# 2017 AIME I Problems/Problem 13

## Problem 13

For every $m \geq 2$, let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$, there is always a perfect cube $k^3$ in the range $n < k^3 \leq m \cdot n$. Find the remainder when $$\sum_{m = 2}^{2017} Q(m)$$is divided by 1000.

## Solution 1

Lemma 1: The ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases.

Lemma 2: If the range $(n,mn]$ includes $y$ cubes, $(p,mp]$ will always contain at least $y-1$ cubes for all $p$ in $[n,+\infty)$.

If $m=14$, the range $(1,14]$ includes one cube. The range $(2,28]$ includes 2 cubes, which fulfills the Lemma. Since $n=1$ also included a cube, we can assume that $Q(m)=1$ for all $m>14$. Two groups of 1000 are included in the sum modulo 1000. They do not count since $Q(m)=1$ for all of them, therefore $$\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000$$

Now that we know this we will find the smallest $n$ that causes $(n,mn]$ to contain two cubes and work backwards (recursion) until there is no cube in $(n,mn]$.

For $m=2$ there are two cubes in $(n,2n]$ for $n=63$. There are no cubes in $(31,62]$ but there is one in $(32,64]$. Therefore $Q(2)=32$.

For $m=3$ there are two cubes in $(n,3n]$ for $n=22$. There are no cubes in $(8,24]$ but there is one in $(9,27]$. Therefore $Q(3)=9$.

For $m$ in $\{4,5,6,7\}$ there are two cubes in $(n,4n]$ for $n=7$. There are no cubes in $(1,4]$ but there is one in $(2,8]$. Therefore $Q(4)=2$, and the same for $Q(5)$, $Q(6)$, and $Q(7)$ for a sum of $8$.

For all other $m$ there is one cube in $(1,8]$, $(2,16]$, $(3,24]$, and there are two in $(4,32]$. Therefore, since there are 10 values of $m$ in the sum, this part sums to $10$.

When the partial sums are added, we get $\boxed{059}\hspace{2 mm}QED\hspace{2 mm} \blacksquare$

This solution is brought to you by a1b2

## Solution 2

We claim that $Q(m) = 1$ when $m \ge 8$.

When $m \ge 8$, for every $n \ge Q(m) = 1$, we need to prove there exists an integer $k$, such that $n < k^3 \le m*n$.

That because $\sqrt{m*n} - \sqrt{n} \ge 2\sqrt{n} - \sqrt{n} = \sqrt{n} \ge 1$, so k exists between $\sqrt{m*n}$ and $\sqrt{n}$ $\sqrt{n} < k \le \sqrt{m*n}$.

We can then hand evaluate $Q(m)$ for $m = 2,3,4,5,6,7$, and get $Q(2) = 32$, $Q(3) = 9$, and all the others equal 2.

There are a total of 2010 integers from 8 to 2017. $$\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{7} Q(m) + 2010 \equiv 32+9+2+2+2+2+10 = \boxed{059} \mod 1000$$

-AlexLikeMath

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