1. Do there exist three different digits $A, B, C$, such that $\overline{A B C}, \overline{C B A}, \overline{C A B}$ are squares of natural numbers? (The bar over the digits denotes the number formed by these digits in the given order.)
Solution. Yes. These are the digits $A=9, B=6, C=1$. In this case $961=31^{2}, 169=13^{2}, 196=14^{2}$.