Number of proper divisors of `3^(2)5^(3)` which are odd, is `p` then find the value of `(p-5)`

Options

A. `6`
B. `5`
C. `7`
D. `8`

Explanation

To find the number of proper odd divisors of \(3^2 \cdot 5^3\), we can follow these steps:

### Step 1: Identify the prime factorization
The given number is \(n = 3^2 \cdot 5^3\).

### Step 2: Use the formula for the number of divisors
The formula to find the total number of divisors of a number \(n = p_1^{e_1} \cdot p_2^{e_2}\) is given by:
\[ d(n) = (e_1 + 1)(e_2 + 1) \]
where \(e_1\) and \(e_2\) are the exponents in the prime factorization.

### Step 3: Apply the formula
For our number:
- The exponent of \(3\) (which is \(e_1\)) is \(2\).
- The exponent of \(5\) (which is \(e_2\)) is \(3\).

Using the formula:
\[ d(n) = (2 + 1)(3 + 1) = 3 \cdot 4 = 12 \]

### Step 4: Calculate the number of proper divisors
The number of proper divisors is given by:
\[ \text{Number of proper divisors} = d(n) - 1 \]
So,
\[ \text{Number of proper divisors} = 12 - 1 = 11 \]

### Step 5: Assign the value to \(p\)
Let \(p\) be the number of proper odd divisors. Thus, we have:
\[ p = 11 \]

### Step 6: Find \(p - 5\)
Now, we need to find \(p - 5\):
\[ p - 5 = 11 - 5 = 6 \]

### Final Answer
Thus, the value of \(p - 5\) is \(6\).

---