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Divisible

This topic has 2 expert replies and 2 member replies
akhilsuhag Master | Next Rank: 500 Posts
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Divisible

Post Wed Nov 19, 2014 12:39 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If a, b and c are integers, is (a·b) a multiple of 18?

    (1) 2a=3b

    (2) 2b=3c

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    Post Wed Nov 19, 2014 1:03 pm
    Hi akhilsuhag,

    This question is perfect for TESTing VALUES.

    We're told that A, B and C are INTEGERS. We're asked if (A)(B) is a multiple of 18? This is a YES/NO question.

    Fact 1: 2A = 3B

    If…
    A = 3
    B = 2
    AB = 6 and the answer to the question is NO

    A = 9
    B = 6
    AB = 54 and the answer to the question is YES
    Fact 1 is INSUFFICIENT

    Fact 2: 2B = 3C

    We know NOTHING about A, so this Fact is likely to be INSUFFICIENT, but here's how you can prove it:

    If…
    B = 3
    C = 2
    A = 1
    AB = 3 and the answer to the question is NO

    B = 3
    C = 2
    A = 6
    AB = 18 and the answer to the question is YES
    Fact 2 is INSUFFICIENT

    Combined, we know…
    2A = 3B
    2B = 3C

    This tells us that B MUST be a multiple of 2 (from Fact 1) AND a multiple of 3 (From Fact 2), so B MUST be a multiple of 6…By extension, this means that A MUST be a multiple of 3. TESTing VALUES proves it:

    If….
    B = 6
    A = 9
    AB = 54 and the answer to the question is YES

    B = 12
    A = 18
    AB = (18)(12) and the answer to the question is YES
    Combined, SUFFICIENT

    Final Answer: C

    GMAT assassins aren't born, they're made,
    Rich

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    Mathsbuddy Master | Next Rank: 500 Posts Default Avatar
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    Post Thu Nov 20, 2014 5:28 am
    If ab is a multiple of 18, then 18 is a factor of ab.
    To be such, then ab = 2 x 3 x 3 x n, where n is a non-negative integer

    Using Statement 1 we know a = 3b/2, which does not necessarily satisfy ab = 3b^2/2 = 2 x 3 x 3 x n
    In fact, the factor of 3 is guaranteed only once. NOT SUFFICIENT

    Using Statement 2 we know b = 3c/2, which does not necessarily satisfy ab = 3ac/2 = 2 x 3 x 3 x n
    In fact, the factor of 3 is guaranteed only once. NOT SUFFICIENT

    Combined: (2) ab = 3b/2 x 3c/2 = 9bc/4. Three is 2 of the factors here. In addition, the fact that a = 3b/2 means that for a to be an integer, then 3b must be a multiple of 2. Hence factors 2, 3 and 3 are guaranteed. This suggests: SUFFICIENT

    However, do not ignore what happens if a and/or b equal zero (because zero is a multiple of all numbers). Quick test: if a and or b = 0, then ab = 0. Also SUFFICIENT.

    ANSWER = C

    Post Sat Nov 22, 2014 7:13 am
    akhilsuhag wrote:
    If a, b and c are integers, is ab a multiple of 18?

    (1) 2a = 3b
    (2) 2b = 3c
    Target question: Is ab a multiple of 18?

    Given: a, b and c are INTEGERS

    Statement 1: 2a = 3b
    This statement doesn't FEEL sufficient, so I'm going to test some values.

    Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.beatthegmat.com/mba/2013/10/25/data-sufficiency-when-to-plug-in-values

    There are several values of a and b that satisfy this condition. Here are two:
    Case a: a = 3 and b = 2, in which case ab = 6, and 6 is NOT a multiple of 18
    Case b: a = 9 and b = 6, in which case ab = 54, and 54 IS a multiple of 18
    Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

