MCQ
The remainder when $(2021)^{2022}+(2022)^{ 2021 }$ is divided by $7$ is.
- ✓$0$
- B$1$
- C$2$
- D$6$
$=(2023-2)^{\text {an2 }}+(2023-1)^{3031}$
$=7 n_{1}+2^{m 2 n 2}+7 n _{2}-1$
$=7\left( n _{1}+ n _{2}\right)+8^{674}-1$
$=7\left(n_{1}+n_{2}\right)+(7-1)^{674}-1$
$=7\left(n_{1}+n_{2}\right)+7 n_{3}+1-1$
$=7\left(n_{1}+n_{2}+n_{3}\right)$
$\therefore$ Given number is divisible by $7$ hence remainder is zero
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