3-3. A two-digit number, when added to the number written with the same digits but in reverse order, gives a perfect square. Find all such numbers.
Solve 3. Answer: $29,38,47,56,65,74,83,92$.

Let $10a+b$ be the desired number. According to the condition, the number $(10a+b)+(10b+a)=11(a+b)$ is a square of some number $k$. Then $k$ is divisible by 11, which means $a+b$ is also divisible by 11. But $a+b \leqslant 18$, so $a+b=11$.