If the remainder when x is divided by 4 is 3... | Binomial Theorem

JEE MAIN 2021

25-02-2021 S2

Question

If the remainder when x is divided by 4 is 3 , then the remainder when $(2020+\mathrm{x})^{2022}$ is divided by 8 is $\_\_\_\_$ .

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Solution

$$ \begin{aligned} & x=4 k+3 \\ & \therefore(2020+x)^{2022}=(2020+4 k+3)^{2022} \\ & =(4(505+k)+3)^{2022} \\ & =(4 \lambda+3)^{2022}=\left(16 \lambda^2+24 \lambda+9\right)^{1011} \\ & =\left(8\left(2 \lambda^2+3 \lambda+1\right)+1\right)^{1011} \\ & =(8 p+1)^{1011} \end{aligned} $$ ∴ Remainder when divided by $8=1$

Question Tags

JEE Main

Mathematics

Medium

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