If u is a primitive element of Z_{37}, then:
u^(36/2) = -1
We thus have that:
36! = Product from k = 0 to 36 of u^k = u^(36*37/2) = (-1)^37 = -1.
So, we have:
33! = -34^(-1) 35^(-1)36^(-1) =
2^(-1) 3^(-1)
2^(-1) = 1/2 = 38/2 = 19
3^(-1) = 1/3 = 75/3 = 25
------>
33! = 19*25 = 31
Find the remainder when 33! is divided by 37.
Please give full solution.
1 answer