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

Question

Accepted Answer

To find the minimum possible value of $$xyz$$, we solve the given equations step by step:

1. **Equations:**  
   $$
   xy + yz = 19 \quad 	ext{(1)}
   $$
   $$
   yz + xz = 51 \quad 	ext{(2)}
   $$

2. **Rewriting Equation (1):**  
   Factorize $$y(x + z) = 19$$. Since $$19$$ is prime, $$y = 1$$ and $$x + z = 19$$ (as $$x, y, z$$ are natural numbers).

3. **Substitute $$y = 1$$ into Equation (2):**  
   $$
   z(x + 1) = 51
   $$
   Factorize $$51 = 17 	imes 3$$. Two cases arise:  
   - Case 1: $$z = 17$$, $$x + 1 = 3$$.  
     Solve for $$x = 2, y = 1, z = 17$$.  
     $$
     xyz = 2 	imes 1 	imes 17 = 34
     $$

- Case 2: $$z = 3$$, $$x + 1 = 17$$.  
     Solve for $$x = 16, y = 1, z = 3$$.  
     $$
     xyz = 16 	imes 1 	imes 3 = 48
     $$

4. **Minimum Value:**  
   The minimum possible value of $$xyz$$ is **34** (from Case 1).

Thus, the correct answer is **34**.