What is the remainder when 11^8577314 is divided by 97?
Since 97 is prime use Fermat's little theorem. It states that
$\displaystyle a^{p-1} \equiv 1 \text{mod}p$ so this gives that
$\displaystyle 11^{96} \equiv 1 \text{mod}97$
So $\displaystyle 8577314 \div 96 = 89347 R 2$
This gives that
$\displaystyle 8577314=96 \cdot 89347 + 2$
So
$\displaystyle 11^{8577314} \equiv 11^{96 \cdot 89347 + 2} \equiv 11^2 \text{mod}97$
It is all down hill from here