So far this is what I can do:
Using Fermat's Little Theorem I know that 3^{16}\equiv 1 \pmod {17}
Also: 3^{2012} = (3^{16})^{125}*3^{12} \pmod{17}
So I am left with 3^{12}\pmod{17}.
Again I'm going to use fermat's theorem so: 3^{12} = \frac{3^{16}}{3^{4}} \pmod{17}
Here I am stuck because I get 3^{-4} \pmod{17} and I don't know how to calculate this because I don't know what \frac{1}{81} \pmod{17} is.
I know 81 = 13 \pmod{17}
But I know the answer is 4. What did I do wrong?
Answer
3^{12}=(3^3)^4=10^4 (mod 17), so we have to find 10000 (mod 17), which is evidently 4 (mod 17).
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