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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

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{If xy is a positive 2-digit number and u, v, x, y are digits,}\hspace{.33em}\\~\\ & \normalsize \text{ then find the number of solutions of the question}\hspace{.33em}\\~\\ & (xy)^{2}=u!+v \hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    i would say at most 7, since 8! >>> 99^2 ^^

  3. anonymous
    • one year ago
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    hey, I found one 71^2 = 7! + 1 =]]

  4. anonymous
    • one year ago
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    oh hey I found yet another one 27^2 = 6! + 9 =]]

  5. anonymous
    • one year ago
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    12^2 = 5! + 24 =]]]

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

  7. mathmath333
    • one year ago
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    how did u find that

  8. anonymous
    • one year ago
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    trial and error XD. I started with 7! and added numbers to see if I have a perfect square.

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