The product and HCF of two numbers are 2160 and 12 respectively. Accordingly, find the number of such pairs possbile?

Options

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

Answer

To solve the problem step by step, we need to find the number of pairs of two numbers whose product is 2160 and whose highest common factor (HCF) is 12.

### Step 1: Understand the relationship between HCF and LCM
The relationship between the HCF (Highest Common Factor) and LCM (Least Common Multiple) of two numbers \( a \) and \( b \) can be expressed as:
\[ \text{HCF} \times \text{LCM} = a \times b \]
Given that the HCF is 12 and the product (which is \( a \times b \)) is 2160, we can find the LCM.

### Step 2: Calculate the LCM
Using the relationship:
\[ 12 \times \text{LCM} = 2160 \]
To find the LCM, we rearrange the equation:
\[ \text{LCM} = \frac{2160}{12} \]
Calculating this gives:
\[ \text{LCM} = 180 \]

### Step 3: Express the numbers in terms of their HCF
Let the two numbers be \( a = 12x \) and \( b = 12y \), where \( x \) and \( y \) are co-prime numbers (since the HCF is 12).

### Step 4: Set up the equation for the product
From the product of the numbers:
\[ a \times b = 12x \times 12y = 144xy \]
We know from the problem that:
\[ 144xy = 2160 \]
Dividing both sides by 144 gives:
\[ xy = \frac{2160}{144} = 15 \]

### Step 5: Find pairs of co-prime factors of 15
Now we need to find pairs of co-prime integers \( (x, y) \) such that \( xy = 15 \). The pairs of factors of 15 are:
1. \( (1, 15) \)
2. \( (3, 5) \)

Both pairs are co-prime.

### Step 6: Calculate the corresponding numbers
For each pair \( (x, y) \):
1. For \( (1, 15) \):
- \( a = 12 \times 1 = 12 \)
- \( b = 12 \times 15 = 180 \)
- Pair: \( (12, 180) \)

2. For \( (3, 5) \):
- \( a = 12 \times 3 = 36 \)
- \( b = 12 \times 5 = 60 \)
- Pair: \( (36, 60) \)

### Step 7: Conclusion
Thus, the possible pairs of numbers are \( (12, 180) \) and \( (36, 60) \). Therefore, the number of such pairs possible is **2**.