What will be the remainder when 7^2020 is divided by 4?

Question

Problem: "What will be the remainder when $7^{2020}$ is divided by $4$?" I can't get a step to approach such type of question but all I know is the answer is $1$.

Answer 1

$7\equiv -1\pmod{4}$

So, $7^{2020}\equiv (-1)^{2020}\equiv 1\pmod{4}$

Answer 2

Hint: $7$ is congruent to $3$ modulo $4$ and powers of $3$ modulo $4$ are just $1$ and $3$

Answer 3

Binomial expansion of $(8-1)^{2020}:$

All terms except the very last one have a factor $8$.

The last term is $(-1)^{2020}$, which is the remainder.

Recall: $(8-1)^{2020}=$

$\displaystyle{\sum}_{k=0}^{2020}$$\binom{2020}{k}8^{2020-k}(-1)^k$.