# Lesson: Least Common Multiple (LCM)

## Comment on Least Common Multiple (LCM)

### I am having difficulty with

I am having difficulty with the third practice problem. After getting the prime factorization of all 3 doctors appointment intervals, I get 5 as a common factor. But then when I take that 5 and multiply it by the remaining prime factors of 2x3x5x3x5 I get 1500. Good question.

Our goal is to find the LCM of 25, 30 and 50

25 = (5)(5)
30 = (2)(3)(5)
50 = (2)(5)(5)

Let's say N = the LCM of 25, 30 and 50

First off, all 3 number share 5 as a common prime.

So, N = (5)(other numbers)

Since N is divisible by 25 (which equals 5 x 5), we can see that N needs a second 5 in its prime factorization.

So, N = (5)(5)(?)(?)

So, now we can be sure that N is a multiple of 25

Next, since N is divisible by 30 (which equals 2 x 3 x 5), we can see that N needs a 2, a 3 and a 5 in its prime factorization. It already has a 5, we need only add a 2 and a 3 to the prime factorization.

So, N = (2)(3)(5)(5)(?)(?)

So, now we can be sure that N is a multiple of 30

Finally, since N is divisible by 50 (which equals 2 x 5 x 5), we can see that N needs a 2 and two 5's in its prime factorization. Well, the prime factorization already has a 2 and two 5's so, as it stands, N is ALREADY is a multiple of 50

So, the LCM of 25, 30 and 50 = (2)(3)(5)(5) = 150

Does that help?

Cheers,
Brent

### VERY tedious... better option

VERY tedious... better option would be to list multiples and finish the question in 30 seconds... ### Agreed.

Agreed.
That said, listing multiples could take a while for numbers with very little in common.
For example, finding the LCM of 45, 81 and 120 would take a while with this technique.

Cheers,
Brent

### If LCM of x and y is 24.

If LCM of x and y is 24.

Besides saying x & y equals 1 and 24, is there an easy approach to finding other possible values of x & y? Particularly if I'm not great at calculating in my head. ### If you're trying to think of

If you're trying to think of two numbers that have an LCM of N, then some number pairs that always work are:
1 and N
N and N
N and any factor of N
That's a start anyway.

### Hi Brent. What if there is no

Hi Brent. What if there is no common prime factors? For example, what is the least common multiple of 2, 3, 15 and 28? ### Here's one approach:

Here's one approach:

Let N = the least common multiple of 2, 3, 15 and 28

This means N is divisible by 2, 3, 15 and 28
So, there must be a 2, a 3, a 15 and a 28 "hiding" within the prime factorization of N.

Let's deal with each number individually.

If N is divisible by 2, then a 2 is "hiding" within the prime factorization of N.
So, N = (2)(?)(?)(?)... [there's our required 2 "hiding"]

If N is divisible by 3, then a 3 is "hiding" within the prime factorization of N.
So, N = (2)(3)(?)(?)(?)...

15 = (3)(5)
If N is divisible by 15, then a 3 AND a 5 must be "hiding" within the prime factorization of N.
We ALREADY have a 3 hiding in the prime factorization of N, but we still need a 5.
So, N = (2)(3)(5)(?)(?)(?)...

28 = (2)(2)(7)
If N is divisible by 28, then two 2's AND one 7 must be "hiding" within the prime factorization of N.
We ALREADY have ONE 2 hiding in the prime factorization of N, so let's add one more to get...
So, N = (2)(3)(5)(2)(?)(?)(?)...
We still need one 7 in the prime factorization of N, so let's add it to get...
So, N = (2)(3)(5)(2)(7)(?)(?)(?)...

Since we're looking for the LEAST common multiple, we can say N = (2)(3)(5)(2)(7) = 420

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,
I understood both of your approach (above mentioned in comment and finding via prime factor). But they both should be giving us the same results , when applied they tend to give us different result . Eg LCM using prime factor of number 30, 25 ,50 is 1500. and LCM using n method mentioned in above example is 150. which method should i use in exam. Please help and thank you in advance ### The video only explains how

The video only explains how to find the LCM of 2 numbers.
For more than two numbers, it's best to use the strategy I describe in my post above (posted on November 22, 2018)

Cheers,
Brent