Remainders JEE

$23^2≡ (-1) \mod 53$ $23^{20} ≡ 1 \mod 53$ $23^3 \mod 53$ $-23 \mod 53=30 \mod 53$

We have $\frac{23.(23^{2})^{11}}{53}$

$23.(530-1)^{11}$

$^{11}C_{0}(530)^{11}+^{11}C_{1}(530)^{10}.(-1).................^{11}C_{11}(-1)^{11}$

$23(53k)-23$

[where $k$ is a positive integer, take $53$ common from the terms leaving $-1$ ]

remainder cannot be negative so add and subtract $53$

$53(23k)-53+53-23$

$53(23k-1)+30$

on diving by 53 we get remainder $\boxed{30}$

$\begin{aligned} 23^{23} & \equiv 23(23^2)^{11} \text{ (mod 53)} & \small \color{#3D99F6} \text{Note that }23^2 = 529 \\ & \equiv 23(530-1)^{11} \text{ (mod 53)} \\ & \equiv 23(-1)^{11} \text{ (mod 53)} \\ & \equiv -23 \text{ (mod 53)} \\ & \equiv \boxed{30} \text{ (mod 53)} \end{aligned}$

20880467999847912034355032910567 MOD 53 = 30

A question of calculator.