smc

We prime factorize $72,600,$ and $900$. The prime factorizations are $2^3\times 3^2$, $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$, respectively. Let $x=2^a\times 3^b\times 5^c$, $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$. We know that $$\max (a,d)=3$$ $$\max (b,e)=2$$ $$\max (a,g)=3$$ $$\max (b,h)=1$$ $$\max (c,i)=2$$ $$\max (d,g)=2$$ $$\max (e,h)=2$$ and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max (d,g)=2$ we know that $a=3$. We also know that since $\max (b,h)=1$ that $e=2$. So now some equations have become useless to us...let's take them out. $$\max (b,h)=1$$ $$\max (d,g)=2$$ are the only two important ones left. We do casework on each now. If $\max (b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$. Similarly if $\max (d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$. Thus our answer is $5\times 3=\boxed{15}.$