# 1987 AIME Problems/Problem 12

## Problem

Let $m$ be the smallest integer whose cube root is of the form $n+r$, where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$. Find $n$.

## Solution 1

In order to keep $m$ as small as possible, we need to make $n$ as small as possible.

$m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3$. Since $r < \frac{1}{1000}$ and $m - n^3 = r(3n^2 + 3nr + r^2)$ is an integer, we must have that $3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000$. This means that the smallest possible $n$ should be quite a bit smaller than 1000. In particular, $3nr + r^2$ should be less than 1, so $3n^2 > 999$ and $n > \sqrt{333}$. $18^2 = 324 < 333 < 361 = 19^2$, so we must have $n \geq 19$. Since we want to minimize $n$, we take $n = 19$. Then for any positive value of $r$, $3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000$, so it is possible for $r$ to be less than $\frac{1}{1000}$. However, we still have to make sure a sufficiently small $r$ exists.

In light of the equation $m - n^3 = r(3n^2 + 3nr + r^2)$, we need to choose $m - n^3$ as small as possible to ensure a small enough $r$. The smallest possible value for $m - n^3$ is 1, when $m = 19^3 + 1$. Then for this value of $m$, $r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}$, and we're set. The answer is $019$.

## Solution 2

To minimize $m$, we should minimize $n$. We have that $(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}$. For a given value of $n$, if $(n + \frac{1}{1000})^3 - n^3 > 1$, there exists an integer between $(n + \frac{1}{1000})^3$ and $n^3$, and the cube root of this integer would be between $n$ and $n + \frac{1}{1000}$ as desired. We seek the smallest $n$ such that $(n + \frac{1}{1000})^3 - n^3 > 1$.

$$(n + \frac{1}{1000})^3 - n^3 > 1$$ $$\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1$$ $$3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3$$

Trying values of $n$, we see that the smallest value of $n$ that works is $\boxed{019}$.