Animated Solution for Mathematics - Binomial Theorem: The remainder when $(2021{)}^{2022}+(2022{)}^{2021}$ is divided by 7 is

Imagine you are standing before a massive, imposing wall of numbers: $(2021{)}^{2022}+(2022{)}^{2021}$. It looks terrifying, doesn't it?

In the world of JEE Advanced, examiners love to throw these gargantuan expressions at you, hoping you will freeze. But here is the secret: they are not testing your ability to calculate; they are testing your ability to see the structure.

We are going to dismantle this wall, brick by brick, using the elegant lens of modular arithmetic.

Our first task is to simplify the bases. We are working modulo $7$.

Let us look at $2021$ first. If we perform a simple division: This tells us that $2021\equiv 5(\mathrm{mod}7)$.

Now, look at $2022$. It is just one greater than $2021$, so: We have already reduced the problem from massive numbers to simple digits. But we can go further.

Working with $5$ and $6$ as bases is fine, but we can make it even easier. In modular arithmetic, a remainder of $5$ is the same as a remainder of $5-7=-2$.

Similarly, a remainder of $6$ is the same as $6-7=-1$. Why do we do this? Because powers of $-1$ and $-2$ are incredibly easy to handle.

Our expression now looks like this: See how the fear factor has vanished?

Let us tackle the second term first: $(-1{)}^{2021}$. Since $2021$ is an odd number, the negative sign survives, giving us $-1$.

Now for the first term: $(-2{)}^{2022}$. Since $2022$ is an even number, the negative sign disappears, leaving us with $2^{2022}$.

Now, we need to find $2^{2022}(\mathrm{mod}7)$. We look for a cycle: This is our golden key! Since $2^3\equiv 1(\mathrm{mod}7)$, we can write $2^{2022}$ as:

We are at the finish line. We have $1$ from the first term and $-1$ from the second term.

Adding them together: The remainder is $0$. The entire expression is perfectly divisible by 7.

You see? By breaking the problem down and using the properties of numbers, we turned a mountain into a molehill. Keep this mindset, and no problem will ever be too big for you.