Problem 979

Prove using mathematical induction.
For each positive integer $n$, there exists a unique $n$-digit number $A_{n}$, which is divisible by $5^{n}$, and its digits all belong to the set $\{1,2,3,4,5\}$.
Clearly, $A_{1}=5, A_{2}=25$.
Assume $A_{n}$ is determined, let $B_{n}=\frac{A_{n}}{5^{n}}$, then the $n+1$-digit number $\overline{c_{n+1} c_{n} \cdots c_{1}}=c_{n+1} 10^{n}+\overline{c_{n} c_{n-1} \cdots c_{1}}$ is divisible by $5^{n}$ if and only if $\overline{c_{n} c_{n-1} \cdots c_{1}}$ is divisible by $5^{n}$.
By the induction hypothesis,
$\overline{c_{n} c_{n-1} \cdots_{c_{1}}}=A_{n}=5^{n} B_{n}$.
Therefore, $\overline{c_{n+1} c_{n} \cdots c_{1}}=5^{n}\left(2^{n} c_{n+1}+B_{n}\right)$.
It is divisible by $5^{n+1}$ if and only if $2^{n} c_{n+1}+B_{n}$ is divisible by 5.
Since $\left(2^{n}, 5\right)=1$, the equation $2^{n} x+b \equiv 0(\bmod 5)$ has a unique solution in the set $\{1,2,3,4,5\}$, where $n \in \mathbf{N}_{+}, b \in \mathbf{N}$.