SOLUTION: find the least number exactly devidable by 27,45,60,72 and 96 ?

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Question 215244: find the least number exactly devidable by 27,45,60,72 and 96 ?
Answer by Edwin McCravy(20065)   (Show Source): You can put this solution on YOUR website!

Let's break all those down into prime factors:

27 = 3*3*3
45 = 3*3*5
60 = 2*2*3*5
72 = 2*2*2*3*3
96 = 2*2*2*2*2*3

Let's check to see which of those has the most 2 factors:

27 has 0 2's
45 has 0 2's
60 has 2 2's
72 has 3 2's
96 has 5 2's

The most in that list is 5 2's.  So the LCM has 5 2's

27 = 3*3*3
45 = 3*3*5
60 = 2*2*3*5
72 = 2*2*2*3*3
96 = 2*2*2*2*2*3

Let's check to see which of those has the most 3 factors:

27 has 3 3's
45 has 2 3's
60 has 1 3
72 has 2 3's
96 has 1 3's

The most in that list is 3 3's.  So the LCM has 5 2's and 3 3's

27 = 3*3*3
45 = 3*3*5
60 = 2*2*3*5
72 = 2*2*2*3*3
96 = 2*2*2*2*2*3

Let's check to see which of those has the most 5 factors:

27 has 0 5's
45 has 1 5
60 has 1 5
72 has 0 5's
96 has 0 5's

The most in that list is 1 5.  So the LCM has 5 2's, 3 3's, and 1 5.

So LCM = 2*2*2*2*2*3*3*3*5 = 4320

Edwin
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