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