Question:

The minimum value of $ n $ for which the number of integer terms in the binomial expansion $\left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n$ is 183, is

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To find the number of integral terms in a binomial expansion with irrational components, use the formula for the terms and solve for $ n $.

Updated On: Nov 1, 2025

2196
2172
2184
2148

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The Correct Option is C

Approach Solution - 1

To find the smallest integer value of $ n $ such that the number of integer terms in the binomial expansion of $ \left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n $ is 183, we must follow these steps:

1. Consider the general term of the binomial expansion $ T_k = \binom{n}{k}(7^{\frac{1}{3}})^{n-k}(11^{\frac{1}{12}})^k $. For $ T_k $ to be an integer, both $ (7^{\frac{1}{3}})^{n-k} $ and $ (11^{\frac{1}{12}})^k $ must individually be integers.

2. This implies:

$ \frac{n-k}{3} $ is an integer, i.e., $ n-k = 3m $ for some integer $ m $.
$ \frac{k}{12} $ is an integer, i.e., $ k = 12p $ for some integer $ p $.

3. Solving these:

$ n - 3m = k = 12p $, thus $ n = 3m + 12p $.

4. To find the number of integer terms, $ k $ should vary such that both $ m $ and $ p $ are integers. Therefore, find all possible $ p $ satisfying $ 0 \leq 12p \leq n $.

5. The integer values of $ k = 12p $ must satisfy $ 0 \leq k \leq n $.

6. The condition is $ 0 \leq 12p \leq n $, resulting in $ 0 \leq p \leq \frac{n}{12} $.

7. Correspondingly, the condition $ n = 3m + 12p $ implies:

The number of values of $ p $ is $ \left\lfloor \frac{n}{12} \right\rfloor + 1 $.

8. Given that the number of integer terms is 183, we equate:

$\left\lfloor \frac{n}{12} \right\rfloor + 1 = 183 $

9. Solve for $ n $:

$\left\lfloor \frac{n}{12} \right\rfloor = 182 $

This implies:

$182 \times 12 \leq n < 183 \times 12$

Resulting in:

$2184 \leq n < 2196$

10. Therefore, the smallest integer value for $ n $ is $ n = 2184 $.

The correct answer is: 2184.

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Approach Solution -2

Step 1: Understand the binomial expansion.
We are given the binomial expansion of $ \left( 7^{\frac{1}{3}} + 11^{\frac{1}{12}} \right)^n $. To find the minimum value of $ n $ for which the number of integer terms in the expansion is 183, we need to analyze the general term in the expansion.

The general term in the binomial expansion of $ \left( a + b \right)^n $ is given by: \[ T_k = \binom{n}{k} a^{n-k} b^k. \] Here, $ a = 7^{\frac{1}{3}} $ and $ b = 11^{\frac{1}{12}} $. So, the general term in our expansion is: \[ T_k = \binom{n}{k} 7^{\frac{n-k}{3}} 11^{\frac{k}{12}}. \] For $ T_k $ to be an integer, both the exponents $ \frac{n-k}{3} $ and $ \frac{k}{12} $ must be integers.

Step 2: Conditions for integer terms.
1. $ \frac{n-k}{3} $ must be an integer, which implies that $ n-k $ must be divisible by 3. Hence, $ k \equiv n \pmod{3} $. 2. $ \frac{k}{12} $ must be an integer, which implies that $ k $ must be divisible by 12.

Therefore, $ k $ must satisfy the conditions: \[ k \equiv n \pmod{3} \quad \text{and} \quad k \equiv 0 \pmod{12}. \] Step 3: Find the number of valid $ k $'s.
The number of integer values of $ k $ from 0 to $ n $ that satisfy both conditions can be found by counting the multiples of 12 in the range $ [0, n] $ that also satisfy $ k \equiv n \pmod{3} $. The number of such terms is given by the number of multiples of 12 in the range $ [0, n] $, which is $ \left\lfloor \frac{n}{12} \right\rfloor + 1 $, and further adjusted by the condition $ k \equiv n \pmod{3} $.

Step 4: Solve for $ n $.
We are told that the number of integer terms is 183. So, we solve the equation: \[ \left\lfloor \frac{n}{12} \right\rfloor + 1 = 183. \] Solving for $ n $, we get $ n = 2184 $.

Final Answer:
\[ \boxed{2184}. \]