The remainder when 7^{2022} + 3^{2022} is div | 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 $7^{2022} + 3^{2022}$ when divided by $5$.Note that $7 \equiv 2 \pmod{5}$ and $3 \equiv -2 \pmod{5}$.Thus,$7^{202

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) We need to find the remainder of $7^{2022} + 3^{2022}$ when divided by $5$.Note that $7 \equiv 2 \pmod{5}$ and $3 \equiv -2 \pmod{5}$.Thus,$7^{2022} + 3^{2022} \equiv 2^{2022} + (-2)^{2022} \pmod{5}$.Since $2022$ is an even number,$(-2)^{2022} = 2^{2022}$.So,$7^{2022} + 3^{2022} \equiv 2^{2022} + 2^{2022} = 2 	imes 2^{2022} = 2^{2023} \pmod{5}$.We know that $2^4 = 16 \equiv 1 \pmod{5}$.Therefore,$2^{2023} = 2^{4 	imes 505 + 3} = (2^4)^{505} 	imes 2^3 \equiv 1^{505} 	imes 8 \equiv 8 \pmod{5}$.Finally,$8 \equiv 3 \pmod{5}$.The remainder is $3$.

The remainder when $7^{2022} + 3^{2022}$ is divided by $5$ is:

Similar Questions

If $a, b$ and $n$ are natural numbers,then $a^{2n-1} + b^{2n-1}$ is divisible by

If $k \in N$,then $3^{3k} - 26k - 1$ is divisible by

If $(27)^{999}$ is divided by $7$,then the remainder is

For a positive integer $n$,$n^{3} + 2n$ is always divisible by

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module