Key Concept: General Term of a Binomial Expansion

For a binomial expansion of the form $(A+B)^n$, the general term, denoted as $T_{r+1}$, is given by the formula: $T_{r+1} = \binom{n}{r} A^{n-r} B^r$ where $r$ is an integer ranging from $0$ to $n$ ($0 \le r \le n$), and $\binom{n}{r}$ is the binomial coefficient, which is always an integer.

Step-by-step Derivation of Integral Terms

1. Identify $A$ and $B$ from the given expression: The given binomial expression is $(\sqrt[3]{7}+\sqrt[12]{11})^n$. We can rewrite the terms with fractional exponents: $A = \sqrt[3]{7} = 7^{1/3}$ $B = \sqrt[12]{11} = 11^{1/12}$

2. Substitute $A$ and $B$ into the general term formula: $T_{r+1} = \binom{n}{r} (7^{1/3})^{n-r} (11^{1/12})^r$ Applying the exponent rule $(x^a)^b = x^{ab}$, we get: $T_{r+1} = \binom{n}{r} 7^{\frac{n-r}{3}} 11^{\frac{r}{12}}$

3. Establish conditions for $T_{r+1}$ to be an integral term: For $T_{r+1}$ to be an integral term (i.e., its value is an integer), two conditions must be met:

   • The binomial coefficient $\binom{n}{r}$ is always an integer.
   • The powers of the prime bases (7 and 11) must be non-negative integers. This is crucial because if the exponents are fractional, the terms $7^{\text{fraction}}$ or $11^{\text{fraction}}$ would result in irrational numbers, making the entire term non-integral. Therefore, we require:
   • $\frac{n-r}{3}$ must be a non-negative integer. This implies that $(n-r)$ must be a multiple of 3.
   • $\frac{r}{12}$ must be a non-negative integer. This implies that $r$ must be a multiple of 12.

4. Determine the possible values of $r$ based on the second condition: Since $r$ must be a multiple of 12 and $r \ge 0$, the possible values for $r$ can be expressed as $r = 12k$, where $k$ is a non-negative integer ($k \in \{0, 1, 2, \dots\}$). These values of $r$ will be $0, 12, 24, 36, \dots$.

5. Relate the number of integral terms to the range of $r$: We are given that there are 183 integral terms in the expansion. Each integral term corresponds to a unique value of $r$ that satisfies the conditions. If $r = 12k$, then the values of $k$ would be $0, 1, 2, \dots, k_{max}$. The number of such values of $k$ is $k_{max} - 0 + 1 = k_{max} + 1$. Given that the number of integral terms is 183, we have: $k_{max} + 1 = 183$ $k_{max} = 182$

6. Calculate the maximum value of $r$ that yields an integral term: Using $k_{max} = 182$, the largest possible value of $r$ that results in an integral term is: $r_{max} = 12 \times k_{max} = 12 \times 182 = 2184$.

7. Determine the least value of $n$: For any term $T_{r+1}$ to exist in the binomial expansion, its index $r$ must be less than or equal to $n$. That is, $0 \le r \le n$. Since the maximum value of $r$ for an integral term is $2184$, it must be true that $n \ge r_{max}$. So, $n \ge 2184$.

   Now, let's revisit the first condition for integral terms: $(n-r)$ must be a multiple of 3. We know that $r = 12k$. Since $12k$ is always a multiple of 3 (as $12$ is a multiple of $3$), for $(n-r)$ to be a multiple of 3, $n$ must also be a multiple of 3. (Because if $n$ is not a multiple of 3, then $n-12k$ would also not be a multiple of 3).

   Therefore, we need to find the least value of $n$ such that:

   • $n \ge 2184$
   • $n$ is a multiple of 3.

   Let's check if 2184 is a multiple of 3: $2+1+8+4 = 15$. Since 15 is a multiple of 3, 2184 is also a multiple of 3. Thus, the smallest integer value of $n$ that satisfies both conditions is 2184.

Tips and Common Mistakes:

• Don't forget $r=0$: The value $r=0$ (which corresponds to $k=0$) gives the first term ($T_1$) and is usually an integral term, provided $n$ is a multiple of the denominator of the first term's exponent. Always include $r=0$ in your count of integral terms.
• Check both exponent conditions: Ensure both $\frac{n-r}{p}$ and $\frac{r}{q}$ (where $p$ and $q$ are denominators of the exponents) are integers. Sometimes one condition implies the other, but it's essential to verify. In this case, $r=12k$ automatically makes $\frac{r}{12}$ an integer. Then $n-r$ being a multiple of 3, combined with $r$ being a multiple of 3, implies $n$ must be a multiple of 3.
• Range of $r$: Remember that $0 \le r \le n$. This constraint is vital for determining the minimum value of $n$.

Summary

To find the number of integral terms in a binomial expansion with fractional exponents, we set up the general term and identify the conditions on $r$ (the index) that make the exponents integers. We then use the given number of integral terms to find the maximum possible value of $r$. Finally, knowing that $r$ cannot exceed $n$, and satisfying any other conditions on $n$ derived from the exponents, we can determine the least value of $n$. In this problem, $r$ had to be a multiple of 12, and $n$ had to be a multiple of 3. The maximum $r$ was 2184, which is already a multiple of 3, making it the least possible value for $n$.

The final answer is $\boxed{\text{2184}}$.