8.2 Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself.
Solution. Let a, b, c be the digits of the number N: $\mathrm{N}=100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}$, where $1 \leq \mathrm{a} \leq 9$, $0 \leq \mathrm{b}, \mathrm{c} \leq 9$. According to the condition, $100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}=11(\mathrm{a}+\mathrm{b}+\mathrm{c})$, i.e., $89 \mathrm{a}=\mathrm{b}+10 \mathrm{c}$.

Therefore, $89 \mathrm{a} \leq 9+10 \cdot 9=99$ and, thus, $\mathrm{a}=1$. Then $10 \mathrm{c} \leq 89$, i.e., $\mathrm{c} \leq 8.9$, but since c is an integer, then c $\leq 8$. In addition, $10 \mathrm{c}+9 \geq 89$, i.e., c $\geq 8$. Therefore, $\mathrm{c}=8$. It remains to find $\mathrm{b}: 89=\mathrm{b}+10 \cdot 8, \mathrm{~b}=9$.