Binomial Theorem Question 8
Question 8 - 25 January - Shift 2
The remainder when $(2023)^{2023}$ is divided by 35
Show Answer
Answer: 7
Solution:
Formula: Binomial Theorem
$(2023)^{2023}$
$=(2030-7)^{2023}$
$=(35 K-7)^{2023}$
$={ }^{2023} C_0(35 K)^{2023}(-7)^{0}+{ }^{2023} C_1(35 K)^{2022}(-7)+\ldots . .+$
…… $+{ }^{2023} C _{2023}(-7)^{2023}$
$=35 N-7^{2023}$.
Now, $-7^{2023}=-7 \times 7^{2022}=-7(7^{2})^{1011}$
$=-7(50-1)^{1011}$
$=-7({ }^{1011} C_0 50^{1011}-{ }^{1011} C_1(50)^{1010}+\ldots \ldots .{ }^{1011} C _{1011})$
$=-7(5 \lambda-1)$
$=-35 \lambda+7$
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7
