If (2021)^{3762} is divided by 17,then the re | vedclass

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(C) We need to find the remainder when $(2021)^{3762}$ is divided by $17$.First,note that $2021 = 17 	imes 118 + 15$,or more conveniently,$2021 = 17

Vedclass · Accepted Answer

$4$

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(C) We need to find the remainder when $(2021)^{3762}$ is divided by $17$.First,note that $2021 = 17 	imes 118 + 15$,or more conveniently,$2021 = 17 	imes 119 - 2 = 2023 - 2$.So,$(2021)^{3762} = (2023 - 2)^{3762}$.Using the Binomial Theorem,$(2023 - 2)^{3762} = \sum_{k=0}^{3762} \binom{3762}{k} (2023)^{3762-k} (-2)^k$.Since $2023 = 17 	imes 119$,all terms containing $2023$ are divisible by $17$.Thus,$(2021)^{3762} \equiv (-2)^{3762} \pmod{17}$.$(-2)^{3762} = 2^{3762} = (2^4)^{940} 	imes 2^2 = 16^{940} 	imes 4$.Since $16 \equiv -1 \pmod{17}$,we have $16^{940} \equiv (-1)^{940} \equiv 1 \pmod{17}$.Therefore,$2^{3762} \equiv 1 	imes 4 \equiv 4 \pmod{17}$.The remainder is $4$.

If $(2021)^{3762}$ is divided by $17$,then the remainder is ........

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