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Am I missing something?

This topic has 1 expert reply and 4 member replies

Am I missing something?

Post Sun Jan 28, 2007 6:13 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the remainder when the positive integer n is divided by the positive integer k, where k>1?

    (1) n = (k+1)^3
    (2) k = 5

    GMAT prep says (1) is sufficent to answer the question.

    I wonder how????

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    jayhawk2001 Community Manager
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    Post Sun Jan 28, 2007 7:35 pm
    N = (k+1)^3 = k^3 + 3k^2 + 3k + 1

    So, N/K always yields the remainder 1

    Hence (A)

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Sun Jan 28, 2007 7:50 pm
    jayhawk2001 wrote:
    N = (k+1)^3 = k^3 + 3k^2 + 3k + 1

    So, N/K always yields the remainder 1

    Hence (A)
    yep, i got the same approach as jayhawk.

    aim-wsc Legendary Member
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    Post Sun Jan 28, 2007 9:08 pm
    good analytical approach FULL MARKS 8)

    sigh
    i hv forgotten all .. i used other approach like any number that is just greater than one than the other number is not divisible by it and the remainder is always one.
    then we have condition A

    good

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    rnschmidt Newbie | Next Rank: 10 Posts Default Avatar
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    Post Mon Jan 29, 2007 12:33 pm
    will you please explain the math for your solution jay?

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    Stacey Koprince GMAT Instructor
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    Post Wed Jan 31, 2007 5:25 pm
    To distribute:
    (k+1)^3 = (k+1)(k+1)(k+1)
    let's take the first two sets of parentheses first:
    (k+1)(k+1) = k^2 + 2k + 1
    now, multiple in our third (k+1):
    (k+1)(k^2 + 2k + 1) =
    k^3 + 2k^2 + k + k^2 + 2k + 1) =
    k^3 + 3k^2 + 3k + 1 = n

    They ask about n/k, so:
    (k^3 + 3k^2 + 3k + 1)
    k

    = k^2 + 3k + 3 + 1/k. In this algebraic representation of division, the numerator of the fraction is the remainder. So the remainder is always 1.

    You can also try some real numbers and see if there's a pattern. We know they're both positive integers and k>1.
    Try k = 2. Then n = (2+1)^3 = 27. 27/2 = 13 R1
    Try k = 3. Then n = (3+1)^3 = 64. 64/3 = 21 R1
    So the first thing I notice is that, when k is even, n is odd, and vice versa. The second thing I notice is that I've gotten R1 for both. Interesting.
    k = 4. Then n = (4+1)^3 = 125. 125/4 = 31 R1. R1 again.
    k = 5. Then n = (5+1)^3 = 216. 216/5 = 43 R1. R1 yet again!
    At this point I'm feeling pretty comfortable that there's a pattern. I'm going to get R1 no matter what I try for k.

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