Question

The remainder when \((2023)^{2023}\) is divided by \(35\) is _____

Answer 1

Correct Answer: 7

Using modular arithmetic:

\( 2023 = 2030 - 7 \implies 2023 \equiv -7 \pmod{35}. \)

Thus:

\( (2023)^{2023} \equiv (-7)^{2023} \pmod{35}. \)

Using properties of modular arithmetic:

\( (-7)^{2023} = (-7) \cdot (-7)^{2022}. \)

Simplify further using binomial theorem, and compute the remainder as:

\( \boxed{7}. \)

Answer 2

The correct answer is 7
$(2023{)}^{2023}$
$=(2030-7{)}^{2023}$
$=(35K-7{)}^{2023}$
$={}^{2023}{C}_0(35K{)}^{2023}(-7{)}^0+{}^{2023}{C}_1(35K{)}^{2022}(-7)+\dots .+$
$\dots \cdots +{}^{2023}{C}_{2022}(-7{)}^{2023}$
$=35N-7^{2023}.$
$\text{Now},-7^{2023}=-7\times 7^{2022}=-7{\left(7^2\right)}^{1011}$
$=-7(50-1{)}^{1011}$
$=-7\left({}^{1011}{C}_{0}50^{1011}-{}^{1011}{C}_1(50{)}^{1010}+\dots \dots \cdot {}^{1011}{C}_{1011}\right)$
$=-7(5\lambda -1)$
$=-35\lambda +7$
$\therefore$ when $(2023{)}^{2023}$ is divided by 35 remainder is 7