SOLUTION: I need some help with a problem where one time I come up with the answer 140 and the other is 240. Can you help me? How do you find the least common multiple (LCM) for

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: I need some help with a problem where one time I come up with the answer 140 and the other is 240. Can you help me? How do you find the least common multiple (LCM) for      Log On


   



Question 39090: I need some help with a problem where one time I come up with the answer 140 and the other is 240. Can you help me?
How do you find the least common multiple (LCM) for the following group of numbers. 12, 20, and 35. Can you explain how you come up with your answer.
Thanks allot.

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
I need some help with a problem where one time I come up
with the answer 140 and the other is 240. Can you help me?
How do you find the least common multiple (LCM) for the 
following group of numbers. 12, 20, and 35. Can you explain 
how you come up with your answer

First prime factor all the numbers

12 = 2·2·3  <---  two 2's and one 3
20 = 2·2·5  <---  two 2's and one 5
35 = 5·7    <---  one 5 and one 7

There are four prime numbers used.  These are 2, 3, 5 and 7. 

For the factor 2:
The largest number of 2's used in some member of the group is TWO.
So the LCM must contain TWO 2's as factors

So let's write LCM = 2·2·_______

For the factor 3:
The largest number of 3's used in some member of the group is ONE.
So the LCM must contain ONE 3 as a factor

So now we can tack on ONE 3.  

Now we have LCM = 2·2·3·_______ 

For the factor 5:
The largest number of 5's used in some member of the group is ONE.
So the LCM must contain ONE 5 as a factor

So now we can tack on ONE 5.  

Now we have LCM = 2·2·3·5_______ 

For the factor 7:
The largest number of 7's used in some member of the group is ONE.
So the LCM must contain ONE 7 as a factor

So now we can tack on ONE 7.  

Now we have LCM = 2·2·3·5·7

That takes care of all factors used, so we are done and

the LCM = 2·2·3·5·7 = 420

Edwin
AnlytcPhil@aol.com