Find the remainder when $$ 47^{121}$$ is divided by 31.

$$47^{121}$$ can be written as $$47^{30}\times\ 47^{30}\times\ 47^{30}\times\ 47^{30}\times\ 47^{1}\times\ 47$$

We know by Fermat's theorem that

when $$a^{p-1}$$ is divide by $$p$$ where $$p$$ is prime and both $$a$$ and $$p$$ are co-prime, the remainder is 1.

Using the theorem in our expression, we get

$$47^{30}mod31=1$$

So the remainder when $$47^{121}$$ is divided by 31 is

$$1\times\ 1\times\ 1\times\ 1\times\ 1\times\ 47mod31=16$$

Hence, the answer is 16.