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

Let $x=4k+3$
$(2020+x{)}^{2022}$
$=(2020+4k+3{)}^{2022}$
$=(2024+4k-1{)}^{2022}$
$=(4A-1{)}^{2022}$
$={}^{2022}{C}_0\left(4A{)}^{2022}\right(-1{)}^0+{}^{2022}{C}_1\left(4A\right){}^{2021}(-1{)}^1+.....+{}^{2022}{C}_{2021}\left(4A{)}^1\right(-1{)}^{2021}+{}^{2022}{C}_{2022}\left(4A{)}^0\right(-1{)}^{2022}$
Which will be of the form $8\lambda +1$
So, Remainder is 1.