What is the remainder when 100! is divided by 97^2?

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100! mod $$97^2$$ = 97*(100*99*98*96! mod 97)

97 is prime. So, 96! mod 97 = 96

=> 100*99*98*96! mod 97 = 3*2*1*96 mod 97 = 3*2*1*(-1) = -6 = 91

So, required remainder = 91*97 = 8827