If HCF and LCM of two number are 17 and 408 How many numbers are possible for such type ?

Answer

To solve the problem, we need to find how many pairs of numbers can be formed given that their HCF is 17 and their LCM is 408.

### Step-by-step Solution:

1. **Understanding HCF and LCM**:
We know that for any two numbers \( a \) and \( b \):
\[ a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b) \]
Given HCF = 17 and LCM = 408, we can substitute these values into the equation.

2. **Setting up the equation**:
\[ a \times b = 17 \times 408 \]
Calculate \( 17 \times 408 \):
\[ 17 \times 408 = 6936 \]
So, we have:
\[ a \times b = 6936 \]

3. **Expressing the numbers**:
Since the HCF is 17, we can express the two numbers as:
\[ a = 17A \quad \text{and} \quad b = 17B \]
where \( A \) and \( B \) are coprime (i.e., they have no common factors other than 1).

4. **Substituting into the equation**:
Substitute \( a \) and \( b \) into the product:
\[ (17A) \times (17B) = 6936 \]
Simplifying this gives:
\[ 289AB = 6936 \]
Dividing both sides by 289:
\[ AB = \frac{6936}{289} = 24 \]

5. **Finding pairs of \( A \) and \( B \)**:
Now we need to find pairs of integers \( (A, B) \) such that:
\[ A \times B = 24 \]
The pairs of factors of 24 are:
- (1, 24)
- (2, 12)
- (3, 8)
- (4, 6)

6. **Checking for coprimeness**:
We need to check which of these pairs are coprime:
- (1, 24): Coprime
- (2, 12): Not coprime (common factor 2)
- (3, 8): Coprime
- (4, 6): Not coprime (common factor 2)

The valid coprime pairs are:
- (1, 24)
- (3, 8)

7. **Conclusion**:
Therefore, there are 2 valid pairs \( (A, B) \), which correspond to 2 pairs of numbers \( (a, b) \).

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