Find two pairs of numbers such that each pair has 16 as the HCF and 3840 as their product.

Answer

To solve the problem of finding two pairs of numbers such that each pair has 16 as the HCF and 3840 as their product, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between HCF, LCM, and product of two numbers**:
The relationship is given by the formula:
\[ \text{Product of two numbers} = \text{HCF} \times \text{LCM} \]

2. **Assign the known values**:
We know:
- HCF = 16
- Product = 3840

So, we can write:
\[ A \times B = 16 \times \text{LCM} \]
Therefore:
\[ 3840 = 16 \times \text{LCM} \]

3. **Calculate the LCM**:
To find the LCM, we rearrange the equation:
\[ \text{LCM} = \frac{3840}{16} \]
Performing the division:
\[ \text{LCM} = 240 \]

4. **Express the numbers in terms of their HCF**:
Since both numbers have 16 as their HCF, we can express them as:
\[ A = 16R \quad \text{and} \quad B = 16S \]
where R and S are co-prime integers (i.e., their HCF is 1).

5. **Set up the equation for the product**:
Substituting A and B into the product equation:
\[ A \times B = (16R) \times (16S) = 3840 \]
This simplifies to:
\[ 256RS = 3840 \]

6. **Solve for RS**:
Dividing both sides by 256:
\[ RS = \frac{3840}{256} \]
Performing the division:
\[ RS = 15 \]

7. **Find pairs of co-prime integers (R, S)**:
Now we need to find pairs of integers R and S such that their product is 15 and they are co-prime. The pairs of factors of 15 are:
- (1, 15)
- (3, 5)

8. **Calculate the corresponding pairs of A and B**:
For each pair (R, S):
- For (1, 15):
\[ A = 16 \times 1 = 16, \quad B = 16 \times 15 = 240 \]
- For (3, 5):
\[ A = 16 \times 3 = 48, \quad B = 16 \times 5 = 80 \]

### Final Answer:
The two pairs of numbers are:
1. (16, 240)
2. (48, 80)