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Factors

This topic has 2 expert replies and 1 member reply
binaras Senior | Next Rank: 100 Posts Default Avatar
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Posted:
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Factors

Post Sat Mar 21, 2015 11:05 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Hi

    Need assistance in figuring out why the answer to the DS question (below) is option (D) Each statement alone is sufficient

    "The positive integers x, y & z are such that x is a factor of y & y is a factor of z.
    Is z even?

    1) xz is even
    2) y is even

    Thanks

    Thanks

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    Post Sat Mar 21, 2015 11:35 am
    Quote:
    The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?

    (1) xz is even
    (2) y is even.

    Here's a more formal approach.

    Target question: Is z even?

    Given: x is a factor of y, and y is a factor of z.

    There's a nice rule that says, "If D is a factor (divisor) of N, then N = kD for some integer k"
    So, if x is a factor of y, then y = kx for some integer k.
    Also, if y is a factor of z, then z = jy for some integer j

    Statement 1: xz is even
    This sets up two possible cases (x is even or z is even). We'll examine both:
    case a: x is even.
    If x is even, then kx is even, which means y is even (since y=kx).
    If y is even, then jy is even, which means z is even (since z=jy).
    case b: z is even
    Since both possible cases yield the same answer to the target question, statement 1 is SUFFICIENT

    Statement 2: y is even
    If y is even, then jy is even, which means z is even (since z=jy).
    Since we can answer the target question with certainty, statement 2 is SUFFICIENT

    Answer = D

    Cheers,
    Brent

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    Post Sat Mar 21, 2015 12:21 pm
    Quote:
    The positive integers x, y and z are such that x is a factor of y and y is a factor of z. Is z even.

    1) xz is even
    2) y is even.
    Let's reverse the language:

    z is a multiple of y, and y is a multiple of x.
    Thus, z is a multiple of x.

    Statement 1: xz is even.
    If x = even, then z must be even since it's a multiple of an even value.
    If x = odd, then z must be even in order for xz to be even.
    Since in either case z is even, sufficient.

    Statement 2: y is even.
    If z is a multiple of an even number, then z too must be even.
    Sufficient.

    The correct answer is D.

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    binaras Senior | Next Rank: 100 Posts Default Avatar
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    Post Sat Mar 21, 2015 7:37 pm
    Thanks for the quick response. Appreciate it.

    Binara

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