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Difference between revisions of "2007 AIME II Problems/Problem 7"
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==Solution 2==
 
==Solution 2==
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Obviously <math>k</math> is positive. Then, <math>n_1</math> clearly equals <math>k^3</math> and we can let <math>n_i</math> equal <math>k^3 + (i - 1)k</math>.  
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Obviously <math>k</math> is positive. Then, we can let <math>n_1</math> equal <math>k^3</math> and similarly let <math>n_i</math> equal <math>k^3 + (i - 1)k</math>.  
  
 
The wording of this problem (which uses "exactly") tells us that <math>k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k</math>. Taking away <math>k^3</math> from our inequality results in <math>69k<3k^2+3k+1\leq 70k</math>. Since <math>69k</math>, <math>3k^2+3k+1</math>, and <math>70k</math> are all integers, this inequality is equivalent to <math>69k\leq 3k^2+3k<70k</math>. Since <math>k</math> is positive, we can divide the inequality by <math>k</math> to get <math>69 \leq 3k+3 < 70</math>. Clearly the only <math>k</math> that satisfies is <math>k=22</math>.  
 
The wording of this problem (which uses "exactly") tells us that <math>k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k</math>. Taking away <math>k^3</math> from our inequality results in <math>69k<3k^2+3k+1\leq 70k</math>. Since <math>69k</math>, <math>3k^2+3k+1</math>, and <math>70k</math> are all integers, this inequality is equivalent to <math>69k\leq 3k^2+3k<70k</math>. Since <math>k</math> is positive, we can divide the inequality by <math>k</math> to get <math>69 \leq 3k+3 < 70</math>. Clearly the only <math>k</math> that satisfies is <math>k=22</math>.  

Revision as of 13:09, 13 March 2015

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 1

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 $\frac{23^3 - 1}{22} = \frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553$.

Solution 2

Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$.

The wording of this problem (which uses "exactly") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k$. Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\leq 70k$. Since $69k$, $3k^2+3k+1$, and $70k$ are all integers, this inequality is equivalent to $69k\leq 3k^2+3k<70k$. Since $k$ is positive, we can divide the inequality by $k$ to get $69 \leq 3k+3 < 70$. Clearly the only $k$ that satisfies is $k=22$.

Then, $\frac{n_{70}}{k}=k^2+69=484+69=\boxed{553}$ is the maximum value of $\frac{n_i}{k}$. (Remember we set $n_i$ equal to $k^3 + (i - 1)k$!)

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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