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Difference between revisions of "2017 AIME I Problems/Problem 13"

m (Added a QED and a square to the end of the proof)
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Lemma 1: The ratio between <math>k^3</math> and <math>(k+1)^3</math> decreases as <math>k</math> increases.
 
Lemma 1: The ratio between <math>k^3</math> and <math>(k+1)^3</math> decreases as <math>k</math> increases.
  
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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 <math>p</math> in <math>[n,+\infty)</math>.
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Lemma 2: If the range <math>(n,mn]</math> includes <math>y</math> cubes, <math>(p,mp]</math> will always contain at least <math>y-1</math> cubes for all <math>p</math> 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 <cmath>\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000</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) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000</cmath>
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When the partial sums are added, we get <math>\boxed{059}\hspace{2 mm}QED\hspace{2 mm} \blacksquare</math>
 
When the partial sums are added, we get <math>\boxed{059}\hspace{2 mm}QED\hspace{2 mm} \blacksquare</math>
  
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This solution is brought to you by a1b2
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This solution is brought to you by [[User:a1b2|a1b2]]
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2017|n=I|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:56, 11 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 ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases.

Lemma 2: If the range $(n,mn]$ includes $y$ cubes, $(p,mp]$ will always contain at least $y-1$ cubes for all $p$ 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 \[\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000\]

Now that we know this we will find the smallest $n$ that causes $(n,mn]$ to contain two cubes and work backwards (recursion) until there is no cube in $(n,mn]$.

For $m=2$ there are two cubes in $(n,2n]$ for $n=63$. There are no cubes in $(31,62]$ but there is one in $(32,64]$. Therefore $Q(2)=32$.

For $m=3$ there are two cubes in $(n,3n]$ for $n=22$. There are no cubes in $(8,24]$ but there is one in $(9,27]$. Therefore $Q(3)=9$.

For $m$ in $\{4,5,6,7\}$ there are two cubes in $(n,4n]$ for $n=7$. There are no cubes in $(1,4]$ but there is one in $(2,8]$. Therefore $Q(4)=2$, and the same for $Q(5)$, $Q(6)$, and $Q(7)$ for a sum of $8$.

For all other $m$ there is one cube in $(1,8]$, $(2,16]$, $(3,24]$, and there are two in $(4,32]$. Therefore, since there are 10 values of $m$ in the sum, this part sums to $10$.

When the partial sums are added, we get $\boxed{059}\hspace{2 mm}QED\hspace{2 mm} \blacksquare$

This solution is brought to you by a1b2

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

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