elementary number theory - How to find last two digits of $2^{2016}$

What should the 'efficient' way of finding the last two digits of $2^{2016}$ be? The way I found them was by multiplying the powers of $2$ because $2016=1024+512+256+128+64+32$. I heard that one way would be with the Chinese Remainder Lemma but I don't really know how I should start?

Answer

Essentially we need $2^{2016}\pmod{100}$

As $(2^{2016},100)=4$

let us find $2^{2016-2}\pmod{100/4}$

Now as $2^{10}\equiv-1\pmod{25}$

$2^{2014}=2^{201\cdot10+4}=(2^{10})^{201}\cdot2^4\equiv(-1)^{201}\cdot2^4\equiv9\pmod{25}$

$$\implies2^2\cdot2^{2014}\equiv2^2\cdot9\pmod{2^2\cdot25}$$