Question

What is the remainder when 100! Is divided by ${97}^2$.

Answer 1

$\frac{100!}{{97}^2}=\frac{100\times 99\times 98\times 97\times 96!}{97\times 97}=\frac{100\times 99\times 98\times 96!}{97}$
From Wilson Theorem,
Remainder $\left[\frac{(p-1)!}{p}\right]=(p-1)$, if p is a prime number.
Therefore
Remainder of $\left[\frac{100\times 99\times 98\times 96!}{97}\right]$ by Wilson Theorem
$=[3\times 2\times 1\times 96][\because \frac{(97-1)!}{97}=\frac{96!}{97}=Remainderof(97-1)=96]$
= 576 mod 97
= 91 mod 97
= 91