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

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

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. $[\textbf{\emph{insert diagram here}}]$

Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and} 5$ . $[\textbf{\emph{insert diagram here for powers of 2}}]$

Analyzing for powers of $2$ , is is quite obvious that $x$ must have $2^3$ as one of it's factors, since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y \text{and} z$ , 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 $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$ .

Now we can move on to powers of $3$ . $[\textbf{\emph{insert diagram here for powers of 3}}]$ Using the same logic as we did for powers of $2$ , is becomes quite easy to note that $y$ must have $3^2$ as one of it's 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 $5$ . $[\textbf{\emph{insert diagram here for powers of 5}}]$ 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: $5\cdot3\cdot1=\boxed{\textbf{(A) }15}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search