Macedonian Mathematical Olympiad

At most $n$ consecutive redistributions. We will start with an example showing that $n$ consecutive redistributions are possible. Let $2^n$ coins be distributed among $3$ children initially as follows: $1$, $2^{n-1}+2^{n-2}+\cdots +2$, $1$. The successive redistributions (a total of $n$) will be: $$2^1,2^{n-1}+2^{n-2}+\cdots +2^2,2^1$$ $$2^2,2^{n-1}+2^{n-2}+\cdots +2^3,2^2$$ $$2^{n-2},2^{n-1},2^{n-2}$$ $$2^{n-1},0,2^{n-1}$$ $$0,0,2^n$$ Let us show that $n$ is the maximal number of consecutive redistributions. We start by assuming that there is an initial distribution for which there are at least $n+1$ possible redistributions and we seek contradiction: let us note that after one redistribution the number of coins that each child has is divisible by $2$ (each child to whom coins have been added has a doubled number of coins which is an even number, and the child from whom coins have been taken has again an even number of coins since there is a total of $2^n$ coins i.e. an even number of them). Analogously after the second redistribution each child has a number of coins that is divisible by $4$, and so on, until the $n$-th redistribution in which the number of coins each child has is divisible by $2^n$; taking into account that the total number of coins is constant and equals $2^n$, the only possible distribution (in some order) is $2^n,0,0,\dots ,0$. But then there cannot be a next redistribution. Contradiction!