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Let us place the numbers $1,2,3....,10$ in a random order on a circular table with 10 places.

The question is: prove that there are three consecutive numbers with a sum of 17 or more.

I know that we need to use the "Generalization of the pigeonhole principle" to solve that problem, I just don't know how to use it.

Any help will be appreciated!

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Proof is the noun, prove is the verb. –  Pedro Tamaroff Sep 28 '13 at 15:35
    
I would like to point out that the true minimum is 18. –  Calvin Lin Sep 28 '13 at 16:27
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1 Answer

up vote 4 down vote accepted

Note that there are 10 different $3-tuples$, and every number is included in 3 such $3-tuples$. That means that the sum of all 10 3-tuples is:

$$3(1+2+3+...+9+10) = 165$$

That means that the average sum of a $3-tuple$ is $16.5$. But in a set, at least of the number is no smaller the average of the set, we also know that the all sums are integers. So this implies that there's at least one 3-tuple has sum of 17 or more.

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Thank you , I have forgot that all the numbers are integer. –  Gil Sep 28 '13 at 16:03
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