Find the remainder when x^100 + 2x + 10 is divided by x − 11 in Z17[x].
Simplify the result using Fermat’s theorem.
I started to do long division, but it was not going well for me. I never have done Fermat's Theorem with this kind of problem before, so if someone could kind of tell me how to start this problem, I'd really appreciate it!
Oh, alright. Thank you.
This is what I did:
Before solving 11^100 + 2*11 + 10, I had figured out earlier that x^100 (mod 17) = x^4 (mod 17)
so, 11^100 (mod 17) = 11^4 (mod 17)
inserting back into the equation, I got:
11^4 + 22 + 10 = 14641 + 22 + 10 = 14673 = 2 (mod 17)