There are two numbers which are greater than 21 and their LCM and HCF are 3003 and 21 respectively. What is the sum of these numbers?

Options

A. 504
B. 508
C. 514
D. 528

Answer

To solve the problem, we need to find two numbers greater than 21, given their LCM (Least Common Multiple) is 3003 and their HCF (Highest Common Factor) is 21. We will denote these two numbers as \( 21A \) and \( 21B \), where \( A \) and \( B \) are co-prime (since their HCF is 21).

### Step-by-Step Solution:

1. **Understanding the relationship between LCM, HCF, and the two numbers:**
We know that:
\[ \text{LCM} \times \text{HCF} = \text{Number 1} \times \text{Number 2} \]
Substituting the values we have:
\[ 3003 \times 21 = (21A) \times (21B) \]

2. **Calculating the product of the two numbers:**
Simplifying the equation:
\[ 3003 \times 21 = 21^2 \times A \times B \]
This can be rewritten as:
\[ 3003 \times 21 = 441 \times A \times B \]
Dividing both sides by 441:
\[ A \times B = \frac{3003 \times 21}{441} \]

3. **Performing the division:**
First, calculate \( 3003 \div 441 \):
\[ 3003 \div 21 = 143 \]
So,
\[ A \times B = 143 \]

4. **Finding co-prime pairs of \( A \) and \( B \):**
The pairs of co-prime integers that multiply to 143 are:
- \( (1, 143) \)
- \( (11, 13) \)

5. **Calculating the corresponding numbers:**
Now, we will calculate the actual numbers:
- For \( A = 1 \) and \( B = 143 \):
\[ 21A = 21 \times 1 = 21 \quad (\text{not greater than 21}) \]
- For \( A = 11 \) and \( B = 13 \):
\[ 21A = 21 \times 11 = 231 \]
\[ 21B = 21 \times 13 = 273 \]

6. **Calculating the sum of the two numbers:**
Now, we find the sum:
\[ 231 + 273 = 504 \]

### Final Answer:
The sum of the two numbers is **504**.