2. A three-digit number is 33 times greater than the sum of its digits. Prove that this number is divisible by 9, and then determine this three-digit number.
Solution. Let $100a + 10b + c$ be the desired number. Then

$$
100a + 10b + c = 33(a + b + c)
$$

The right side of the above equation is divisible by 3, so the number $100a + 10b + c$ is also divisible by 3. This means that $a + b + c = 3k$. Now we have $100a + 10b + c = 99k$, which means the desired number is divisible by 9. Therefore, $a + b + c = 9$ or $a + b + c = 18$ or $a + b + c = 27$. Since

$$
\begin{aligned}
& 33 \cdot 9 = 297 \neq 33(2 + 9 + 7) \\
& 33 \cdot 18 = 594 = 33(5 + 9 + 4) \text{ and } \\
& 33 \cdot 27 = 891 \neq 33(8 + 9 + 1)
\end{aligned}
$$

we conclude that the desired number is 594.