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

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m (Solution)
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<math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be quite a bit smaller than 1000, so in particular <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   
 
<math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be quite a bit smaller than 1000, so in particular <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   
  
In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to insure a small-enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>019</math>.
+
In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to insure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>019</math>.
 
 
  
 
== See also ==
 
== See also ==

Revision as of 13:17, 24 October 2007

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 $3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000$. 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 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions
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