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

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==Solution 2==
 
==Solution 2==
  
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It is well known that if the <math>\text{lcm}a,b=c</math> and <math>c</math> can be written as <math>p_1^ap_2^bp_3^c\dots</math>, then the highest power of all prime numbers <math>p_1,p_2,p_3\dots</math> must divide into either <math>a</math> and/or <math>b</math>. Or else a lower <math>c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots</math> is the <math>\text{lcm}</math>.
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It is well known that if the <math>\text{lcm}(a,b)=c</math> and <math>c</math> can be written as <math>p_1^ap_2^bp_3^c\dots</math>, then the highest power of all prime numbers <math>p_1,p_2,p_3\dots</math> must divide into either <math>a</math> and/or <math>b</math>. Or else a lower <math>c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots</math> is the <math>\text{lcm}</math>.
  
 
Start from <math>x</math>:<math>\text{lcm}(x,y)=72</math> so <math>8\mid x</math> or <math>9\mid x</math> or both. But <math>9\nmid x</math> because <math>\text{lcm}(x,z}=600</math> and <math>9\nmid 600</math>.
 
Start from <math>x</math>:<math>\text{lcm}(x,y)=72</math> so <math>8\mid x</math> or <math>9\mid x</math> or both. But <math>9\nmid x</math> because <math>\text{lcm}(x,z}=600</math> and <math>9\nmid 600</math>.

Revision as of 21:18, 5 February 2016

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

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 $\text{lcm}(x,z}=600$ (Error compiling LaTeX. Unknown error_msg) and $9\nmid 600$. So $x=8,24$.

$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\lcm{y,z}=900$ (Error compiling LaTeX. Unknown error_msg) 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 $5$. $z\in\{25,50,75,100,150,300\}$. $z\nin\{200,225\}$ (Error compiling LaTeX. Unknown error_msg) because of $\text{lcm}$ restrictions.

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

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

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

See Also

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

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