The product to two 20 digit numbers is 2160 and their H.C.F. is 12 . The numbers are

Options

A. (12, 60)
B. (72, 30)
C. (36, 60)
D. (60, 72)

Answer

To solve the problem, we need to find two 20-digit numbers whose product is 2160 and whose H.C.F. (Highest Common Factor) is 12. We can use the relationship between H.C.F., L.C.M. (Least Common Multiple), and the product of two numbers.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Product of the two numbers (X and Y) = 2160
- H.C.F. of the two numbers = 12

2. **Use the Relationship between H.C.F., L.C.M., and the Product:**
- The relationship is given by:
\[ \text{H.C.F.} \times \text{L.C.M.} = X \times Y \]
- Substituting the known values:
\[ 12 \times \text{L.C.M.} = 2160 \]

3. **Calculate the L.C.M.:**
- Rearranging the equation to find L.C.M.:
\[ \text{L.C.M.} = \frac{2160}{12} \]
- Performing the division:
\[ \text{L.C.M.} = 180 \]

4. **Express the Numbers in Terms of H.C.F.:**
- Let the two numbers be:
\[ X = 12a \quad \text{and} \quad Y = 12b \]
- Here, \(a\) and \(b\) are co-prime (since the H.C.F. is 12).

5. **Substitute into the Product Equation:**
- The product of the numbers can be expressed as:
\[ X \times Y = (12a) \times (12b) = 144ab \]
- Setting this equal to the known product:
\[ 144ab = 2160 \]

6. **Solve for \(ab\):**
- Dividing both sides by 144:
\[ ab = \frac{2160}{144} \]
- Performing the division:
\[ ab = 15 \]

7. **Find Pairs of Co-prime Factors of 15:**
- The pairs of co-prime factors of 15 are:
- (1, 15)
- (3, 5)

8. **Calculate the Two Numbers:**
- For the pair (1, 15):
\[ X = 12 \times 1 = 12 \quad \text{and} \quad Y = 12 \times 15 = 180 \]
- For the pair (3, 5):
\[ X = 12 \times 3 = 36 \quad \text{and} \quad Y = 12 \times 5 = 60 \]

9. **Final Numbers:**
- The two pairs of numbers that satisfy the conditions are:
- (12, 180)
- (36, 60)

### Conclusion:
The two numbers are **12 and 180** or **36 and 60**.

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