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Difference between revisions of "2007 AIME I Problems/Problem 4"
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== Problem ==
 
== Problem ==
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Three planets orbit a star circularly in the same plane.  Each moves in the same direction and moves at constant speed.  Their periods are <math>60</math>,<math>84</math>, and <math>140</math>.  The three planets and the star are currently collinear.  What is the fewest number of years from now that they will all be collinear again?
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Three planets orbit a star circularly in the same plane.  Each moves in the same direction and moves at [[constant]] speed.  Their periods are <math>60</math>,<math>84</math>, and <math>140</math>.  The three planets and the star are currently [[collinear]].  What is the fewest number of years from now that they will all be collinear again?
  
 
== Solution ==
 
== Solution ==
−
First you prime factor all of the numbers: <math>\displaystyle 60=2^2*3*5</math>, <math>\displaystyle 84=2^2*3*7</math>, and <math>\displaystyle 140=2^2*5*7</math>. To find out when they are next collinear, you take <math>\displaystyle \frac{lcm}{gcf}</math>, which is <math>\displaystyle \frac{2^2*3*5*7}{2^2}=\frac{420}{4}=105</math>
+
First, take the [[prime factorization]] of all the numbers: <math>\displaystyle 60=2^2*3*5</math>, <math>\displaystyle 84=2^2*3*7</math>, and <math>\displaystyle 140=2^2*5*7</math>. To find out when they are next collinear, you take <math>\displaystyle \frac{lcm}{gcf}</math>, which is <math>\displaystyle \frac{2^2*3*5*7}{2^2}=\frac{420}{4}=105</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2007|n=I|num-b=4|num-a=6}}

Revision as of 19:24, 15 March 2007

Problem

Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are $60$,$84$, and $140$. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?

Solution

First, take the prime factorization of all the numbers: $\displaystyle 60=2^2*3*5$, $\displaystyle 84=2^2*3*7$, and $\displaystyle 140=2^2*5*7$. To find out when they are next collinear, you take $\displaystyle \frac{lcm}{gcf}$, which is $\displaystyle \frac{2^2*3*5*7}{2^2}=\frac{420}{4}=105$

See also

2007 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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