From Wilson's Theorem, we know that $ \scriptstyle100!\ mod\ 101\ =\ 100. $
Then dividing by $ \scriptstyle100 $, $ \scriptstyle99!\ mod\ 101\ =\ 98!\cdot99\ mod\ 101\ =\ 1 $.
We want to get rid of the $ \scriptstyle99 $, that is, reduce it somehow to a unity. We know that $ \scriptstyle99\ \equiv\ -2 $ and $ \scriptstyle-100\ \equiv\ 1 $, so the easiest course would be to simply multiply both sides by 50:
$ \scriptstyle98!*-2*50\ mod\ 101\ =\ 50 $
$ \scriptstyle\Rightarrow98!*-100\ mod\ 101\ =\ 50 $
$ \scriptstyle\Rightarrow98!*1\ mod\ 101\ =\ 50 $
$ \scriptstyle\Rightarrow98!\ mod\ 101\ =\ 50 $. $ \scriptstyle\checkmark $
- --Nick Rupley 06:55, 2 April 2009 (UTC)
