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question from Gmatprep test1

This topic has 3 member replies
banona Senior | Next Rank: 100 Posts Default Avatar
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question from Gmatprep test1

Post Sat Mar 03, 2007 11:22 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Can somebody tell me what the most efficient way to solve this kind of DS GMAT questions ;

    the question says , if the remainder of the division of (n-1)*(n+1) by 24 is r, what's the value of r ?
    1- n is not divisible by 2
    2- n is not divisible by3

    I appologize if the question rose somewhere else here in this forum while,butI didn't see it.

    Thanks

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    jayhawk2001 Community Manager
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    Post Sun Mar 04, 2007 6:36 pm
    Not sure if this is the most efficient way but here goes...

    1 - insufficient. Take n = 1 and n = 3 for example.
    They yield remainders of 0 and 8 resp. No unique value. So rule out 1.

    2 - insufficient. Take n = 1 and n = 2.
    They yield remainders of 0 and 3 resp. No unique value. So, rule out 2

    A number that is not divisible by 2 AND not divisible by 3 can be
    written as n = 6x + 1 or n = 6x - 1

    For n = 6x+1
    (n-1)*(n+1) / 24
    = 6x * (6x+2) / 24
    = 12x * (x+1) / 24

    We know that x*(x+1) will be even or 0, so the above equation will
    always yield a remainder of 0.

    Similarly for n=6x-1
    (n-1)*(n+1) / 24
    = (6x - 2)*6x
    = 12x*(x-1) / 24

    Again, x*x(-1) will be even or 0 and hence remainder r will be 0.

    So, C looks like the correct answer.

    On the actual GMAT, in the interest of time, it might be worthwhile
    trying n = 1, 3, 5, 7, 9 etc. We can see that the remainder is 8 when n
    is not divisible by 3 and remainder is 0 when n is divisible by 3 (for
    all values n > 0). Using this pattern, I guess we can settle on C.

    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Mon Mar 05, 2007 5:50 am
    thank you,
    I would go for picking numbers on the real GMAT too.

    rajesh_ctm Master | Next Rank: 500 Posts Default Avatar
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    Post Tue Apr 03, 2007 6:44 pm
    jayhawk2001, that is a wonderful explanation! Thank you. Gives us a valuable technique, we can use for similar questions.

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