Question

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

Answer 1

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