2017 AIME I Problem 13


For every m2m \geq 2, let Q(m)Q(m) be the least positive integer with the following property: For every nQ(m)n \geq Q(m), there is always a perfect cube k3k^3 in the range n<k3mnn < k^3 \leq m \cdot n. Find the remainder when
m=22017Q(m) \sum_{m = 2}^{2017} Q(m) is divided by 1000.


Leading zeroes must be inputted, so if your answer is 34, then input 034. Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.

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Category: AIME I
Points: 6
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