Problem:
At each of the sixteen circles in the network below stands a student. A total of coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.
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Solution:
Let , and be the sums of the numbers of coins held by all the students standing at all the circles in the diagram below labeled , and , respectively. Using the numbers of connections in the diagram, it follows that , and . From the first equation it follows that . The second equation then gives . Finally, both the third and the fourth equation give . The total number of coins is . Thus, .
Note that if each student with neighbors starts with coins, the conditions of the problem are satisfied.
Query: Are there other initial distributions of the coins which satisfy the conditions of the problem?
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The problems on this page are the property of the MAA's American Mathematics Competitions