# 2007 AIME II Problems/Problem 7

## 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 $k=\lfloor\sqrt{n_{1}}\rfloor = \lfloor\sqrt{n_{2}}\rfloor = \cdots = \lfloor\sqrt{n_{70}}\rfloor$ 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

### Solution 1

For $x = 1$, we see that $\sqrt{1} \ldots \sqrt{7}$ all work, giving 7 integers. For $x=2$, we see that in $\sqrt{8} \ldots \sqrt{26}$, 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 $\sqrt{x^3} \ldots \sqrt{(x+1)^3 - 1}$ 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 $k^3 + (i - 1)k$!)

## Video Solution by the Beauty of Math

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