4. Determine all three-digit numbers that are divisible by 11, and the sum of their digits is 10.
Solution: Let's denote with $\overline{\mathrm{abc}}$ (where $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$ are digits) the required three-digit numbers. For them, the following holds: $11 \mid \overline{a b c}$ and $a+b+c=10$.

$\overline{a b c}=100 a+10 b+c=99 a+9 b+(a+b+c)$ Since $a+b+c=10$, it follows that $\overline{a b c}=$ $99 a+9 b+10$, or $\overline{a b c}=99 a+11 b+11-(2 b+1)$. It is clear that $99 a+11 b+11$ is divisible by 11. Since $11 \mid \overline{a b c}$, it follows that $11 \mid 2 b+1$, and this is satisfied only if $2 b+1=0$ or $2 b+1=11$. Since $\mathrm{b}$ is a digit, it follows that $2 b+1 \neq 0$, and from $2 b+1=11$ it follows that $\mathrm{b}=5$. From the condition, it follows that $\mathrm{a} c=5$, so the required three-digit numbers are: 154, 253, 352, 451, and 550.