The product of two numbers is 6760 and their HGF is 13. How many such ...
Understanding the Problem
To determine how many pairs of numbers can be formed given that their product is 6760 and their HCF (Highest Common Factor) is 13, we can follow these steps:
Step 1: Determine the Numbers
- The two numbers can be expressed as 13a and 13b, where a and b are coprime (i.e., their HCF is 1).
- This means that 13a * 13b = 6760.
Step 2: Simplifying the Equation
- Simplifying the equation gives us: 169ab = 6760.
- Dividing both sides by 169 leads to: ab = 40.
Step 3: Finding Pairs of (a, b)
- Now, we need to find all pairs of positive integers (a, b) such that their product is 40 and they are coprime.
- The factor pairs of 40 are: (1, 40), (2, 20), (4, 10), (5, 8).
Step 4: Identifying Coprime Pairs
- Check which pairs are coprime:
- (1, 40): Coprime
- (2, 20): Not coprime
- (4, 10): Not coprime
- (5, 8): Coprime
Step 5: Count Valid Pairs
- The valid coprime pairs are: (1, 40) and (5, 8).
- Each pair (a, b) can give two combinations: (a, b) and (b, a).
Conclusion
- Therefore, the valid pairs of (a, b) are (1, 40) and (5, 8), leading to the combinations:
- (13*1, 13*40) = (13, 520)
- (13*40, 13*1) = (520, 13)
- (13*5, 13*8) = (65, 104)
- (13*8, 13*5) = (104, 65)
Total Pairs
- In total, there are 4 pairs of numbers that can be formed. The answer is 4.