5
$\begingroup$

Write the numbers from $1$ to $10$ in a circle. Consider all the groups of three consecutive numbers and their sums. Show that we can find a group with sum at least $18$.

Approach 1 ( not usefull ): The arithmetic mean of all sums of the groups is $\frac{3(1+2+3+\dots+10)}{10} = \frac{165}{10} = 16.5$ so, at least one group has sum $17$. But that doesn't help.

Approach 2: Suppose that there exist a configuration in which all groups have sums less that $18$. Therefore, $10$ and $9$, $10$ and $8$, $8$ and $9$, $10$ and $7$ can not be in the same group.

From the first three observations between $8$, $9$ and $10$ there are group of numbers of size, two, two and three.

Here I am stuck. I suppose looking at where 7 can be placed and with lots of case work you can get to a contradiction. Is this a good approach ? Do you have a better alternative ?

Also feel free to change the tags, I do not know which tags are appropriate

$\endgroup$
2
  • 1
    $\begingroup$ I don’t like statistics, but I love the probabilistic method. It would have been so nice if your first approach had worked out... $\endgroup$ Mar 15, 2020 at 12:20
  • $\begingroup$ Approach 1: If $a_1+a_2+a_3=a_2+a_3+a_4$ then $a_1=a_4$. So to avoid $18$s, the sums must alternate $16$ and $17$... $\endgroup$
    – Empy2
    Mar 15, 2020 at 12:53

1 Answer 1

11
$\begingroup$

Let us say the numbers in clockwise direction are $a_{i},i=1,2,...,10$ with $a_{1}=1$.

$$ \begin{aligned} \sum_{i=2}^{10}{a_{i}}&=2+3+...+10\\ &=3\times 18 \end{aligned} $$

At least one of $(a_2+a_3+a_4)$, $(a_5+a_6+a_7)$, $(a_8+a_9+a_{10})$ is greater than or equal to $18$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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