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Difference between revisions of "2017 AIME I Problems/Problem 13"
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Lemma 2: If the range <math>(n,mn]</math> includes two cubes, <math>(p,mp]</math> will always contain at least one cube for all integers in <math>[n,+\infty)</math>.
 
Lemma 2: If the range <math>(n,mn]</math> includes two cubes, <math>(p,mp]</math> will always contain at least one cube for all integers in <math>[n,+\infty)</math>.
  
−
If <math>m=14</math>, the range <math>(1,14]</math> includes one cube. The range <math>(2,28]</math> includes 2 cubes, which fulfills the Lemma. Since <math>n=1</math> also included a cube, we can assume that <math>Q(m)=1</math> for all <math>m>14</math>. Two groups of 1000 are included in the sum modulo 1000. They do not count since <math>Q(m)=1</math> for all of them, therefore <math>\sum_{m = 2}^{2017} Q(m) = \sum_{m = 2}^{17} Q(m)</math>.
+
If <math>m=14</math>, the range <math>(1,14]</math> includes one cube. The range <math>(2,28]</math> includes 2 cubes, which fulfills the Lemma. Since <math>n=1</math> also included a cube, we can assume that <math>Q(m)=1</math> for all <math>m>14</math>. Two groups of 1000 are included in the sum modulo 1000. They do not count since <math>Q(m)=1</math> for all of them, therefore <cmath>\sum_{m = 2}^{2017} Q(m) = \sum_{m = 2}^{17} Q(m)</cmath>

Revision as of 18:39, 8 March 2017

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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