Current time:0:00Total duration:7:55

# Common divisibility examples

## 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.