# Lesson: Least Common Multiple (LCM)

## Comment on Least Common Multiple (LCM)

### Hi Brent, I have difficulties

Hi Brent, I have difficulties with the skill of ¨selecting numbers with a certain least common multiple¨, could you explain it a bit further, and help me with this question please? Thank you x

If x is a positive integer, what is the least common multiple of x, 6, and 9 ?
(1) The least common multiple of x and 6 is 30.
(2) The least common multiple of x and 9 is 45. ### Great question. Here are two

Great question. Here are two different approaches:

Please let me know if either of those approaches help.

Cheers,
Brent

### Thank you thank you, now I

Thank you thank you, now I fully understand everything!! x

### Hey Brent,

Hey Brent,

https://gmatclub.com/forum/the-number-of-students-who-attend-a-school-could-be-divided-among-273174.html
QQ, why doesn't using the Prime Factorization trick you taught earlier give us 240 but instead 480 as the LCM?

I understand how 240 is the LCM but wanted to understand the application of that rule here. The technique for finding the LCM of 3 numbers is a little bit different from finding the LCM of 2 numbers.
10 = (2)(5)
12 = (2)(2)(3)
16 = (2)(2)(2)(2)

The three numbers share ONE 2 in their prime factorizations.
So far, LCM = (2)(some other stuff)

Ignore those three shared 2's to get:
10 = (-)(5)
12 = (-)(2)(3)
16 = (-)(2)(2)(2)

Among the remaining primes, two numbers share ONE 2
So, we now have: LCM = (2)(2)(some other stuff)

Ignore those two shared 2's to get:
10 = (-)(5)
12 = (-)(-)(3)
16 = (-)(-)(2)(2)

Among the remaining primes, none are shared.
So we take these remaining primes and add them to our calculations to get:
LCM = (2)(2)(5)(3)(2)(2) = 240

-------ALTERNATE APPROACH----------------
When it comes to finding the LCM of three integers, I prefer to just "build" the LCM by examining one number at a time.
Let N = the LCM of 10, 12 and 16
This means N is divisible by 10, 12 and 16

10 = (2)(5)
So, if N is divisible by 10, there must be one 2 and one 5 in the prime factorization of N.
N = (2)(5)(more possible primes)

12 = (2)(2)(3)
So, if N is divisible by 12, there must be two 2's and one 3 in the prime factorization of N.
We already have ONE 2 in the prime factorization of N. So, we need only add one more 2 (and one 3)
We get: N = (2)(5)(2)(3)(more possible primes)

16 = (2)(2)(2)(2)
So, if N is divisible by 16, there must be four 2's in the prime factorization of N.
We already have TWO 2's in the prime factorization of N, we need to add TWO more 2's
We get: N = (2)(5)(2)(3)(2)(2)

Done!

### I'm still confused on how to

I'm still confused on how to apply the LCM in

2, 3, 4, 5, 6, 7, 8, and 9 these numbers.

say if a number is shared in prime factorization then it doesn't add to the multiplication of LCM? ### Here's another way to look at

Here's another way to look at it:
If some integer N is the LCM of 2, 3, 4, 5, 6, 7, 8, and 9, then the prime factorization of N must include 2, 3, 4, 5, 6, 7, 8, and 9

So, for example, if and is divisible by 2, then there must be a 2 hiding in the prime factorization of N.
In other words, we now know that N = (2)(?)(?)(?)(?), where the other "?" symbols represent other possible values in the prime factorization of N.

Next, there must be a 3 hiding in the prime factorization of N.
So we now know that N = (2)(3)(?)(?)(?)(?)

Important: In order for N to be divisible by 4, there must be two 2's hiding in the prime factorization of N.
Since we already have one 2 in the prime factorization of N, we need only add one more 2.
In other words: N = (2)(3)(2)(?)(?)(?)(?)
As we can see, we can't be certain that N is divisible by 2, 3 and 4

Next, there must be a 5 hiding in the prime factorization of N.
So we now know that N = (2)(3)(2)(5)(?)(?)(?)

6 = (2)(3)
So, in order for N to be divisible by 6, there must be one 2 and one 3 hiding in the prime factorization of N.
Since our prime factorization of N already includes one 2 and one 3, we need not add any numbers to our prime factorization.
In other words, if N = (2)(3)(2)(5), then we can be certain that N is divisible by 6.

Next, there must be a 7 hiding in the prime factorization of N.
So we now know that N = (2)(3)(2)(5)(7)(?)(?)

8 = (2)(2)(2)
So, in order for N to be divisible by 8, there must be three 2's hiding in the prime factorization of N.
Since our prime factorization of N already includes two 2's, we need to add one more 2.
In other words, if N = (2)(3)(2)(5)(7)(2)(?)(?), then we can be certain that N is divisible by 8.

9 = (3)(3)
So, in order for N to be divisible by 9, there must be two 3's hiding in the prime factorization of N.
Since our prime factorization of N already includes one 3, we need to add one more 3.
In other words, if N = (2)(3)(2)(5)(7)(2)(3), then we can be certain that N is divisible by 9.

So, if N = (2)(3)(2)(5)(7)(2)(3), then we can be certain that N is divisible by 2, 3, 4, 5, 6, 7, 8, and 9.
Similarly, if N = (2)(3)(2)(5)(7)(2)(3), then we can be certain that N is a MULTIPLE OF 2, 3, 4, 5, 6, 7, 8, and 9.

Does that help?