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

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

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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