Question:
If the remainder when $x$ is divided by 4 is 3 , then the remainder when $(2020+x)^{2022}$ is devided by 8 is_______.
Solution:
$x=4 k+3$
$\therefore(2020+\mathrm{x})^{2022}=(2020+4 \mathrm{k}+3)^{2022}$
$=(4(505+\mathrm{k})+3)^{2022}$
$=(4 \lambda+3)^{2022}=\left(16 \lambda^{2}+24 \lambda+9\right)^{1011}$
$=\left(8\left(2 \lambda^{2}+3 \lambda+1\right)+1\right)^{1011}$
$=(8 \mathrm{p}+1)^{1011}$
$\therefore$ Remainder when divided by $8=1$