# Difference between revisions of "2016 AMC 10A Problems/Problem 25"

## Problem

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600$ and $\text{lcm}(y,z)=900$?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$

## Solution 1

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 $5\times 3=\boxed{\textbf{(A)15}}$.

## Solution 2

It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$, then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$. Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$.

Start from $x$:$\text{lcm}(x,y)=72$ so $8\mid x$ or $9\mid x$ or both. But $9\nmid x$ because $\text{lcm}(x,z)=600$ and $9\nmid 600$. So $x=8,24$.

$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\text{lcm}(y,z)=900$ and $72\nmid 900$.

So there are six sets of $x,y$ and we will list all possible values of $z$ based on those.

$25\mid z$ because $z$ must source all powers of $5$. $z\in\{25,50,75,100,150,300\}$. $z\ne\{200,225\}$ because of $\text{lcm}$ restrictions.

By different sourcing of powers of $2$ and $3$,

$$(8,9):z=300$$ $$(8,18):z=300$$ $$(8,36):z=75,150,300$$ $$(24,9):z=100,300$$ $$(24,18):z=100,300$$ $$(24,36):z=25,50,75,100,150,300$$

$z=100$ is "enabled" by $x$ sourcing the power of $3$. $z=75,150$ is uncovered by $y$ sourcing all powers of $2$. And $z=25,50$ is uncovered by $x$ and $y$ both at full power capacity.

Counting the cases, $1+1+3+2+2+6=\boxed{\textbf{(A) }15}.$

## Solution 3 (Less Casework!)

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: $$5\cdot3\cdot1=\boxed{\textbf{(A) }15}$$

(Diagrams by ColtsFan10)