[ $\left.\frac{\text { Evenness and Oddness }}{\text { Case Enumeration }}\right]$

Find such a three-digit number $A^{2}$, which is a perfect square, that the product of its digits equals $A-1$.

#
Let $A^{2}=100 x+10 y+z$. Since $100 \leq A^{2} \leq 599$.

In the second case, we check the squares of the numbers 19 and 29; only 19 fits.

## Answer

361.