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 from each of the integers at two neighboring vertices and adding to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount 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.
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Note that interchanging the order of the turns does not affect anything. Thus we can consider just the 5 distinct possible moves. Let be the initial values of the 5 vertices, and let be the total values of directed at the five vertices. Consider the case where is the vertex that reaches We then get the equations,


We have is a multiple of This means that for a set of to exist, must be a multiple of Now note that Therefore, all the are different values mod 5, so there can be at most 1 vertex that reaches for each arrangement.

Now it remains to prove that there always exists a set of Without loss of generality let be the vertex that reaches Consider the following values of


There values are all integers considering is. They satisfy the aforementioned equations and thus provide a valid way to reach the end position from any given arrangement. We are done.

Main Point(s): Once we have equations for the values of try to find an equation relating the values so that we can single out just one vertex. In this situation, always try multiplying by different numbers and taking mod 5.