Question

The sum of two numbers is 135 and their HCF is 9. How many such pairs of numbers can be formed?

• A. 6
• B. 2
• C. 5
• D. 4

Answer 1

The Correct Option is D

To solve the problem, we need to find pairs of numbers whose sum is 135 and have a Highest Common Factor (HCF) of 9.

1. The sum of two numbers is 135. Let's denote these numbers as \(9a\) and \(9b\) where \(a\) and \(b\) are co-prime (i.e., their HCF is 1) since the HCF of the original numbers is 9.
2. So, we set up the equation based on their sum: \(9a + 9b = 135\)
3. Factor out the common factor 9: \(9(a + b) = 135\)
4. Divide both sides by 9 to simplify: \(a + b = 15\) 
5. Now, we look for pairs of positive integers \((a, b)\) that are coprime and satisfy \(a + b = 15\).

Let's find such pairs:

1. If \(a = 1\), then \(b = 14\). Here, 1 and 14 are coprime.
2. If \(a = 2\), then \(b = 13\). Here, 2 and 13 are coprime.
3. If \(a = 4\), then \(b = 11\). Here, 4 and 11 are coprime.
4. If \(a = 5\), then \(b = 10\). Here, 5 and 10 are not coprime as they share a factor of 5.
5. If \(a = 7\), then \(b = 8\). Here, 7 and 8 are coprime.

Thus, the valid coprime pairs are: \((1, 14)\), \((2, 13)\), \((4, 11)\), and \((7, 8)\).

Corresponding to each pair \((a, b)\), the numbers are \(9a\) and \(9b\):

• For \((1, 14)\): \(9 \times 1\) and \(9 \times 14\)
• For \((2, 13)\): \(9 \times 2\) and \(9 \times 13\)
• For \((4, 11)\): \(9 \times 4\) and \(9 \times 11\)
• For \((7, 8)\): \(9 \times 7\) and \(9 \times 8\)

Thus, the number of such pairs is 4.