The remainder when $$428^{2024}$$ is divided by 21 is __________

Firstly, we need to find the remainder when 428 is divided by 21.

$$428 = (21 \times 20) + 8$$

So, $$428 \equiv 8 \pmod{21}$$.

Hence, we can say: $$428^{2024} \equiv 8^{2024} \pmod{21}$$

Now, we can find the remainders for the first few powers of 8 so as to find any recurring pattern of the remainders. 

$$8^1\ ≡\ 8$$ (mod 21)

$$8^2=24$$ (mod 21)

Now, find the remainder of 64 divided by 21: $$64 = (21 \times 3) + 1$$

So, $$8^2 \equiv 1 \pmod{21}$$.

Now, because $$8^2 \equiv 1 \pmod{21}$$, we can express the large exponent, 2024, in terms of 2. We divide 2024 by 2: 

$$2024 = 2 \times 1012$$

So, we can say, $$8^{2024}=\left(8^2\right)^{2024}$$

Substitute 1 for $$8^2$$,

$$(8^2)^{1012} \equiv 1^{1012} \pmod{21}$$

Hence, the remainder wil be 1.