AMC10 2016 A

Answer (A): Because $\mathrm{lcm}(x,y)=2^3\cdot 3^2$ and $\mathrm{lcm}(x,z)=2^3\cdot 3\cdot 5^2$, it follows that $5^2$ divides $z$, but neither $x$ nor $y$ is divisible by 5. Furthermore, $y$ is divisible by $3^2$, and neither $x$ nor $z$ is divisible by $3^2$, but at least one of $x$ or $z$ is divisible by 3. Finally, because $\mathrm{lcm}(y,z)=2^2\cdot 3^2\cdot 5^2$, at least one of $y$ or $z$ is divisible by $2^2$, but neither is divisible by $2^3$. However, $x$ must be divisible by $2^3$. Thus $x=2^3\cdot 3^j$, $y=2^k\cdot 3^2$, and $z=2^m\cdot 3^n\cdot 5^2$, where $\max(j,n)=1$ and $\max(k,m)=2$. There are 3 choices for $(j,n)$ and 5 choices for $(k,m)$, so there are 15 possible ordered triples $(x,y,z)$.