## mathmath333 one year ago 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

1. mathmath333

\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}}

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

12^2 = 5! + 24 =]]]

6. anonymous

7^2 = (4! + 25), but 7 is a one digit number. So not part of the solution Answer: 3 solutions ^^

7. mathmath333

how did u find that

8. anonymous

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