Animated Solution for Mathematics - Binomial Theorem: The remainder when $19^{200}+23^{200}$ is divided by 49, is ______.

Imagine standing before a mountain of a problem: finding the remainder of $19^{200}+23^{200}$ when divided by $49$. At first glance, your intuition might suggest this is impossible. But in the world of JEE Advanced, we do not calculate; we observe.

We look for the hidden architecture beneath the numbers. The first step is to recognize the symmetry. Look at the bases $19$ and $23$; they are perfectly balanced around the number $21$.

Because $21$ is a multiple of $7$ and our divisor is $49$ (which is $7^2$), this symmetry is the key that unlocks the door. We rewrite our expression as:

Now, we invoke the Binomial Theorem. When we expand $(21-2{)}^{200}+(21+2{)}^{200}$, we are looking at a sum of two expansions. Because the power $200$ is even, the odd-powered terms in the expansion of $(21-2{)}^{200}$ will perfectly cancel out the corresponding terms in $(21+2{)}^{200}$.

We are left with twice the sum of the even terms:

This looks intimidating, but here is where the magic happens. We are working modulo $49$. Notice that $21^2=441$, and since $441=9\times 49$, $21^2$ is a multiple of $49$.

Consequently, any term in our expansion that contains $21^k$ where $k\ge 2$ becomes $0$ modulo $49$. The entire series collapses! Every single term vanishes except for the very last one, which is $2\times 2^{200}$. We have simplified a monstrous expression down to $2^{201}$.

Conquering $2^{201}$

We are now left with finding the remainder of $2^{201}$ when divided by $49$. We need to bring this closer to a multiple of $7$. We know that $2^3=8$, and $8=1+7$. This is our bridge.

We rewrite $2^{201}$ as $(2^3{)}^{67}$, which is $8^{67}$. Now, we apply the Binomial Theorem one last time:

Again, look at the terms. Any term containing $7^2$ or higher is a multiple of $49$ and vanishes. We are left with only the first two terms:

Calculating this, we get $1+469=470$. Finally, we divide $470$ by $49$. Since $49\times 9=441$, the remainder is $470-441=29$.

The final remainder is $29$.