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## Class 6 math (India)

### Unit 3: Lesson 5

Some more divisibility rules

# Common divisibility examples

Common Divisibility Examples. Created by Sal Khan.

## Want to join the conversation?

• On the second problem, the prime factorization for 9 is 3 and 3. So why do you only use one of the threes from that side? The prime factorization for 24 is 2, 2, 2, and 3 and all of those were used. This confused me and I hope I can get an answer! •   So the prime factors of 9 are 3 and 3, so your final factors must contain at least two 3s.
There is already a three from the prime factors of 24, so it is shared.

For example, if instead of 9 it was 18, you would get 2x2x2x3x3. The prime factorisation of 18 is 2x3x3, so one 3 and one 2 is shared with the 24.

I hope this helps, difficult to explain!
• What are you supposed to do when one of the numbers is prime? •   It took me several viewings over several days to figure out what to do when one of the numbers is prime.

When one number is prime, nothing is done to it but it's a good clue because if any other primes larger than it show up when you're breaking each answer (on the list of answers) into their own primes, that's not a correct answer. So move on to breaking down the next answer in the list into factors even if you haven't completely factored that answer. It's good to make the habit of checking at each level of the factor "tree" to see if either one is a prime (from a list of primes) before trying to divide them out for the next level. That also helps learn the more common primes.
• Why are some of the prime factors not listed or "SHARED" like others? For example the prime factorization of 12 + 20 = 2x2x3 + 2x2x5 but is only listed as 2x2x3x5. Why are two of the 2's shown instead of 4? If its to not repeat, then shouldn't it just be 2x3x5? I just don't understand why the sharing is selective and what designated which numbers to be "Shared" or not. Also should it be originally displayed as (12, 24) and (9, 20) instead of the + sign, because its proposed as a list not an addition problem? If I am wrong, i would appreciate someone explaining to me why •   Suppose I have two numbers A and B,
If I get the prime factorization of each, the prime factorization of the least common multiple has to contain the maximum number of 2s from A and B, max number of 3s from A and B, max number of 5s from A and B, etc. for all the prime factors.
e.g. A=12=2*2*3, B=20=2*2*5
A has two 2s, and B has two 2s. The max number of 2s is 2
A has one 3 , and B has zero 3s. The max number of 3s is 1.
A has zero 5s and B has one 5. The max number of 5s is 1.
So our least common multiple (LCM) has two 2s, one 3, and one 5.
LCM=2*2*3*5=60

Why is this true ? For us to be able to divide the LCM by A it must at least have all the prime factors that A has. For us to be able to divide the LCM by B it must have at least the prime factors that B has. Thus it must have, at least, for each of the prime factors the max number of that factor from A and B.

60/A=60/12=(2*2*3*5)/(2*2*3)=5
60/B=60/20=(2*2*3*5)/(2*2*5)=3

Another example:
Least common multiple for 21 and 49
A=21=3*7
B=49=7*7

A has one 3, B has zero 3s, max number of 3s is one
A has one 7, B has two 7s, max number of 7s is two
our least common multiple (LCM) must have one 3 and two 7s
LCM=3*7*7=147

Hope this helps
• What does prime factorization mean? I saw the prime factorization video and i know how to do it but, I dont understand what is it? Can anyone help??? •   Prime Factorization means breaking down a number into all of the PRIME numbers that make up a number.

So the PRIME factors of 5 are (5 and 1). The prime factors of 10 are (5 and 2 and 1). Frequently, 1 is not listed as a prime factor because all whole numbers have 1 in addition to their other prime factors.

Let's give another example: The number 8 has a prime factorization of (2 and 2 and 2); its only prime factor (besides 1) is 2. Even though 8 has other factors (such as 4 and 8), these numbers are not PRIME. 2 is the only PRIME factor of 8, and in order to make 8, you need 3 2's (2x2x2).

Does that help?
• • Holly, we're not looking for a number that is divisible by 24 & 21. The wording is hard to make clear, but let's say we have some number, A, and A is divisible by both 24 & 21. What we're looking for is the other numbers, B, C, D, and so on, that A will ALSO be divisible by. Since A is divisible by both 24 & 21, then A will also be divisible by any number that is made by multiplying any combination of the prime factors of 24 & 21, which are 2x2x2x3x7. (There is only one 3, because 24 & 21 only have one 3 each, so they "share" it.)

I hope that helps.
• After watching the "Common Divisibility Examples" video, it is still very confusing. I think there needs to be another video on this that explains it differently. • At how does 4 work? I thought that for the number to work it had to have the same number of numbers in it as the example Sal gave
e.g. 9 = 3*3 or 8 = 2*2*2 • 4 works because its prime factorization is 2*2. 2*2 is part of 2*2*2*3*3, so all numbers divible by 9 and 24 are also divisible by 4.

In other words, we're not looking for an exact match. There is no 3 in 4, but we don't really care, because we just want to know if 9 and 24 combined have at least two 2's. Maybe they also have lots of numbers in them, but that doesn't matter, we just want two 2's for 4 to work.

Does it help?   