Difference between revisions of "2007 AIME II Problems/Problem 7"

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== Problem ==
 
== Problem ==
Given a real number <math>x,</math> let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x.</math> For a certain integer <math>k,</math> there are exactly <math>70</math> positive integers <math>n_{1}, n_{2}, \ldots, n_{70}</math> such that <math>k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor</math> and <math>k</math> divides <math>n_{i}</math> for all <math>i</math> such that <math>1 \leq i \leq 70.</math>
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Given a [[real number]] <math>x,</math> let <math>\lfloor x \rfloor</math> denote the [[floor function|greatest integer]] less than or equal to <math>x.</math> For a certain [[integer]] <math>k,</math> there are exactly <math>70</math> positive integers <math>n_{1}, n_{2}, \ldots, n_{70}</math> such that <math>k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor</math> and <math>k</math> divides <math>n_{i}</math> for all <math>i</math> such that <math>1 \leq i \leq 70.</math>
  
 
Find the maximum value of <math>\frac{n_{i}}{k}</math> for <math>1\leq i \leq 70.</math>
 
Find the maximum value of <math>\frac{n_{i}}{k}</math> for <math>1\leq i \leq 70.</math>
  
 
== Solution ==
 
== Solution ==
For <math>x = 1</math>, we see that <math>\sqrt[3]{1} \ldots \sqrt[3]{7}</math> all work, giving 7 integers. For <math>x=2</math>, we see that in <math>\sqrt[3]{8} \ldots \sqrt[3]{26}</math>, all of the even numbers work, giving 10 integers. For <math>x = 3</math>, we get 13, and so on. We can predict that at <math>x = 22</math> we get 70.
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For <math>x = 1</math>, we see that <math>\sqrt[3]{1} \ldots \sqrt[3]{7}</math> all work, giving 7 integers. For <math>x=2</math>, we see that in <math>\sqrt[3]{8} \ldots \sqrt[3]{26}</math>, all of the [[even]] numbers work, giving 10 integers. For <math>x = 3</math>, we get 13, and so on. We can predict that at <math>x = 22</math> we get 70.
  
 
To prove this, note that all of the numbers from <math>\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}</math> divisible by <math>x</math> work. Thus, <math>\frac{(x+1)^3 - 1 - x^3}{x} + 1  = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4</math> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>.
 
To prove this, note that all of the numbers from <math>\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}</math> divisible by <math>x</math> work. Thus, <math>\frac{(x+1)^3 - 1 - x^3}{x} + 1  = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4</math> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>.

Revision as of 19:50, 29 March 2007

Problem

Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$

Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$

Solution

For $x = 1$, we see that $\sqrt[3]{1} \ldots \sqrt[3]{7}$ all work, giving 7 integers. For $x=2$, we see that in $\sqrt[3]{8} \ldots \sqrt[3]{26}$, all of the even numbers work, giving 10 integers. For $x = 3$, we get 13, and so on. We can predict that at $x = 22$ we get 70.

To prove this, note that all of the numbers from $\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}$ divisible by $x$ work. Thus, $\frac{(x+1)^3 - 1 - x^3}{x} + 1  = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4$ (the one to be inclusive) integers will fit the conditions. $3k + 4 = 70 \Longrightarrow k = 22$.

The maximum value of $\displaystyle n_i = (x + 1)^3 - 1$. Therefore, the solution is $\frac{23^3 - 1}{22} = \frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553$.

See also

2007 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions