67th Romanian Mathematical Olympiad

Let $\overline{abc}$ be such a number. The condition is $\overline{abc}=13\cdot \overline{ac}$, that is $100a+10b+c=130a+13c$, same as $10b=30a+12c$. We notice that $5$ divides both $10b$ and $30a$, so $12c$ is divisible by $5$, hence $c\in \{0,5\}$.

If $c=0$, then $b=3a$. Since $a$ is a non-null digit, $(a=1,b=3)$, $(a=2,b=6)$ and $(a=3,b=9)$, which lead to the solutions $130$, $260$ and $390$.

If $c=5$, then $b=3a+6$. Since $a$ is a non-null digit, the only possibility is $(a=1,b=9)$, which leads to the solution $195$.