Lesson: Least Common Multiple (LCM)

Question link: https://gre.myprepclub.com/forum/in-a-certain-medical-group-dr-schwartz-...

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

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.

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

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