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    This is the question and how far I've gotten;

    With k being a natural number such that  p =  2^k - 1 is a prime number (i.e 3, 7, 31) Define and list all positive integers which divide N;

     N = 2^{k-1}p


    So the positive integers that divide N must divide  2^{k-1} or p. Now p is prime, so the only thing that can divide it is itself and 1. The only things that can divide  2^{k-1} are multiples of two.

    So divisors of  2^{k-1} are  2^{k-1}, 2^{k-2}, 2^{k-3}, 2^{k-4} ...  2^{k-(k-2)}, 2^{k-(k-1)}, 2^{k-(k-0)}

    Thus this tells me the positive integers which divide N (I'm assuming there's an infinite number of possibilities because I'm not sure if there is or isn't) are;

     2^{k} -1, 2^{k-1}, 2^{k-2}, 2^{k-3}, 2^{k-4} ...  2^{k-(k-2)}, 2^{k-(k-1)}, 2^{k-(k-0)}


    Now it asks prove that the sum of all its divisors (including 1 but not N) is equal to N , this bit I can't do.

    It's saying


     2^{k} -1 + 2^{k-1} + 2^{k-2} + 2^{k-3} + 2^{k-4} + ... + 2^{k-(k-2)} + 2^{k-(k-1)} + 2^{k-(k-0)} =  (2^{k-1})(2^k -1)

    Basically simplified (since  2^{k-(k-0)} = 1 )

     2^{k} + 2^{k-1} + 2^{k-2} + 2^{k-3} + 2^{k-4} + ... + 2^{2} + 2^{1} =  (2^{k-1})(2^k -1)

    Now if I take a factor of  2^{k-1} out I get

     (2^{k-1})(2^{1} + 2^{0} + 2^{-1} + 2^{-2} + 2^{-3} + ... + 2^{3-k} + 2^{2-k}) =  (2^{k-1})(2^k -1)

    Which basically means I have to show that

     2^{1} + 2^{0} + 2^{-1} + 2^{-2} + 2^{-3} + ... + 2^{3-k} + 2^{2-k} =  2^k -1


    Sorry for the long explanation and I hope you understand. Let me know if I've gone wrong somewhere. I can't do the bolded bit, how exactly do I prove that last statement? Any help/hints appreciated.
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    \sum_{i=0}^{n-1} {2^i} = 2^{n}-1

    It's probably (slightly) easier to try to show that the sum of all the factors, including N, sum to 2N.
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    (Original post by Hopple)
    \sum_{i=0}^{n-1} {2^i} = 2^{n}-1

    It's probably (slightly) easier to try to show that the sum of all the factors, including N, sum to 2N.
    Hey thanks. Is that just a general equation I'm meant to know or something? And I'm not sure if I can do what you proposed as the question says without including N as a divisor.

    I'm not exactly sure how I can use your useful piece of information. I know that

     2^{k} -1 + 2^{k-1} + 2^{k-2} + 2^{k-3} + 2^{k-4} + ... + 2^{k-(k-2)} + 2^{k-(k-1)} + 2^{k-(k-0)} =  (2^{k-1})(2^k -1)

    And using what you've said;

     2^{k-1} + 2^{k-2} + 2^{k-3} + 2^{k-4} + ... + 2^{k-(k-2)} + 2^{k-(k-1)} + 2^{k-(k-0)} =  2^{k}-1

    But then I just get

     2^{k} -1 + 2^{k} -1 = 2^{k+1} - 2 = 2(2^{k} -1)

    What have I done wrong? :|
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    Let's take a concrete example: k=3, so p = 2^k - 1 = 7. Then N = 4.7 = 28.

    If you list all the factors of 28, you'll see that you've missed (almost) a whole category in your earlier analysis.
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    (Original post by Anti Elephant Mine)

    Thus this tells me the positive integers which divide N (I'm assuming there's an infinite number of possibilities because I'm not sure if there is or isn't) are;

     2^{k} -1, 2^{k-1}, 2^{k-2}, 2^{k-3}, 2^{k-4} ...  2^{k-(k-2)}, 2^{k-(k-1)}, 2^{k-(k-0)}

    i think that you have overlooked factors made by combining the powers of 2 with the prime number p...

    the bear

    dfranklin beat me to it
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    (Original post by Anti Elephant Mine)
    Hey thanks. Is that just a general equation I'm meant to know or something? And I'm not sure if I can do what you proposed as the question says without including N as a divisor.
    It'd be handy to know, yes I'm sure you can do it that way, as long as you subtract off N at the end.

    Does 2p divide N?
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    Ah my mistake, I've done it now, can't believe I didn't notice that. Thanks for the help, rep for all of you!
 
 
 
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