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

(Solution 2)
(Solution 2)
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<cmath>(24,18):z=100,300</cmath>
 
<cmath>(24,18):z=100,300</cmath>
 
<cmath>(24,36):z=25,50,75,100,150,300</cmath>
 
<cmath>(24,36):z=25,50,75,100,150,300</cmath>
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 +
<math>z=100</math> is "enabled" by <math>x</math> sourcing the power of <math>3</math>. <math>z=75,150</math> is uncovered by <math>y</math> sourcing all powers of <math>2</math>. And <math>z=25,50</math> is uncovered by <math>x</math> and <math>y</math> both at full power capacity.
  
 
Counting the cases, <math>1+1+3+2+2+6=\boxed{\textbf{(A) }15}.</math>
 
Counting the cases, <math>1+1+3+2+2+6=\boxed{\textbf{(A) }15}.</math>

Revision as of 22:21, 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. ! Extra }, or forgotten $.) 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. ! Undefined control sequence.) 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. ! Undefined control sequence.) 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, $1+1+3+2+2+6=\boxed{\textbf{(A) }15}.$

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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