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

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

$$ 2023=289 \times 7
2023 is a multiple of 7
\mathrm{n}=(2023)^{2023} is multiple of 7
$$ \text { and } \begin{aligned} (2023)^{2023}=(-2)^{2023} & =-2\left(2^2\right)^{1011} \\ & =-2(5-1)^{1011} \\ & =-2\left[{ }^5 \mathrm{C}_0 5^{1011}-{ }^5 \mathrm{C}_1 5^{1010}+\ldots .-{ }^{1011} \mathrm{C}_{1011}\right] \end{aligned}
(2023)^{2023} when divided by 5

gives remainder 2

If \mathrm{n}=(2023)^{2023} divided by 35=7 \times 5
\mathrm{n}=7 \mathrm{k}
\mathrm{n}-7=7(\mathrm{k}-1) \rightarrow \mathrm{n} 7 is multiple of 7
and \mathrm{n}=5 \mathrm{~m}+2
so n-7=5 \mathrm{~m}-5= multiple of 5
so n-7 is multiple of 35 so when n is divided by 35 , reminder =7

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chirag

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