If $x, y$ and $z$ are positive integers with $x y=18, x z=3$ and $y z=6$, what is the value of $x+y+z$ ?
(A) 6
(B) 10
(C) 25
(D) 11
(E) 8
Since $x$ and $z$ are positive integers and $x z=3$, the only possibilities are $x=1$ and $z=3$ or $x=3$ and $z=1$.

Assuming that $x=1$ and $z=3, y z=6$ implies $3 y=6$ or $y=2$.

Thus, $x=1$ and $y=2$ and $x y=2$.

This contradicts the first equation $x y=18$.

Therefore, our assumption was incorrect and it must be true that $x=3$ and $z=1$.

Then $y z=6$ and $z=1$ implies $y=6$.

Checking, $x=3$ and $y=6$ also satisfies $x y=18$, the first equation.

Therefore, the required sum is $x+y+z=3+6+1=10$.

ANSWER: (B)