The remainder when 20232023 is divided by 35 is | (Mathematics)

$\begin{aligned} & (2023)^{2023} \\\\ & =(2030-7)^{2023} \\\\ & =(35 \mathrm{~K}-7)^{2023} \\\\ & ={ }^{2023} \mathrm{C}_0(35 \mathrm{~K})^{2023}(-7)^0+{ }^{2023} \mathrm{C}_1(35 \mathrm{~K})^{2022}(-7)+ \\\\ & \ldots . .+\ldots \ldots .+{ }^{2023} \mathrm{C}_{2023}(-7)^{2023} \\\\ & =35 \mathrm{~N}-7^{2023} \\\\ & \text { Now },-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011} \\\\ & =-7(50-1)^{1011} \\\\ & =-7\left({ }^{1011} \mathrm{C}_0 50^{1011}-{ }^{1011} \mathrm{C}_1(50)^{1010}+\ldots \ldots{ }^{1011} \mathrm{C}_{1011}\right) \\\\ & =-7(5 \lambda-1) \\\\ & =-35 \lambda+7 \end{aligned}$

$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7

Explanation

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