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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.

The seventeen numbers are arranged like the numbers on an analogue clock, but placed in a seemingly random order: 13, 4, 14, 10, 5, and so on.

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.

The variables a subscript 1 through a subscript 17 are arranged in order in the clockwise direction around a circle. Variable a subscript 1 is between a subscript 17 and a subscript 2.

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 \(27\).