0:00:00.000,0:00:00.700


0:00:00.700,0:00:03.130
William and Luis are in[br]different physics classes

0:00:03.130,0:00:04.370
at Santa Rita.

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Luis's teacher always gives[br]exams with 30 questions

0:00:07.750,0:00:10.870
on them, while William's[br]teacher gives more

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frequent exams with[br]only 24 questions.

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Luis's teacher also assigns[br]three projects per year.

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Even though the two classes[br]have to take a different number

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of exams, their[br]teachers have told them

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that both classes-- let me[br]underline-- both classes will

0:00:25.250,0:00:29.040
get the same total number[br]of exam questions each year.

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What is the minimum[br]number of exam questions

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William's or Luis's class can[br]expect to get in a given year?

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So let's think about[br]what's happening.

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So if we think about[br]Luis's teacher who

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gives 30 questions per test,[br]so after the first test,

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he would have done 30 questions.

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So this is 0 right over here.

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Then after the second test,[br]he would have done 60.

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Then after the third test,[br]he would have done 90.

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And after the fourth test,[br]he would have done 120.

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And after the fifth test,[br]if there is a fifth test,

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he would do-- so this is if[br]they have that many tests-- he

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would get to 150[br]total questions.

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And we could keep[br]going on and on looking

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at all the multiples of 30.

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So this is probably a hint[br]of what we're thinking about.

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We're looking at[br]multiples of the numbers.

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We want the minimum multiples[br]or the least multiple.

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So that's with Luis.

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Well what's going[br]on with William?

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Will William's teacher,[br]after the first test,

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they're going to[br]get to 24 questions.

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Then they're going to get[br]to 48 after the second test.

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Then they're going to get[br]to 72 after the third test.

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Then they're going to get to 96.

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I'm just taking multiples of 24.

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They're going to get to[br]96 after the fourth test.

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And then after the fifth test,[br]they're going to get to 120.

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And if there's a sixth test,[br]then they would get to 144.

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And we could keep going[br]on and on in there.

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But let's see what[br]they're asking us.

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What is the minimum[br]number of exam questions

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William's or Luis's class[br]can expect to get in a year?

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Well the minimum[br]number is the point

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at which they've gotten the[br]same number of exam questions,

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despite the fact[br]that the tests had

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a different number of items.

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And you see the point at which[br]they have the same number

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is at 120.

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This happens at 120.

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They both could have[br]exactly 120 questions

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even though Luis's teacher[br]is giving 30 at a time

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and even though William's[br]teacher is giving 24 at a time.

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And so the answer is 120.

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And notice, they had a[br]different number of exams.

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Luis had one, two,[br]three, four exams

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while William would have to[br]have one, two, three, four,

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five exams.

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But that gets them both[br]to 120 total questions.

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Now thinking of it in terms[br]of some of the math notation

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or the least common multiple[br]notation we've seen before,

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this is really asking us what is[br]the least common multiple of 30

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and 24.

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And that least common[br]multiple is equal to 120.

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Now there's other[br]ways that you can

0:03:04.150,0:03:06.399
find the least common multiple[br]other than just looking

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at the multiples like this.

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You could look at it[br]through prime factorization.

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30 is 2 times 15,[br]which is 3 times 5.

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So we could say that 30 is[br]equal to 2 times 3 times 5.

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And 24-- that's a different[br]color than that blue-- 24

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is equal to 2 times 12.

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12 is equal to 2 times 6.

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6 is equal to 2 times 3.

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So 24 is equal to 2[br]times 2 times 2 times 3.

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So another way to come up with[br]the least common multiple,

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if we didn't even do this[br]exercise up here, says, look,

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the number has to be[br]divisible by both 30 and 24.

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If it's going to[br]be divisible by 30,

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it's going to have to[br]have 2 times 3 times 5

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in its prime factorization.

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That is essentially 30.

0:04:03.420,0:04:05.830
So this makes it[br]divisible by 30.

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And say, well in order[br]to be divisible by 24,

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its prime factorization is[br]going to need 3 twos and a 3.

