2007 AIME II Problems/Problem 7

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

2007 AIME II (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
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2007 AIME II Problems/Problem 7

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Problem

Solution

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