The product of two numbers is 6300 and their HCF is 15. How many pairs of such numbers are there?

Answer

To solve the problem, we need to find how many pairs of numbers \( x \) and \( y \) exist such that their product is 6300 and their highest common factor (HCF) is 15.

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that the product of two numbers \( x \) and \( y \) is given as:
\[ xy = 6300 \]
and their HCF is:
\[ \text{HCF}(x, y) = 15 \]

2. **Expressing the Numbers**:
Since the HCF of \( x \) and \( y \) is 15, we can express \( x \) and \( y \) in terms of two new variables \( a \) and \( b \):
\[ x = 15a \quad \text{and} \quad y = 15b \]
where \( a \) and \( b \) are co-prime (i.e., their HCF is 1).

3. **Substituting into the Product**:
Substituting \( x \) and \( y \) into the product equation:
\[ (15a)(15b) = 6300 \]
This simplifies to:
\[ 225ab = 6300 \]

4. **Solving for \( ab \)**:
To find \( ab \), we divide both sides by 225:
\[ ab = \frac{6300}{225} \]
Simplifying the right-hand side:
\[ ab = 28 \]

5. **Finding Pairs of \( (a, b) \)**:
We need to find pairs of integers \( (a, b) \) such that \( ab = 28 \). The pairs of factors of 28 are:
- \( (1, 28) \)
- \( (2, 14) \)
- \( (4, 7) \)

Since \( a \) and \( b \) must be co-prime, we check which pairs are co-prime:
- \( (1, 28) \) are co-prime.
- \( (2, 14) \) are not co-prime (common factor is 2).
- \( (4, 7) \) are co-prime.

Thus, the valid pairs of \( (a, b) \) are:
- \( (1, 28) \)
- \( (4, 7) \)

6. **Calculating Corresponding \( (x, y) \) Pairs**:
Now we can find the corresponding \( (x, y) \) pairs:
- For \( (1, 28) \):
\[ x = 15 \times 1 = 15, \quad y = 15 \times 28 = 420 \quad \Rightarrow \quad (15, 420) \]
- For \( (4, 7) \):
\[ x = 15 \times 4 = 60, \quad y = 15 \times 7 = 105 \quad \Rightarrow \quad (60, 105) \]

7. **Conclusion**:
The pairs of numbers \( (x, y) \) that satisfy the conditions are:
- \( (15, 420) \)
- \( (60, 105) \)

Therefore, there are **2 pairs** of such numbers.