Question

If xy is a positive 2-digit number and u, v, x, y are digits,
then find the number of solutions of the question
& (xy)^{2}=u!+v

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{If xy is a positive 2-digit number and u, v, x, y are digits,}\hspace{.33em}\\~\\
& \normalsize \text{ then find the number of solutions of the question}\hspace{.33em}\\~\\
& (xy)^{2}=u!+v \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

i would say at most 7, since 8! >>> 99^2

^^

Answer 3

hey, I found one
71^2 = 7! + 1

=]]

Answer 4

oh hey I found yet another one
27^2 = 6! + 9

=]]

Answer 5

12^2 = 5! + 24

=]]]

Answer 6

7^2 = (4! + 25), but 7 is a one digit number. So not part of the solution

Answer: 3 solutions

^^

Answer 7

how did u find that

Answer 8

trial and error XD. I started with 7! and added numbers to see if I have a perfect square.