Let r be the remainder when 2021^{2020} is di | vedclass

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(A) We have $(2021)^{2020} = (1 + 2020)^{2020}$.Using the Binomial Theorem,$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$Here,$x = 2020$ and $n =

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$0$ and $5$

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(A) We have $(2021)^{2020} = (1 + 2020)^{2020}$.Using the Binomial Theorem,$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$Here,$x = 2020$ and $n = 2020$.$(1 + 2020)^{2020} = 1 + (2020)(2020) + \frac{2020 	imes 2019}{2} 	imes (2020)^2 + \dots$$(1 + 2020)^{2020} = 1 + (2020)^2 + 1010 	imes 2019 	imes (2020)^2 + \dots$$(1 + 2020)^{2020} = 1 + (2020)^2 	imes [1 + 1010 	imes 2019 + \dots]$Since all terms from the second term onwards are multiples of $(2020)^2$,the remainder $r$ when divided by $(2020)^2$ is $1$.Since $1$ lies between $0$ and $5$,the correct option is $A$.

Let $r$ be the remainder when $2021^{2020}$ is divided by $2020^2$. Then $r$ lies between

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