If the number of terms free from radicals in the expansion of `(7^((1)/(3))+11^((1)/(9)))^(6561)` is k, then the value of `(k)/(100)` is equal to

Answer

To solve the problem, we need to find the number of terms free from radicals in the expansion of \((7^{\frac{1}{3}} + 11^{\frac{1}{9}})^{6561}\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = 7^{\frac{1}{3}}\), \(b = 11^{\frac{1}{9}}\), and \(n = 6561\).

2. **Write the General Term for Our Case**:
Substituting the values, we have:
\[ T_{r+1} = \binom{6561}{r} (7^{\frac{1}{3}})^{6561-r} (11^{\frac{1}{9}})^r \]
Simplifying this, we get:
\[ T_{r+1} = \binom{6561}{r} 7^{\frac{6561 - r}{3}} 11^{\frac{r}{9}} \]

3. **Determine Conditions for Integer Terms**:
For \(T_{r+1}\) to be an integer, both exponents \(\frac{6561 - r}{3}\) and \(\frac{r}{9}\) must be non-negative integers. This leads us to the following conditions:
- \(6561 - r \equiv 0 \mod 3\)
- \(r \equiv 0 \mod 9\)

4. **Find Values of \(r\)**:
From the first condition, \(r\) must be congruent to \(0 \mod 3\). From the second condition, \(r\) must be a multiple of \(9\). Therefore, \(r\) can take values of \(0, 9, 18, \ldots, 6561\).

5. **Count the Valid Values of \(r\)**:
The values of \(r\) that satisfy both conditions can be expressed as:
\[ r = 9k \quad \text{where } k = 0, 1, 2, \ldots, \frac{6561}{9} \]
The maximum value of \(k\) is:
\[ k = \frac{6561}{9} = 729 \]
Thus, \(k\) can take values \(0, 1, 2, \ldots, 729\), which gives us a total of \(729 + 1 = 730\) values (including \(r = 0\)).

6. **Final Calculation**:
The number of terms free from radicals is \(k = 730\). We need to find \(\frac{k}{100}\):
\[ \frac{k}{100} = \frac{730}{100} = 7.3 \]

### Conclusion:
The value of \(\frac{k}{100}\) is \(7.3\).