# A Sum for 2015

Algebra Level pending

Let $$x_1, x_2, x_3, \dots, x_{2015}$$ be real numbers ranging from $$-1$$ to $$1$$ such that $$\sum_{i=1}^{2015}x_i^2=1$$.

Find the remainder when the absolute value of the minimum possible value of $$\sum_{j=1}^{2015}x_j\sqrt{j^3}$$ is divided by $$1000$$.

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