SAUDI ARABIAN IMO Booklet 2023

The answer is: all distributions where ${\sum }_{i=1}^{n}i{c}_i=\frac{n(n+1)}{2}(\mathrm{mod}n)$, where $c_i$ denotes the number of coins the $i$-th child starts with.

Encode the sequence $c_i$ as polynomial $p(x)={\sum }_i{a}_i{x}_i$. The cyclic nature of the problem makes it natural to work modulo $x^n-1$. Child $i$ performing a step is equivalent to adding $x^i(x-1{)}^2$ to the polynomial, and we want to reach the polynomial $q(x)=1+x+\cdots +x^{n-1}$.

Since we only add multiples of $(x-1{)}^2$, this is only possible if $p(x)=q(x)$ modulo the ideal generated by $x^n-1$ and $(x-1{)}^2$, i.e. $$(x^n-1,(x-1{)}^2)=(x-1)\left(\frac{x^n-1}{x-1},(x-1)\right)=(x-1)\cdot (n,(x-1))$$ This is equivalent to $p(1)=q(1)$ (which simply translates to the condition that there are $n$ coins) and $p^{\prime }(1)=q^{\prime }(1)(\mathrm{mod}n)$, which translates to the invariant. We also could show that this condition is also sufficient. $\square$