Difference between revisions of "2007 AIME I Problems/Problem 4"
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| + | == 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 <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 == | ||
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 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> | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2007|n=I|num-b=4|num-a=6}} | ||
Revision as of 16:46, 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 
,
, and 
. 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 you prime factor all of the numbers: 
, 
, and 
. To find out when they are next collinear, you take 
, which is 
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 | ||
