Problem 1. Determine the digits $a, b, c, d$ knowing that $\overline{\frac{a b c d}{8625}}=\overline{1,0 a 6}$.

Costel Anghel, Gazeta Matematică nr. 9/2014
Solution and grading criteria

From the statement, we get $8 \cdot \overline{a b c d}=69 \cdot \overline{10 a 6}$

$8 \mid \overline{10 a 6} \Rightarrow a \in\{1,5,9\}$

For $a=1$ we get $\overline{a b c d}=8763$, so $a$ is simultaneously 1 and 8, which is absurd

For $a=5$ we get $\overline{a b c d}=9108$, so $a$ is simultaneously 5 and 9, which is absurd

For $a=9$ we get $\overline{a b c d}=9453$, which works, so $a=9, b=4, c=5, d=3$