What is the three-digit (integer) number that is 12 times the sum of its digits?
I. solution. Let the number we are looking for be $A$, and the sum of its digits be $S$, then $A=12 S$. Therefore, $A$ is divisible by 3, but we know that in this case $S$ is also divisible by 3, so $A$ is also divisible by 9. However, this implies that $S$ is also divisible by 9. On the other hand, the sum of the digits of a three-digit number is at most $9+9+9=27$. However, $A$ is even, its last digit is not 9, so $S<27$. Thus, only 9 and 18 can be considered for $S$, from which $A$ can only be $12 \cdot 9=108$, or $12 \cdot 18=216$. In both obtained numbers, $S=9$, so only $A=108$ fits, this is the number we are looking for.

Mariann Szalay (Kunszentmárton, Mátyás kir. u. ált. isk. 8. o. t.)

II. solution. Denoting the digits of the number we are looking for by $x$, $y$, and $z$ in order, according to the problem

$$
\begin{aligned}
100 x+10 y+z & =12(x+y+z), \text { i.e., } \\
88 x-11 z & =11(8 x-z)=2 y
\end{aligned}
$$

The left side is divisible by 11, so $2 y$ and thus $y$ must also be, because 2 is not divisible by 11. Since $0<y \leq 9$, this only holds for $y=0$.

Then $8 x-z=0$, so $z=8 x$, meaning $z$ is a digit divisible by 8. The only such digit is $z=8$, so $x=1$, and the number we are looking for is 108. This indeed fits, because $108=12 \cdot 9=12(1+0+8)$.

Mária Galambos (Zirc, Reguly A. g. II. o. t.)