, 2012

Let us write $n = \overline{abc}$. The condition of the problem says that $\overline{bc}$ divides $\overline{abc} = 100a + \overline{bc}$, hence it also divides $100a$. First suppose that $\overline{bc}$ is not divisible by $5$. Then $\overline{bc}$ must divide $4a$. Because $\overline{bc}$ is a 2-digit natural number, $4a$ must also be a 2-digit natural number, hence $3 \le a \le 9$. The number $4a$ can thus take values $12$, $16$, $20$, $24$, $28$, $32$ or $36$. 2-digit divisors of these numbers are $12$, $16$, $20$ and $10$, $24$ and $12$, $28$ and $14$, $32$ and $16$, $36$ and $18$ and $12$, respectively. Allowing only non-zero digits, we get the following solutions: $312$, $416$, $624$, $612$, $728$, $714$, $832$, $816$, $936$, $918$ and $912$.

If, on the other hand, $\overline{bc}$ is divisible by $5$, we get $c = 5$, because $c$ must be a non-zero digit. Hence $\overline{bc} = 10b + 5 = 5(2b + 1)$, from which we conclude that $2b + 1$ must divide $20a$. Because $2b + 1$ is odd, it must divide $5a$. If $2b + 1$ is divisible by $5$, we get $b = 2$ (hence $2b + 1 = 5$) and $a$ is arbitrary or $b = 7$ (hence $2b + 1 = 15$) and $a$ is equal to $3$, $6$ or $9$. We get the following solutions: $125$, $225$, $325$, $425$, $525$, $625$, $725$, $825$ and $925$ as well as $375$, $675$ and $975$.

If, on the other hand, $2b + 1 \ne 5$, it is coprime to $5$ and hence a divisor of $a$. It must hold $2b + 1 \le 9$ and therefore $b \le 4$ and $b \ne 2$ (otherwise we would have $2b + 1 = 5$). If $b$ is successively equal to $4$, $3$ and $1$, then $a$ can be successively equal to $9$, $7$ and $3$ or $6$ or $9$. We get the following solutions: $945$, $735$, $315$, $615$ and $915$.

To summarize, the solutions are: $125$, $225$, $312$, $315$, $325$, $375$, $416$, $425$, $525$, $612$, $615$, $624$, $625$, $675$, $714$, $725$, $728$, $735$, $816$, $825$, $832$, $912$, $915$, $918$, $925$, $936$, $945$, $975$.