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Math Help - problem involving sum of divisors

  1. #1
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    Post problem involving sum of divisors

    studying for finals and enountered this problem on a practice test.
    If n is a natural number such that n = 2^k(2^(k+1) - 1) with 2^(k+1) -1 prime then 2n = sum of the divisors.
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  2. #2
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    Quote Originally Posted by ChrisBickle View Post
    studying for finals and enountered this problem on a practice test.
    If n is a natural number such that n = 2^k(2^(k+1) - 1) with 2^(k+1) -1 prime then 2n = sum of the divisors.

    What are the divisors of n=2^k\left(2^{k+1}-1\right)? Let's write them down:

    ** 1,2,2^2,\ldots,2^k

    ** 2^{k+1}-1

    ** 2^{k+2}-2,2^{k+3}-2^2,\ldots,2^{2k+1}-2^k

    Well, let's sum up the bastards above:

    \sum\limits_{i=0}^k2^i+\sum\limits_{j=0}^k\left(2^  {k+1+j}-2^k\right) =2^{k+1}\sum\limits_{j=0}^k2^j=2^{k+1}\,\frac{2^{k  +1}-1}{2-1}=2^{k+1}(2^{k+1}-1) ...

    Tonio
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  3. #3
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    awesome..everytime you see a solution it looks so easy...im convinced math is to make smart people feel stupid
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  4. #4
    Senior Member Shanks's Avatar
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    that is all the even perfect number of the form 2^{n-1}(2^n-1) \text{provide that} 2^n-1 \text{is prime}.
    Last edited by Shanks; December 13th 2009 at 08:41 PM.
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    Quote Originally Posted by Shanks View Post
    that is all the even perfect number of the form 2^n(2^n-1) \text{provide that} 2^n-1 \text{is prime}.

    Perhaps you meant to say: an even natural number is perfect iff it is of the form 2^n\left(2^{n+1}-1\right) AND 2^{n+1}-1 is prime .

    Tonio
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  6. #6
    Senior Member Shanks's Avatar
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    oooops,,,sorry, type error...
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