Positive Unit Fractions
Date: 10/02/2002 at 17:35:35 From: Jessica Subject: Positive unit fractions Hi! Here is the problem: Find five different positive unit fractions whose sum is 1. (A unit fraction is a fraction whose numerator is 1. All denominators must also be natural numbers.) Find all possibilities. I initially had no idea where to start with this, so I randomly tried adding different unit fractions together until I finally got five of them to equal 1. I found two sets of five fractions that equal 1. But I don't see any similarities between these solutions so I don't know how I can find a common formula. I noticed that in both sets that I found, two denominators were divisible by 5, and the other three were divisible by 2. I thought this might be something, but when i tried other random numbers with that pattern, it didn't equal 1. When I asked the teacher, she suggested looking at it like 1/a+1/b+1/c+1/d+1/e and putting them all over the common denominator of abcde. Then the object would be to get the numerator to equal the denominator. But when I tried to do it this way, I got stuck with just a lot of letters that I don't know how to solve. If you have any ideas or suggestions, it would be greatly appreciated! Thanks! -Jessica
Date: 10/02/2002 at 22:29:08 From: Doctor Greenie Subject: Re: Positive unit fractions Hi, Jessica - Here is a link to a page in the Dr. Math archives where you can find several suggestions for ways to attack this problem: Unit Fractions Summing to 1 http://mathforum.org/library/drmath/view/56821.html I'm not sure about your instructions to find "all" the possibilities. I haven't tried to work the problem completely, but I think there are a huge number of answers; and I don't know of any way to make sure you have found them all - or, rather, I can imagine that an exhaustive investigation to find all the answers would involve a huge amount of effort. So the rest of my response to your inquiry will be devoted to methods for finding solutions, without worrying about the question of finding ALL solutions. The referenced page in the archives describes three different methods for solving a problem like yours: I. brute force, or the "greedy" algorithm Basically, this method just says to choose the unit fractions one at a time, always choosing the largest one you can without going over the desired total. The arithmetic with this method is often very messy, and the answer will seldom be one of the simpler ones. II. using powers of 2 for the denominators of all but the last two fractions For your example, this method would say to use 1/2, 1/4, and 1/8 as the first 3 fractions; the sum of the last two fractions must then be 1/8; we can easily write this as the sum of two different unit fractions as follows: 1/8=3/24 = 1/24 + 2/24 = 1/12 + 1/24. So using this method we would have the solution 1/2 + 1/4 + 1/8 + 1/12 + 1/24. III. choose a number with many divisors and find five of those divisors whose sum is the number For your example, you might try a common denominator of 60; the proper divisors are 1 2 3 4 5 6 10 12 15 20 30 (a) If we choose the two largest divisors first, we have 30+20=50, so we need 3 more divisors whose sum is 10: 6,3,1; or 5,4,1; or 5,3,2. This gives us 3 solutions: (30+20+6+3+1)/60 = 1/2 + 1/3 + 1/10 + 1/20 + 1/60 (30+20+5+4+1)/60 = 1/2 + 1/3 + 1/12 + 1/15 + 1/60 (30+20+5+3+2)/60 = 1/2 + 1/3 + 1/12 + 1/20 + 1/30 (b) 30+15=45, so we need 3 more divisors whose sum is 15: 10,3,2; or 10,4,1; or 6,5,4. This gives us 3 more solutions: (30+15+10+3+2)/60 = 1/2 + 1/4 + 1/6 + 1/20 + 1/30 (30+15+10+4+1)/60 = 1/2 + 1/4 + 1/6 + 1/15 + 1/60 (30+15+6+5+4)/60 = 1/2 + 1/4 + 1/10 + 1/12 + 1/15 And just looking at this list of divisors of 60, I can see that there will be at least one solution if the two largest fractions are 1/2 and 1/5.... And you can probably find many other solutions by using the same method with another common denominator with many divisors, such as 72, whose divisors are 1 2 3 4 6 8 9 12 18 24 36 IV. and yet another method... Yet another approach to this problem, not described on the reference page in the archives, is essentially a variation of the "greedy" algorithm. With this method, you just choose, at each stage, any fraction that is CLOSE TO the largest one you can still choose without going over the desired total. You use this process to find all but the last two fractions and then solve an equation to find the last two fractions. With your problem, we might start by choosing 1/2 for our first fraction. With the "greedy" algorithm, we would next choose 1/3; but let's choose 1/4 instead. Then let's choose 1/5 for our third fraction. The sum of these fractions is (1/2)+(1/4)+(1/5) = (10+5+4)/20 = 19/20 so the sum of our last two fractions must be 1/20. The easiest way I know of to write 1/20 as the sum of two unit fractions is to say (1/20) = (3/60) = (2/60)+(1/60) = (1/30)+(1/60) or (1/20) = (4/80) = (3/80)+(1/80) = ... OOPS! This one won't work, because 3/80 does not reduce to a unit fraction or (1/20) = (5/100) = (4/100)+1/100) = (1/25)+(1/100) or ... By rewriting 1/20 as other equivalent fractions with even larger denominators, you might possibly find many other solutions using (1/2), (1/4), and (1/5) as the first three fractions. Or, another way to finish this example - which will find you ALL the solutions using (1/2), (1/4), and (1/5) as the first three fractions - is to solve the equation (1/x)+(1/y)=(1/20) for all possible integer values of x and y. This is an example of a diophantine equation, where we solve the equation for one variable in terms of the other and search for all integer solutions of the resulting equation. We have (1/y) = (1/20)-(1/x) = (x-20)/20x and so (taking the reciprocal of each side of the equation) y = (20x)/(x-20) Searching for all values of x which make y an integer, we find x y -------------------- 21 420/1 = 420 22 440/2 = 220 24 480/4 = 120 25 500/5 = 100 28 560/8 = 70 30 600/10 = 60 36 720/16 = 45 40 800/20 = 40 The values of x between 21 and 40 not shown in the table do not produce integer values for y. Also, since the equation we are solving is symmetrical in the two variables, continuing with larger values of x will just find us the solutions we already have. So this table shows us all the solutions to your problem in which the first three denominators are 2, 4, and 5: 1/2 + 1/4 + 1/5 + 1/21 + 1/420 1/2 + 1/4 + 1/5 + 1/22 + 1/220 1/2 + 1/4 + 1/5 + 1/24 + 1/120 1/2 + 1/4 + 1/5 + 1/25 + 1/100 1/2 + 1/4 + 1/5 + 1/28 + 1/70 1/2 + 1/4 + 1/5 + 1/30 + 1/60 1/2 + 1/4 + 1/5 + 1/36 + 1/45 I hope all this helps rather than overwhelms.... As you can see, in my partial analysis of your problem I have already found several different solutions; but I have no idea if I am anywhere close to finding the complete set of solutions.... - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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