Question

The HCF of two numbers is 5 and their LCM is 60. How many pairs of such numbers are possible?

• A. 0
• B. 1
• C. 2
• D. 3
• E. 4

Hint:

Always remember the fundamental relationship: Product of two numbers = Product of their HCF and LCM. When given the HCF, immediately represent the numbers as multiples of the HCF with co-prime factors. This simplifies the problem significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves the relationship between the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two numbers.
Step 2: Key Formula or Approach:
There are two key properties we will use: 1. For any two positive integers \(a\) and \(b\), \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \).
2. If the HCF of two numbers is \(h\), the numbers can be expressed as \(hx\) and \(hy\), where \(x\) and \(y\) are co-prime integers (i.e., their HCF is 1).
Step 3: Detailed Explanation:
We are given HCF = 5 and LCM = 60.
Using the second property, let the two numbers be \(5x\) and \(5y\), where \(x\) and \(y\) are co-prime.
Now, using the first property: \[ (5x) \times (5y) = 5 \times 60 \] \[ 25xy = 300 \] \[ xy = \frac{300}{25} \] \[ xy = 12 \] We need to find pairs of co-prime integers (\(x, y\)) whose product is 12. The possible pairs of integers whose product is 12 are: (1, 12), (2, 6), and (3, 4).
Now we check which of these pairs are co-prime:

Pair (1, 12): HCF(1, 12) = 1. They are co-prime. This is a valid pair.
Pair (2, 6): HCF(2, 6) = 2. They are not co-prime. This is not a valid pair.
Pair (3, 4): HCF(3, 4) = 1. They are co-prime. This is a valid pair.
We have found two valid pairs for \((x, y)\). Let's find the corresponding pairs of numbers:

For \((x, y) = (1, 12)\), the numbers are \( (5 \times 1, 5 \times 12) = (5, 60) \).
For \((x, y) = (3, 4)\), the numbers are \( (5 \times 3, 5 \times 4) = (15, 20) \).
There are 2 such pairs of numbers.
Step 4: Final Answer:
The two possible pairs are (5, 60) and (15, 20). Therefore, there are 2 possible pairs.