Problem: How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$

Solution: As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri;  //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle;  //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W); label("$2^33^15^2$",X--Z,2E); label("$2^23^25^2$",Y--Z,2S); [/asy]
Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri;  //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle;  //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^3$",X--Y,2W); label("$2^3$",X--Z,2E); label("$2^2$",Y--Z,2S); [/asy]
Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y$ and $z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri;  //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle;  //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$3^2$",X--Y,2W); label("$3^1$",X--Z,2E); label("$3^2$",Y--Z,2S); [/asy]
Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$
The final and last case is the powers of $5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri;  //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle;  //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$5^0$",X--Y,2W); label("$5^2$",X--Z,2E); label("$5^2$",Y--Z,2S); [/asy]
This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: \[5\cdot3\cdot1=\boxed{15}\]

Answer: A 15