The remainder when $$(2021)^{2023}$$ is divided by $$7$$ is

We need to find the remainder when $$(2021)^{2023}$$ is divided by 7.

Find $$2021 \mod 7$$.

$$2021 = 288 \times 7 + 5$$

So $$2021 \equiv 5 \pmod{7}$$

Find the pattern of powers of 5 modulo 7.

$$5^1 \equiv 5 \pmod{7}$$

$$5^2 \equiv 25 \equiv 4 \pmod{7}$$

$$5^3 \equiv 20 \equiv 6 \pmod{7}$$

$$5^4 \equiv 30 \equiv 2 \pmod{7}$$

$$5^5 \equiv 10 \equiv 3 \pmod{7}$$

$$5^6 \equiv 15 \equiv 1 \pmod{7}$$

The powers of 5 repeat with a cycle of length 6 (by Fermat's Little Theorem).

Find $$2023 \mod 6$$.

$$2023 = 337 \times 6 + 1$$

So $$2023 \equiv 1 \pmod{6}$$

Compute the remainder.

$$(2021)^{2023} \equiv 5^{2023} \equiv 5^1 \equiv 5 \pmod{7}$$

The remainder is $$\textbf{5}$$.

The correct answer is Option D.