If the coefficient of `x^(6)` in the expansion of `(2+x)^(3)(3+x)^(2)(5+x)^(3)` is K, then the value of `(K)/(100)` is

Answer

To find the coefficient \( K \) of \( x^6 \) in the expansion of \( (2+x)^3 (3+x)^2 (5+x)^3 \), we will follow these steps:

### Step 1: Identify the contributions to \( x^6 \)
The expression can be expanded as follows:
- \( (2+x)^3 \) contributes terms of the form \( 2^{3-k} x^k \) for \( k = 0, 1, 2, 3 \).
- \( (3+x)^2 \) contributes terms of the form \( 3^{2-m} x^m \) for \( m = 0, 1, 2 \).
- \( (5+x)^3 \) contributes terms of the form \( 5^{3-n} x^n \) for \( n = 0, 1, 2, 3 \).

We need to find combinations of \( k \), \( m \), and \( n \) such that \( k + m + n = 6 \).

### Step 2: Determine valid combinations of \( k, m, n \)
The possible combinations of \( (k, m, n) \) that satisfy \( k + m + n = 6 \) are:
1. \( (3, 2, 1) \)
2. \( (3, 1, 2) \)
3. \( (2, 2, 2) \)
4. \( (2, 3, 1) \)
5. \( (1, 2, 3) \)

### Step 3: Calculate the contributions for each combination
1. For \( (3, 2, 1) \):
- Coefficient: \( 2^{3-3} \cdot 3^{2-2} \cdot 5^{3-1} = 1 \cdot 1 \cdot 25 = 25 \)

2. For \( (3, 1, 2) \):
- Coefficient: \( 2^{3-3} \cdot 3^{2-1} \cdot 5^{3-2} = 1 \cdot 3 \cdot 5 = 15 \)

3. For \( (2, 2, 2) \):
- Coefficient: \( 2^{3-2} \cdot 3^{2-2} \cdot 5^{3-2} = 2 \cdot 1 \cdot 5 = 10 \)

4. For \( (2, 3, 1) \):
- Coefficient: \( 2^{3-2} \cdot 3^{2-3} \cdot 5^{3-1} = 2 \cdot \frac{1}{3} \cdot 25 = \frac{50}{3} \)

5. For \( (1, 2, 3) \):
- Coefficient: \( 2^{3-1} \cdot 3^{2-2} \cdot 5^{3-3} = 4 \cdot 1 \cdot 1 = 4 \)

### Step 4: Sum the contributions
Now, we sum all the contributions:
\[ K = 25 + 15 + 10 + \frac{50}{3} + 4 \]
Converting to a common denominator (which is 3):
\[ K = \frac{75}{3} + \frac{45}{3} + \frac{30}{3} + \frac{50}{3} + \frac{12}{3} = \frac{212}{3} \]

### Step 5: Calculate \( \frac{K}{100} \)
Now, we find \( \frac{K}{100} \):
\[ \frac{K}{100} = \frac{\frac{212}{3}}{100} = \frac{212}{300} = \frac{53}{75} \]

Thus, the final answer is:
\[ \frac{K}{100} = \frac{53}{75} \]