## Task B-1.5.

Determine all three-digit numbers that are three times greater than the square of the sum of their digits.
## Second solution.

Let $x=\overline{a b c}$ be the desired number.

According to the condition of the problem, $\overline{a b c}=3(a+b+c)^{2}$.

Obviously, the desired number is divisible by 3, and then the sum of its digits is also divisible by 3, i.e., $a+b+c=3 k, k \in \mathbb{N}$.

Then $\overline{a b c}=3(a+b+c)^{2}=27 k^{2}$.

This will be a three-digit number if $k^{2} \in\{4,9,16,25,36\}$. 1 point

It follows that $\overline{a b c} \in\{108,243,432,675,972\}$.

By direct verification, we conclude that the numbers satisfying the conditions of the problem are

$243=3(2+4+3)^{2}$ and $972=3(9+7+2)^{2}$.