Question 165102: Planets A, B, and C orbit a specific star once every 3, 7, and 18 months. If they are all lined up now in the same stright line, what is the smallest number of months that must pass before they line up again?
I took 3, 7, and 18 and found out they were all divisible by 72, yet now I don't know how to figure what the smallest number of months will be when A B and C can all line uip again.
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! You need to find the Least Common Multiple of these numbers
3*7 = 21
I would find the LCM of 18 and 21
2*21 = 42 and 42/18 = 2.333
3*21 = 63 and 63/18 = 3.5
4*21 = 84 and 84/18 = 4.666
5*21 = 105 and 105/18 = 5.833
6*21 = 126 and 126/18= 7
So far 126 looks like the answer
126/3 = 42
126/7 = 18
126/18 = 7
What about 42?
42/3 = 14
42/7 = 6
42/18 = 2.333
Doesn't work
What about 63?
63/3 = 21
63/7 = 9
63/18 = 3.5
It appears they all line up again after 126 revolutions
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