The LCM and HCF of two numbers are `1530` and `51` . Find how many such pairs are possible .

Options

A. 2
B. 3
C. 4
D. 1

Answer

To find how many pairs of numbers have an LCM of 1530 and an HCF of 51, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between LCM, HCF, and the numbers**:
The relationship between the LCM (Least Common Multiple), HCF (Highest Common Factor), and the two numbers \( a \) and \( b \) is given by:
\[ \text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b \]

2. **Substitute the known values**:
Given that LCM = 1530 and HCF = 51, we can substitute these values into the equation:
\[ 1530 \times 51 = a \times b \]

3. **Calculate the product of the numbers**:
Calculate \( 1530 \times 51 \):
\[ 1530 \times 51 = 78030 \]
So, we have:
\[ a \times b = 78030 \]

4. **Express the numbers in terms of HCF**:
Since the HCF of the two numbers is 51, we can express the numbers as:
\[ a = 51x \quad \text{and} \quad b = 51y \]
where \( x \) and \( y \) are co-prime numbers.

5. **Substitute \( a \) and \( b \) into the product equation**:
Substitute into the equation \( a \times b = 78030 \):
\[ (51x) \times (51y) = 78030 \]
This simplifies to:
\[ 2601xy = 78030 \]

6. **Solve for \( xy \)**:
Divide both sides by 2601:
\[ xy = \frac{78030}{2601} = 30 \]

7. **Find pairs of co-prime numbers whose product is 30**:
Now we need to find pairs \( (x, y) \) such that \( x \times y = 30 \) and \( x \) and \( y \) are co-prime. The pairs of factors of 30 are:
- (1, 30)
- (2, 15)
- (3, 10)
- (5, 6)

Now we check which of these pairs are co-prime:
- (1, 30) → Co-prime
- (2, 15) → Co-prime
- (3, 10) → Co-prime
- (5, 6) → Co-prime

8. **Count the valid pairs**:
All four pairs are co-prime. Therefore, the total number of pairs \( (x, y) \) is 4.

### Final Answer:
Thus, the number of such pairs \( (a, b) \) is **4**.

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