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Positive/Negative: If x, y, and z are positive integers, is

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Positive/Negative: If x, y, and z are positive integers, is Post Sun Apr 13, 2008 3:16 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If x, y, and z are positive integers, is x-y odd ?

    (1) x = z^2
    (2) y = (z - 1)^2

    Thanks.



    Last edited by II on Mon May 05, 2008 1:41 am; edited 1 time in total

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    Post Sun Apr 13, 2008 5:14 am
    II wrote:
    If x, y, and z are positive integeres, is x-y odd ?

    (1) x = z^2
    (2) y = (z - 1)^2

    Thanks.
    Interesting question II.

    This is the very essence of Data Sufficiency. Within 10-20 seconds you should be able to eliminte A / B and D as possible answers.

    Why?

    Since the relationship being tested requires us to look for the outcome of X - Y, it becomes obvious that Statement 1 and Statement 2 by themselves are not going to pull it off (they only have X and Z, and Y and Z respectively.

    If on test-da your time is running out (or even if it is not), something like this is really going to help you.

    Now to decide between C and E:

    Together:
    X - Y = Z^2 - (Z-1)^2.................(This is the classic difference of squares rule)

    Therfore, X- Y = ( Z - (Z-1)) ( Z+(Z+1))
    which is then = (Z - Z +1) (2Z+1)
    which simplifies as = (1) (2Z +1) = 2Z +1

    The odd number kingdom is defined as 2(integer) + 1 OR 2(Integer) - 1.

    Since Z is an integer, 2Z+1 MUST be odd. This answers the question with an ALWAYS YES, and hence sufficiency.

    C is the answer.

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    Post Sun Apr 13, 2008 8:19 am
    I will go with C, too.

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    Post Wed Jun 17, 2015 12:14 am
    It should be 2z-1

    Answer is C

    Post Wed Jun 17, 2015 7:33 am
    Quote:
    If x, y, and z are positive integers, is x-y odd?

    1) x = z²
    2) y = (z-1)²
    Here's an algebraic approach:

    Target question: Is x-y odd?

    Given: x, y, and z are positive integers

    Statement 1: x = z²
    There's no information about y, so there's no way to determine whether or not x-y is odd.
    Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

    Statement 2: y = (z-1)²
    There's no information about x, so there's no way to determine whether or not x-y is odd.
    Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

    Statements 1 and 2 combined
    Statement 1: x = z²
    Statement 2: y = (z-1)²
    Subtract equations to get: x-y = z² - (z-1)²
    Expand to get: x-y = z² - [z² - 2z + 1]
    Simplify to get: x-y = 2z - 1
    Since z is a positive integer, we know that 2z is EVEN, which means 2z-1 is ODD.
    If 2z-1 is ODD, we can conclude that x-y is definitely ODD
    Since we can answer the target question with certainty, the combined statements are SUFFICIENT

    Answer = C

    Cheers,
    Brent

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    Post Wed Jun 17, 2015 9:41 am
    II wrote:
    If x, y, and z are positive integers, is x-y odd ?

    (1) x = z^2
    (2) y = (z - 1)^2
    Clearly, neither statement alone is sufficient.

    Statements combined:
    Since x and y are both positive integers in terms of positive integer z -- and neither statement involves division -- we can determine whether x-y must be ODD by testing two cases:
    z = EVEN and z = ODD.

    Case 1: z=2
    Here, x = 2² = 4 and y = (2-1)² = 1.
    In this case, x-y = 4-1 = 3, which is ODD.

    Case 2: z=3
    Here, x = 3² = 9 and y = (3-1)² = 4.
    In this case, x-y = 9-4 = 5, which is ODD.

    Since x-y is ODD in both cases, the answer to the question stem is YES.
    Thus, the two statements combined are SUFFICIENT.

    The correct answer is C.

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    Post Fri Jun 26, 2015 2:54 am
    Answer is C
    2z-1 is always ODD

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    Post Mon Jun 29, 2015 4:29 pm
    Another approach:

    z² - (z-1)² =
    z² - (z² - 2z + 1) =
    2z - 1

    For any integer z, 2z is even, so 2z - 1 is odd. We're set!

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