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LCM and GCF greatest common factor) word problems

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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.
Title:
LCM and GCF greatest common factor) word problems
Description:

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Video Language:
English
Team:
Khan Academy
Duration:
08:34

English subtitles

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