The number of proper divisors of `2^(p)*6^(q)*21^(r),AA p,q,r in N`, is

Options

A. (a) (p+q+1)(q+r+1)(r+1)
B. (b) (p+q+1)(q+r+1)(r+1)-2
C. (c) (p+q)(q+r)r-2
D. (d) (p+q)(q+r)r

Answer

To find the number of proper divisors of the expression \(2^p \cdot 6^q \cdot 21^r\), where \(p, q, r\) are natural numbers, we can follow these steps:

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

Thus, we can rewrite the expression:
\[ 6^q = (2 \cdot 3)^q = 2^q \cdot 3^q \]
\[ 21^r = (3 \cdot 7)^r = 3^r \cdot 7^r \]

### Step 2: Combine the Factors
Now, substituting back into the original expression:
\[ 2^p \cdot 6^q \cdot 21^r = 2^p \cdot (2^q \cdot 3^q) \cdot (3^r \cdot 7^r) \]
This simplifies to:
\[ 2^{p+q} \cdot 3^{q+r} \cdot 7^r \]

### Step 3: Count Total Divisors
To find the total number of divisors of a number expressed in the form \(p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3}\), the formula is:
\[ (e_1 + 1)(e_2 + 1)(e_3 + 1) \]
Applying this to our expression:
- For \(2^{p+q}\), the exponent is \(p+q\), so we have \(p+q + 1\).
- For \(3^{q+r}\), the exponent is \(q+r\), so we have \(q+r + 1\).
- For \(7^r\), the exponent is \(r\), so we have \(r + 1\).

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

### Step 4: Count Proper Divisors
Proper divisors are all divisors except the number itself. Therefore, we subtract \(1\) (the number itself) from the total number of divisors:
\[ \text{Number of proper divisors} = D - 1 = (p + q + 1)(q + r + 1)(r + 1) - 1 \]

### Final Expression
Thus, the number of proper divisors of \(2^p \cdot 6^q \cdot 21^r\) is:
\[ (p + q + 1)(q + r + 1)(r + 1) - 1 \]