Finding remainder of dividing and moding large exponents

Question

Can anyone explain how to calculate the remainder for types of problems like this:

$2^{2131312213123}$ divided by 100

$13^{6601}$ mod 77

Answer 1

For the first one, note that:

$$2^2\equiv 4\mod 100$$

We also see that:

$$2^{22}\equiv 4\mod 100$$

This can be easily found by hand by starting with $2^1$ and continuously multiplying by $2$, only remembering the last two digits in each step.

Now since $2131312213120$ is divisible by $20$, we must have:

$$2^{2131312213122}\equiv 4\mod 100$$

Hence

$$2^{2131312213123}\equiv 8\mod 100$$

For the second one, we will use Euler's theorem ($\varphi$ is the Euler totient function; http://mathworld.wolfram.com/TotientFunction.html):

$$\varphi(77)=\varphi(7)\varphi(11)=60$$

So

$$13^{60}\equiv 1\mod 77$$

Hence

$$13^{6600}=(13^{60})^{110}\equiv 1\mod 77$$

So

$$13^{6601}\equiv 13 \mod 77$$