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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
    
Associated Topics:
High School Discrete Mathematics

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