Difference between revisions of "2017 AIME I Problems/Problem 13"
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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> | 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> | ||
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+ | ==See Also== | ||
+ | {{AIME box|year=2017|n=I|num-b=12|num-a=14}} | ||
+ | {{MAA Notice}} |
Revision as of 19:31, 8 March 2017
Problem 13
For every , let be the least positive integer with the following property: For every , there is always a perfect cube in the range . Find the remainder when is divided by 1000.
Solution
Lemma 1: The ratios between and decreases as increases.
Lemma 2: If the range includes two cubes, will always contain at least one cube for all integers in .
If , the range includes one cube. The range includes 2 cubes, which fulfills the Lemma. Since also included a cube, we can assume that for all . Two groups of 1000 are included in the sum modulo 1000. They do not count since for all of them, therefore
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.