# Difference between revisions of "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 $\boxed{019}$.

## 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 $\boxed{019}$.

Why is it $(n + \frac{1}{1000})^3 - n^3 > 1$ and not greater than or equal to? - awesomediabrine

Because if its equal to, then there is no integer in between the two values. - resources

## Solution 3 (Similar to Solution 2)

Since $r$ is less than $1/1000$, we have $\sqrt[3]{m} < n + \frac{1}{1000}$. Notice that since we want $m$ minimized, $n$ should also be minimized. Also, $n^3$ should be as close as possible, but not exceeding $m$. This means $m$ should be set to $n^3+1$. Substituting and simplifying, we get $$\sqrt[3]{n^3+1} < n + \frac{1}{1000}$$ $$n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}$$ The last two terms in the right side can be ignored in the calculation because they are too small. This results in $1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}$. The minimum positive integer $n$ that satisfies this is $\boxed{019}$. ~ Hb10

## Solution 4 (Calculus)

Note that the cube root is increasing for positive reals while its derivative is decreasing, so linear approximation gives $$\sqrt[3]{n^3+1} - n < \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3} = \frac{1}{3n^2}$$ and $$\sqrt[3]{n^3+1} - n > \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3+1} = \frac{1}{3\sqrt[3]{(n^3+1)^2}}$$ From this, it is clear that $n = \boxed{019}$ is the smallest $n$ for which LHS will be less than $\frac{1}{1000}$. ~ Hyprox1413