Main content

## Pre-algebra

### Course: Pre-algebra > Unit 1

Lesson 3: Prime factorization# 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!(207 votes)
- 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!(162 votes)

- What are you supposed to do when one of the numbers is prime?(89 votes)
- 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.(52 votes)

- 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(36 votes)
- 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(103 votes)

- What is Divisible by 24 and 21?(7 votes)
- 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.(9 votes)

- 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.(14 votes)
- At7:33how 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

Please explain someone....(7 votes)- 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?(14 votes)

- I know this isn't directly related to the video, but if 1 is neither prime nor composite, what word would be used to describe it?(6 votes)
- Unique.

Zero and One are neither composite nor prime numbers.

"There can be only one." - Highlander(6 votes)

- This is beginning to sink in - it has taken a while. I do not remember anything like this in grade or high school, nor on any math test that I've taken since (admittedly, it's has been a while). With that said, I'd like to get an idea of the usefulness or usability of this mathematical concept. Where in higher math will this concept be useful? I'm grasping for some logic behind it. Thanks.(5 votes)
- Prime factorization (and the previous lesson of the fundamental theorem of arithmetic - that every natural number is either Prime or a composite) is used every single day by nearly everyone in the world, though they may not know it.

The fact that it is so very hard to factor out huge numbers is the driving concept behind the encryption technology in your emails - something under discussion this very week in the U.S. Congress. Even the best sub-quantum computers can't perform prime factorization at anything like a practical rate, which makes it a great check against fraud. That few-digit pin# on your credit card is nothing more than a check on whether your card number matches it's unique combination of factors. Using factoring to derive GCF and LCD comes up directly in the algebra section on this site of which this lesson is a part. Those concepts both have daily real world application. (Hotdogs come in packs of 10, buns in packs of 8; the least common multiple tells you how many of each you would need to buy to have an even split with no bun-less wieners.)

As for higher math, there's lots of kooky stuff (Gauss, Eisenstein, rings, reciprocity, quadratics) that touches back on this simple concept.(6 votes)

- in the first equation Sal uses 12 as one of the two numbers and one of the multiple choice numbers so shouldn't it be a different because 12 is always divisible by 12(5 votes)
- at2:16I'm not sure what he is trying to say does anyone know a better way to explain it? please help.(5 votes)

## Video transcript

- [Voiceover] In this video I wanna do a bunch of example problems that show up on standardized exams and definitely will help you
with our divisibility module because it's asking you
questions like this. And this is just one of the examples. All numbers divisible by both 12 and 20 are also divisible by: And the trick here is to realize that if a number is
divisible by both 12 and 20, it has to be divisible by each
of these guy's prime factors. So let's take the prime factorization, the prime factorization of 12, let's see, 12 is 2 times 6. 6 isn't prime yet so 6 is 2 times 3. So that is prime. So any number divisible by 12 needs to be divisible
by 2 times 2 times 3. So its prime factorization needs to have a 2 times a 2 times a 3 in it, any number that's divisible by 12. Now any number that's divisible by 20 needs to be divisible by, let's take it's prime factorization. 2 times 10 10 is 2 times 5. So any number divisible by 20 needs to also be divisible by 2 times 2 times 5. Or another way of thinking about it, it needs to have two 2's and a
5 in its prime factorization. If you're divisible by both, you have to have two 2's, a 3, and a 5, two 2's and a 3 for 12, and then two 2's and a 5 for 20, and you can verify this for yourself that this is divisible by both. Obviously if you divide it by 20, let me do it this way. Dividing it by 20 is the same thing as dividing by 2 times 2 times 5, so you're going to have the 2's are going to cancel out, the 5's are going to cancel out. You're just going to have a 3 left over. So it's clearly divisible by 20, and if you were to divide it by 12, you'd divide it by 2 times 2 times 3. This is the same thing as 12. And so these guys would cancel out and you would just have a 5 left over so it's clearly divisible by both, and this number right here is 60. It's 4 times 3, which is 12, times 5, it's 60. This right here is actually
the least common multiple of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's
answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's
or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that
we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2
is divisible into it, cuz we have the 2 times 2 over there. We know that 3 is divisible into it. We know that 2 times 3
is divisible into it. So that's 6. Let me write these. This is 4. This is 6. We know that 2 times 2 times 3 is divisible into it. I could go through every combination of these numbers right here. We know that 3 times 5
is divisible into it. We know that 2 times 3 times 5 is divisible into it. So in general you can look
at these prime factors and any combination of these prime factors is divisible into any
number that's divisible by both 12 and 20, so if this was a multiple choice question, and the choices were 7 and 9 and 12 and 8. You would say, well let's see, 7 is not one of these
prime factors over here. 9 is 3 times 3 so I need to have two 3's here. I only have one 3 here so 9 doesn't work. 7 doesn't work, 9 doesn't work. 12 is 4 times 3 or another way to think about it, 12 is 2 times 2 times 3. Well there is a 2 times 2 times 3 in the prime factorization, of this least common multiple
of these two numbers, so this is a 12 so 12 would work. 8 is 2 times 2 times 2. You would need three 2's
in the prime factorization. We don't have three 2's, so this doesn't work. Let's try another example just so that we make sure that we understand this fairly well. So let's say we wanna know, we ask the same question. All numbers divisible by and let me think of two
interesting numbers, all numbers divisible by 12 and let's say 9, and I don't know, let's
make it more interesting, 9 and 24 are also divisible by are also divisible by And once again we just do
the prime factorization. We essentially think about
the least common multiple of 9 and 24. You take the prime factorization of 9, it's 3 times 3 and we're done. Prime factorization of 24 is 2 times 12. 12 is 2 times 6, 6 is 2 times 3. So anything that's divisible by 9 has to have a 9 in it's factorization or if you did its prime factorization would have to be a 3 times 3, anything divisible by 24 has got to have three 2's in it, so it's gotta have a
2 times a 2 times a 2, and it's got to have at lease one 3 here. And we already have at
least one 3 from the 9, so we have that. So this number right here is divisible by both 9 and 24. And this number right here is actually 72. This is 8 times 9, which is 72. So if the choices for this question, let's assume that it was multiple choice. Let's say the choices here were 16 27 5 11 and 9. So 16, if you were to do
its prime factorization, is 2 times 2 times 2 times 2. It's 2 to the 4th power. So you would need four 2's here. We don't have four 2's over here. I mean there could be
some other numbers here but we don't know what they are. These are the only numbers that we can assume are in
the prime factorization of something divisible by both 9 and 24. So we can rule out 16. We don't have four 2's here. 27 is equal to 3 times 3 times 3, so you need three 3's in
the prime factorization. We don't have three 3's. We only have two of them. So once again, cancel that out. 5's a prime number. There are no 5's here. Rule that out. 11, once again, prime number. No 11's here. Rule that out. 9 is equal to 3 times 3. And actually I just realized
that this is a silly answer because obviously all numbers divisible by 9 and 24 are also divisible by 9. So obviously 9 is going to work but I shouldn't have made that a choice cuz that's in the
problem, but 9 would work, and what also would work if we had a, if 8 was one of the choices, because 8 is equal to 2 times 2 times 2, and we have a 2 times 2 times 2 here. 4 would also work. That's 2 times 2. That's 2 times 2. 6 would work since that's 2 times 3. 18 would work cuz that's 2 times 3 times 3. So anything that's made
up of a combination of these prime factors will be divisible into something divisible by both 9 and 24. Hopefully that doesn't
confuse you too much.