Animated Solution for Mathematics - Binomial Theorem: The number of positive integers $k$ such that the constant term in the binomial expansion of ${\left(2x^3+\frac{3}{x^k}\right)}^{12}$, $x\ne 0$ is $2^8\cdot l$, where $l$ is an odd integer, is ____.

Imagine you are standing on the precipice of a complex binomial expansion: ${\left(2x^3+\frac{3}{x^k}\right)}^{12}$. At first glance, it looks like a daunting algebraic expression.

But in the world of JEE Advanced, we don't just solve; we uncover the hidden structure. Our mission is to find the number of positive integers $k$ such that the constant term in this expansion takes the form $2^8\cdot l$, where $l$ is an odd integer.

This is not just algebra; it is a beautiful dance between combinatorics and number theory.

The Anatomy of the General Term

To find the constant term, we must first summon the general term of the binomial expansion. The Binomial Theorem tells us that for any expansion $(a+b{)}^n$, the general term $T_{r+1}$ is given by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^r{b}^{n-r}$.

Substituting our specific values, $a=2x^3$ and $b=3x^{-k}$, we get:

Now, let's peel back the layers. We separate the constants from the variables to see the true nature of the term:

Combining the powers of $x$, the net exponent becomes $3r-k(12-r)$. This is the heartbeat of the problem.

For the term to be constant, it must be independent of $x$, meaning the exponent must vanish into thin air. Thus, we set $3r-k(12-r)=0$.

The Constraint of the Constant Term

Solving for $k$, we find the relationship that governs our search:

To make this more manageable, let's perform a little algebraic manipulation:

Since $k$ must be a positive integer, $(12-r)$ must be a positive divisor of $36$. Given that $0\le r\le 12$, the possible values for $(12-r)$ are $1,2,3,4,6,9$.

This leads us to a set of candidate values for $r$: $11,10,9,8,6,3$.

The Number Theory Twist

Now, we enter the final arena. We need the coefficient $C=\left(\genfrac{}{}{0ex}{}{12}{r}\right)2^r{3}^{12-r}$ to have exactly $2^8$ as its highest power of 2.

As we discussed, $3^{12-r}$ is always odd, so it contributes nothing to the power of 2. We are left with the condition:

Let's test our candidates:

For $r=3$: $\left(\genfrac{}{}{0ex}{}{12}{3}\right)=220=2^2\cdot 55$. So, $v_2 = 2 + 3 = 5 eq 8$. For $r=6$: $\left(\genfrac{}{}{0ex}{}{12}{6}\right)=924=2^2\cdot 231$. So, $v_2=2+6=8$. (Valid!) For $r=8$: $\left(\genfrac{}{}{0ex}{}{12}{8}\right)=495=2^0\cdot 495$. So, $v_2=0+8=8$. (Valid!) For $r=9$: $\left(\genfrac{}{}{0ex}{}{12}{9}\right)=220=2^2\cdot 55$. So, $v_2 = 2 + 9 = 11 eq 8$. For $r=10$: $\left(\genfrac{}{}{0ex}{}{12}{10}\right)=66=2^1\cdot 33$. So, $v_2 = 1 + 10 = 11 eq 8$. For $r=11$: $\left(\genfrac{}{}{0ex}{}{12}{11}\right)=12=2^2\cdot 3$. So, $v_2 = 2 + 11 = 13 eq 8$.

Only $r=6$ and $r=8$ satisfy our rigorous conditions. For $r=6$, we find $k=3$. For $r=8$, we find $k=6$.

Both are positive integers. Thus, there are exactly 2 such values of $k$.

We have navigated the binomial expansion, tamed the algebraic constraints, and solved the number theory puzzle. Remember, in JEE Advanced, the path is just as important as the destination. Keep exploring!