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

Answer: 1

$x=4k+3$
$\therefore (2020+x{)}^{2022}=(2020+4k+3{)}^{2022}$
$=(4(505+k)+3{)}^{2022}$
$=(4\lambda +3{)}^{2022}={\left(16{\lambda }^2+24\lambda +9\right)}^{1011}$
$={\left(8\left(2{\lambda }^2+3\lambda +1\right)+1\right)}^{1011}$
$=(8p+1{)}^{1011}$
$\therefore$ Remainder when divided by $8=1$