If $$(2021)^{3762}$$ is divided by 17, then the remainder is ________.

We need the remainder when $$(2021)^{3762}$$ is divided by 17.

First, $$2021 \mod 17$$: since $$17 \times 118 = 2006$$, we get $$2021 - 2006 = 15$$, so $$2021 \equiv 15 \equiv -2 \pmod{17}$$.

Since the exponent 3762 is even, $$(2021)^{3762} \equiv (-2)^{3762} = 2^{3762} \pmod{17}$$.

By Fermat's Little Theorem (17 is prime, $$\gcd(2,17)=1$$): $$2^{16} \equiv 1 \pmod{17}$$.

Dividing: $$3762 = 16 \times 235 + 2$$, so $$2^{3762} = (2^{16})^{235} \cdot 2^2 \equiv 1 \cdot 4 = 4 \pmod{17}$$.

The remainder is 4.