Animated Solution for Mathematics - Binomial Theorem: The least value of $n$ for which the number of integral terms in the binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11}{)}^n$ is 183, is:

To solve this problem, we examine the binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11}{)}^n$. We seek the smallest integer $n$ such that there are exactly $183$ rational terms in the expansion.

Our Key to the Kingdom

We invoke the general term formula for the expansion $(a+b{)}^n$, where the $(r+1)$-th term is given by:

Substituting $a=7^{1/3}$ and $b=11^{1/12}$, we obtain:

Simplifying the exponents, we arrive at the expression:

Here, $r$ is an integer such that $0\le r\le n$.

For $T_{r+1}$ to be an integer, the exponents of $7$ and $11$ must be non-negative integers. First, consider the exponent of $11$: $\frac{r}{12}$. For this to be an integer, $r$ must be a multiple of $12$.

Next, consider the exponent of $7$: $\frac{n-r}{3}$. For this to be an integer, $n-r$ must be a multiple of $3$. Since $r$ is a multiple of $12$ (and thus a multiple of $3$), $n$ must also be a multiple of $3$ for the expression to be an integer.

The valid values for $r$ form an arithmetic progression: $0,12,24,\dots ,r_{max}$. We are given that there are exactly $183$ such terms.

Using the formula for the number of terms in an arithmetic progression:

Substituting the known values:

Solving for $r_{max}$:

Since $r$ ranges from $0$ to $n$, the maximum value $r_{max}$ must be less than or equal to $n$. To find the smallest $n$, we set $n=r_{max}$.

Thus, $n=2184$.

We verify that $2184$ is divisible by $3$ (sum of digits $2+1+8+4=15$), which satisfies our earlier constraint. The smallest value of $n$ is $2184$.