Can two numbers have 18 as their HCF and 146 as their LCM? Give reasons.

Answer

To determine if two numbers can have 18 as their HCF (Highest Common Factor) and 146 as their LCM (Lowest Common Multiple), we can follow these steps:

### Step 1: Understand the relationship between HCF and LCM
The relationship between the HCF and LCM of two numbers can be expressed with the formula:
\[ \text{HCF} \times \text{LCM} = \text{Product of the two numbers} \]

### Step 2: Apply the values given in the question
Given:
- HCF = 18
- LCM = 146

Using the formula:
\[ 18 \times 146 = \text{Product of the two numbers} \]

### Step 3: Calculate the product
Now, we calculate the product:
\[ 18 \times 146 = 2628 \]

### Step 4: Analyze the factors of the product
Next, we need to find two numbers whose product is 2628 and whose HCF is 18. To do this, we can express the two numbers in terms of their HCF:
Let the two numbers be:
- \( a = 18m \)
- \( b = 18n \)

Where \( m \) and \( n \) are coprime (i.e., their HCF is 1).

### Step 5: Substitute into the product equation
Substituting \( a \) and \( b \) into the product equation:
\[ (18m) \times (18n) = 2628 \]
\[ 324mn = 2628 \]

### Step 6: Solve for \( mn \)
Now, divide both sides by 324:
\[ mn = \frac{2628}{324} \]
Calculating this gives:
\[ mn = 8.125 \]

### Step 7: Check if \( mn \) is an integer
Since \( mn \) must be an integer (as \( m \) and \( n \) are whole numbers), and 8.125 is not an integer, it indicates that there are no such integers \( m \) and \( n \) that satisfy the condition.

### Conclusion
Therefore, it is not possible for two numbers to have 18 as their HCF and 146 as their LCM.

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