If HCF and LCM of two numbers are 29 and 4147. Find the sum of both numbers if both numbers are greater than 29.

Answer

To solve the problem, we need to find two numbers based on their HCF (Highest Common Factor) and LCM (Least Common Multiple). Given that the HCF is 29 and the LCM is 4147, we can use the relationship between HCF, LCM, and the two numbers.

### Step-by-Step Solution:

1. **Use the relationship between HCF, LCM, and the numbers**:
The relationship states that:
\[ \text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2} \]
Here, we have:
\[ 29 \times 4147 = \text{Number 1} \times \text{Number 2} \]

2. **Calculate the product of HCF and LCM**:
\[ 29 \times 4147 = 120263 \]
Therefore, we have:
\[ \text{Number 1} \times \text{Number 2} = 120263 \]

3. **Express the numbers in terms of HCF**:
Let:
\[ \text{Number 1} = 29x \quad \text{and} \quad \text{Number 2} = 29y \]
where \(x\) and \(y\) are co-prime integers (since HCF is 29).

4. **Substituting into the product equation**:
\[ 29x \times 29y = 120263 \]
Simplifying gives:
\[ 841xy = 120263 \]

5. **Solve for \(xy\)**:
\[ xy = \frac{120263}{841} = 143 \]

6. **Find pairs of integers \(x\) and \(y\)**:
The pairs of integers that multiply to 143 are:
- (1, 143)
- (11, 13)

7. **Choose the valid pair**:
Since both numbers must be greater than 29, we discard (1, 143) because:
\[ 29 \times 1 = 29 \quad \text{(not greater than 29)} \]
We take the pair (11, 13).

8. **Calculate the two numbers**:
\[ \text{Number 1} = 29 \times 11 = 319 \]
\[ \text{Number 2} = 29 \times 13 = 377 \]

9. **Calculate the sum of the two numbers**:
\[ \text{Sum} = 319 + 377 = 696 \]

### Final Answer:
The sum of both numbers is **696**.