Question

The remainder when $(2021{)}^{2022}+(2022{)}^{2021}$ is divided by 7 is

• A. 0
• B. 1
• C. 2
• D. 6

Answer 1

Answer: (A)

$(2021{)}^{2022}+(2022{)}^{2021}$
$=(2023-2{)}^{2022}+(2023-1{)}^{2021}$
$=7n_1+2^{2022}+7n_2-1$
$=7\left(n_1+n_2\right)+8^{674}-1$
$=7\left(n_1+n_2\right)+(7-1{)}^{674}-1$
$=7\left(n_1+n_2\right)+7n_3+1-1$
$=7\left(n_1+n_2+n_3\right)$
$\therefore$ Given number is divisible by 7 hence remainder is zero