Q: Find the remainder upon dividing 98! by 101.
By Wilson's theorem (problem 34) we know that for every integer n>1, (n-1)! mod n = n -1 which is equivalent to the statement for every integer n>1, (n-1)! mod n = -1 mod n. (FOR N PRIME!)
We know that 100 mod 101 = -1 and 99 mod 101 = -2 just from basic modular arithmetic. We also know from Wilson's theorem that (100)! mod 101 = -1 mod 101.
The key here is that from what we have above we can rewrite 98!. Since 100! = 100*99*(98!) then 98! = (100!)/ (100*99) and thus:
98! mod 101 = (100!)/ (100*99) mod 101 = (100! mod 101)/ [(100 mod 101) *(99 mod 101)] by distribution and thus:
98! mod 101 = (-1)/(-1 * -2) = -1/2
Which is actually the remainder upon diving 98! by 101.
Moreover under mod 101 we can rewrite -1/2 as 50 since:
-1/2 = x mod 101 -1 = 2x mod 101 Since x=50 can satisfy this equation we are set. --Bakey 11:39, 29 October 2012 (UTC)
