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We begin by noting that solutions to these puzzles are not unique. In particular, doing the `lending' action from each of the vertices once brings us back to where we started. Moreover, the act of doing the `borrowing' action from one vertex is equivalent to doing the`lending' action from each of the other vertices. In particular, without loss of generality one can assume that there is (at least) one vertex for which you do neither action and for all other vertices you do the `lending' action a nonnegative number of times. Below we give possible solutions to four of the puzzles by showing the number of times one lends from each vertex in order to eliminate all debt.

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Mathematics Commons