The remainder when (2023)$^{2023}$ is divided by 35 is __________.
Answer (integer)
7
Solution
$$
\begin{aligned}
& (2023)^{2023} \\\\
& =(2030-7)^{2023} \\\\
& =(35 \mathrm{~K}-7)^{2023} \\\\
& ={ }^{2023} \mathrm{C}_0(35 \mathrm{~K})^{2023}(-7)^0+{ }^{2023} \mathrm{C}_1(35 \mathrm{~K})^{2022}(-7)+ \\\\
& \ldots . .+\ldots \ldots .+{ }^{2023} \mathrm{C}_{2023}(-7)^{2023} \\\\
& =35 \mathrm{~N}-7^{2023} \\\\
& \text { Now },-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011} \\\\
& =-7(50-1)^{1011} \\\\
& =-7\left({ }^{1011} \mathrm{C}_0 50^{1011}-{ }^{1011} \mathrm{C}_1(50)^{1010}+\ldots \ldots{ }^{1011} \mathrm{C}_{1011}\right) \\\\
& =-7(5 \lambda-1) \\\\
& =-35 \lambda+7
\end{aligned}
$$<br/><br/>
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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