If the remainder when $$x$$ is divided by 4 is 3, then the remainder when $$(2020 + x)^{2022}$$ is divided by 8 is ______.

We are given that when $$x$$ is divided by 4, the remainder is 3, so $$x = 4k + 3$$ for some non-negative integer $$k$$.

We need the remainder when $$(2020 + x)^{2022}$$ is divided by 8. Substituting $$x = 4k + 3$$, we get $$2020 + x = 2020 + 4k + 3 = 2023 + 4k$$.

Now $$2023 + 4k \pmod{8}$$: since $$2023 = 252 \times 8 + 7$$, we have $$2023 \equiv 7 \pmod{8}$$, so $$2023 + 4k \equiv 7 + 4k \pmod{8}$$. Depending on whether $$k$$ is even or odd, this is either $$7$$ or $$3 \pmod{8}$$. In both cases, the number is odd.

Let $$m = 2023 + 4k$$. Since $$m$$ is odd, we can write $$m = 2q + 1$$. Then $$m^2 = 4q^2 + 4q + 1 = 4q(q+1) + 1$$. Since one of $$q, q+1$$ is even, $$q(q+1)$$ is even, so $$4q(q+1) \equiv 0 \pmod{8}$$. Thus $$m^2 \equiv 1 \pmod{8}$$.

Therefore $$m^{2022} = (m^2)^{1011} \equiv 1^{1011} = 1 \pmod{8}$$.

The remainder when $$(2020 + x)^{2022}$$ is divided by 8 is $$1$$.