The product of two numbers is 11830 and their HCF is 13. How many pairs of such number can be formed?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem step by step, we need to find how many pairs of numbers can be formed given that their product is 11830 and their HCF is 13.

### Step 1: Understand the relationship between the numbers and their HCF
Let the two numbers be \( x \) and \( y \). Since the HCF of \( x \) and \( y \) is 13, we can express the numbers as:
\[ x = 13a \quad \text{and} \quad y = 13b \]
where \( a \) and \( b \) are coprime (i.e., their HCF is 1).

### Step 2: Set up the equation using the product
Given that the product of the two numbers is:
\[ x \cdot y = 11830 \]
Substituting the expressions for \( x \) and \( y \):
\[ (13a)(13b) = 11830 \]
This simplifies to:
\[ 169ab = 11830 \]

### Step 3: Solve for \( ab \)
To find \( ab \), we divide both sides by 169:
\[ ab = \frac{11830}{169} \]
Calculating this gives:
\[ ab = 70 \]

### Step 4: Find pairs of \( (a, b) \) such that \( ab = 70 \)
Now, we need to find pairs of integers \( (a, b) \) such that their product is 70. We will list the factor pairs of 70:
1. \( 1 \times 70 \)
2. \( 2 \times 35 \)
3. \( 5 \times 14 \)
4. \( 7 \times 10 \)

### Step 5: Check the HCF of each pair
We need to check if \( a \) and \( b \) are coprime (i.e., their HCF is 1):
- For \( (1, 70) \): HCF is 1 (valid)
- For \( (2, 35) \): HCF is 1 (valid)
- For \( (5, 14) \): HCF is 1 (valid)
- For \( (7, 10) \): HCF is 1 (valid)

### Step 6: Count the valid pairs
All the pairs we found are valid. Thus, the pairs \( (a, b) \) that satisfy the conditions are:
1. \( (1, 70) \)
2. \( (2, 35) \)
3. \( (5, 14) \)
4. \( (7, 10) \)

### Conclusion
The total number of pairs \( (x, y) \) that can be formed is 4.

### Final Answer
The number of pairs of such numbers is **4**.

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