The remainder when (2021)^{2022} + (2022)^{20 | vedclass

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Question

(A) We need to find the remainder of $(2021)^{2022} + (2022)^{2021}$ when divided by $7$.First,note that $2021 = 7 	imes 288 + 5 \equiv -2 \pmod{7}$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We need to find the remainder of $(2021)^{2022} + (2022)^{2021}$ when divided by $7$.First,note that $2021 = 7 	imes 288 + 5 \equiv -2 \pmod{7}$ and $2022 = 7 	imes 288 + 6 \equiv -1 \pmod{7}$.So,$(2021)^{2022} + (2022)^{2021} \equiv (-2)^{2022} + (-1)^{2021} \pmod{7}$.Since $2022$ is even,$(-2)^{2022} = 2^{2022} = (2^3)^{674} = 8^{674} \equiv 1^{674} = 1 \pmod{7}$.Since $2021$ is odd,$(-1)^{2021} = -1$.Thus,$(2021)^{2022} + (2022)^{2021} \equiv 1 - 1 = 0 \pmod{7}$.Therefore,the remainder is $0$.

The remainder when $(2021)^{2022} + (2022)^{2021}$ is divided by $7$ is:

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