Find the number of odd proper divisors of `3^pxx6^mxx21^ndot`

Explanation

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

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

Thus, we can rewrite the expression:
\[ 6^m = (2 \times 3)^m = 2^m \times 3^m \]
\[ 21^n = (3 \times 7)^n = 3^n \times 7^n \]

Now, substituting these back into the original expression:
\[ 3^p \times 6^m \times 21^n = 3^p \times (2^m \times 3^m) \times (3^n \times 7^n) \]
Combining the powers of \(3\):
\[ = 2^m \times 3^{p+m+n} \times 7^n \]

### Step 2: Identify the odd part of the expression
To find the odd proper divisors, we only consider the odd factors, which means we ignore the factor of \(2^m\). Thus, we focus on:
\[ 3^{p+m+n} \times 7^n \]

### Step 3: Calculate the total number of odd divisors
The number of divisors of a number in the form \(a^x \times b^y\) is given by:
\[ (x+1)(y+1) \]
For our odd part \(3^{p+m+n} \times 7^n\):
- The exponent of \(3\) is \(p + m + n\), so the number of choices for \(3\) is \(p + m + n + 1\).
- The exponent of \(7\) is \(n\), so the number of choices for \(7\) is \(n + 1\).

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

### Step 4: Exclude the number 1 to find proper odd divisors
Since we need the number of proper odd divisors, we must exclude the divisor \(1\). Therefore, we subtract \(1\) from the total count:
\[ \text{Number of odd proper divisors} = (p + m + n + 1)(n + 1) - 1 \]

### Final Answer
Thus, the number of odd proper divisors of \(3^p \times 6^m \times 21^n\) is:
\[ (p + m + n + 1)(n + 1) - 1 \]

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