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

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== Solution 2 ==
 
== Solution 2 ==
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We have that <math>(n + 1/1000)^3 = n^3 + 3/10^3 n^2 + 3/10^6 n + 1/10^9</math>. If <math>(n + 1/1000)^3 - n^3 > 1</math>, there exists an integer between <math>(n + 1/1000)^3</math> and <math>n^3</math>, and the cube root of this integer would be between <math>n</math> and <math>n + 1/1000</math>. We seek the smallest <math>n</math> such that <math>(n + 1/1000)^3 - n^3 > 1</math>.  
+
To minimize <math>m</math>, we should minimize <math>n</math>. We have that <math>(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}</math>. For a given value of <math>n</math>, if <math>(n + \frac{1}{1000})^3 - n^3 > 1</math>, there exists an integer between <math>(n + \frac{1}{1000})^3</math> and <math>n^3</math>, and the cube root of this integer would be between <math>n</math> and <math>n + \frac{1}{1000}</math> as desired. We seek the smallest <math>n</math> such that <math>(n + \frac{1}{1000})^3 - n^3 > 1</math>.  
  
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<cmath>(n + 1/1000)^3 - n^3 > 1</cmath>
+
<cmath>(n + \frac{1}{1000})^3 - n^3 > 1</cmath>
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<cmath>3/10^3 n^2 + 3/10^6 n + 1/10^9 > 1</cmath>
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<cmath>\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1</cmath>
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<cmath>3n^2 + 3/10^3 n + 1/10^6 > 10^3</cmath>.
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<cmath>3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3</cmath>
  
 
Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</math>.
 
Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</math>.

Revision as of 14:00, 23 March 2018

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 $019$.

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 $\boxed{019}$.

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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