What is $26^{32}\bmod 12$? What is the correct answer to this expression:  
$26^{32}  \pmod  {12}$ 
When I tried in Wolfram Alpha the answer is $4$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $0.$

First, note that $26 \equiv 2 \pmod {12}$, so $26^{32} \equiv 2^{32} \pmod {12}$.
Next, note that $2^4 \equiv 16 \equiv 4 \pmod {12}$, so $2^{32} \equiv \left(2^4\right)^8 \equiv 4 ^8 \pmod {12}$, and $4^2 \equiv 4 \pmod {12}$.
Finally, $4^8 \equiv \left(4^2\right)^4 \equiv 4^4 \equiv \left(4^2\right)^2 \equiv 4^2 \equiv 4 \pmod {12}$.
Then we get the result. 
There are slicker solutions with just a few results from elementary number theory, but this is a very basic method which should be easy enough to follow.
