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

  • 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


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 $18$


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