$10 \cdot 41$ Find such a three-digit number, which is 12 times the sum of its digits.
(Kyiv Mathematical Olympiad, 1964)
[Solution] Let the required three-digit number be
$$
\overline{a b c}, \quad a, b, c \in\{0,1,2, \cdots, 9\}, \quad a \neq 0 .
$$

According to the problem, we have
$$
\begin{array}{c}
100 a+10 b+c=12(a+b+c) . \\
2 b=88 a-11 c=11(8 a-c) .
\end{array}
$$

Thus,
$$
11 \mid b \text {. }
$$

Since $0 \leqslant b \leqslant 9$, then $b=0$.
At this point, we get
$$
c=8 a .
$$

Furthermore, by $1 \leqslant a \leqslant 9,0 \leqslant c \leqslant 9$, then
$$
a=1, \quad c=8 .
$$

Therefore, the three-digit number that meets the requirements is 108.