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### 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.

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 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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High School Basic Algebra
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