Triples of NumbersDate: 7/23/96 at 10:24:53 From: Anonymous Subject: Triples of Numbers Find two triples of positive integers such that their sum equals their product. I found one right away (1,2,3). I insist that there is no way there can be another one. Is there? And if there is not, can you prove it? Date: 7/26/96 at 15:41:36 From: Doctor Ceeks Subject: Re: your response Hi, There are more: (0,A,-A) works...both product and sum equal 0. Are you restricting yourself to integers? If not, then there are many, many, many solutions. If you are considering only positive integers, then you've found the only one. To see this: you want a,b,c such that a+b+c=abc. Rearranging, get c=(a+b)/(ab-1). For c to be an integer, we must have ab-1<= a+b. If both a and b are bigger than 2, then ab > max(a,b)*2 >= a+b, so we would need equality to occur in the second inequality...that is a=b, but there is no solution with a=b (check that 2a/(a^2-1) is never an integer). So one of a or b is equal to 1 or 2. Case 1: a = 1. c =(b+1)/(b-1) is an integer only when b=2 or 3, in which cases c = 3 or 2, respectively. Case 2: a = 2. c =(b+2)/(2b-1) is an integer only when b=1 or 3, in which cases, c=3 or 1, respectively. (There are some small things to check above, but this outline is correct.) -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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