Two numbers , both greater than 29, have HCF 29 and LCM 4147 . The sum of the numbers is

Options

A. 966
B. 696
C. 669
D. 666

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript:

### Step 1: Understand the relationship between HCF and LCM
Given that the two numbers have an HCF of 29 and an LCM of 4147, we can use the relationship between HCF, LCM, and the two numbers. The product of the two numbers can be expressed as:
\[ \text{Product of the two numbers} = \text{HCF} \times \text{LCM} \]

### Step 2: Set up the equation
Let the two numbers be \(29x\) and \(29y\) where \(x\) and \(y\) are integers. Then, we can write:
\[ 29x \times 29y = 29 \times 4147 \]

### Step 3: Simplify the equation
Now, simplifying the left side:
\[ 29^2 \cdot xy = 29 \cdot 4147 \]
Dividing both sides by 29:
\[ 29xy = 4147 \]
Now, divide both sides by 29:
\[ xy = \frac{4147}{29} \]

### Step 4: Calculate \(xy\)
Calculating \(4147 \div 29\):
\[ xy = 143 \]

### Step 5: Find pairs of factors of 143
Next, we need to find pairs of integers \(x\) and \(y\) such that their product is 143. The factor pairs of 143 are:
1. \(1 \times 143\)
2. \(11 \times 13\)

### Step 6: Determine valid pairs
Since both numbers must be greater than 29, we check each pair:
- For \(x = 1\) and \(y = 143\):
- The numbers would be \(29 \times 1 = 29\) and \(29 \times 143 = 4147\). The first number is not greater than 29, so this pair is invalid.

- For \(x = 11\) and \(y = 13\):
- The numbers would be \(29 \times 11 = 319\) and \(29 \times 13 = 377\). Both numbers are greater than 29, so this pair is valid.

### Step 7: Calculate the sum of the valid numbers
Now, we find the sum of the valid numbers:
\[ 319 + 377 = 696 \]

### Final Answer
Thus, the sum of the two numbers is:
\[ \boxed{696} \]

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