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### Unit Fractions Summing to 1

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Date: 07/15/2001 at 16:33:37
From: Pamela
Subject: Unit Fractions

Find seven different unit fractions whose sum is 1, i.e.
1/a+1/b+1/c+1/d+1/f+1/g = 1.

I have worked on this problem for over a week and still cannot come
up with the correct answer. I came up with fractions; however, they
had zero as the numerator. The fractions did add up to 1. My teacher
said this is not correct. Even my dad could not help me. HELP!

My teacher said there is an infinite number of answers. Do you think
maybe she needs a vacation?
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Date: 07/16/2001 at 13:39:59
From: Doctor Greenie
Subject: Re: Unit Fractions

Hi, Pamela -

Your teacher does not need a vacation; there is an infinite number of
answers, but the arithmetic required to find most of them is
unmanageable except for a computer.

Here are some different approaches to this problem....

I. "Brute force"

One way to attack this problem is by using the "greedy" algorithm.
With this method, you simply choose the largest unit fractions you can
without any regard for how ugly the arithmetic gets. This method will
always work, but the arithemetic usually becomes quite unpleasant.

Here is how the beginning of your problem would be handled with this
method:

(1) The largest unit fraction less than 1 is 1/2, so we have

1 = 1/2 + ...

(2) The sum of the remaining unit fractions needs to be 1/2; the
largest unit fraction less than 1/2 is 1/3. So we have

1 = 1/2 + 1/3 + ... = 5/6 + ...

(3) The sum of the remaining unit fraction needs to be 1/6; the
largest unit fraction less than 1/6 is 1/7. So we have

1 = 1/2 + 1/3 + 1/7 + ... = 41/42 + ...

(4) The sum of the remaining unit fractions needs to be 1/42; the
largest unit fraction less than 1/42 is 1/43. So we have

1 = 1/2 + 1/3 + 1/7 + 1/43 + ... = ???

Let's stop at this point, because the arithmetic is getting quite
ugly.

We want to find a different approach for solving this problem, using
unit fractions that can be added more easily.

II. Unit fractions with denominators equal to powers of 2

One group of different unit fractions that can always be added fairly
easily is those whose denominators are powers of 2 - 1/2, 1/4, 1/8,
and so on:

1/2 + 1/4 = 3/4
1/2 + 1/4 + 1/8 = 7/8
1/2 + 1/4 + 1/8 + 1/16 = 15/16
...

We know that we can always make a sum of 1 using different numbers of
these unit fractions:

1/2 + 1/2 = 1
1/2 + 1/4 + 1/4 = 1
1/2 + 1/4 + 1/8 + 1/8 = 1
1/2 + 1/4 + 1/8 + 1/16 + 1/16 = 1
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/32 = 1  (1)
...

In the last sum above, the number 1 has been written as the sum of 6
unit fractions, but the last two unit fractions are the same.  If we
can write the second "1/32" as the sum of two unit fractions, then we
will be done with the problem.

Whenever we have a unit fraction 1/x where the denominator is an even
number, the fraction 1/x can be written as the sum of two different
unit fractions, one of which is twice the other.  For example

1/2  = 3/6   = 1/6   + 2/6   = 1/6   + 1/3
1/10 = 3/30  = 1/30  + 2/30  = 1/30  + 1/15
1/66 = 3/198 = 1/198 + 2/198 = 1/198 + 1/99

As long as the denominator is an even number, the second fraction in
the sum will be able to be reduced to give a unit fraction.

So we can write the second "1/32" in (1) above as a sum of two
different unit fractions:

1/32 = 3/96 = 1/96 + 2/96 = 1/96 + 1/48  (2)

Substituting (2) in (1) above, we have our first solution to the
problem:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/48 + 1/96 = 1

we have

1     1     1     1     1     1     1
--- + --- + --- + --- + --- + --- + ---
2     4     8    16    32    48    96

48    24    12     6     3     2     1
= --- + --- + --- + --- + --- + --- + ---
96    96    96    96    96    96    96

48 + 24 + 12 + 6 + 3 + 2 + 1     96
= ------------------------------ = ---- = 1
96                   96

This arithemetic suggests that we might search for "nice" answers
(consisting of unit fractions with relatively small denominators) by
looking for a common denominator which has many divisors. We can then
try to find a case where seven of the divisors add up to the common
denominator.

III. A common denominator with many divisors

Here is another solution I found using this approach....

Common denominator = 48:  The divisors of 48 are

1, 2, 3, 4, 6, 8, 12, 16, and 24

Seven of these whose sum is 48 are the following:

24 + 8 + 6 + 4 + 3 + 2 + 1 = 48

This gives me the solution

24/48 + 8/48 + 6/48 + 4/48 + 3/48 + 2/48 + 1/48 = 1
or
1/2 + 1/6 + 1/8 + 1/12 + 1/16 + 1/24 + 1/48 = 1

IV. Some more examples for you to try...

Solutions can also be found using this method using common
denominators of 60 or 72. See if you can find these solutions. I will
get you started.

Common denominator = 60:  The divisors of 60 are

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, and 30

Find seven of these divisors whose sum is 60 and use them to find
another solution.

Common denominator = 72:  The divisors of 72 are

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, and 36

Find seven of these divisors whose sum is 72 and use them to find yet
another solution.

I hope all this helps and that you are able to find the last two
solutions I suggested.

Write back if you have further questions on this type of problem.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Puzzles
Middle School Fractions
Middle School Puzzles

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