number theory - How can I calculate the remainder of $3^{2012}$ modulo 17?

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 $17$).