Animated Solution for Mathematics - Binomial Theorem: If $\{p\}$ denotes the fractional part of the number $p$, then $\{\frac{3^{200}}{8}\}$ is equal to:

Imagine you are standing before a mountain of a number: $3^{200}$. If you were asked to find the fractional part of this number divided by $8$, your first instinct might be panic.

How could anyone possibly calculate $3^{200}$? But here is the secret of the JEE Advanced: the problem is not about calculation; it is about structure. We are not here to compute the number; we are here to understand its soul.

The fractional part of a number $p$, denoted as $\{p\}$, is defined as $\{p\}=p-[p]$, where $[p]$ is the greatest integer less than or equal to $p$. Think of it as the 'leftover' after you have stripped away all the whole numbers.

When we look at $\{\frac{3^{200}}{8}\}$, we are essentially asking: "What is the remainder when $3^{200}$ is divided by $8$, and how does that remainder behave when divided by $8$?"

Whenever you see a large power, look for a connection between the base and the divisor. We have a base of $3$ and a divisor of $8$. Is there a relationship? Indeed! $3^2=9$, and $9$ is just $1$ more than $8$.

This is our golden key. We can rewrite $3^{200}$ as $(3^2{)}^{100}$, which is $9^{100}$. Now, we can express $9$ as $(1+8)$. Our expression becomes:

Now, we invoke the Binomial Theorem, the ultimate tool for expanding powers of sums. The expansion of $(1+x{)}^n$ is $\left(\genfrac{}{}{0ex}{}{n}{0}\right)+\left(\genfrac{}{}{0ex}{}{n}{1}\right)x+\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^2+\cdots +\left(\genfrac{}{}{0ex}{}{n}{n}\right)x^n$. Substituting $x=8$ and $n=100$, we get:

Look closely at this expansion. The first term is $\left(\genfrac{}{}{0ex}{}{100}{0}\right)=1$. Every single term that follows contains at least one factor of $8$.

This is beautiful! We can write this as $1+8K$, where $K$ is the sum of all the subsequent terms. Since every term in $K$ is an integer, $K$ itself must be an integer.

Now, let's return to our original fraction:

By splitting this fraction, we get $\frac{1}{8}+K$. We are now looking for the fractional part of $(\frac{1}{8}+K)$.

Because $K$ is an integer, it vanishes when we apply the fractional part operator, leaving us with $\{\frac{1}{8}\}$. Since $\frac{1}{8}$ is a positive fraction less than $1$, its fractional part is simply $\frac{1}{8}$.

We started with a terrifyingly large number and, through the elegance of the Binomial Theorem, reduced it to a simple fraction. This is the essence of mathematics—finding the underlying simplicity in what appears to be complex.

You didn't need a calculator; you needed insight. Keep that curiosity alive, and you will conquer any problem that comes your way.