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

Revision as of 17:49, 7 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

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

2016 AMC 10A (Problems • Answer Key • Resources)
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2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 4: / Line 4: @@
 <math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math>
-==Solution 1==
+== Solution ==
+===Solution 1===
 We prime factorize <math>72,600,</math> and <math>900</math>. The prime factorizations are <math>2^3\times 3^2</math>, <math>2^3\times 3\times 5^2</math> and <math>2^2\times 3^2\times 5^2</math>, respectively. Let <math>x=2^a\times 3^b\times 5^c</math>, <math>y=2^d\times 3^e\times 5^f</math> and <math>z=2^g\times 3^h\times 5^i</math>. We know that <cmath>\max(a,d)=3</cmath> <cmath>\max(b,e)=2</cmath> <cmath>\max(a,g)=3</cmath> <cmath>\max(b,h)=1</cmath> <cmath>\max(c,i)=2</cmath> <cmath>\max(d,g)=2</cmath> <cmath>\max(e,h)=2</cmath> and <math>c=f=0</math> since <math>\text{lcm}(x,y)</math> isn't a multiple of 5. Since <math>\max(d,g)=2</math> we know that <math>a=3</math>. We also know that since <math>\max(b,h)=1</math> that <math>e=2</math>. So now some equations have become useless to us...let's take them out. <cmath>\max(b,h)=1</cmath> <cmath>\max(d,g)=2</cmath> are the only two important ones left. We do casework on each now. If <math>\max(b,h)=1</math> then <math>(b,h)=(1,0),(0,1)</math> or <math>(1,1)</math>. Similarly if <math>\max(d,g)=2</math> then <math>(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)</math>. Thus our answer is <math>5\times 3=\boxed{15 \text{(A)}}</math>.
-==Solution 2==
+===Solution 2===
 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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