The remainder when $$(2021)^{2022} + (2022)^{2021}$$ is divided by $$7$$ is

1. Reduce the bases modulo 7

First, divide the base numbers by 7 to find their remainders:

$$2021 = 7 \times 288 + 5 \implies 2021 \equiv 5 \equiv -2 \pmod 7$$

$$2022 = 7 \times 288 + 6 \implies 2022 \equiv 6 \equiv -1 \pmod 7$$

2. Evaluate the first term

Substitute the reduced base into the first term:

$$(2021)^{2022} \equiv (-2)^{2022} \pmod 7$$

Since the exponent is even, the negative sign disappears:

$$(-2)^{2022} = 2^{2022}$$

By Fermat's Little Theorem, since 7 is a prime number:

$$2^6 \equiv 1 \pmod 7$$

Now, divide the exponent 2022 by 6:

$$2022 = 6 \times 337 + 0$$

Therefore:

$$2^{2022} = (2^6)^{337} \equiv 1^{337} \equiv 1 \pmod 7$$

3. Evaluate the second term

Substitute the reduced base into the second term:

$$(2022)^{2021} \equiv (-1)^{2021} \pmod 7$$

Since the exponent 2021 is an odd number, any negative base raised to an odd power remains negative:

$$(-1)^{2021} = -1 \equiv 6 \pmod 7$$

4. Combine the results

Add the two evaluated remainders together:

$$(2021)^{2022} + (2022)^{2021} \equiv 1 + (-1) \pmod 7$$

$$1 - 1 = 0$$