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Combinatorics Question

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data

    The sum of 5 positive real numbers is 100. Prove that there are two numbers among them whose difference is at most 10.

    2. Relevant equations

    Nothing really...

    3. The attempt at a solution

    The biggest problem I'm running into is that I can think of specific examples, but translating that into an algebraic argument has always been my weak area. Getting started is where I struggle the most... but I'm thinking the following:

    Let's assume there are no two positive real numbers whose difference is at most 10.
    Let [tex]a_{1}, a_{2}, a_{3}, a_{4}, a_{5}[/tex] each represent some positive real number whose sum equal 100. Given that [tex]a_{1}[/tex] is the smallest real number, [tex]a_{2} = 10 + a_{1}[/tex], [tex]a_{3} = 10 + a_{2}[/tex] and so on...

    Don't really know where to go from here! Suggestions?
     
    Last edited: Aug 23, 2009
  2. jcsd
  3. Aug 23, 2009 #2

    Tom Mattson

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    Gold Member

    Well you've got a finite set of numbers, so you know it has a minimum. Let [itex]x_1[/itex] be that minimum. Furthermore they're real numbers so they're ordered. Let [itex]x_1\leq x_2 \leq x_3 \leq x_4 \leq x_5[/itex]. Now suppose that there do not exist a pair of numbers such that their difference is at most 10.

    Try to show that [itex]x_1+x_2+x_3+x_4+x_5>100[/itex].
     
  4. Aug 23, 2009 #3
    Wow, that was straight forward... Totally figured it out...
    I had an epiphany right before you responded, so I should be good now. I was definitely thinking the same thing, and ended up solving in terms of [tex]a_{1}[/tex]. Regardless of how small [tex]a_{1}[/tex] is, as long as it's not 0, you will receive some value that is greater than 100, thus we have a contradiction, which means that there must be at least two values who have a difference of at most 10!

    Thanks for the help!
     
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