If the remainder when x is divided by 4 is 3 ,

Question:

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

Solution:

Let $x=4 k+3$

$(2020+x)^{2022}$

$=(2020+4 k+3)^{2022}$

$=(4(505)+4 k+3)^{2022}$

$=(4 P+3)^{2022}$

$=(4 P+4-1)^{2022}$

$=(4 A-1)^{2022}$

${ }^{2022} \mathrm{C}_{0}(4 \mathrm{~A})^{0}(-1)^{2022}+{ }^{2022} \mathrm{C}_{1}(4 \mathrm{~A})^{1}(-1)^{2021}+\ldots \ldots$

$1+8 \lambda$

Reminder is 1

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