Place the numbers $1,2,3....,10$ in a random order on a circular table with 10 places.Prove there are three consecutive numbers with a sum of $\ge17$.

Question

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!

Very minor nitpick for a decade-old question: "Random" should probably be "arbitrary." "Random" isn't wrong so much as it's irrelevant—we're not after a probability or anything else that randomness might bear on. — Brian Tung, Jul 28 '23 at 06:38

score 4 · Accepted Answer · answered Sep 28 '13 at 15:43

4

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.

answered Sep 28 '13 at 15:43

Stefan4024

35,843

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

Place the numbers $1,2,3....,10$ in a random order on a circular table with 10 places.Prove there are three consecutive numbers with a sum of $\ge17$.

1 Answers1