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Well we already have 1 three.

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And we already have 1 two,[br]so we just need 2 more twos.

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So 2 times 2.

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So this makes it--[br]let me scroll up

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a little bit-- this right over[br]here makes it divisible by 24.

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And so this is essentially[br]the prime factorization

0:04:32.030,0:04:34.920
of the least common[br]multiple of 30 and 24.

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You take any one of[br]these numbers away,

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you are no longer going to be[br]divisible by one of these two

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numbers.

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If you take a two away, you're[br]not going to be divisible by 24

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anymore.

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If you take a two[br]or a three away.

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If you take a three[br]or a five away,

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you're not going to be[br]divisible by 30 anymore.

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And so if you were to[br]multiply all these out,

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this is 2 times 2 times 2 is 8[br]times 3 is 24 times 5 is 120.

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Now let's do one more of these.

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Umama just bought one[br]package of 21 binders.

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Let me write that number down.

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21 binders.

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She also bought a[br]package of 30 pencils.

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She wants to use all of[br]the binders and pencils

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to create identical[br]sets of office supplies

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for her classmates.

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What is the greatest[br]number of identical sets

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Umama can make using[br]all the supplies?

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So the fact that we're[br]talking about greatest

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is clue that it's probably[br]going to be dealing

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with greatest common divisors.

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And it's also dealing with[br]dividing these things.

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We want to divide these[br]both into the greatest

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number of identical sets.

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So there's a couple of ways[br]we could think about it.

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Let's think about what the[br]greatest common divisor

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of both these numbers are.

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Or I could even say the[br]greatest common factor.

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The greatest common[br]divisor of 21 and 30.

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So what's the largest number[br]that divides into both of them?

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So we could go with[br]the prime factor.

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We could list all of[br]their normal factors

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and see what is the[br]greatest common one.

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Or we could look at the[br]prime factorization.

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So let's just do the prime[br]factorization method.

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So 21 is the same[br]thing as 3 times 7.

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These are both prime numbers.

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30 is, let's see,[br]it's 3-- actually,

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I could write it this[br]way-- it is 2 times 15.

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We already did it[br]actually just now.

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And 15 is 3 times 5.

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So what's the largest[br]number of prime numbers that

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are common to both[br]factorizations?

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Well you only have a[br]three right over here.

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Then you don't have a[br]three times anything else.

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So this is just going[br]to be equal to 3.

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So this is essentially[br]telling us,

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look, we can divide both[br]of these numbers into 3

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and that will give[br]us the largest

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number of identical sets.

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So just to be clear[br]of what we're doing.

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So we've answered[br]the question is 3,

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but just to visualize[br]it for this question,

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let's actually draw 21 binders.

0:07:07.070,0:07:13.728
So let's say the 21 binders so[br]1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

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11, 12, 13, 14, 15,[br]16, 17, 18, 19, 20, 21.

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And then 30 pencils, so[br]I'll just do those in green.

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So 1, 2, 3, 4, 5,[br]6, 7, 8, 9, 10.

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Let me just copy and paste that.

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This is getting tedious.

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So copy and paste.

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So that's 20 and then[br]paste that is 30.

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Now, we figured out that 3[br]is the largest number that

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divides into both[br]of these evenly.

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So I can divide both of[br]these into groups of 3.

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So for the binders, I could[br]do it into three groups of 7.

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And then for the[br]pencils, I could do it

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into three groups of 10.

0:08:01.320,0:08:03.050
So if there are[br]three people that

0:08:03.050,0:08:05.710
are coming into[br]this classroom, I

0:08:05.710,0:08:11.640
could give them each seven[br]binders and 10 pencils.

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But that's the greatest[br]number of identical sets

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Umama can make.

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I would have three sets.

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Each set would have seven[br]binders and 10 pencils.

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And we essentially[br]are just thinking

0:08:23.500,0:08:27.960
about what's the number that we[br]can divide both of these sets

0:08:27.960,0:08:30.050
into evenly, the[br]largest number that we

0:08:30.050,0:08:33.263
can divide both of[br]these sets into evenly.

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