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I want to find the least residue of $95!$ mod $97$. So I wanna solve $$95! \equiv_{97} x$$ where $x \in \mathbb{Z}^+$ is minimum, but Wilsons theorem

$$96! \equiv_{97} -1$$

so multiplying the top equation by 96 I obtain

$$-1 \equiv_{97} 96x$$

which forces $x=1$. Did I do this correctly using Wilsons theorem?

anonymous
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    Just to be sure, the $19$'s are meant to be $97,$ right? If so then I agree, but if not then it's clearly false because $19 \leq 95 \Rightarrow 19 | 95! \Rightarrow 95! \equiv_{19} 0$ – Stephen Donovan Jun 15 '22 at 05:31
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    Does this answer your question? [Showing $p$ divides $(p-2)!-1$ when $p$ is prime](https://math.stackexchange.com/questions/102378/showing-p-divides-p-2-1-when-p-is-prime) (note that $97$ is prime) - found using an [Approach0 search](https://approach0.xyz/search/?q=OR%20content%3A%24(p-2)!%5Cequiv%201%5Cpmod%7Bp%7D%24&p=1). – John Omielan Jun 15 '22 at 05:56
  • @StephenDonovan yes edit made. Does it work ? – anonymous Jun 15 '22 at 16:02

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