Find the sum of three numbers which are prime to one another such that the product of the first two is 437 and that of the last two is 551-

Options

A. 91
B. 81
C. 71
D. 70

Answer

To solve the problem, we need to find three numbers \( x, y, z \) that are coprime (prime to one another) such that:

1. The product of the first two numbers \( x \) and \( y \) is 437.
2. The product of the last two numbers \( y \) and \( z \) is 551.

Let's break down the steps to find these numbers and their sum.

### Step 1: Factor the products

First, we need to factor the given products to find possible values for \( x, y, z \).

**For \( xy = 437 \)**:
- The prime factorization of 437 can be found by checking divisibility by prime numbers:
- 437 is not divisible by 2 (it's odd).
- Check 3: \( 4 + 3 + 7 = 14 \) (not divisible by 3).
- Check 5: does not end in 0 or 5.
- Check 7: \( 437 \div 7 \approx 62.43 \) (not divisible).
- Check 11: \( 437 \div 11 \approx 39.73 \) (not divisible).
- Check 19: \( 437 \div 19 = 23 \) (exactly).

Thus, the factors of 437 are \( 19 \) and \( 23 \). Therefore, we can set:
- \( x = 19 \)
- \( y = 23 \)

### Step 2: Find \( z \)

Next, we need to find \( z \) using the second product \( yz = 551 \).

**For \( yz = 551 \)**:
- We already have \( y = 23 \), so we can find \( z \):
\[ z = \frac{551}{y} = \frac{551}{23} = 24 \]

### Step 3: Verify coprimeness

Now we have \( x = 19 \), \( y = 23 \), and \( z = 24 \). We need to check if these numbers are coprime:
- The GCD of \( 19 \) and \( 23 \) is \( 1 \) (both are prime).
- The GCD of \( 19 \) and \( 24 \) is \( 1 \) (19 is prime and does not divide 24).
- The GCD of \( 23 \) and \( 24 \) is \( 1 \) (23 is prime and does not divide 24).

Since all pairs are coprime, we can proceed.

### Step 4: Calculate the sum

Finally, we calculate the sum of the three numbers:
\[ \text{Sum} = x + y + z = 19 + 23 + 24 = 66 \]

### Final Answer

The sum of the three numbers is \( 66 \).

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