Animated Solution for Mathematics - Binomial Theorem: If $(2021{)}^{3762}$ is divided by 17, then the remainder is ____.

Imagine you are standing at the base of a mountain. The mountain is labeled $2021^{3762}$. It is massive, imposing, and frankly, terrifying.

If you try to calculate this number directly, you would run out of paper, ink, and time before you even finished the first few steps. In the world of JEE Advanced, these problems are not tests of your ability to calculate; they are tests of your ability to see the structure hidden beneath the chaos.

Today, we are going to dismantle this monster using the elegant tools of Number Theory.

When you see a large base like $2021$ and a divisor like $17$, your first instinct should be to simplify. We are looking for the remainder, which means we are working in the realm of modular arithmetic. We want to find a number $x$ such that $2021\equiv x(\mathrm{mod}17)$.

Let us perform a quick division. If we divide $2021$ by $17$, we get $\mathrm{118.88...}$. This tells us that $17\times 118$ is close, but let us look at $17\times 119$. That gives us $2023$.

Look at that! $2023$ is just $2$ units away from $2021$. So, we can write:

Since $2023$ is a perfect multiple of $17$, it vanishes into the ether of modular arithmetic (it becomes $0$). Thus, we are left with:

Suddenly, our monster $2021^{3762}$ has shrunk into the much more manageable $(-2{)}^{3762}$. We have successfully tamed the base.

Now we face the exponent: $3762$. It is even, which is a gift. A negative number raised to an even power becomes positive. So, $(-2{)}^{3762}$ is exactly the same as $2^{3762}$.

Now the problem is simply: What is $2^{3762}(\mathrm{mod}17)$? We need a cycle. We need to find a power of $2$ that lands us near a multiple of $17$.

Let us test the waters: - $2^1=2$ - $2^2=4$ - $2^3=8$ - $2^4=16$

Stop right there. $16$ is the magic number. Why? Because $16\equiv -1(\mathrm{mod}17)$.

This is the Golden Key we were looking for. Whenever you find a power that results in $\pm 1$, you have found the rhythm of the modular cycle. We know that $2^4\equiv -1(\mathrm{mod}17)$.

Now, we must break down the giant exponent $3762$ to fit our cycle of $4$. We divide $3762$ by $4$:

This means we can rewrite our expression using the laws of indices:

Now, substitute our Golden Key, $2^4\equiv -1(\mathrm{mod}17)$:

Look at the exponent $940$. It is even! Therefore, $(-1{)}^{940}=1$. The entire complex term collapses into simplicity:

And there it is. The remainder is $4$.

Think about what we just did. We took a number so large it would fill a book, and through the lens of modular arithmetic, we reduced it to a simple multiplication of $1\times 4$.

This is the beauty of mathematics. It is not about brute force; it is about finding the pattern, identifying the cycle, and letting the structure do the work for you. Whenever you face a problem like this in the future, do not panic. Find the base reduction, find the cycle, and watch the monster vanish.