Math/Logic Puzzle

We can describe a situation by three numbers $(M,m,r)$, where $2M$ is the maximal amount a player has, $m$ is the minimal amount, and $r$ is the number of players with $2m$.

If a player as $2a$ pieces and their left neighbour has $2b$ pieces, in the next round, this player will have $ (2a-a)+b= a+b\le 2M$ pieces. Even if $a+b$ is odd, it is $\le 2M-1$ and so after refill from the jar, it is still $\le 2M$. Likewise, the new amount is $\ge 2m$ with equality if and only if $a=b=m$. We conclude that a situation $(M,m,r)$ turns into $(M',m',r')$ with $m\le m'\le M'\le M$. Moreover, if $m'=m$, then $r'\le r$.

But it cannot happen that $(M',m',r')=(M,m,r)$ unless $M=m$. Indeed, if $m<M$, then there must be some person with $2m$ pieces while their left neighbour has more. Then this person will have $>2m$ pieces in the next round, meaning that either $m'>m$ or at least $r'<r$.

At any rate, if $M>m$, then it takes only finitely mann steps until $M-m$ decreases by at least one. Then still after finitely many steps, we reach $M=m$.

Tags:

Puzzle

Induction

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