it is the smallest number that all the numbers being compared goes into...
such as: the LCM of 2 and 3 is 6, because both 2 and 3 go into 6, and 6 is the smallest number both go into.
Or: the LCM of 4 and 6 is 12
There is a way to find the LCM if you would like to know it.
I'll demonstrate one process. I'm sure there are others.
Take 60 and 630. First break both of these down into their prime factorizations:
60 = 2^2 * 3 * 5
630 = 2 * 3^2 * 5 * 7
Now, the LCM is the smallest number that is divisible by both.
Look at each prime factor. How many of each will the LCM contain?
Example, 2: 60 is divisible by 2^2, and 630 is divisible by 2. The LCM MUST be divisible by 2^2, right? Else it won't be divisible by 60.
So what we are going to do is look at the list of prime factors and choose the greatest exponent from the two lists.
So we need 2 2's, 2 3's, 1 5 and 1 7.
Thus LCM(60, 630) = 2^2 * 3^2 * 5 * 7 = 1260.
Look again at the example in the previous post. What is the LCM of 4 and 6?
4 = 2^2
6 = 2 * 3
Thus LCM(4, 6) = 2^2 * 3 = 12.
-Dan
Don't ask me how it works (I have not yet examined it enough to know).
ANyway, the method is this:
pick any amount of numbers, let's do 18; 32; and 3
Now to find the LCM of all of them we write out there prime factors:
now this will sound confusing...Code:Number Prime Factors 18 2^1 x 3^2 32 2^5 x 3^0 3 2^0 x 3^1
take a look at those prime factorizations, the only prime factors for those three numbers are 2's and 3's. Therefore our LCM has prime factors of 2's and 3's
LCM = 2^a x 3^b
now (here's the part that get's everybody), find the biggest group for each prime factor, In this case the biggest group for 2 is 2^5, and the biggest group for 3 is 3^2. Now put that into your equation to get:
LCM = 2^5 x 3^2 = 32 x 9 = 288
so the lowest common multiple for 18, 32, and 3 is 288
Now you try to find the LCM of 3; 4; and 10; and I'll correct your work.