The product of two numbers is 2160 and their HCF is 12. Number of such possible pairs are:

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the relationship between product, HCF, and LCM**:
The relationship between the product of two numbers (n1 and n2), their HCF (H), and their LCM (L) is given by:
\[ n1 \times n2 = H \times L \]
Here, \( n1 \times n2 = 2160 \) and \( H = 12 \).

2. **Calculate the LCM**:
We can rearrange the formula to find the LCM:
\[ L = \frac{n1 \times n2}{H} \]
Substituting the known values:
\[ L = \frac{2160}{12} = 180 \]

3. **Express the numbers in terms of HCF**:
Let’s denote the two numbers as:
\[ n1 = H \times A \quad \text{and} \quad n2 = H \times B \]
where A and B are coprime numbers (i.e., their HCF is 1).

4. **Substituting into the product equation**:
Substituting \( n1 \) and \( n2 \) into the product equation:
\[ (H \times A) \times (H \times B) = H^2 \times A \times B \]
This gives us:
\[ 12^2 \times A \times B = 2160 \]
Simplifying this:
\[ 144 \times A \times B = 2160 \]
Dividing both sides by 144:
\[ A \times B = \frac{2160}{144} = 15 \]

5. **Finding coprime pairs (A, B)**:
Now we need to find pairs of coprime numbers (A, B) such that their product is 15. The factor pairs of 15 are:
- (1, 15)
- (3, 5)

Both pairs (1, 15) and (3, 5) are coprime.

6. **Count the pairs**:
We have found two pairs of coprime numbers:
- (1, 15)
- (3, 5)

Therefore, the total number of such possible pairs is **2**.

### Final Answer:
The number of such possible pairs is **2**.

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