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

Correct Answer: 34

Given:

\[ y(x + z) = 19 \quad \text{and} \quad z(x + y) = 51 \]

From the first equation: \[ y(x + z) = 19 \] Since \( y \neq 19 \), the simplest solution to consider is:

\[ y = 1 \Rightarrow x + z = 19 \]

Now we substitute \( y = 1 \) into the second equation:

\[ z(x + 1) = 51 \]

Since \( x + z = 19 \Rightarrow x = 19 - z \), we plug this into the second equation:

\[ z(19 - z + 1) = 51 \Rightarrow z(20 - z) = 51 \]

Let's solve the equation:

\[ z(20 - z) = 51 \Rightarrow 20z - z^2 = 51 \Rightarrow z^2 - 20z + 51 = 0 \]

Solve the quadratic equation:

\[ z = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 51}}{2} = \frac{20 \pm \sqrt{400 - 204}}{2} = \frac{20 \pm \sqrt{196}}{2} = \frac{20 \pm 14}{2} \]

So the two possible values of \( z \) are:

• \( z = \frac{20 + 14}{2} = 17 \)
• \( z = \frac{20 - 14}{2} = 3 \)

Now compute \( xyz \) in both cases:

Case 1: \( z = 3 \)

Then \( x = 19 - z = 16 \), and \( y = 1 \) \[ xyz = 1 \cdot 16 \cdot 3 = 48 \]

Case 2: \( z = 17 \)

Then \( x = 2 \), and \( y = 1 \) \[ xyz = 1 \cdot 2 \cdot 17 = 34 \]

Conclusion:

Minimum value of \( xyz \) is: \[ \boxed{34} \]