# Problem of the Week Problem E and Solution Stand in a Circle

## Problem

The numbers from $$1$$ to $$17$$ are arranged around a circle. One such arrangement is shown.

Explain why every possible arrangement of these numbers around a circle must have at least one group of three adjacent numbers whose sum is at least $$27$$.

Note:
In solving the above problem, it may be helpful to use the fact that the sum of the first $$n$$ positive integers is equal to $$\tfrac{n(n+1)}{2}$$. That is, $1 + 2 + 3 + … + n = \frac{n(n+1)}{2}$

## Solution

We will use a proof by contradiction to explain why every possible arrangement of these numbers around a circle must have at least one group of three adjacent numbers whose sum is at least 27.

In general, to prove that a statement is true using a proof by contradiction, we first assume the statement is false. We then show this leads to a contradiction, which proves that our original assumption was wrong, and therefore the statement must be true.

First, we will assume that there exists an arrangement of the numbers from $$1$$ to $$17$$ around a circle where the sums of all groups of three adjacent numbers are less than $$27$$. This arrangement is shown, where the variables $$a_1, a_2, a_3, \ldots, a_{17}$$ represent the numbers from $$1$$ to $$17$$, in some order, for this particular arrangement.

Now we will add up the sums of all groups of three adjacent numbers and call this value $$S$$. \begin{aligned} %S=&(a_1+a_2+a_3) + (a_2+a_3+a_4) + (a_3+a_4+a_5) + \cdots + (a_{15}+a_{16}+a_{17}) + \\&(a_{16}+a_{17}+a_1) + (a_{17}+a_1+a_2)\\ S=&(a_1+a_2+a_3) + (a_2+a_3+a_4) + (a_3+a_4+a_5)\\ & ~+ (a_4+a_5+a_6) +(a_5+a_6+a_7) + (a_6+a_7+a_8)\\ & ~+ (a_7+a_8+a_9) + (a_8+a_9+a_{10}) +(a_9+a_{10}+a_{11}) \\ & ~+ (a_{10}+a_{11}+a_{12})+ (a_{11}+a_{12}+a_{13})+ (a_{12}+a_{13}+a_{14})\\ & ~ +(a_{13}+a_{14}+a_{15})+ (a_{14}+a_{15}+a_{16}) + (a_{15}+a_{16}+a_{17}) \\ & ~+ (a_{16}+a_{17}+a_1) +(a_{17}+a_1+a_2)\end{aligned} We can see that there are $$17$$ groups of three adjacent numbers around the circle. Since each of these groups has a sum that is less than $$27$$, we can conclude that $$S$$ must be less than $$17 \times 27=459$$. So, $$S<459$$.

Looking again at the value of $$S$$, we can see that each of $$a_1, a_2, a_3, \ldots, a_{17}$$ appears exactly three times. So, \begin{aligned} S &= 3(a_1) + 3(a_2) + 3(a_3) + \cdots + 3(a_{17})\\ &= 3(a_1+a_2+a_3+\cdots+a_{17})\end{aligned} However, we know that $$a_1+a_2+a_3+\cdots+a_{17}$$ is equal to the sum of all the numbers from $$1$$ to $$17$$, which is $$\frac{17(18)}{2}=153$$. Therefore, $$S=3(153)=459$$.

But this is a contradiction, since we stated earlier that $$S<459$$. It can’t be possible that $$S<459$$ and $$S=459$$. Therefore, our original assumption that there exists an arrangement of the numbers from $$1$$ to $$17$$ around a circle where the sums of all groups of three adjacent numbers are less than $$27$$ must be false. Thus, it follows that every possible arrangement of the numbers from $$1$$ to $$17$$ around a circle must have at least one group of three adjacent numbers whose sum is at least $$27$$.