We prove that a nice number has at most 20 digits which proves that there are finitely many nice numbers. Assume that $n = \overline{abc}$ where $a, c$ can have more than one digit and $b$ is a digit such that $\overline{ac} = nk$ for some natural number $k$. If $c$ has $t$ digits, then $n = \overline{ac}$ (mod $10^{t-1}$) therefore we have $kn = n$ (mod $10^{t-1}$). Because there is no zero digit in $n$, we know that $\text{gcd}(n, 2) = 1$ or $\text{gcd}(n, 5) = 1$. If $\text{gcd}(n, 2) = 1$ then we have $k = 1$ (mod $2^{t-1}$), on the other hand $n < 100\overline{ac}$ so $k < 7$ which implies that $t < 9$. Similarly if $\text{gcd}(n, 2) = 5$ we have $t < 4$.

Now assume that $a$ has more than 10 digits, we have $\overline{ac} \mid \overline{ac0}$, $\overline{ac} \mid \overline{abc}$ therefore $\overline{ac} \mid \overline{abc} - \overline{ac0}$. But the previous paragraph implies that $\overline{|abc - ac0|}$ has at most 10 digits which implies that $\overline{abc} - \overline{ac0} \mid \overline{ac}$. So we should have $n = \overline{ac0}$ which is in contradiction with the fact that $n$ is an interesting number.
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