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Sum of Reciprocals

Date: 08/05/2008 at 23:28:56
From: Cynthia
Subject: Sum of reciprocals

The 3 positive integers 2, 3, and 6 have reciprocals whose sum equals
1 since 1/2 + 1/3 + 1/6 = 1.

What are all the sets of 7 different positive integers (a, b, c, d,
e, f, g) so that the sum of their reciprocals equals 1?

I can't find even one set which equals 1.



Date: 08/11/2008 at 23:05:31
From: Doctor Greenie
Subject: Re: Sum of reciprocals

Hi, Cynthia --

One tool we can use is the fact that

  1/2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2

In other words, to produce two fractions that add to the starting
fraction (with numerator 1 and even denominator), first triple the
denominator, then take half of that new denominator, and add the two
resulting fractions.  Another example is 

  1/10 = 1/30 + 1/15 = 1/30 + 2/30 = 3/30 = 1/10

Hold onto that that idea for a moment as we will use it to help find
solutions to your problem.

For the easiest solution I can think of, we can use the familiar 
series

  1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + ...

The sum of any finite number of terms of this sequence is equal to 1 
minus the last term we added:

  1/2 + 1/4 = 3/4 = 1 - 1/4
  1/2 + 1/4 + 1/8 = 7/8 = 1 - 1/8

and so on.

Let's use this to find our first "simple" solution to your problem.  
Our first five numbers are 2, 4, 8, 16, and 32; the sum of the 
reciprocals of these five is 31/32.  We need two more unit fractions 
whose sum is 1/32.

We can do that using the tool noted earlier:

  1/32 = 1/96 + 1/48

And so our first solution to your problem is using the seven numbers

  2, 4, 8, 16, 32, 48, 96

A general strategy for finding solutions is to use a "greedy" 
algorithm, which means each new unit fraction we add is the largest 
possible one we can add and keep the sum under 1.  Here is one
relatively simple solution that uses this method, except that we start
with 1/3 instead of 1/2 as our first unit fraction:

  1/3 + 1/4 = 7/12
  7/12 + 1/5 = 47/60
  47/60 + 1/6 = 57/60
  57/60 + 1/30  59/60

We now have five unit fractions whose sum is 59/60; we need two unit 
fractions whose sum is 1/60.  Again we use the same tool as before to 
find the last two unit fractions:

  1/2 = 1/6 + 1/3

so

  1/60 = 1/180 + 1/90

And we have another solution to the problem:

  3, 4, 5, 6, 30, 90, 180

And here is a really ugly one that we can get, using the greedy rule 
at its most greedy....

  1/2 + 1/3 = 5/6

The largest unit fraction we can add and stay under a sum of 1 is
1/7:

  5/6 + 1/7 = 41/42

The largest unit fraction we can add and stay under a sum of 1 is 
1/43:

  41/42 + 1/43 = 1805/1806

The largest unit fraction we can add and stay under a sum of 1 is 
1/1807; however, at this point, in order to use our standard tool to 
get the last two unit fractions, we have to have a denominator here 
which is an even number--so our next fraction is 1/1808:

  1805/1806 + 1/1808 = 3265246/3265248 = 1632623/1632624

Our last two fractions have to have a sum of 1/1632624, so

  1/2 = 1/6 + 1/3

  1/1632624 = 1/4897872 + 1/2448936

And our third--and very ugly--solution to the problem is with the 
seven numbers

  2, 3, 7, 43, 1808, 2448936, and 4897872

I hope this is of some help....

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Algorithms
College Discrete Math
High School Discrete Mathematics

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