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Current time:0:00Total duration:8:34

CCSS.Math:

William and Louise are in different physics classes at Santa Rita Louise's teacher always gives exams with 30 questions on them while Williams teacher gives more frequent exams with only 24 questions Louise's teacher also assigns three projects per year even though the two classes have to take a different number of exams their teachers have told them that both classes let me underline both classes will get the same total number of exam questions each year what is the minimum number of exam questions Williams or Lewis's class can expect to get in a given a year so let's think about what's happening so if we think about Louis's teacher who gives 30 questions per test so after the first test he would have done 30 questions then after so this is zero right over here then after the second test he would have done 60 then after the third test he would have done 90 and after the fourth test he would have done 120 and after the fifth test if there is a fifth test he would do so this is if they had that made test he would get to 150 total questions and we could keep going on and on looking at all the multiples of 30 so this is probably a hint of what we're thinking about working at multiples of the numbers we want the minimum multiples or the least multiple so that's what Louise well what's going on with William well William's teacher the first after the first test they're going to get 224 questions 24 then they're going to get 248 after the second test then they're going to get to 72 72 after the third test then they're going to get to 96 I'm just taking multiples of 24 they're going to get to 96 after the fourth test and then after the fifth test they're going to get to 120 and if there's a sixth test then they would get to 144 they would get to 144 and we could keep going on and on and there let's see what they're asking us what is the minimum number of exam questions Williams or Lewis's class can expect to get in a year well the minimum number is the point at which they've gotten the same number of exam questions despite the fact that the tests had different number of items and you see the point at which they have the same number is that a and 20 this happens at 120 they both can have exactly 120 questions even though Lewis's teachers giving 30 at a time and even though Williams teachers giving 24 at a time and so the answer is 120 and notice they had a different number of exams Lewis had one two three four exams while William would have to have one two three four five exams but that gets them both to 120 total questions now thinking of it in terms of some of the math notation or the or the least common multiple notation we've seen before this is really asking us what is the least common multiple of 30 and of 30 and 24 and that least common multiple is equal to is equal to 120 now there's other ways that you can find the least common multiple other than just looking at the multiples like this you can look at it through prime factorization 30 is 2 times 15 which is 3 times 5 so we could say that 30 is equal to 2 times 3 times 5 and 24 24 that's a different color in that blue 24 is equal to 2 times 12 12 is equal to 2 times 6 6 is equal to 2 times 3 so 24 24 is equal to 2 times 2 times 2 times 3 so another way to come up with the least common multiple if we didn't even do this exercise up here it says look the number has to be divisible by both 30 and 24 if it's going to be divisible by 30 it's going to have to have 2 it's going to have to have 2 times 3 times 5 in its prime factorization that is essentially 30 so this makes it divisible by 30 and say well in order to be divisible by 24 it's going to need to its prime factorization is going to need 3 2s and a 3 well we already have one 3 and we already have 1 2 so we just need 2 more twos so 2 times 2 so this makes it so this makes it let me score a little bit this right over here makes it divisible by 24 and so this is essentially the prime factorization of the least common multiple of 30 and 24 you take any one of these numbers away you are no longer going to be divisible by one of these two numbers if you take a 2 away you're not going to be divisible by 24 anymore if you take a 2 or 3 away if you take a if you take a if you take a 3 or a 5 away you're not going to be divisible by 30 anymore and so if you were to multiply all these out this is 2 times 2 times 2 is 8 times 3 is 24 times 5 is it is 120 now let's do one more of these oh mama just bought one package of 21 binders we write that number down 21 binders she also bought a package of 30 pencils 30 pencils she wants to use all of the binders and pencils to create identical sets of office supplies for her classmates what is the greatest number of identical sets well mam what can make using all of the supplies so the fact that we're talking about greatest is clue that it's probably going to be dealing with greatest common divisors it's also dealing with dividing these things we want to divide these both into the greatest number of identical the greatest number of identical sets so there's a couple of ways we could think about let's think about what the greatest common divisor of both these numbers are or I could even say the greatest common factor the greatest common divisor of 21 21 and 30 so what's the largest number that divides into both of them so we could go with the prime factor we could list all of their normal factors and see what is the greatest common one or or we could say we could we could look at the prime factorization so let's just do the prime factorization method so 21 is the same thing as 3 times 7 these are both prime numbers 30 is let's see it's 3 I check it write it this way it is 2 times 15 we already did it actually just now and 15 is 3 times 5 so what's the greatest what's the largest number of prime numbers that are into both factorizations well you only have a three right over here there you'll have a three times anything else so this is just going to be equal to three so this is essentially telling us like let's divide both of them in the let we can divide both of these numbers into three and that will give us the largest number of identical sets so just to be clear of what we're doing so we've answered the question it's three but just to visualize it for this question let's actually draw 21 21 binders so let's say the 21 binders so 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 and then 30 pencils so I'll just do those in green so 1 2 3 4 5 6 7 8 9 10 let me just copy and paste that lets getting tedious so copy and paste so that's 20 and then paste that is 30 now we figured out that 3 is the largest number that divides into both of these evenly so I can divide both of these into groups of 3 so for the binders I can do it into 3 groups of 7 and then for the for the pencils I could do it into 3 groups of 10 so if there are 3 people that are coming into this classroom I could give them each 3 I can just give them each 7 binders and 10 pencils but that's the greatest number of identical sets who mamuh can make I would have 3 sets each set would have 7 binders and 10 and 10 pencils and we essentially just thinking about what's the greatest what's the great the number that we can divide both of these sets into evenly the largest number that we can divide both of these sets into evenly