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

Correct answer: 0 x = 4k + 3 ∴ (2020 + x) ^ 2022 = (2020 + 4k + 3) ^ 2022 = (4(505 + k) + 3) ^ 2022 = (4λ + 3) ^ 2022 = (16 λ ^ 2 + 24λ + 9) ^ 1011 = (8(2 λ ^ 2 + 3λ + 1) + 1) ^ 1011 = (8p + 1) ^ 1011 ∴ Remainder when divided by 8 = 1 .

JEE Main 202125 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Binomial Theorem

If the remainder when x is divided by 4 is 3 , then the remainder...

If the remainder when

x

is divided by

4

3

, then the remainder when

{\left(2020 + x\right)}^{2022} is divided by

8 is ___ .

Correct answer

Step-by-step solution

$x = 4k + 3$

$∴ {\left(2020 + x\right)}^{2022} = {\left(2020 + 4k + 3\right)}^{2022}$

$= {\left(4\left(505 + k\right) + 3\right)}^{2022}$

$= {\left(4\lambda + 3\right)}^{2022} = {\left(16{\lambda}^{2} + 24\lambda + 9\right)}^{1011}$

$= {\left(8\left(2{\lambda}^{2} + 3\lambda + 1\right) + 1\right)}^{1011}$

$= {\left(8p + 1\right)}^{1011}$

$∴$ Remainder when divided by $8 = 1$ .

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