Question

For natural numbers x,y, and z, if xy+yz=19 and yz+xz=51, then the minimum possible value of xyz is

Answer 1

1. Given Equations

We are given: \[ y(x + z) = 19 \tag{1} \] \[ z(x + y) = 51 \tag{2} \]

2. Deduction from Equation (1)

From equation (1): \[ y(x + z) = 19 \] Since 19 is a prime number, and \( y \) is a positive integer, the only integer values possible for \( y \) are 1 and 19. 
If \( y = 1 \), then: \[ x + z = 19 \] 
Try this value in equation (2): \[ z(x + 1) = 51 \tag{3} \] From equation (1), we now explore pairs \((x, z)\) such that \( x + z = 19 \)

3. Exploring Cases

Case 1: \( z = 3 \Rightarrow x = 16 \)

Plug into equation (3): \[ z(x + 1) = 3 \times (16 + 1) = 3 \times 17 = 51 \quad \text{✔️ Satisfies} \] So, \[ xyz = x \cdot y \cdot z = 16 \cdot 1 \cdot 3 = 48 \]

Case 2: \( z = 17 \Rightarrow x = 2 \)

Check equation (3): \[ z(x + 1) = 17 \cdot (2 + 1) = 17 \cdot 3 = 51 \quad \text{✔️ Satisfies} \] So, \[ xyz = 2 \cdot 1 \cdot 17 = 34 \]

4. Final Answer

The two valid solutions yield: \[ xyz = 48 \quad \text{and} \quad xyz = 34 \] 
The minimum value is: \[ \boxed{34} \]