Animated Solution for Mathematics - Binomial Theorem: The total number of two digit numbers 'n', such that $3^n+7^n$ is a multiple of 10, is

Imagine you are standing before a complex, intimidating expression: $3^n+7^n$. The problem asks us to find all two-digit numbers $n$ such that this sum is a multiple of $10$.

At first glance, you might be tempted to start calculating powers of $3$ and $7$ for every $n$ from $10$ to $99$. But stop! That is a trap. In JEE Advanced, we don't calculate; we observe. We look for the hidden symmetry.

The secret lies in the relationship between the bases. We are working modulo $10$. Notice that $7\equiv -3(\mathrm{mod}10)$.

This is the spark! By rewriting $7$ as $10-3$, we transform our expression into $3^n+(10-3{)}^n$. This simple substitution is the key that unlocks the entire problem.

We have moved from a brute-force calculation to a structural analysis.

Now, let us apply the Binomial Theorem to the term $(10-3{)}^n$. The expansion is:

Look closely at this expansion. Every single term, except for the very last one, contains a power of $10$.

In the world of modulo $10$, any term that is a multiple of $10$ is effectively zero. So, the entire expansion collapses! We are left with just the last term: $(-3{)}^n$.

Our original expression, $3^n+7^n$, now simplifies beautifully to $3^n+(-3{)}^n(\mathrm{mod}10)$.

Now we face a fork in the road, determined by the parity of $n$.

Case 1: $n$ is even. If $n$ is even, then $(-3{)}^n=3^n$. Our expression becomes $3^n+3^n=2\cdot 3^n$.

Since $3^n$ is always odd, $2\cdot 3^n$ is $2$ times an odd number. This will always end in $2,4,6,$ or $8$. It will never be a multiple of $10$. So, even numbers are out.

Case 2: $n$ is odd. If $n$ is odd, then $(-3{)}^n=-3^n$. Our expression becomes $3^n-3^n=0$.

And $0$ is, by definition, a multiple of $10$. This means the condition is satisfied for every odd number!

We have discovered that $n$ must be an odd two-digit number. The range is $10\le n\le 99$.

The odd numbers in this range are $11,13,15,\dots ,99$. This is an arithmetic progression where the first term $a=11$, the last term $l=99$, and the common difference $d=2$.

To find the total count, we use the formula $N=\frac{l-a}{d}+1$. Substituting our values:

There are exactly $45$ such numbers. We didn't just solve a problem; we navigated a mathematical landscape, using the Binomial Theorem as our compass and parity as our guide.