The number of terms in the expansion of `(5^((1)/(6))+7^((1)/(9)))^(1824)` which are integers is

Options

A. 100
B. 101
C. 102
D. 103

Answer

To find the number of integer terms in the expansion of \((5^{\frac{1}{6}} + 7^{\frac{1}{9}})^{1824}\), we will use the binomial theorem.

### Step-by-Step Solution:

1. **Identify the General Term:**
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = 5^{\frac{1}{6}}\), \(b = 7^{\frac{1}{9}}\), and \(n = 1824\).

2. **Write the General Term for Our Expansion:**
Substituting the values into the formula, we get:
\[ T_{r+1} = \binom{1824}{r} (5^{\frac{1}{6}})^{1824 - r} (7^{\frac{1}{9}})^r \]
This simplifies to:
\[ T_{r+1} = \binom{1824}{r} 5^{\frac{1824 - r}{6}} 7^{\frac{r}{9}} \]

3. **Determine When the Term is an Integer:**
For \(T_{r+1}\) to be an integer, both exponents \(\frac{1824 - r}{6}\) and \(\frac{r}{9}\) must be integers. This means:
- \(1824 - r\) must be divisible by 6.
- \(r\) must be divisible by 9.

4. **Set Up the Conditions:**
From \(1824 - r \equiv 0 \mod 6\), we can express this as:
\[ r \equiv 1824 \mod 6 \]
Since \(1824 \div 6 = 304\), we have \(1824 \equiv 0 \mod 6\). Thus, \(r \equiv 0 \mod 6\).

From \(r \equiv 0 \mod 9\), we can express this as:
\[ r \equiv 0 \mod 9 \]

5. **Find the Common Multiples:**
We need \(r\) to be a common multiple of both 6 and 9. The least common multiple (LCM) of 6 and 9 is 18. Thus, \(r\) must be a multiple of 18.

6. **Determine the Range of \(r\):**
The possible values of \(r\) range from 0 to 1824. We can find the multiples of 18 within this range:
\[ r = 0, 18, 36, \ldots, 1824 \]

7. **Count the Number of Terms:**
The multiples of 18 can be expressed as:
\[ r = 18k \quad \text{where } k = 0, 1, 2, \ldots, \frac{1824}{18} \]
Calculating \(\frac{1824}{18}\):
\[ \frac{1824}{18} = 101.333 \quad \Rightarrow \quad k \text{ can take values from } 0 \text{ to } 101 \]
Thus, \(k\) can take 102 values (including 0).

### Final Answer:
The number of integer terms in the expansion is **102**.