For abstract algebra I have to find the last two digits of $27^{2018}$, without the use of a calculator, and as a hint it says you should work in $\mathbb{Z}/100\mathbb{Z}$. 

I thought breaking up the problem into $\mod(100)$ arguments. Thus:
$27^{2}=729\equiv 29 \mod (100)$, and

$27^{4}=(27^{2})^{2} \equiv 29^{2}=861\equiv 61 \mod (100)$ and

$27^{8}=(27^{4})^{2} \equiv 61^{2}=3421 \equiv 21 \mod(100)$ and so on until I would find something that repeated itself. But I've done quite some terms no and I've not seen any iteration yet. So I'm thinking this is the wrong way.

Any suggestions?