Finding Sets of IntegersDate: 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 already a unit fraction] 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/