2017 AIME I Problems/Problem 13

Problem 13

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

Solution

Lemma 1: The ratios between $k^3$ and $(k+1)^3$ decreases as $k$ increases.

Lemma 2: If the range $(n,mn]$ includes two cubes, $(p,mp]$ will always contain at least one cube for all integers in $[n,+\infty)$ .

If $m=14$ , the range $(1,14]$ includes one cube. The range $(2,28]$ includes 2 cubes, which fulfills the Lemma. Since $n=1$ also included a cube, we can assume that $Q(m)=1$ for all $m>14$ . Two groups of 1000 are included in the sum modulo 1000. They do not count since $Q(m)=1$ for all of them, therefore $\sum_{m = 2}^{2017} Q(m) = \sum_{m = 2}^{17} Q(m)$

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2017 AIME I Problems/Problem 13

Problem 13

Solution