Prove that 61! = -1mod71

#HelpMe! #MathProblem #Math

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

This is a simple application of Wilson's Theorem.

First, note that $9! \equiv -1$ (mod $71$ ).

Then, since $71$ is a prime, by Wilson's Theorem,

$70! \equiv -1$ (mod $71$ ).

But $70! \equiv 61! (62)(63)(64)...(70) \equiv 61! (-9)(-8)(-7)...(-1)$ $\equiv -61!(9!) \equiv 61!$ (mod $71$ ).

But $70! \equiv -1$ (mod $71$ ).

So $61! \equiv 70! \equiv -1$ (mod $71$ ).

Zi Song Yeoh - 8 years, 5 months ago

I'm not that great at modulus arithmetic but how can 61!= -1 mod(7) ? Doesn't that mean that it is smaller than 7 ?

johnson adeleke - 8 years, 5 months ago

Oh I see why I was looking at it the wrong way. Nice proof Zi song :)

johnson adeleke - 8 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$