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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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Charlie-Boo

Posts: 1,585
Registered: 2/27/06
Re: Reciprocals of integers summing to 1
Posted: Nov 17, 2012 2:44 PM
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On Nov 17, 11:52 am, luir...@yahoo.com wrote:
> El viernes, 16 de noviembre de 2012 00:39:24 UTC-4:30, Charlie-Boo  escribió:
>

> > For each n, what are the solutions in positive integers (or in
> > integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?

>
> There are infinitely many different solutions. But I am not sure that it is valid for all n.
> Take k primes and do the following sum: 1/p1 + 1/p2 + 1/p3 + ...+ 1/pk = S < 1
> Now make Q = 1 - S.
> By the theorem of the Egyptian fractions, Q always can be decomposed as:
> Q = 1/x1 + 1/x2 + 1/x3 +....+ 1/xj.
> I am not sure that it is possible, ever, that k+j = n.
> Ludovicus


There are an infinite number of sets of positive integers whose
reciprocals sum to 1:

N = # of terms. Term # M where M = 1 to N = ( If M < N then (M+1)! /
M Else M ! )

C-B



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