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 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.
Solution:
Let and represent the integers at vertices to (in order around the pentagon) at the start of the game. We will first show that the game can be won at only one of the vertices. Observe that the quantity is an invariant of the game. For instance, one move involves replacing and by and . Thus the quantity becomes
which is unchanged mod 5 . The other moves may be checked similarly. Now suppose that the game may be won at vertex . The value of the invariant at the winning position is . If the initial value of the invariant is , then we must have , or . Hence the game may only be won at vertex , where is the least positive residue of .
By renumbering the vertices, we may assume without loss of generality that the winning vertex is . We will show that the game can be won in four moves by adding a suitable amount at vertex (and subtracting from the opposite vertices) on the th turn for . The net change at vertex after these four moves is , which must equal if we are to finish with 0 at . In this fashion we find that
The sum of the first four equations is the negative of the fifth equation, so it is redundant. Multiplying the first four equations by and adding them yields . But
since we are assuming is the winning vertex. Therefore we may divide by 5 to obtain . We also know that , and one easily confirms that the right-hand sides of these equations are integers with the same parity. Hence the system admits an integral solution for and . The second and third equations then quickly give integer values for and as well, so it is indeed possible to win the game at vertex .
The problems on this page are the property of the MAA's American Mathematics Competitions