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.