Find the remainder when 33! is divided by 37.

Please give full solution.

1 answer

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

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33! = 19*25 = 31