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

Revision as of 21:21, 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. 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 $3$ ,

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

$z=100$ 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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 29: / Line 29: @@
 <cmath>(24,18):z=100,300</cmath>
 <cmath>(24,36):z=25,50,75,100,150,300</cmath>
+<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>

Page

Toolbox

Search

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

Revision as of 21:21, 5 February 2016

Contents

Problem

Solution 1

Solution 2

See Also