= Discrete prize-count optimization for a power-valued contest
{title2=$x_*=(1-\alpha)/(2-\alpha)$}
For prize scale $w(x)=x^{-\alpha}$, total effort is proportional to $h(x)=x^{1-\alpha}(1-x)$ on the feasible grid $x=m/n$. If $\alpha\geq1$, one prize is optimal. For $0<\alpha<1$, $h$ increases up to $x_*=(1-\alpha)/(2-\alpha)$ and then decreases, so the optimal feasible integer is among $\lfloor nx_*\rfloor$ and $\lfloor nx_*\rfloor+1$. Remove infeasible candidates and compare their objective values. No definition at $x=0$ is needed.
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