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

The remainder when

{\left(2021\right)}^{2022} + {\left(2022\right)}^{2021} is divided by

7$ is

Options

A $0$
B $1$
C $2$
D $6$

Correct answer

A. $0$

Step-by-step solution

We know $2023 = 7 \times 289$
So ${\left(2021\right)}^{2022} + {\left(2022\right)}^{2021}$
$= {\left(2023 - 2\right)}^{2022} + {\left(2023 - 1\right)}^{2021}$
$= 7{m}_{1} + {2}^{2022} + 7{m}_{2} + {\left(- 1\right)}^{2021}$
$= 7\left({m}_{1} + {m}_{2}\right) + {8}^{674} - 1$
$= 7\left({m}_{1} + {m}_{2}\right) + ( 7 + 1{)}^{674} - 1$
$= 7\left({m}_{1} + {m}_{2}\right) + 7{m}_{3} + 1 - 1$
$= 7\left({m}_{1} + {m}_{2} + {m}_{3}\right)$
This is a multiple of $7$
Hence, remainder is $0$

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