Animated Solution for Mathematics - Binomial Theorem: Remainder when $64^{32^{32}}$ is divided by 9 is equal to ______.

Imagine you are standing at the base of a colossal mountain. The problem before you, $64^{32^{32}}$, is not just a number; it is a monolith of mathematical complexity that seems designed to intimidate you.

When you see an exponent like $32^{32}$, your brain naturally wants to retreat, to find a calculator, or to assume that the problem is impossible. But in the world of JEE Advanced, intimidation is just a mask for a hidden simplicity. Let us peel back that mask together and see the elegant structure beneath.

The first step in our journey is to change how we see the world. We are not dealing with standard arithmetic here; we are in the realm of modular arithmetic.

Think of modular arithmetic as looking at the world through a lens that only sees the remainder. It is the study of cycles, of rhythms, and of the patterns that repeat when we divide by a number.

When we divide by $9$, we are essentially looking at the world through a lens that only cares about the remainder. This lens simplifies everything, turning the infinite sea of integers into a small, manageable set of possibilities.

Consider our base, $64$. In the standard world, $64$ is a large, cumbersome number. But in the world of modulo $9$, $64$ is just a neighbor of $63$.

And $63$ is a multiple of $9$. This is the spark, the moment of clarity that changes everything. We can write $64$ as $63+1$, which is $9\times 7+1$.

This means that $64\equiv 1(\mathrm{mod}9)$. This is the foundational insight. We have transformed a complex base into a simple one; we are no longer dealing with $64$, we are dealing with $1$.

Now, we must address the exponent, $32^{32}$. It looks terrifying, doesn't it? But here is the beauty of the power property in modular arithmetic: if $a\equiv b(\mathrm{mod}n)$, then $a^k\equiv b^k(\mathrm{mod}n)$.

This property is our engine. It allows us to raise both sides of our congruence to the power of $32^{32}$. So, we have:

This is the moment of collapse. The complex exponent, which seemed so daunting, suddenly becomes irrelevant because $1$ raised to any positive integer power is still $1$. The massive power of $32^{32}$ is completely neutralized by the simplicity of the base $1$.

This is the essence of the JEE mindset. It is not about brute force; it is about pattern recognition. It is about finding the elegant path through the complexity.

It is about seeing the structure beneath the surface. When you face a problem like this, do not be intimidated by the size of the numbers. Look for the pattern, look for the remainder, and look for the simplicity that lies beneath the complexity.

The goal is to reach a state where you can look at $64^{32^{32}}$ and immediately see the $1$ hiding inside it.

As you continue your journey through the world of mathematics, remember this problem. Remember the power of modular arithmetic. Remember that even the most complex problems can be broken down into simple, elegant steps.

The remainder is $1$. It is a simple, clean, and beautiful answer to a problem that looked like a monster. Keep practicing, keep exploring, and keep falling in love with the beauty of the math.