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Problem involving prime factors

This topic has 6 member replies
wlvoh Newbie | Next Rank: 10 Posts Default Avatar
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Problem involving prime factors

Post Sat Feb 10, 2007 12:23 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    The following is a doozy of a PS problem from a practice test on the Official GMAT Prep Software from GMAC. If anyone can help explain how to solve this one, I'd greatly appreciate it.

    "For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is:

    A.) between 2 and 10

    B.) between 10 and 20

    C.) between 20 and 30

    D.) between 30 and 40

    E.) greater than 40

    The official answer is: E

    (highlight to see the OA)

    Thanks.

    wlvoh

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    aim-wsc Legendary Member
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    Post Sat Feb 10, 2007 1:12 pm
    difficult indeed.
    or maybe i lost number properties basics. Confused

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    jayhawk2001 Community Manager
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    Post Sat Feb 10, 2007 1:20 pm
    I posted this on an earlier thread. Hope this helps.

    h(100) + 1
    = 2 * 4 * ... * 100 + 1
    = 2*50! + 1

    2*50! can be expressed as a multiple of X (2 <= X <= 50) and so
    (2*50! + 1) cannot be a multiple of of that same number X.

    Since p is a factor of h(100) +1, p should be > X and so option
    "E" looks like the correct answer.

    aim-wsc Legendary Member
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    Post Sat Feb 10, 2007 1:36 pm
    yeah thanks for the quick reply.
    i knew that answer is E.

    but i started thinking of the value of p Rolling Eyes
    yeah! now thats really a wrong approach for GMAT i know... & i still fall in the trap. Sad
    it doesnt expect the value but the answer choice.... same applies for DS.... Confused

    thanks once again.
    the value of p by the way would be far much higher than 40 for sure.

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    wlvoh Newbie | Next Rank: 10 Posts Default Avatar
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    Post Mon Feb 12, 2007 5:16 am
    jayhawk2001 wrote:
    I posted this on an earlier thread. Hope this helps.

    h(100) + 1
    = 2 * 4 * ... * 100 + 1
    = 2*50! + 1

    2*50! can be expressed as a multiple of X (2 <= X <= 50) and so
    (2*50! + 1) cannot be a multiple of of that same number X.

    Since p is a factor of h(100) +1, p should be > X and so option
    "E" looks like the correct answer.
    Thanks for the help. But, in your explanation, one thing isn't quite right. The product 2*4*6*...*100 cannot be expressed as 2*50! It isn't a sum, so you can't simply factor out a single 2. Instead, you must factor out 2 from each element in the sum. So, actually, 2*4*6*...*100 is equal to:

    (2^50) * 50!

    I'm not sure if this makes any difference in the reasoning. Any thoughts?

    Thanks.

    wlvoh

    jayhawk2001 Community Manager
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    Post Tue Feb 13, 2007 10:36 pm
    wlvoh, Yes, that is correct. The value is (2^50 * 50!) not 2*50!.

    In anycase, the element to focus on is 50! and the rest of the logic should
    follow as before.

    Thanks.

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