The remainder when 2023 2023 is divided by 35 is __________ . [2023]

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Solution

Q.
The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

Ans.
(7)

${(2023)}^{2023} = {(2030 - 7)}^{2023} = {(35 K - 7)}^{2023}$

$= {}^{2023}{C_{0}} {(35 K)}^{2023} {(- 7)}^{0} + {}^{2023}{C_{1}} {(35 K)}^{2022} {(- 7)}^{1} + \dots + {}^{2023}{C_{2023}} {(- 7)}^{2023}$

$= 35 N - 7^{2023}$

Now, $- 7^{2023} = - 7 \times 7^{2022} = - 7 {(7^{2})}^{1011} = - 7 {(50 - 1)}^{1011}$

$= - 7 [{}^{1011}{C_{0} (50)}^{1011} - {}^{1011}{C_{1}} {(50)}^{1010} + \dots + {}^{1011}{C_{1011}}]$

$= - 7 (5 λ - 1) = 35 λ + 7$

$∴ When {(2023)}^{2023} is divided by 35, remainder is 7 .$

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