WEBVTT

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William and Luis are in
different physics classes

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at Santa Rita.

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Luis's teacher always gives
exams with 30 questions

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on them, while William's
teacher gives more

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frequent exams with
only 24 questions.

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Luis's teacher also assigns
three projects per year.

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Even though the two classes
have to take a different number

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of exams, their
teachers have told them

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that both classes-- let me
underline-- both classes will

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get the same total number
of exam questions each year.

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What is the minimum
number of exam questions

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

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So let's think about
what's happening.

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So if we think about
Luis's teacher who

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gives 30 questions per test,
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,
he would have done 60.

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

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

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And after the fifth test,
if there is a fifth test,

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

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would get to 150
total questions.

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And we could keep
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
of what we're thinking about.

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We're looking at
multiples of the numbers.

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We want the minimum multiples
or the least multiple.

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So that's with Luis.

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Well what's going
on with William?

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Will William's teacher,
after the first test,

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they're going to
get to 24 questions.

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

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Then they're going to get
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
96 after the fourth test.

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

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

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And we could keep going
on and on in there.

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But let's see what
they're asking us.

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What is the minimum
number of exam questions

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

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Well the minimum
number is the point

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

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despite the fact
that the tests had

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a different number of items.

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And you see the point at which
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
exactly 120 questions

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

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and even though William's
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
different number of exams.

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Luis had one, two,
three, four exams

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while William would have to
have one, two, three, four,

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five exams.

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But that gets them both
to 120 total questions.

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Now thinking of it in terms
of some of the math notation

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

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

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and 24.

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And that least common
multiple is equal to 120.

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Now there's other
ways that you can

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find the least common multiple
other than just looking

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at the multiples like this.

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You could look at it
through prime factorization.

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30 is 2 times 15,
which is 3 times 5.

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

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And 24-- that's a different
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
times 2 times 2 times 3.

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

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

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

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If it's going to
be divisible by 30,

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it's going to have to
have 2 times 3 times 5

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in its prime factorization.

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That is essentially 30.

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So this makes it
divisible by 30.

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And say, well in order
to be divisible by 24,

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its prime factorization is
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,
so we just need 2 more twos.

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So 2 times 2.

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So this makes it--
let me scroll up

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

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And so this is essentially
the prime factorization

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of the least common
multiple of 30 and 24.

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You take any one of
these numbers away,

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

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numbers.

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

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anymore.

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If you take a two
or a three away.

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If you take a three
or a five away,

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

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And so if you were to
multiply all these out,

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this is 2 times 2 times 2 is 8
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
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
package of 30 pencils.

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She wants to use all of
the binders and pencils

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to create identical
sets of office supplies

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for her classmates.

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What is the greatest
number of identical sets

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Umama can make using
all the supplies?

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So the fact that we're
talking about greatest

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is clue that it's probably
going to be dealing

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with greatest common divisors.

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And it's also dealing with
dividing these things.

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We want to divide these
both into the greatest

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number of identical sets.

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

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Let's think about what the
greatest common divisor

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of both these numbers are.

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Or I could even say the
greatest common factor.

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The greatest common
divisor of 21 and 30.

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

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So we could go with
the prime factor.

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We could list all of
their normal factors

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and see what is the
greatest common one.

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Or we could look at the
prime factorization.

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So let's just do the prime
factorization method.

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So 21 is the same
thing as 3 times 7.

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These are both prime numbers.

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30 is, let's see,
it's 3-- actually,

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

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We already did it
actually just now.

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And 15 is 3 times 5.

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So what's the largest
number of prime numbers that

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are common to both
factorizations?

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Well you only have a
three right over here.

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

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So this is just going
to be equal to 3.

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So this is essentially
telling us,

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look, we can divide both
of these numbers into 3

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and that will give
us the largest

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number of identical sets.

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

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So we've answered
the question is 3,

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but just to visualize
it for this question,

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let's actually draw 21 binders.

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So let's say the 21 binders so
1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

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

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

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So 1, 2, 3, 4, 5,
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
paste that is 30.

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Now, we figured out that 3
is the largest number that

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divides into both
of these evenly.

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So I can divide both of
these into groups of 3.

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

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And then for the
pencils, I could do it

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into three groups of 10.

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So if there are
three people that

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are coming into
this classroom, I

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could give them each seven
binders and 10 pencils.

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But that's the greatest
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
binders and 10 pencils.

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And we essentially
are just thinking

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about what's the number that we
can divide both of these sets

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into evenly, the
largest number that we

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can divide both of
these sets into evenly.

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