If the remainder when x is divided by 4 is 3 , then the remainder when 2020 + x 2022 is divided by 8 is ___ .

If the remainder when x is divided by 4 is 3, then the remainder when 2020+x2022 is divided by 8 is ___ .

Solution

x=4k+3

∴2020+x2022=2020+4k+32022

=4505+k+32022

=4λ+32022=16λ2+24λ+91011

=82λ2+3λ+1+11011

=8p+11011

∴ Remainder when divided by 8=1.

Asked in: JEE Main 2021 (25 Feb Shift 2)

Practice more Binomial Theorem questions on Aicharya