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Topic: Reciprocals of integers summing to 1
Replies: 21   Last Post: Nov 23, 2012 8:44 PM

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billh04

Posts: 7
Registered: 1/27/11
Re: Reciprocals of integers summing to 1
Posted: Nov 16, 2012 10:07 AM
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On Nov 16, 8:57 am, gus gassmann <g...@nospam.com> wrote:
> On 16/11/2012 10:10 AM, redm...@siu.edu wrote:
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> > On Thursday, November 15, 2012 11:39:57 PM UTC-6, Charlie-Boo wrote:
> >> On Nov 16, 12:21 am, William Elliot <ma...@panix.com> wrote:
>
> >>> On Thu, 15 Nov 2012, Charlie-Boo wrote:
>
> >>>> For each n, what are the solutions in positive integers (or in
>
> >>>> integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?
>
> >>> x1 = x2 =..= x_n = n
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> >> But there is also 1/2 + 1/3 + 1/6 = 1.  I am asking for all solutions.
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> >> C-B
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> > Here's another
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> > 1/2 + 1/3 + 1/7 + 1/42 = 1.
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> > don
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> What about 1/1 + 1/1 + 1/(-1)?
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> Seems to me that the original question was not very well stated.
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> Maybe you want x_i =/= x_j for i =/= j? That is still quite a bit more
> work than I am willing to invest.
>
> On the other hand,
>
> M = {n | exist 0 < x_1 < x_2 < ... < x_n with sum 1/x_i = 1}
>
> would be interesting. I offer k=2 as one positive integer that is not
> contained in M. Are there others? Is there a nice characterization?


Since 1 = 1/2 + 1/3 + 1/6, then 1/2 = 1/4 + 1/6 + 1/12. Hence 1 = 1/2
+ 1/2 + 1/4 + 1/6 + 1/12. We probably want to rule out cases like this
also. More generally, 1/n = 1/2n + 1/3n + 1/6n and then 1 = 1/n + 1/n
+ 1/n + ... + 1/2n + 1/3n + 1/6n.



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