Problem: Given a sequence of real numbers,
a move consists of choosing two terms
and replacing each with their arithmetic mean.
Show that there exists a sequence of $2015$ distinct real numbers
such that after one initial move
is applied to the sequence --- no matter what move --- there
is always a way to continue with a finite sequence of moves
so as to obtain in the end a constant sequence.

Solution: One valid example of a sequence is $0$, $1$, \dots, $2014$.
We will show how to achieve the all-$1007$ sequence
based on the first move.

Say two numbers are \emph{opposites} if their average is $1007$.
We consider $1007$ as its own opposite.

We consider two cases:
\begin{itemize}
  \ii First, suppose the first initial move did
  \emph{not} involve the number $1007$.
  Suppose the two numbers changed were $a$ and $b$,
  replaced by $c = \half(a+b)$ twice.
  \begin{itemize}
    \ii If $a$ and $b$ are opposites, we
    simply operate on all the other pairs of opposites.
    \ii Otherwise let $a'$ and $b'$ be the opposites of $a$ and $b$,
    so all four of $a$, $b$, $a'$, $b'$ are distinct.
    Then operate on $a'$ and $b'$ to get $c' = 2014-c$.
    We work with only these four numbers ande replace them as
    follows:
    \[
      \begin{array}{cccc}
        \half(a+b) & \half(a+b) & a' & b' \\
        \half(a+b) & \half(a+b) & \half(a'+b') & \half(a'+b') \\
        1007 & \half(a+b) & 1007 & \half(a'+b') \\
        1007 & 1007 & 1007 & 1007
      \end{array}
    \]
    Finally, we operate on the remaining $1005$ pairs of opposites.
  \end{itemize}

  \ii Now suppose the first initial move involved the number $1007$
  and some $a$.
  Let $k$ be any number other than $a$ or its opposite,
  and let $a'$, $k'$ be the opposites of $a$ and $k$.
  We work with only these five numbers:
  and replace them in the following way:
  \[
    \begin{array}{ccccc}
      \half(a+1007) & \half(a+1007) & a' & k & k' \\
      \half(a+1007) & \half(a+1007) & a' & 1007 & 1007 \\
      \half(a+1007) & \half(a+1007) & \half(a'+1007) &
        \half(a'+1007) & 1007 \\
      1007 & \half(a+1007) & 1007 &
        \half(a'+1007) & 1007 \\
      1007 & 1007 & 1007 & 1007 & 1007 \\
    \end{array}
  \]
  Finally, we operate on the remaining $1005$ pairs of opposites.
\end{itemize}


\begin{remark*}
  In fact, the same proof basically works for any sequence
  with average $m$ such that $m$ is in the sequence,
  and every term has an opposite.

  However for ``most'' sequences one expects
  the result to not be possible.
  As a simple example,
  the goal is impossible for $(0, 1, \dots, 2013, 2015)$
  since the average of the terms is $1007 + \frac{1}{2015}$,
  but in the process the only denominators
  ever generated are powers of $2$.
  This narrows the search somewhat.
\end{remark*}