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

Given:
$y(x+z)=19$
$z(x+y)=51$
From the first equation, we can deduce that $y$ is equal to 1, as $y≠19.$ .
So, $y=1$ and $x+z=19$
Now, let's explore the two possible cases for $z$.

**Case 1: **$z=3$
If $z=3$, then $x=16$
So, $xyz$=$3\times1\times16$
$xyz=48$

Case 2: $z=17$
If $z=17$, then $x=2$
So, $xyz=17\times1\times2$
$xyz=34$
The minimum value of $xyz$ among the two cases is 34.

Therefore, the answer is 34.