The remainder when $$(2023)^{2023}$$ is divided by 35 is

We need to find the remainder when $$(2023)^{2023}$$ is divided by $$35$$.

Since $$35 = 5 \times 7$$ and $$\gcd(5, 7) = 1$$, we use the Chinese Remainder Theorem.

Find $$(2023)^{2023} \pmod{5}$$.

$$2023 = 404 \times 5 + 3$$, so $$2023 \equiv 3 \pmod{5}$$.

By Fermat's Little Theorem, $$3^4 \equiv 1 \pmod{5}$$.

$$2023 = 4 \times 505 + 3$$, so $$2023 \equiv 3 \pmod{4}$$.

Therefore: $$3^{2023} \equiv 3^3 = 27 \equiv 2 \pmod{5}$$.

Find $$(2023)^{2023} \pmod{7}$$.

$$2023 = 289 \times 7$$, so $$2023 \equiv 0 \pmod{7}$$.

Therefore: $$(2023)^{2023} \equiv 0 \pmod{7}$$.

Apply the Chinese Remainder Theorem.

We need $$x$$ such that:

$$x \equiv 2 \pmod{5}$$ and $$x \equiv 0 \pmod{7}$$.

From the second condition, $$x = 7k$$ for some integer $$k$$.

Substituting: $$7k \equiv 2 \pmod{5} \implies 2k \equiv 2 \pmod{5} \implies k \equiv 1 \pmod{5}$$.

So $$k = 5m + 1$$ and $$x = 7(5m + 1) = 35m + 7$$.

The remainder when $$(2023)^{2023}$$ is divided by $$35$$ is $$\boxed{7}$$.