If the remainder when x is divided by 4 is 3, then the remainder when (2020+x)^2022 is div…

Question

If the remainder when $x$ is divided by 4 is 3, then the remainder when $(2020+x)^{2022}$ is divided by 8 is

Answer: 1

Accepted Answer

Write $x=4p+3$. Then $2020+x=2020+4p+3=2024+4p-1=(4k-1)$, where $k$ is an integer (since $2024$ is divisible by 4).

$$(2020+x)^{2022}=(4k-1)^{2022}=\sum_{i=0}^{2022}\binom{2022}{i}(4k)^{2022-i}(-1)^i$$

Since $2022$ is even, the last term is $+1$. All terms with $i\le2020$ carry $4^2$ at minimum times $\binom{2022}{i}$ where $\binom{2022}{2021}\cdot4k$ contributes $2022\times4k$, which is $8\times(1011k)$ --- a multiple of 8. So the whole expression $\equiv1\pmod8$.

Remainder = 1

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