# Closure regarding arithmetic mean

Say I have n integers bigger than zero with $n > 3$. I need to choose them in a way that leads to the arithmetic mean of any three of these integers to be one of the integers as well. I figured that for the arithmetic mean of any three integers to be an integer again, they all need to be modulo 3 equivalent. This means that every of these integers can be described as $3i+k; i \in \mathbb{N}$ with $k \in \{0, 1, 2\}$ and the same for all numbers.

The airthmetic mean of three of these numbers is $\frac{(3i_1+k)+(3i_2+k)+(3i_3+k)}{3} = i_1+i_2+i_3+k$. From the previous assumption of how the numbers must be able to be represented follows that $i_1+i_2+i_3$ needs to be a multiple of 3 as well. Which is the exact same problem I was trying to solve with examining the modulo. I was hoping to reduce the set of numbers that might possibly be the integers in question by solving a smaller part of the problem first.

Which ultimately leads me to my question: Is my idea usable in its basics (and I made 'minor' mistakes) or completely flawed?

Say you have a set of $n$ integers fulfilling your condition, with $n>3$. Further let $$\{a_1, a_2, \ldots, a_n\}$$ be a sorted list of your numbers. Then we must have $$\frac{a_1 + a_2 + a_3}{3} = a_2$$ (It cannot be neither $a_1$ nor $a_3$ since the numbers are distinct.) We also have that $\frac{a_1 + a_2 + a_4}{3}$ is strictly larger than $a_2$, and it must be smaller than $a_4$, so it has to be $a_3$. Now we run into a problem with $\frac{a_1 + a_3 + a_4}{3}$, which is again strictly larger than $a_3$, but still cannot be as large as $a_4$.

The conclusion is that you cannot have such a set with more than $3$ elements. For the record, $\{1, 2, 3\}$ fulfills the condition, so there are sets with three elements. Any set with less than $3$ elements will vacuously statisfy your condition as well.

• Nice solution, except that one may interpret the question to allow two of the three chosen numbers to be the same. In this case a set with 2 elements no longer works, but a set with 1 element still does. – vadim123 Jan 19 '14 at 14:25
• I must admit I forgot to state that $n > 3$. However the numbers need not be distinct, this is why I tried to avoid the word 'set' in my description, but I used it once anyway - I'll correct that. I understand that all numbers will need to be equal to satisfy the condition, right? – Bk1ng Jan 19 '14 at 14:35
• @Bk1ng As long as there are distinct values present, it matters little whether there are more of any single one. I've proven that there cannot be more than three values present (if you tweak my wording, at least), and the "two-of-one-and-one-of-another" average shows that there cannot be three or two istinct values. So they all have to be equal, yes. – Arthur Jan 19 '14 at 14:39

If your numbers are $a_1<a_2<\cdots<a_n$, then:

${1\over 3}(a_1+a_2+a_3)=a_2,$

$a_1+a_2+a_3=3a_2,$

$a_3=2a_2-a_1.$

By the same reason, $a_4$ is determined by $a_2$ and $a_3$... and your finite sequence is determined by $a_1$ and $a_2$.

Continue from here and ask again if required.

• You've only mentioned consecutive averages. Once you consider averages that are not of three consecutive elements, you can see that it is indeed impossible to form such a sequence (see my answer). – Arthur Jan 19 '14 at 14:42