Asked by Anonymous
Plantets A, B, and C orbit a certain star once every 2, 5, and 12 months. If the three plantes are now in the same straight line, what is the smallest number of months that must pass before they line up again?
I do not want the answer, I just do not understand how to set up the equation.
I do not want the answer, I just do not understand how to set up the equation.
Answers
Answered by
Anonymous
I think I figured it out...If I go through all the multiples of
2 5 and 12, the LCM would be 60, so wouldn't that be the number of months?
2 5 and 12, the LCM would be 60, so wouldn't that be the number of months?
Answered by
Reiny
yes
Answered by
MathMate
I believe the LCM solution is most probably the correct one.
However, if we interpret the expression "in the same straight line" in the general sense, it means that any planet could be on the same side OR on the other side of the sun. In which case the LCM would be for half the period, time it takes to go across the other side of the sun.
If the general interpretation is accepted, we can say that the planets will be on a straight line again in LCM(1, 2.5, 6)=15 months.
However, if we interpret the expression "in the same straight line" in the general sense, it means that any planet could be on the same side OR on the other side of the sun. In which case the LCM would be for half the period, time it takes to go across the other side of the sun.
If the general interpretation is accepted, we can say that the planets will be on a straight line again in LCM(1, 2.5, 6)=15 months.
Answered by
MathMate
change "sun" for "star".
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