Slovenia — National Math Olympiad, 2012

Suppose the decimal notation of a natural number $n$ is equal to $\overline{a_k a_{k-1} \dots a_2 a_1}$. The main condition of the problem says that the number $\overline{a_k a_{k-1} \dots a_2}$ divides the number $\overline{a_k a_{k-1} \dots a_2 a_1} = 10 \cdot \overline{a_k a_{k-1} \dots a_2} + a_1$, hence it also divides the number $a_1$. Since $a_1 \neq 0$, the number $\overline{a_k a_{k-1} \dots a_2}$ can contain at most one digit, hence $k=2$.

If now $n = \overline{a_2 a_1}$, then $a_2$ must divide $a_1$, from which we conclude $a_2 = a_1$ or $a_2 \le 4$. The number $n$ can thus only be equal to 99, 88, 77, 66, 55, 48, 44, 39, 36, 33, 28, 26, 24, 22, 19, 18, 17, 16, 15, 14, 13, 12 or 11. Among these numbers only 99, 88, 77, 66, 55, 48, 44, 36, 33, 24, 22, 15, 12 and 11 fulfill the condition of the problem.

To summarize, the solutions are 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99.