
00.2. The persons $P_{1}, P_{1}, \ldots, P_{n-1}, P_{n}$ sit around a table, in this order, and each one of them has a number of coins. In the start, $P_{1}$ has one coin more than $P_{2}, P_{2}$ has one coin more than $P_{3}$, etc., up to $P_{n-1}$ who has one coin more than $P_{n}$. Now $P_{1}$ gives one coin to $P_{2}$, who in turn gives two coins to $P_{3}$ etc., up to $P_{n}$ who gives $n$ coins to $P_{1}$. Now the process continues in the same way: $P_{1}$ gives $n+1$ coins to $P_{2}, P_{2}$ gives $n+2$ coins to $P_{3}$; in this way the transactions go on until someone has not enough coins, i.e. a person no more can give away one coin more than he just received. At the moment when the process comes to an end in this manner, it turns out that there are to neighbours at the table such that one of them has exactly five times as many coins as the other. Determine the number of persons and the number of coins circulating around the table.


Solution. Assume that $P_{n}$ has $m$ coins in the start. Then $P_{n-1}$ has $m+1$ coins, ... and $P_{1}$ has $m+n-1$ coins. In every move a player receives $k$ coins and gives $k+1$ coins away, so her net loss is one coin. After the first round, when $P_{n}$ has given $n$ coins to $P_{1}$, $P_{n}$ has $m-1$ coins, $P_{n-1}$ has $m$ coins etc., after two rounds $P_{n}$ has $m-2$ coins, $P_{n-1}$ has $m-1$ coins etc. This can go on during $m$ rounds, after which $P_{n}$ has no money, $P_{n-1}$ has one coin etc. On round $m+1$ each player still in possession of money can receive and give away coins as before. The penniless $P_{n}$ can no more give away coins according to the rule. She receives $n(m+1)-1$ coins from $P_{n-1}$, but is unable to give $n(m+1)$ coins to $P_{1}$. So when the game ends, $P_{n-1}$ has no coins and $P_{1}$ has $n-2$ coins. The only pair of neighbours such that one has 5 times as many coins as the other can be $\left(P_{1}, P_{n}\right)$. Because $n-21$, the possibilities are $n=3, m=1$ or $n=9, m=3$. Both are indeed possible. In the first case the number of coins is $3+2+1=6$, in the second $11+10+\cdots+3=63$.
