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Difference between revisions of "1987 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
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Let <math>\displaystyle m</math> be the smallest [[integer]] whose [[cube root]] is of the form <math>\displaystyle n+r</math>, where <math>\displaystyle n</math> is a [[positive integer]] and <math>\displaystyle r</math> is a [[positive]] [[real number]] less than <math>\displaystyle 1/1000</math>. Find <math>\displaystyle n</math>.
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Let <math>m</math> be the smallest [[integer]] whose [[cube root]] is of the form <math>n+r</math>, where <math>n</math> is a [[positive integer]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>.
 
== Solution ==
 
== Solution ==
 
In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.
 
In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Revision as of 18:09, 4 July 2013

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

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, 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 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 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 $019$.

See also

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
All AIME Problems and Solutions

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