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# Common divisibility examples

## Video transcript

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 you 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 divisible by both 12 and 20 it has to be divisible by each of these guys prime factors so let's take their prime factorization the prime factorization of 12c 12 is 2 times 6 6 is a 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 the 2 times a 3 in it any number that's divisible by 12 now any number that's divisible by 20 leads to be divisible by let's take its prime factorization 2 times 10 10 is 2 times 5 so any number divisible by 20 means to be 2 needs to also be divisible by 2 times 2 times 5 or another way of thinking about it it needs to have two 2s and a 5 in its prime factorization now if you're divisible by both you have to have two twos a 3 and a 5 two 2s and a 3 4 12 and then two 2s and a 5 4 20 and you can verify this for yourself that this is divisible by both obviously if you divide it by 20 if you divide it by 20 if you let me do it this way if you dividing it by 20 is the same thing as dividing it by 2 times 2 times 5 so you're going to have the twos are going to cancel out the 5s are going to cancel out it's going that you're just going to have a 3 left over and if you were to divide it so it's clearly divisible by 20 and if you were to divide it by 12 if you divide it by 12 you divided 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 other 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 as 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 we can't really address it this they might just be ones 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 two can be we know that two is legitimate answer two is obviously divisible into two times two times three times five we know that two times two is divisible into it because we have the two times two over there we know that three is divisible into it we know that two times three is divisible into it so that's six let me write these this is 4 this is 6 we know that 2 times 2 times 3 is divisible into it I could do it I could go through every combination of these numbers right here we know that 3 we know that 3 times 5 is divisible into it we know that 2 times 3 times 5 is a visible 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 & 9 and why let me and 12 and let me and 8 you would say well let's see 7 is not one of these prime factors over here 9 9 is 3 times 3 so I'd need to have two threes 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 kind of 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 twos in the prime factorization we don't have three twos so this doesn't 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 want to know we want we asked the same question all numbers all numbers divisible by let me think of two interesting numbers all numbers divisible by 12 and let's say let's say 9 9 and 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 its factorization or if you did its prime factorization would have to be a 3 times 3 anything divisible by 24 has got to have a three 2's in it so it's going to have a 2 times a 2 times a 2 and it's got to have at least 1 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 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 16 27 511 and 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 fourth power so you would need four 2's here we don't have four twos over here I mean there 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 in 24 so we can rule out 16 we don't have four twos here 27 is equal to 3 times 3 times 3 so we need three 3s in the prime factorization we don't have three threes we only have two of them so once again cancel that out 5 5 is a prime number there are no 5s here rule that out 11 once again prime number no 11s here rule that out 9 is equal to 3 times 3 and actually I just realize 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 because that's that's in the problem but 9 would work and what also would work is 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 for would also work that's 2 times 2 6 would work since 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 divisible by both 9 and 24 hopefully that doesn't confuse you too much