2. Determine the two-digit number that is equal to the sum of the cube of the value of the tens digit and the square of the value of the units digit of that two-digit number.
Solution. Let the desired number be $\overline{a b}(a \neq 0)$. Then $10 a+b=a^{3}+b^{2}$, so $10 a-a^{3}=b^{2}-b$, i.e., $a\left(10-a^{2}\right)=b(b-1)$. Clearly, $a \in\{1,2,3\}$. For $a=1$ we get $9=b(b-1)$ and for $a=3$ we get $3=b(b-1)$, which is not possible (Why?). For $a=2$ we get $12=b(b-1)$, from which we find $b=4$. Finally, the desired number is 24.