    Statement 2: 2b = 3c
    There's no information about the variable a, so a can have ANY value. So, this statement SEEMS/FEELS insufficient. Let's test some values.
    There are several values of a, b and c that satisfy this condition (keeping in mind that variable a can have ANY value). Here are two possible cases:
    Case a: a = 1, b = 3 and c = 2, in which case ab = 3, and 3 is NOT a multiple of 18
    Case b: a = 6, b = 3 and c = 2, in which case ab = 18, and 18 IS a multiple of 18
    Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

    Statements 1 and 2 combined
    Statement 1 tells us that 2a = 3b
    Divide both sides by 2 to get: a = 3b/2
    Since a is an INTEGER, we know that 3b/2 is an INTEGER
    If 3b/2 is an INTEGER, then b must be divisible by 2

    Statement 2 tells us that 2b = 3c
    Divide both sides by 3 to get: c = 2b/3
    Rewrite as c = (2/3)b
    Let's also take the statement 1 equation (2a = 3b) and divide both sides by 3 to get: b = 2a/3
    Now take c = (2/3)b and replace b with 2a/3
    We get: c = (2/3)(2a/3)
    Simplify to get: c = 4a/9
    Since c is an INTEGER, we know that 4a/9 is an INTEGER
    If 4a/9 is an INTEGER then a must be divisible by 9

    We now know that b must be divisible by 2 and a must be divisible by 9.
    So, we can conclude that ab is divisible by (2)(9)
    In other words, ab is a multiple of 18
    Since we can answer the target question with certainty, the combined statements are SUFFICIENT

    Answer = C

    Cheers,
    Brent

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    Gurpreet singh Senior | Next Rank: 100 Posts Default Avatar
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    Post Sun Jun 19, 2016 10:29 pm
    Having knowledge is one thing but putting it in a language which can be understood by all is another skill.
    I was lost in this question
    Brilliant explanation Brent.
    Regards


    Brent@GMATPrepNow wrote:
    akhilsuhag wrote:
    If a, b and c are integers, is ab a multiple of 18?

    (1) 2a = 3b
    (2) 2b = 3c
    Target question: Is ab a multiple of 18?

    Given: a, b and c are INTEGERS

    Statement 1: 2a = 3b
    This statement doesn't FEEL sufficient, so I'm going to test some values.

    Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.beatthegmat.com/mba/2013/10/25/data-sufficiency-when-to-plug-in-values

    There are several values of a and b that satisfy this condition. Here are two:
    Case a: a = 3 and b = 2, in which case ab = 6, and 6 is NOT a multiple of 18
    Case b: a = 9 and b = 6, in which case ab = 54, and 54 IS a multiple of 18
    Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

    Statement 2: 2b = 3c
    There's no information about the variable a, so a can have ANY value. So, this statement SEEMS/FEELS insufficient. Let's test some values.
    There are several values of a, b and c that satisfy this condition (keeping in mind that variable a can have ANY value). Here are two possible cases:
    Case a: a = 1, b = 3 and c = 2, in which case ab = 3, and 3 is NOT a multiple of 18
    Case b: a = 6, b = 3 and c = 2, in which case ab = 18, and 18 IS a multiple of 18
    Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

    Statements 1 and 2 combined
    Statement 1 tells us that 2a = 3b
    Divide both sides by 2 to get: a = 3b/2
    Since a is an INTEGER, we know that 3b/2 is an INTEGER
    If 3b/2 is an INTEGER, then b must be divisible by 2

    Statement 2 tells us that 2b = 3c
    Divide both sides by 3 to get: c = 2b/3
    Rewrite as c = (2/3)b
    Let's also take the statement 1 equation (2a = 3b) and divide both sides by 3 to get: b = 2a/3
    Now take c = (2/3)b and replace b with 2a/3
    We get: c = (2/3)(2a/3)
    Simplify to get: c = 4a/9
    Since c is an INTEGER, we know that 4a/9 is an INTEGER
    If 4a/9 is an INTEGER then a must be divisible by 9

    We now know that b must be divisible by 2 and a must be divisible by 9.
    So, we can conclude that ab is divisible by (2)(9)
    In other words, ab is a multiple of 18
    Since we can answer the target question with certainty, the combined statements are SUFFICIENT

    Answer = C

    Cheers,
    Brent

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