Iranian Mathematical Olympiad

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. ■