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Question 120756: three strings of a musical instrument vibrate 6,8 and 12 times per second respectively.If all the three begin to vibrate simultaneously,find the shortest time interval before all three vibrate together again?
Answer by Edwin McCravy(20081)   (Show Source):
You can put this solution on YOUR website! three strings of a musical instrument vibrate 6,8 and 12 times per second respectively.If all the three begin to vibrate simultaneously,find the shortest time interval before all three vibrate together again?
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That's the least common multiple of 6, 8, and 12
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Break each into prime factors:
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6 = 2x3
8 = 2x2x2
12 = 2x2x3
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2 appears 1 time as a factor of 6, 3 times as a factor of 8
and 2 times as a factor of 12. That's at most 3 times.
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Therefore the LCM must contain 2 as a factor the most number of
times it occurs in any one of those, which is 3 times.
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3 appears 1 time as a factor of 6, 0 times as a factor of 8
and 1 time as a factor of 12. That's at most 1 time.
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Therefore the LCM must contain 3 as a factor the most number of
times it occurs in any one of those, which is 1 time.
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Therefore the LCM must contain 2 as a factor 3 times, and
3 as a factor 1 time. So LCM = 2x2x2x3 = 24
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Answer: 24 seconds
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Edwin
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