The number of odd proper divisors of `3^(p)*6^(m)*21^(n)` is

Options

A. `(p+1)(m+1)(n+1)-2`
B. `(p+m+n+1)(n+1)-1`
C. `(p+1)(m+1)(n+1)-1`
D. `(p+m+n+1)(n+1)-2`

Answer

To find the number of odd proper divisors of the expression \(3^p \cdot 6^m \cdot 21^n\), we can follow these steps:

### Step 1: Factor the expression
First, we need to express \(6\) and \(21\) in terms of their prime factors:
- \(6 = 2^1 \cdot 3^1\)
- \(21 = 3^1 \cdot 7^1\)

Now, substituting these into the original expression:
\[ 3^p \cdot 6^m \cdot 21^n = 3^p \cdot (2^1 \cdot 3^1)^m \cdot (3^1 \cdot 7^1)^n \]
This simplifies to:
\[ 3^p \cdot 2^m \cdot 3^m \cdot 3^n \cdot 7^n = 2^m \cdot 3^{p+m+n} \cdot 7^n \]

### Step 2: Identify the odd part
Since we are interested in the odd divisors, we can ignore the factor of \(2^m\) (as it contributes to even divisors). Thus, we focus on the odd part:
\[ 3^{p+m+n} \cdot 7^n \]

### Step 3: Calculate the number of odd divisors
The formula for the number of divisors of a number given its prime factorization \(p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}\) is:
\[ (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \]
For our odd part \(3^{p+m+n} \cdot 7^n\), the exponents are:
- For \(3\): \(p + m + n\)
- For \(7\): \(n\)

Thus, the total number of odd divisors is:
\[ (p + m + n + 1)(n + 1) \]

### Step 4: Calculate the number of proper odd divisors
Proper divisors are all divisors except the number itself. Therefore, we subtract 1 from the total number of odd divisors:
\[ \text{Number of odd proper divisors} = (p + m + n + 1)(n + 1) - 1 \]

### Final Result
The number of odd proper divisors of \(3^p \cdot 6^m \cdot 21^n\) is:
\[ (p + m + n + 1)(n + 1) - 1 \]