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


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


$x=4 k+3$

$\therefore(2020+\mathrm{x})^{2022}=(2020+4 \mathrm{k}+3)^{2022}$


$=(4 \lambda+3)^{2022}=\left(16 \lambda^{2}+24 \lambda+9\right)^{1011}$

$=\left(8\left(2 \lambda^{2}+3 \lambda+1\right)+1\right)^{1011}$

$=(8 \mathrm{p}+1)^{1011}$

$\therefore$ Remainder when divided by $8=1$

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