If the remainder when x is divided by 4 is 3, | vedclass

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

Question

(A) Given that the remainder when $x$ is divided by $4$ is $3$,we can write $x = 4k + 3$ for some integer $k$.We need to find the remainder when $(202

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given that the remainder when $x$ is divided by $4$ is $3$,we can write $x = 4k + 3$ for some integer $k$.We need to find the remainder when $(2020 + x)^{2022}$ is divided by $8$.Substitute $x = 4k + 3$ into the expression:$(2020 + x)^{2022} = (2020 + 4k + 3)^{2022} = (2023 + 4k)^{2022}$.Since $2023 = 8 	imes 252 + 7$,we have $2023 \equiv 7 \equiv -1 \pmod{8}$.Thus,$(2023 + 4k)^{2022} \equiv (-1 + 4k)^{2022} \pmod{8}$.If $k$ is even,let $k = 2m$,then $(-1 + 4(2m))^{2022} = (-1 + 8m)^{2022} \equiv (-1)^{2022} \equiv 1 \pmod{8}$.If $k$ is odd,let $k = 2m + 1$,then $(-1 + 4(2m + 1))^{2022} = (-1 + 8m + 4)^{2022} = (3 + 8m)^{2022} \equiv 3^{2022} \pmod{8}$.Note that $3^2 = 9 \equiv 1 \pmod{8}$,so $3^{2022} = (3^2)^{1011} \equiv 1^{1011} \equiv 1 \pmod{8}$.In both cases,the remainder is $1$.

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

Similar Questions

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

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

For all $n \in N$,if $n(n^2+3)$ is divisible by $k$,then the maximum value of $k$ is

If $n \in N$,then ${7^{2n}} + {2^{3n - 3}} \cdot {3^{n - 1}}$ is always divisible by

If $n \in N$,then ${x^{2n - 1}} + {y^{2n - 1}}$ is divisible by

Vedclass Test Series

Exam Paper Generator

Online Exam Module