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 \((5^{\frac{1}{6}} + 7^{\frac{1}{8}})^{1824}\), we will follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \((x + y)^n\) is given by:
\[ \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r \]
In our case, \(x = 5^{\frac{1}{6}}\), \(y = 7^{\frac{1}{8}}\), and \(n = 1824\).

### Step 2: Write the General Term
The general term \(T_r\) in the expansion can be expressed as:
\[ T_r = \binom{1824}{r} (5^{\frac{1}{6}})^{1824 - r} (7^{\frac{1}{8}})^r = \binom{1824}{r} 5^{\frac{1824 - r}{6}} 7^{\frac{r}{8}} \]

### Step 3: Conditions for Integral Terms
For \(T_r\) to be an integer, both exponents \(\frac{1824 - r}{6}\) and \(\frac{r}{8}\) must be integers. This leads to the following conditions:
1. \(1824 - r\) must be a multiple of 6.
2. \(r\) must be a multiple of 8.

### Step 4: Express Conditions Mathematically
From the first condition:
\[ 1824 - r \equiv 0 \mod 6 \implies r \equiv 1824 \mod 6 \]
Calculating \(1824 \mod 6\):
\[ 1824 \div 6 = 304 \quad \text{(exact division, remainder 0)} \]
Thus, \(r \equiv 0 \mod 6\).

From the second condition:
\[ r \equiv 0 \mod 8 \]

### Step 5: Find Common Multiples
We need \(r\) to be a common multiple of both 6 and 8. The least common multiple (LCM) of 6 and 8 is 24. Therefore, \(r\) must be a multiple of 24:
\[ r = 24k \quad \text{for integers } k \]

### Step 6: Determine the Range of \(r\)
Since \(r\) can take values from 0 to 1824, we need to find the maximum integer \(k\) such that:
\[ 24k \leq 1824 \]
Dividing both sides by 24:
\[ k \leq \frac{1824}{24} = 76 \]
Thus, \(k\) can take values from 0 to 76, giving us:
\[ k = 0, 1, 2, \ldots, 76 \]
This gives us \(77\) possible values for \(k\).

### Step 7: Count the Integral Terms
Since \(k\) can take \(77\) values, there are \(77\) integral terms in the expansion.

### Final Answer
The number of integral terms in the expansion of \((5^{\frac{1}{6}} + 7^{\frac{1}{8}})^{1824}\) is **77**.

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