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

Options

A. 84
B. 96
C. 91
D. 102

Answer

To find the number of integral terms in the expansion of \((5^{\frac{1}{6}} + 7^{\frac{1}{9}})^{1824}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^r b^{n-r} \]
In our case, \(n = 1824\), \(a = 7^{\frac{1}{9}}\), and \(b = 5^{\frac{1}{6}}\). Thus, the general term becomes:
\[ T_{r+1} = \binom{1824}{r} (7^{\frac{1}{9}})^r (5^{\frac{1}{6}})^{1824 - r} \]
This simplifies to:
\[ T_{r+1} = \binom{1824}{r} 7^{\frac{r}{9}} 5^{\frac{1824 - r}{6}} \]

### Step 2: Determine Conditions for Integral Terms
For \(T_{r+1}\) to be an integral term, both \(7^{\frac{r}{9}}\) and \(5^{\frac{1824 - r}{6}}\) must be integers.

1. **Condition for \(7^{\frac{r}{9}}\)**: This term is an integer when \(\frac{r}{9}\) is an integer, meaning \(r\) must be a multiple of \(9\).
2. **Condition for \(5^{\frac{1824 - r}{6}}\)**: This term is an integer when \(\frac{1824 - r}{6}\) is an integer, meaning \(1824 - r\) must be a multiple of \(6\).

### Step 3: Combine Conditions
From the above conditions:
- Let \(r = 9k\) for some integer \(k\).
- Then, \(1824 - r = 1824 - 9k\) must be a multiple of \(6\).

This can be expressed as:
\[ 1824 - 9k \equiv 0 \mod 6 \]
Calculating \(1824 \mod 6\):
\[ 1824 \div 6 = 304 \quad \text{(exact, so } 1824 \equiv 0 \mod 6\text{)} \]
Thus, the equation simplifies to:
\[ -9k \equiv 0 \mod 6 \]
Since \(9 \equiv 3 \mod 6\), we have:
\[ 3k \equiv 0 \mod 6 \]
This implies \(k\) must be a multiple of \(2\).

### Step 4: Find Valid Values of \(r\)
Let \(k = 2m\) for some integer \(m\). Then:
\[ r = 9k = 9(2m) = 18m \]
Now, we need \(r\) to be in the range \(0 \leq r \leq 1824\):
\[ 0 \leq 18m \leq 1824 \]
Dividing the entire inequality by \(18\):
\[ 0 \leq m \leq 101.33 \]
Thus, \(m\) can take integer values from \(0\) to \(101\), which gives us \(102\) possible values for \(m\).

### Conclusion
Therefore, the total number of integral terms in the expansion is:
\[ \boxed{102} \]