Question on a Math Counts Test
Date: 13 Aug 1995 04:05:16 -0400 From: Kara Enzenberger Subject: algebra Question: This question was on a Math Counts test, and I am interested in the quick (80 second) solution (I have the answer but I need to know how to get it). Find the sum of the x-coordinates of all possible positive integral solutions to: 1 1 1 - + - = - x y 7 Can you help me?
Date: 13 Aug 1995 14:04:27 -0400 From: Dr. Ken Subject: Re: algebra Hello there! Well, the first thing I think I would do on this question is solve for y: 1/y = 1/7 - 1/x 1/y = (x-7)/7x y = 7x/(x-7) So if we're looking for positive integral solutions, we have to find a positive integer x for which 7x/(x-7) is also a positive integer. That means that all of the factors of x-7 must cancel with something in the numerator, either the 7 or the x. When can x-7 share factors with x? Well, let's suppose a divisor d divides x and it also divides x-7. Since d divides x, we can write x = dk for some k. Since d divides x-7, d must divide dk-7. But what's dk-7 divided by d? It's k-7/d, which must be an integer, so d divides 7. So d = 1 or 7. What have we shown here? We've shown that if we've got any hope of getting factors in the denominator to cancel with stuff in the numerator, either x and x-7 have to be divisible by 7, or x-7 = 1 (i.e. x=8 and y=56). Right? Supposing x is divisible by 7, let's write x = 7n. Then we have 7*7n/(7n-7) = 7n/(n-1). Now, certainly n and n-1 can't share any factors except 1; if something divides n-1 evenly, it divides n with a remainder of 1. So the only time we'll be able to cancel all the factors in the denominator is when n-1 = 1 or 7, making n = 2 or 8. Since x = 7n, That makes x = 14 or 56, and y= 14 or 8. So the only positive integer solutions we have are (8,56), (14,14) and (56,8), and 8+14+56 = 78. So that's the proof. But if you're just trying to get the answer fast, you might just look at 7x/(x-7) and try to plug in things that cancel. The obvious things to try are to make the bottom 1, and then to make it 7. That gives you two of the answers, and then you can get the third by realizing that the problem is symmetric in x and y. What I mean by this is that in the original equation 1/x + 1/y = 1/7, there was absolutely no difference in the roles of x and y. So since (8,56) is a solution, (56,8) must also be a solution. I really wouldn't feel bad if it took you longer than 80 seconds to do the problem. Unless there's an easier way than the way I found, I think this problem is much harder than that. It's also much cooler. -K
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum