## Problem 2

$100a + 10b + c$ is divisible by 17

$\Rightarrow 15a + 10b + c$ is divisible by 17 (1)

But $12a - 6b + c$ is divisible by 17

Adding the last two relations, we obtain that: $27a + 4b + 2c$ is divisible by 17 $2 p$.

Thus $5a + 2b + c$ is divisible by 17 $1 p$.

Subtracting the last relation from relation (1) we obtain $5a + 4b$ divisible by 17 $1 p$.

Since $a$ and $b$ are digits, $a \neq 0$, we have $5 \leq 5a + 4b \leq 81$, so $5a + 4b \in \{17, 34, 51, 68\}$ $1 p$

Analyzing the possible cases, we obtain the solutions: 136; $612; 748$ $2 p$.
## Problem 2

Determine the natural numbers of the form $\overline{abc}$ divisible by 17, knowing that $12a - 6b + c$ is a natural number divisible by 17.

Selected by Prof. Ursan Rodica