7. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of different digits $\Pi, B$, and $\Gamma$, for which the equality $\overline{\Pi B \Gamma}=(\Pi+B+\Gamma) \times(\Pi+B+\Gamma+1)$ holds.
Answer: 156.

Solution: Note that П+В $\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers

$\overline{\Pi В \Gamma}$ and $(П+B+\Gamma)$ must give the same remainder when divided by 9. This is only possible when П+В $\Gamma) \times(П+B+\Gamma+1)=90$-a two-digit number. By trying $12,15,18,21,24$, we get $\overline{\Pi В \Gamma}=$ $156=12 \times 13$.

## Variant 1-b