Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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.

share|cite|improve this answer
Thank you , I have forgot that all the numbers are integer. – Gil Sep 28 '13 at 16:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.