1. (BUL) Find all the three-digit numbers for which one obtains, when dividing the number by 11, the sum of the squares of the digits of the initial number.
1. Given the number $\overline{a c b}$, since $11 \mid \overline{a c}$, it follows that $c=a+b$ or $c=$ $a+b-11$. In the first case, $a^{2}+b^{2}+(a+b)^{2}=10 a+b$, and in the second case, $a^{2}+b^{2}+(a+b-11)^{2}=10(a-1)+b$. In the first case the LHS is even, and hence $b \in\{0,2,4,6,8\}$, while in the second case it is odd, and hence $b \in\{1,3,5,7,9\}$. Analyzing the 10 quadratic equations for $a$ we obtain that the only valid solutions are 550 and 803 .