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Show: $(27!)^6 \equiv 1 \pmod{899}$. Note: $899=30^2 - 1$.

Could anyone show me how to go through this please? I think I have to use Wilson's theorem which is $(p-1)! \equiv -1 \pmod p$ but not exactly sure how else to tackle this. Thanks!

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    Hints: $30^2-1=(30+1)(30-1)$; $30+1$ and $30-1$ are close enough to $27$ to make Wilson's Theorem an attractive option; $n\equiv-(p-n)\pmod p$.2012-12-18
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    could you expand on that a little please? I still dont quite grasp the concept2012-12-18
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    No, but you could try thinking about it for more than three minutes.2012-12-18

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