Animated Solution for Mathematics - Binomial Theorem: The remainder when $(2023{)}^{2023}$ is divided by 35 is ______.

Imagine you are standing before a mathematical mountain: calculating the remainder of $(2023{)}^{2023}$ when divided by $35$. At first glance, it looks impossible. The number is astronomically large, and your calculator would simply give up.

But in the world of JEE Advanced, we don't use brute force; we use elegance. We use modular arithmetic to dismantle the problem from the inside out.

The Strategy of Coprime Factors

Our divisor is $35$. The secret to unlocking this problem lies in its factors.

Since $35=5\times 7$, and $5$ and $7$ are coprime, we can solve the problem in two parallel universes: one modulo $5$ and one modulo $7$. By finding the remainder in each, we can reconstruct the final answer using the Chinese Remainder Theorem.

This is the beauty of number theory—breaking a complex whole into manageable, bite-sized pieces.

The Modulo $7$ Victory

Let us step into the first universe: modulo $7$. We need to evaluate $(2023{)}^{2023}(\mathrm{mod}7)$.

First, we check the base. Is $2023$ divisible by $7$? A quick division reveals that:

The remainder is exactly $0$! This is a massive stroke of luck. It means $2023\equiv 0(\mathrm{mod}7)$.

Consequently:

We have already conquered half the mountain.

The Modulo $5$ Journey

Now, we enter the second universe: modulo $5$. We need $(2023{)}^{2023}(\mathrm{mod}5)$.

The last digit of $2023$ is $3$, so $2023\equiv 3(\mathrm{mod}5)$. To make our lives easier, we use the negative remainder: $3\equiv -2(\mathrm{mod}5)$.

Now our expression is $(-2{)}^{2023}(\mathrm{mod}5)$. Since $2023$ is odd, the negative sign persists: $-(2^{2023})(\mathrm{mod}5)$.

We need to find the pattern of powers of $2$ modulo $5$: $2^1\equiv 2$ $2^2\equiv 4\equiv -1$ * $2^4\equiv (-1{)}^2\equiv 1$

The cycle length is $4$. Dividing the exponent $2023$ by $4$, we get $2023=4\times 505+3$.

Thus:

Don't forget the negative sign! We have $-(3)=-3\equiv 2(\mathrm{mod}5)$.

The Synthesis

We have our two clues: $x\equiv 0(\mathrm{mod}7)$ and $x\equiv 2(\mathrm{mod}5)$.

From the first, $x$ must be a multiple of $7$, so $x=7k$. Substituting this into the second:

Simplifying $7(\mathrm{mod}5)$ gives $2k\equiv 2(\mathrm{mod}5)$, which leads to $k\equiv 1(\mathrm{mod}5)$.

The smallest positive integer $k$ is $1$, giving us $x=7(1)=7$. The mountain has been climbed. The remainder is $7$.