Animated Solution for Mathematics - Binomial Theorem: If the coefficient of $x^{10}$ in the binomial expansion of ${\left(\frac{\sqrt{x}}{5^{1/4}}+\frac{\sqrt{5}}{x^{1/3}}\right)}^{60}$ is $5^{k}l$, where $l,k\in \mathbb{N}$ and $l$ is coprime to 5, then $k$ is equal to ____.

Unlocking the Power of 5

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of the Binomial Theorem.

When you look at an expression like ${\left(\frac{\sqrt{x}}{5^{1/4}}+\frac{\sqrt{5}}{x^{1/3}}\right)}^{60}$, it is natural to feel a momentary shiver. A power of 60 seems gargantuan, but in the world of JEE Advanced, we do not fear size; we embrace structure. Let us peel back the layers of this problem together.

The General Term—Our North Star

Every binomial expansion is governed by a fundamental law: the general term formula. We know that for any expansion $(a+b{)}^n$, the $(r+1)$-th term is given by:

In our specific case, $n=60$, our first term $a$ is $\frac{x^{1/2}}{5^{1/4}}$, and our second term $b$ is $\frac{5^{1/2}}{x^{1/3}}$.

When we substitute these into our formula, we get:

Do not rush to expand everything. Instead, separate the constants from the variables to organize your workspace. Grouping the powers of 5 together and the powers of $x$ together is the key to clarity.

The Hunt for $r$

We are on a mission to find the coefficient of $x^{10}$. This means we must isolate the variable $x$.

From our first term, we have $x^{1/2}$ raised to the power of $(60-r)$, which gives us $x^{(60-r)/2}$. From our second term, we have $x^{-1/3}$ raised to the power of $r$, which gives us $x^{-r/3}$.

When we multiply these, we add the exponents:

To solve this, we multiply the entire equation by 6 to clear the denominators:

Expanding this, we get $180-3r-2r=60$, which simplifies to $180-5r=60$. A quick rearrangement gives us $5r=120$, and finally, $r=24$. We have found our target; the 25th term is the one we are looking for.

The Hidden Constants

Now that we know $r=24$, we return to our constant terms. We substitute $r=24$ into the constant part of our general term:

Let us simplify these exponents of 5. The first part becomes $(5^{-1/4}{)}^{36}=5^{-9}$, and the second part becomes $(5^{1/2}{)}^{24}=5^{12}$.

Multiplying these, we get $5^{-9}\cdot 5^{12}=5^3$. So, our coefficient is $\left(\genfrac{}{}{0ex}{}{60}{24}\right)\cdot 5^3$.

Legendre's Magic

We are told the coefficient is $5^k\cdot l$, where $l$ is coprime to 5. This means we need to find the total power of 5 in $\left(\genfrac{}{}{0ex}{}{60}{24}\right)\cdot 5^3$.

We already have a $5^3$, but we must check if there are more factors of 5 hidden inside $\left(\genfrac{}{}{0ex}{}{60}{24}\right)=\frac{60!}{24!\cdot 36!}$. This is where Legendre's Formula shines, which states the exponent of a prime $p$ in $n!$ is $\sum \lfloor \frac{n}{p^k}\rfloor$.

For $60!$, we have:

For $24!$, we have $\lfloor \frac{24}{5}\rfloor =4$. For $36!$, we have:

The exponent of 5 in $\left(\genfrac{}{}{0ex}{}{60}{24}\right)$ is $14-(4+8)=2$.

Finally, we combine everything: the $5^2$ from the combination and the $5^3$ we found earlier. The total exponent $k=5$. Through patience and the right tools, we have conquered the problem.