Animated Solution for Mathematics - Binomial Theorem: The remainder left out when $8^{2n}-(62{)}^{2n+1}$ is divided by 9 is

Imagine you are standing in front of a giant, circular clock. But this isn't a normal clock with twelve hours; this is a Modulo 9 Clock.

On this clock, the numbers wrap around every $9$ units. This means that $9$ is equivalent to $0$, $10$ is equivalent to $1$, and so on.

When we face a problem like finding the remainder of $8^{2n}-(62{)}^{2n+1}$ divided by $9$, we are essentially asking: "Where does this massive expression land on our Modulo 9 Clock?"

Large powers like $2n$ and $2n+1$ can look intimidating, but they are actually a gift in disguise. The secret to unlocking this problem lies in the Binomial Theorem.

We want to express our bases, $8$ and $62$, in a form that makes them easy to handle. Notice that $8$ is just $9-1$, and $62$ is $63-1$, which is $9\times 7-1$.

By writing them this way, we can use the binomial expansion: $(9k-1{)}^m$. In this expansion, every single term will contain a factor of $9$, except for the very last term, $(-1{)}^m$.

This is the "magic" of the method—all the complex terms simply vanish into the multiple of $9$, leaving us with a tiny, manageable number.

Let's tackle the first part: $8^{2n}$. We rewrite this as $(9-1{)}^{2n}$.

When we expand this, every term is a multiple of $9$ except for $(-1{)}^{2n}$. Since $2n$ is always an even integer, $(-1{)}^{2n}$ becomes $1$.

So, on our Modulo 9 Clock, $8^{2n}$ lands perfectly on $1$. It is as simple as that!

Now for the second part: $62^{2n+1}$. We know $62=63-1$, and since $63$ is a multiple of $9$, we can treat it as $0$ on our clock.

So, $62^{2n+1}$ becomes $(63-1{)}^{2n+1}$, which simplifies to $(-1{)}^{2n+1}(\mathrm{mod}9)$.

Here, the exponent $2n+1$ is always an odd integer. Therefore, $(-1{)}^{2n+1}$ remains $-1$. On our clock, landing on $-1$ is the same as landing on $8$ (since $-1+9=8$).

Now, we bring it all together. Our original expression is $8^{2n}-62^{2n+1}$.

Substituting our findings, we get:

The remainder is $2$.

It is elegant, it is fast, and it is powerful. By mastering this "clock" logic, you have turned a terrifying algebraic expression into a simple arithmetic walk in the park. Keep practicing this, and you will find that even the most complex JEE problems have a hidden, simple rhythm waiting to be discovered!