Let S be the set of all ordered pairs (x,y) of positive intergers, with HCF (x,y)=16 and LCM (x,y) = 48000. The number of elements in S is

Options

A. 4
B. 8
C. 16
D. 32

Explanation

To solve the problem, we need to find the number of ordered pairs \((x, y)\) of positive integers such that:

1. \( \text{HCF}(x, y) = 16 \)
2. \( \text{LCM}(x, y) = 48000 \)

### Step 1: Use the relationship between HCF and LCM
We know that for any two integers \(x\) and \(y\):

\[ \text{HCF}(x, y) \times \text{LCM}(x, y) = x \times y \]

Substituting the given values:

\[ 16 \times 48000 = x \times y \]

Calculating the left-hand side:

\[ 16 \times 48000 = 768000 \]

Thus, we have:

\[ x \times y = 768000 \]

### Step 2: Express \(x\) and \(y\) in terms of their HCF
Since \( \text{HCF}(x, y) = 16 \), we can express \(x\) and \(y\) as:

\[ x = 16a \quad \text{and} \quad y = 16b \]

where \(a\) and \(b\) are coprime integers (i.e., \( \text{HCF}(a, b) = 1\)).

### Step 3: Substitute \(x\) and \(y\) into the product equation
Substituting \(x\) and \(y\) into the product equation gives:

\[ (16a)(16b) = 768000 \]

This simplifies to:

\[ 256ab = 768000 \]

Dividing both sides by 256:

\[ ab = \frac{768000}{256} = 3000 \]

### Step 4: Factor \(3000\) to find coprime pairs \((a, b)\)
Next, we need to find the pairs \((a, b)\) such that \(ab = 3000\) and \( \text{HCF}(a, b) = 1\).

First, we factor \(3000\):

\[ 3000 = 3 \times 1000 = 3 \times 10^3 = 3 \times (2^3 \times 5^3) = 2^3 \times 3^1 \times 5^3 \]

### Step 5: Count the coprime pairs
To find the coprime pairs, we need to consider the divisors of \(3000\). The number of divisors can be calculated using the formula:

\[ \text{Number of divisors} = (e_1 + 1)(e_2 + 1)(e_3 + 1) \]

where \(e_1, e_2, e_3\) are the powers of the prime factors.

For \(3000 = 2^3 \times 3^1 \times 5^3\):

\[ (3 + 1)(1 + 1)(3 + 1) = 4 \times 2 \times 4 = 32 \]

### Step 6: Identify coprime pairs
Each divisor \(d\) of \(3000\) can be paired with \(\frac{3000}{d}\). However, we need to ensure that \(d\) and \(\frac{3000}{d}\) are coprime.

To find the number of coprime pairs, we can use the property that if \(d\) has a prime factor in common with \(3000\), it cannot be paired with \(\frac{3000}{d}\).

### Step 7: Calculate the pairs
The number of coprime pairs can be calculated as follows:

1. Count the total pairs of divisors: 32
2. Identify pairs that are not coprime.

After checking the pairs, we find that there are \(8\) pairs of \((a, b)\) such that \( \text{HCF}(a, b) = 1\).

### Final Step: Count the ordered pairs
Since each pair \((a, b)\) can be arranged as \((x, y)\) and \((y, x)\), the total number of ordered pairs \((x, y)\) is:

\[ 2 \times 8 = 16 \]

Thus, the number of elements in the set \(S\) is:

\[ \boxed{16} \]