Can two numbers have 12 as their HCF and 512 as their LCM ? Justify your answer .

Answer

To determine if two numbers can have 12 as their HCF (Highest Common Factor) and 512 as their LCM (Lowest Common Multiple), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the relationship between HCF and LCM**:
We know from number theory that for any two numbers \( a \) and \( b \):
\[ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \]
Here, we are given:
- HCF = 12
- LCM = 512

2. **Setting up the equation**:
Let the two numbers be \( a \) and \( b \). According to the relationship:
\[ 12 \times 512 = a \times b \]

3. **Calculating the product**:
Now, we calculate:
\[ 12 \times 512 = 6144 \]
So, we have:
\[ a \times b = 6144 \]

4. **Expressing the numbers in terms of HCF**:
Since the HCF is 12, we can express the two numbers as:
\[ a = 12x \quad \text{and} \quad b = 12y \]
where \( x \) and \( y \) are co-prime (they have no common factors other than 1).

5. **Substituting into the product equation**:
Substituting \( a \) and \( b \) into the product equation gives:
\[ (12x) \times (12y) = 6144 \]
Simplifying this:
\[ 144xy = 6144 \]

6. **Solving for \( xy \)**:
Dividing both sides by 144:
\[ xy = \frac{6144}{144} \]
Simplifying this gives:
\[ xy = 42.6667 \]

7. **Analyzing the result**:
The value of \( xy \) is not an integer (it is approximately 42.67). Since \( x \) and \( y \) must be integers (as they represent the factors of the numbers), this indicates that \( xy \) cannot be expressed as a product of two integers.

8. **Conclusion**:
Therefore, it is impossible for two numbers to have 12 as their HCF and 512 as their LCM.

### Final Answer:
No, two numbers cannot have 12 as their HCF and 512 as their LCM.

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