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

Solution: Note that $P+B+\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers $\overline{P B \Gamma}$ and $(P + B + \Gamma)$ should give the same remainder when divided by 9. This is only possible when $P+B+\Gamma$ is a multiple of 3. Note that $P+B+\Gamma=9$ does not work, since $(P + B + \Gamma) \times (P+B+\Gamma+1)=90$ is a two-digit number. By trying $12,15,18,21,24$, we get $\overline{P B \Gamma}=$ $156=12 \times 13$.