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### Finding Sets of Integers

```
Date: 01/02/2002 at 16:36:20
From: keith
Subject: Algebra

Without computer assistance, find five different sets of three
positive integers such that

k < m < n and 1/k + 1/n + 1/m = 19/84
```

```
Date: 01/03/2002 at 03:17:50
From: Doctor Greenie
Subject: Re: Algebra

Hello, Keith -

The desired sum, 19/84, is less than 1/4 and greater than 1/5, so the
smallest denominator, k, must be greater than 4.

Also, with k < m < n, 1/k > 1/m > 1/n, so

1/k + 1/m + 1/n < 1/k + 1/k + 1/k = 3/k

and so

1/k + 1/m + 1/n = 19/84 < 3/k

which means

19/252 < 1/k

or, since k is an integer,

k < 15

So we can start looking for solutions to the problem, knowing that

4 < k < 15

Let's just plunge in head first, looking first for solution(s) with
k = 5.  We have

1/k + 1/m + 1/n = 1/5 + 1/m + 1/n = 19/84

84/320 + 1/m + 1/n = 95/320

1/m + 1/n = 11/320

We need to try to decompose 11/320 into the sum of two fractions
a/320 and b/320 such that both a/320 and b/320 reduce to unit
fractions. We have the following choices:

6/320 + 5/320  [no good -- 320 is divisible by 5 but not by 6]
7/320 + 4/320  [no good -- 320 is divisible by 4 but not by 7]
8/320 + 3/320  [no good -- 320 is divisible by 8 but not by 3]
9/320 + 2/320  [no good -- 320 is divisible by 2 but not by 9]
10/320 + 1/320  [okay -- 320 is divisible by 10, and 1/320 is

So we have our first solution:

19/84 = 95/320 = 84/320 + 10/320 + 1/320 = 1/5 + 1/32 + 1/320

Now let's try finding solution(s) with k = 6.  We have

1/k + 1/m + 1/n = 1/6 + 1/m + 1/n = 19/84

14/84 + 1/m + 1/n = 19/84

1/m + 1/n = 5/84

We need to try to decompose 5/84 into the sum of two fractions with
denominator 84 that are both equivalent to unit fractions. This time
we have the following choices:

3/84 + 2/84
4/84 + 1/84

In this case, all of these fractions are unit fractions, so we get two
more solutions to the problem:

19/84 = 14/84 + 3/84 + 2/84 = 1/6 + 1/28 + 1/42

19/84 = 14/84 + 4/84 + 1/84 = 1/6 + 1/21 + 1/84

You can continue trying the remaining higher values for k. Working
quickly and without a great deal of care, I found two more solutions
with k = 7 and another with k = 8, making a total of 6 solutions to
the problem. I looked very briefly at the higher possible values of k,
but chances for other solutions appeared slim, and I did not complete
an exhaustive search for solutions.

I hope this helps. Write back if you have any further questions on
this.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Number Theory

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