Two numbers have `16` as their HCF and `146` as their LCM. How many such pairs of numbers are there ?

Options

A. Zero
B. Only I
C. Only 2
D. Many

Answer

To solve the problem of finding how many pairs of numbers have an HCF of 16 and an LCM of 146, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the relationship between HCF and LCM**:
The relationship between two numbers \(a\) and \(b\) in terms of their HCF (Highest Common Factor) and LCM (Lowest Common Multiple) is given by:
\[ \text{HCF} \times \text{LCM} = a \times b \]
Here, we know that HCF = 16 and LCM = 146.

2. **Setting up the equation**:
Using the formula, we can set up the equation:
\[ 16 \times 146 = a \times b \]

3. **Calculating the product**:
Now, we calculate the product:
\[ 16 \times 146 = 2336 \]
So, we have:
\[ a \times b = 2336 \]

4. **Expressing the numbers in terms of HCF**:
Since the HCF is 16, we can express the two numbers as:
\[ a = 16x \quad \text{and} \quad b = 16y \]
where \(x\) and \(y\) are coprime integers.

5. **Substituting into the product equation**:
Substitute \(a\) and \(b\) into the product equation:
\[ (16x) \times (16y) = 2336 \]
Simplifying gives:
\[ 256xy = 2336 \]

6. **Solving for \(xy\)**:
Divide both sides by 256:
\[ xy = \frac{2336}{256} = 9.125 \]

7. **Analyzing the result**:
Since \(x\) and \(y\) must be coprime integers, and their product \(xy\) cannot be a decimal (as integers cannot multiply to give a decimal), we conclude that there are no pairs of integers \(x\) and \(y\) that satisfy this condition.

### Conclusion:
Thus, there are **no pairs of numbers** that have an HCF of 16 and an LCM of 146.

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