Prime factorization
Common Divisibility Examples Common Divisibility Examples
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- In this video I want to 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 questions like this
- all numbers, 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 both divisible by 12 and 20
- it has to be divisible by each of these guy's prime factors
- So let's take their prime factorization.
- The prime factorization of 12 is 2 time 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 it's 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 it's prime factorization
- Now if you're divisible by both, you need 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 if this is divisible by both
- Obviously, if you divide it by 20, is the same thing as dividing it 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
- your just going to have a 3 leftover, 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 leftover
- 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 divisible by 12 and 20
- You could multiply this number here by a whole bunch
- of other factors I could call them a, b, and c.
- But this is kinda the smallest number thats divisible by 12 and 20
- any larger number will also be divisible by the same things as this smaller number
- Now, with that said, lets answer the questions.
- All numbers divisible by 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 ones, or they might not exist
- because the number might be sixty, it might be a hundred twenty
- Who knows what this number is? So the only numbers that we know can be divided in to this number
- well we know two can be, We know that two is a legitimate answer.
- Two is obviously divisible in 2 times 2 times 3 times 5
- We know that 2 times 2 is divisible into it.
- 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 six
- 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
- 7 is not of these prime factors over here,
- 9 is 3 times 3 so I'd need to have two threes over here, so nine doesn't work
- 7 doesn't work, 9 doesn't work,
- 12 is 4 times 3, or another way to divide it,
- 12 is 2 times 2 times 3
- Well there is a 2 times 2 times 3 in this prime factorization
- of this LCM of these two numbers
- So this is a 12. So a 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 twos, so this doesn't work.
- Let's try another example, just so that we understand this fairly well
- So let's say we want to know, we'll ask the same question
- All numbers divisible by 9 and 24 are also divisible by
- and once again we just do the prime factorization
- We essentially think of the least common multiple
- of 9 and 24
- We take the prime factorization of 9
- is 3 times 3
- and we're done
- prime factorization fo 24 is 2 times 12
- 12 is 2 times 6
- 6 is 2 times 3
- So anything divisible by 9 has to have a 9 in it's factorization
- or it's prime factorization would have be a 3 times 3
- anything divisible by 24 has to have three 2's in it
- So it's gotta have a 2 times 2 times 2
- and it's gotta have at least one three and we already have a three from the nine
- So we have that, so this number right here is divisible by both
- 9 and 24. This number right here is actually 72.
- This is 8 times 9 which is 72
- So for 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 it's prime factorization
- is 2 times 2 times 2 times 2, its 2 to the fourth power
- So you would need 4 two's here, we don't have 4 2's here
- I mean there could some other numbers, but we don't know what they are
- These are the only numbers that we can assume are the prime factorization
- of something divisible by both 9 and 24
- So we can rule out 16 because we don't have four 2's in it
- 27 is equal to 3 times 3 times 3.
- We don't have three 3's, we only have two of them
- So once again, cancel that out
- 5, 5's a prime number, no 5 here, so rule that out
- 11, once again prime number, no eleven's here, rule that out
- 9 is equal to 3 times 3
- And I just realized that's a silly answer
- because all numbers divisible by 9 and 24 are divisible
- by 9
- So obviously 9 is gonna work but I shouldn't have made that a choice
- Because that's in the problem
- But 9 would work. And what also would work is 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 have worked. That's 2 times 2
- 6 would work. Because that's 2 times 3
- 18 would work. Because 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 that's divisible by
- both 9 and 24
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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