The remainder when (2021)^{2023} is divided b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We need to find the remainder of $(2021)^{2023}$ when divided by $7$.First,divide $2021$ by $7$: $2021 = 7 	imes 288 + 5$.So,$2021 \equiv 5 \pmod

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) We need to find the remainder of $(2021)^{2023}$ when divided by $7$.First,divide $2021$ by $7$: $2021 = 7 	imes 288 + 5$.So,$2021 \equiv 5 \pmod{7}$,which is equivalent to $2021 \equiv -2 \pmod{7}$.Therefore,$(2021)^{2023} \equiv (-2)^{2023} \pmod{7}$.$(-2)^{2023} = - (2^{2023}) = - (2^3)^{674} 	imes 2^1$.Since $2^3 = 8 \equiv 1 \pmod{7}$,we have $2^3 \equiv 1 \pmod{7}$.Thus,$(2^3)^{674} \equiv 1^{674} \equiv 1 \pmod{7}$.So,$(-2)^{2023} \equiv -(1 	imes 2) \equiv -2 \pmod{7}$.To express the remainder as a positive value,we add $7$: $-2 + 7 = 5$.Thus,the remainder is $5$.

The remainder when $(2021)^{2023}$ is divided by $7$ is

Similar Questions

For every natural number $n$,${3^{2n + 2}} - 8n - 9$ is divisible by

For what values of $m \in \mathbb{N}$,does the divisibility $(x+y) \mid (x^m+y^m)$ hold?

The remainder on dividing $5^{99}$ by $11$ is

The product of the last two digits of $(1919)^{1919}$ is . . . . . . .

The remainder when $3^{37}$ is divided by $80$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module