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
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
Show Hint
- 2196
- 2172
- 2184
- 2148
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.
Approach Solution -2
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:
\[ \boxed{2184}. \]
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