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So I have the following problem:

Let n be a positive integer. Is it possible to arrange the numbers 1, 2, . . . , n in a row so that the arithmetic mean of any two of these numbers is not equal to some number between them?

I'm having some trouble solving it, but what I have done so far is:

  • I think it is possible because I can think of an example. If we took 1,2,3 and 4, I arranged it as 1,3,2,4
  • I thought that by maybe arranging the odds on the left side and evens on the right side could get me somewhere. I tried to look at this post, Arrangement of Numbers but I couldn't quite understand it.
user21820
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user604720
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  • I would suggest you follow through the algorithm in your linked question for some small $n$. You have solutions for $1,2,3$ and $4$, so try $5$ and $8$. I chose $8$ because you split the list in half and you have a solution for $4$. – Ross Millikan Oct 20 '18 at 17:38
  • The statement of the problem is missing. Please revise. – N. F. Taussig Feb 20 '19 at 09:40
  • Please do not vandalize your posts. It disrespects those who have taken the time to answer. – robjohn Jun 25 '20 at 17:13

1 Answers1

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I thought that by maybe arranging the odds on the left side and evens on the right side could get me somewhere.

That's a good idea: the average of an odd number and an even number is not an integer, so you can then consider the two halves separately.

The arithmetic mean is a linear operator, which is to say that $\textrm{avg}(ax+b, ay+b) = a\,\textrm{avg}(x,y)+b$. If we take the odd numbers, $a = \frac12$ and $b=\frac12$ maps them to $1,2,3,\ldots$; if we take the even numbers, $a = \frac12$ and $b=0$ maps them to $1,2,3,\ldots$

Suppose there is a smallest $n$ such that there is no valid arrangement of $1,2,\ldots,n$. We know that $n > 4$, because of the solution you give in the question. But if we separate the odds and evens, we reduce to two halves which are smaller than $n$, and we have valid arrangements for them by using the linearity. By contradiction, there is no $n$ which has no valid arrangement.

Peter Taylor
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