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magical cook GMAT Destroyer! Default Avatar
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DS question Post Wed Feb 14, 2007 9:57 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Hi,

    I am hoping that anyone could explaing this question in an easy way...

    Dose the integer K has a factor p such that 1<p<K?

    1) K>4!
    2) 13!+2<k<13!+13

    The answer is B but I only was able to understand A k>24 would not be sufficient to answer the question but I have no idea as to 2).

    Thanks!
    R.

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    amitamit2020 Just gettin' started! Default Avatar
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    Post Wed Feb 14, 2007 12:02 pm
    Good question.

    The question askes - if we can find an number which is lesser than k and can devide k without remainder.

    Let us examine statement (2), it says that k belongs to [13!+2, 13!+13].
    take k = 13! + 2 -- we are now to find a number (les than k) which can devide this without remainder clearly p = 2 can do that.

    go to k = 13! + 3 -- it is clear that p = 3 can answer the question.

    ... similarly till 13! + 13 ... p = 4,5,6,7,8,9,10,11,12 and 13 can devide k without remainder and obviously it will be lesser than 13! + any positive integer.

    hope it helps...

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    Stacey Koprince GMAT Instructor
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    Post Wed Feb 14, 2007 9:13 pm
    Amit is right about this process of examining statement 2. I just want to take one step back and look at the question itself b/c there's a very easy way to rephrase it. This is basically asking us whether k is prime.

    A prime number has exactly two factors: itself and 1. By definition, it cannot have a factor p that is between itself and 1. A non-prime number, on the other hand, that is greater than 1 will have a factor between itself and 1 (again, by definition).

    So I can answer this question if I know whether k is prime.

    (1) just tells me that k>24. There are prime numbers and non-prime numbers greater than 24, so that's not useful.

    (2) for any sum, if the two numbers in that sum have a common factor, that factor will also be a factor of the sum. E.g., 2 + 4 = 6. 2 is a factor of 2 and 2 is a factor of 4. Therefore, 2 will also be a factor of 6.

    Now, to follow amit's logic:
    13! + 2 = 13*12*11*10*9*8*7*6*5*4*3*2 + 2. 2 is a factor of both of the numbers, so the sum will also have 2 as a factor. Therefore, the first possibility is not prime. Follow this logic all the way through 13! + 13; you will always have at least one factor, so that number is not prime. All of the possibilities given by statement 2 are not prime so I can answer the question: No, the integer k does not have a factor p such that 1<p<k.

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