IMO 1960 - Problem 1

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