Find the remainder when $2^{2016}$ is divided by $9$?

Question

How to find the remainder of $2^{2016}$ divided by $9$?

The possible prime factor of $2^{2016}$ is $2$ and the possible prime factor of $9$ is $3$. So the remainder should be non zero.

Give some hints. Thanks in advance.

Answer 1

Here is a HINT

$2^6=64=1 \mod 9$

Answer 2

By the binomial theorem, $2^{2016} = (3-1)^{2016} = 9a - 2016\cdot 3 +1 = 9b+1$.