9.1. $\left[7-8.1\right.$ (15 points)] Find all solutions to the numerical riddle AB $=$ B $^{V}$ (different letters correspond to different digits; on the left side, there is a two-digit number, not the product of digits A and B).

Answer: $32=2^{5}, 36=6^{2}, 64=4^{3}$.
Solution. If $\mathrm{B}=1$, then Б ${ }^{\mathrm{B}}=$ Б - is not a two-digit number. Let's list the two-digit numbers obtained if $\mathrm{B} \geqslant 2$:

1) $\mathrm{B}=2: 4^{2}=16,5^{2}=25,6^{2}=36,7^{2}=49,8^{2}=64,9^{2}=81$. In this case, the solution to the puzzle is only the equality $36=6^{2}$.
2) $\mathrm{B}=3: 3^{3}=27,4^{3}=64$. In this case, we also get a unique solution $64=4^{3}$.
3) $\mathrm{B}=4: 2^{4}=16,3^{4}=81$. In this case, there are no solutions.
4) $\mathrm{B}=5: 2^{5}=32$. In this case, we have a unique solution $32=2^{5}$.
5) $\mathrm{B}=6: 2^{6}=64-$ is not a solution.

For $\mathrm{B} \geqslant 6$ and $Б \geqslant 2$, the number Б $^{\mathrm{B}}$ will always be greater than 100. Thus, the puzzle has 3 solutions: $32=2^{5}, 36=6^{2}, 64=4^{3}$.