Sum of ReciprocalsDate: 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/ |
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