2007 AIME I Problems/Problem 4

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 you prime factor all of 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 $\displaystyle \frac{2^2*3*5*7}{2^2}=\frac{420}{4}=105$

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2007 AIME I Problems/Problem 4

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