1987 AIME Problems/Problem 12

Problem

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

Solution

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 $3n^2 + 3nr + r^2 \geq \frac{1}{r} > \frac{1000}$ (Error compiling LaTeX. Unknown error_msg). This means that the smallest possible $n$ should be quite a bit smaller than 1000, so 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 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 insure 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 $019$ .

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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1987 AIME Problems/Problem 12

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