The number of odd proper divisors of `3^(p)*6^(q)*15^(r),AA p,q,r, in N`, is

Options

A. `(p+1)(q+1)(r+1)-2`
B. `(p+1)(q+1)(r+1)-1`
C. `(p+q+r+1)(r+1)-2`
D. `(p+q+r+1)(r+1)-1`

Answer

To find the number of odd proper divisors of the expression \(3^p \cdot 6^q \cdot 15^r\), where \(p, q, r \in \mathbb{N}\), we can follow these steps:

### Step 1: Factor the expression into its prime factors
First, we need to express \(6\) and \(15\) in terms of their prime factors:
- \(6 = 2^1 \cdot 3^1\)
- \(15 = 3^1 \cdot 5^1\)

Now we can rewrite the expression:
\[ 3^p \cdot 6^q \cdot 15^r = 3^p \cdot (2^1 \cdot 3^1)^q \cdot (3^1 \cdot 5^1)^r \]
This simplifies to:
\[ 3^p \cdot 2^q \cdot 3^q \cdot 3^r \cdot 5^r = 2^q \cdot 3^{p + q + r} \cdot 5^r \]

### Step 2: Identify the odd part
To find the odd proper divisors, we need to exclude any factors of \(2\). Therefore, we focus on the odd part of the expression:
\[ 3^{p + q + r} \cdot 5^r \]

### Step 3: Determine the number of odd divisors
The formula for finding the number of divisors from the prime factorization \(p_1^{e_1} \cdot p_2^{e_2} \cdots p_n^{e_n}\) is given by:
\[ (e_1 + 1)(e_2 + 1) \cdots (e_n + 1) \]

For our odd part \(3^{p + q + r} \cdot 5^r\):
- The exponent of \(3\) is \(p + q + r\)
- The exponent of \(5\) is \(r\)

Thus, the number of odd divisors is:
\[ (p + q + r + 1)(r + 1) \]

### Step 4: Exclude the number 1
Since we are looking for proper divisors, we need to exclude \(1\) from our count of divisors. Therefore, we subtract \(1\) from the total number of odd divisors:
\[ \text{Number of odd proper divisors} = (p + q + r + 1)(r + 1) - 1 \]

### Final Answer
Thus, the number of odd proper divisors of \(3^p \cdot 6^q \cdot 15^r\) is:
\[ (p + q + r + 1)(r + 1) - 1 \]