Problem: An integer is assigned to each vertex of a regular pentagon
so that the sum of the five integers is $2011$.
A turn of a solitaire game consists of subtracting an integer $m$
(not necessarily positive) from each of the integers at two neighboring vertices
and adding $2m$ to the opposite vertex, which is not adjacent
to either of the first two vertices.
(The amount $m$ and the vertices chosen can vary from turn to turn.)
The game is won at a certain vertex if, after some number of turns,
that vertex has the number $2011$ and the other four vertices have the number $0$.
Prove that for any choice of the initial integers,
there is exactly one vertex at which the game can be won.

Solution: Call the vertices $0$, $1$, $2$, $3$, $4$ in order.
First, notice that the quantity
\[ S \coloneq N_1 + 2N_2 + 3N_3 + 4N_4 \pmod 5 \]
is invariant, where $N_i$ is the amount at vertex $i$.
This immediately implies that at most one vertex can win,
since in a winning situation all $N_i$ are $0$ except for one, which is $2011$.
(For example, if $S \equiv 3 \pmod 5$, any victory must occur at the
vertex $3$, via $N_3 = 2011$, $N_0 = N_1 = N_2 = N_4 = 0$.)

Now we prove we can win on this unique vertex.
Let $a_i$, $x_i$ denote the number initially at $i$
and $x_i$ denote $\sum m$ over all $m$ where vertex $i$ gains $2m$.
WLOG the possible vertex is $0$,
meaning $a_1 + 2a_2 + 3a_3 + 4a_4 \equiv 0 \pmod 5$.
Moreover we want
\begin{align*}
  2011 &= a_0 + 2x_0 - x_2 - x_3 \\
  0 &= a_1 + 2x_1 - x_3 - x_4 \\
  0 &= a_2 + 2x_2 - x_4 - x_0 \\
  0 &= a_3 + 2x_3 - x_0 - x_1 \\
  0 &= a_4 + 2x_4 - x_1 - x_2.
\end{align*}
We can ignore the first equation since its the sum of the other four,
and we can WLOG shift $x_0 \to 0$ by shifting each $x_i$ by a fixed amount.
We will now solve the resulting system of equations.

First, we have \[ x_4 = 2x_2 + a_2 \text{ and } x_1 = 2x_3 + a_3. \]
Using these to remove all instances of $x_1$ and $x_4$ gives
\[ 2x_2-3x_3 = 2a_3+a_1-a_2 \text{ and } 2x_3-3x_2 = 2a_2+a_4-a_3 \]
whence we have a two-variable system of equations!
To verify its solution is integral, note that
\[ x_2-x_3 = \frac{a_1 - 3a_2 + 3a_3 - a_4}{5} \]
is an integer, since
\[ a_1 - 3a_2 + 3a_3 - a_4 \equiv a_1 + 2a_2 + 3a_3 + 4a_4 \equiv 0 \pmod 5. \]
Abbreviating $\frac{a_1 - 3a_2 + 3a_3 - a_4}{5}$ as $k$, we obtain the desired $x_i$:
\begin{align*}
  x_2 &= 2a_3+a_1-a_2 + 2k \\
  x_3 &= x_2 + k \\
  x_1 &= 2x_3 + a_3 \\
  x_4 &= 2x_2 + a_2 \\
  x_0 &= 0.
\end{align*}
This is the desired integer solution.

\begin{remark*}
  In principle, you could unwind all the definitions above to explicitly
  write every $x_i$ as a function of $a_1$, $a_2$, $a_3$, $a_4$.
  If you did this, you could get the long equations
  \begin{align*}
    x_0 &= 0 \\
    x_1 &= -\tfrac15(6a_1 + 2a_2 + 3a_3 + 4a_4) \\
    x_2 &= -\tfrac15(2a_1 + 4a_2 + a_3 + 3a_4) \\
    x_3 &= -\tfrac15(3a_1 + a_2 + 4a_3 + 2a_4) \\
    x_4 &= -\tfrac15(4a_1 + 3a_2 + 2a_3 + 6a_4)
  \end{align*}
  which indeed are all integers whenever $a_1 + 2a_2 + 3a_3 + 4a_4 \equiv 0 \pmod 5$.

  However, this is quite tedious and also unnecessary to solve the problem.
  That's because we only care that the $x_i$ are integers,
  and do not need to actually know the values.
  This lets us work more indirectly to avoid long calculation, as we did above.
\end{remark*}