
  N_5 &= 5 \cdot 10^n + M \\
  N_7 &= 7 \cdot 10^n + M \\
  N_9 &= 9 \cdot 10^n + M
\end{align*}
is divisible by $5^{n+1}$;
as they are all divisible by $5^n$
and $N_k/5^n$ are all distinct.