Find the three prime numbers such that product of the first two is 1147 and the product of the last two numbers is 1517? Options are (a) 21, 47, 61 (b) 31, 37, 61 (c) 21, 37, 41 (d) 31, 37, 41 .

Options

A. 21, 47, 61
B. 31, 37, 61
C. 21, 37, 41
D. 31, 37, 41

Answer

To solve the problem of finding three prime numbers such that the product of the first two is 1147 and the product of the last two is 1517, we can follow these steps:

### Step 1: Identify the Products
We know:
- The product of the first two prime numbers (let's call them \( p_1 \) and \( p_2 \)) is 1147.
- The product of the last two prime numbers (let's call them \( p_2 \) and \( p_3 \)) is 1517.

### Step 2: Factor 1147
To find \( p_1 \) and \( p_2 \), we need to factor 1147. We will check the prime numbers given in the options to see which two multiply to give 1147.

1. **Check 31 and 37**:
- \( 31 \times 37 = 1147 \) (This is correct)

Thus, \( p_1 = 31 \) and \( p_2 = 37 \).

### Step 3: Factor 1517
Next, we need to find \( p_3 \) using the product 1517. We know \( p_2 = 37 \), so we can find \( p_3 \) by dividing 1517 by 37.

2. **Calculate \( p_3 \)**:
- \( p_3 = \frac{1517}{37} \)
- Performing the division:
- \( 1517 \div 37 = 41 \)

Thus, \( p_3 = 41 \).

### Step 4: Verify the Prime Numbers
Now we have:
- \( p_1 = 31 \)
- \( p_2 = 37 \)
- \( p_3 = 41 \)

We should check if these numbers are prime:
- 31 is prime.
- 37 is prime.
- 41 is prime.

### Conclusion
The three prime numbers that satisfy the conditions are 31, 37, and 41. Therefore, the answer is option (d) 31, 37, 41.