# Combinatorics Question

1. Aug 23, 2009

### rbzima

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 $$a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$$ each represent some positive real number whose sum equal 100. Given that $$a_{1}$$ is the smallest real number, $$a_{2} = 10 + a_{1}$$, $$a_{3} = 10 + a_{2}$$ and so on...

Don't really know where to go from here! Suggestions?

Last edited: Aug 23, 2009
2. Aug 23, 2009

### Tom Mattson

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

Try to show that $x_1+x_2+x_3+x_4+x_5>100$.

3. Aug 23, 2009

### rbzima

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 $$a_{1}$$. Regardless of how small $$a_{1}$$ 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!