757. Find all three-digit numbers that are 11 times the sum of their digits.
$\triangle$ Let the desired number be $\overline{a b c}$. We get:

$$
\overline{a b c}=11(a+b+c), \quad 100 a+10 b+c=11 a+11 b+11 c, \quad 89 a=b+10 c
$$

Let's try using divisibility by 10 (but we can also use 11):

$$
90 a-a=b+10 c, \quad 90 a-10 c=a+b .
$$

Therefore, the sum $a+b$ is divisible by 10, i.e., $a+b=10$. Then the last equation becomes:

$$
90 a-10 c=10, \quad 9 a-c=1, \quad 9 a=c+1
$$

From this, $a=1, c=8$. Let's find $b$ using the fact that $a+b=10: \quad b=9$.

Answer: 198.