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If y = 2z, and y and z are both positive integers

This topic has 5 expert replies and 0 member replies

Top Member

If y = 2z, and y and z are both positive integers

Post Sat May 20, 2017 10:19 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

    A) x = w

    B) x > w

    C) x/y is an integer

    D) w/z is an integer

    E) x/z is an integer

    OAC

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    Post Sun May 21, 2017 2:33 am
    rsarashi wrote:
    x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

    A) x = w

    B) x > w

    C) x/y is an integer

    D) w/z is an integer

    E) x/z is an integer
    w, x, y and z are all integers.
    Test the SMALLEST POSSIBLE CASE.
    Let z=1.

    Since z=1, it must be possible that w/z and x/z are integers.
    Eliminate D and E.

    Since z=1, w is the sum of 1 consecutive integer, implying that w can be ANY INTEGER.
    Thus, it must be possible that x=w or that x>w.
    Eliminate A and B.

    The correct answer is C.

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    Thanked by: rsarashi
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    Post Sun May 21, 2017 2:55 pm
    This is testing a very particular rule in consecutive integers: if you have an EVEN number of terms, the sum will never be divisible by the number of terms. If you have an ODD number of terms, the sum *will* be divisible by the number of terms.

    This is because, by definition, SUM = (AVERAGE)*(NUMBER OF TERMS), and in a consecutive set, AVERAGE = MEDIAN.

    If you have an EVEN number of terms, the median is a non-integer:
    [4, 5, 6, 7]
    median = 5.5
    sum = (5.5)(4) = 22

    If we divide the sum by the number of terms, we get a non-integer: the median. Thus, the sum is not divisible by the number of terms.

    If you have an ODD number of terms, the median is an integer:
    [4, 5, 6, 7, 8]
    median = 6
    sum = (6)(5) = 30

    If we divide the sum by the number of terms, we get an integer: the median. Thus, the sum is divisible by the number of terms.

    So in this problem, if y = 2z, then y must be even: there are an EVEN number of terms in the set. The sum cannot possibly be divisible by the number of terms in an even set, so x cannot be divisible by y.

    The answer is C.

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    Thanked by: rsarashi
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    Post Sun May 21, 2017 2:56 pm
    Also, please always POST YOUR SOURCES. It is a copyright violation not to do so.

    This question comes from Manhattan Prep CATs.

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    Manhattan Prep GMAT & GRE instructor
    EdM in Mind, Brain, and Education
    Harvard Graduate School of Education


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    GMAT/MBA Expert

    Post Wed May 24, 2017 5:51 pm
    It seems to me that the question really tests the ability to either quickly pick smart numbers that rule out four of the answers and/or the ability to quickly whip up an equation that will show the impossibility of one of the answers.

    If I take the first approach, some easy sets are:

    {1} vs {0,1} (eliminate A)

    {1} vs {1,2} (eliminate B)

    {1, 2, 3} (eliminate D)

    {1, 2, 3, 4} vs {1, 2} (eliminate E)

    If I take the second approach, I know that

    x = n + (n + 1) + ... + (n + 2z - 1)

    x = 2z * n + (1 + 2 + ... + 2z - 1)

    x = 2zn + (2z - 1)*2z*(1/2) = 2zn + (2z - 1)*z

    Since y = 2z, we know that x/y = (2nz + z * (2z - 1)) / 2z

    Simplified, that becomes 2zn/2z + z*(2z - 1)/2z. The first term is an integer, but the second is not, so we'll never get an integer here for any integer z.

    It's a great Q, with (at least) two clever and generally relevant paths to a solution - don't sell it short! Smile

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    GMAT/MBA Expert

    Post Wed May 24, 2017 5:53 pm
    One other trick that just occurred to me:

    If you're trying to backsolve from the answers, note that C implies E. (If x is divisible by 2z, then x must be divisible by z.)

    With that in mind, you don't even need to bother with E!

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