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Jeremy Ting edited p7.tex
about 10 years ago
Commit id: 17162df76eba74eef6542825580e872eadeeea5e
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Let n=5:
$4! \equiv -1\ (\textrm{mod}\ n)$
$4! \equiv -1\ (\textrm{mod}\ 5)$
$4*3*2*1 \equiv -1\ (\textrm{mod}\ 5)$
$(4*1)(3*2) \equiv -1\ (\textrm{mod}\ 5)$
$(4*1) \equiv -1\ (\textrm{mod}\ 5)$
$(3*2) \equiv 1\ (\textrm{mod}\ 5)$
Then via Wilson's Theorem, it will then equal to $1\ (\textrm{mod}\ 5)$
\end{itemize}