CAT 2022 Slot 1 QA Question 5: For natural numbers $x, y, z$, if $xy + yz = 19$ and $yz + xz = 51$, then the minimum possible value of $xyz$ is

Question

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

Accepted Answer

$xy + yz = y(x+z) = 19$
Since $19$ is prime and $x$ and $z$ are natural numbers, $y$ cannot be $19$ because that would give $x + z = 1$. Hence,
$y = 1$
$x + z = 19$
$yz + xz = z(y+x) = 51$
$\Rightarrow z(x+1) = 51$
The useful positive divisors of $51$ are $3$ and $17$.
If $z = 3$, then $x = 16$ and

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

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