Animated Solution for Mathematics - Binomial Theorem: The remainder when $((64{)}^{(64)}{)}^{(64)}$ is divided by 7 is equal to

A Journey into Modular Elegance

Imagine standing before a mountain. This mountain is not made of rock, but of numbers. You are presented with the expression $((64{)}^{64}{)}^{64}$, and you are asked to find the remainder when this titan is divided by $7$.

At first glance, your instinct might be to panic. How can anyone compute such a number? It is a tower of powers so high that it would dwarf the observable universe if written out in decimal form.

But here is the secret of the JEE Advanced: the more intimidating the expression, the more elegant the shortcut. We are not here to calculate; we are here to observe.

Taming the Exponent

Before we even touch the divisor, let us simplify the structure. We have a base $64$ raised to the power of $64$, and that entire result is raised to the power of $64$ again.

We recall the fundamental law of exponents: $(a^m{)}^n=a^{m\times n}$. Applying this, our expression $N$ becomes $64^{64\times 64}$.

Calculating $64\times 64$, we get $4096$. So, our expression is now $N=64^{4096}$. This is much more manageable. We have reduced a "tower" to a single, albeit large, exponent.

The Modular Epiphany

This is where we bridge the gap between the base and the divisor. We are dividing by $7$. In modular arithmetic, we are interested in the remainder.

We look at our base, $64$, and ask: what is the closest multiple of $7$? We know that $7\times 9=63$. This is the "Aha!" moment.

We can write $64$ as $63+1$. In the language of modular arithmetic, $64\equiv 1(\mathrm{mod}7)$. This means that when we raise $64$ to any power, it will behave very similarly to $1$ raised to that same power.

The Binomial Vanishing Act

Now, let us substitute our discovery into the expression: $N=(63+1{)}^{4096}$. This is a classic setup for the Binomial Theorem.

When we expand $(x+a{)}^n$, we get a series of terms. For our expression, this looks like:

Look at every term except the very last one. Each of these terms contains at least one factor of $63$. Since $63=7\times 9$, every one of these terms is a multiple of $7$.

When we divide these terms by $7$, the remainder is $0$. They effectively vanish into the ether of the quotient.

What are we left with? Only the final term: $\left(\genfrac{}{}{0ex}{}{4096}{4096}\right)(1{)}^{4096}$. Since $\left(\genfrac{}{}{0ex}{}{4096}{4096}\right)=1$ and $1^{4096}=1$, the entire expression simplifies to just $1$.

We started with a terrifying tower of powers and ended with a simple $1$. This is the beauty of mathematics. We didn't need to compute the number; we only needed to understand its structure.

Whenever you face a problem like this in the JEE, remember: look for the multiple of the divisor that is closest to the base. Express the base as $(divisor\times k\pm 1)$, and let the Binomial Theorem do the heavy lifting for you.

You have conquered the tower! The final remainder is $1$.