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

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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>.
  
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The maximum value of <math>\displaystyle n_i = (x + 1)^3 - 1</math>. Therefore, the solution is <math>\frac{23^3 - 1}{22} = \frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553</math>.
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The maximum value of <math>n_i = (x + 1)^3 - 1</math>. Therefore, the solution is <math>\frac{23^3 - 1}{22} = \frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 22:32, 4 July 2013

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 $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ 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 $n_i = (x + 1)^3 - 1$. 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 (Problems • Answer Key • Resources)
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

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