Is there a way to do this problem without multiplying it out or is that the easiest way to do this question? $$\begin{align}6! = 6\times5\times4\times3\times2\times1 &\equiv 720 &\mod 7 \\&\equiv 6 &\mod 7\end{align}$$
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5Wilson's theorem. – Robert Israel Nov 01 '17 at 02:02
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16=-1 mod 7, so without multiplying, you can do -54321, see if you can apply a similar trick again – mdave16 Nov 01 '17 at 02:02
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Motivated by the proof of Wilson's theorem, you might note that $5 \cdot 3 \equiv 1 \mod 7$ and $4 \cdot 2 \equiv 1 \mod 7$.
Robert Israel
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You can make the multiplication easier by reducing the product $\mod7$ as you go along. Thus $6\times5\times4\times3\times2\times1\equiv -1\times-1\times-1\times1\equiv -1 \equiv 6 \mod7$, because $5\times4 \equiv -1\mod7$ and $3\times2 \equiv -1\mod7$.
Stephen Meskin
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One possible method is to try and get squares$$\begin{align}6\times5\times4\times3\times2\times1&\equiv (-1)\times(-2)\times(-3)\times3\times2\times1\mod 7\\&\equiv (-1^2)\times(-2^2)\times(-3^2)\\&\equiv-( 1\times4\times2)\\&\equiv-8\\&\equiv 6\mod 7\end{align}$$
John Doe
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