Problem 4. Find all solutions to the puzzle

$$
\mathrm{G}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}=\mathrm{Mb} .
$$

(Same letters encode the same digits, different letters encode different digits.)

[5 points] (D. E. Schnol)
Answer. $0+12+12+12+12+12+12+12+12=96$.

Solution. Since $8 \cdot \mathrm{OH}<100$, then $\mathrm{OH} \leqslant 12$, which means $\mathrm{O}=1$, and Н is either 0 or 2. But $\mathrm{H}$ cannot be 0, because then Я and Ы would represent the same digit. Therefore, $\mathrm{H}=2$. Thus, $8 \cdot \mathrm{OH}=8 \cdot 12=96$, so МЫ can be 96, 97, or 98. The last two cases do not work, because for them Я would have to be 1 or 2, and these digits are already used. Therefore, МЫ $=96$, and $Я=0$.