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

Latest revision as of 18:48, 28 December 2023

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 less 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

Because if its equal to, then there is no integer in between the two values. - resources

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

@@ Line 1: / Line 1: @@
 == Problem ==
+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 1 ==
+In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.
-== Solution ==
+<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 less than 1000. 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 is 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 ensure 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>\boxed{019}</math>.
+== Solution 2 ==
+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>.
+<cmath>(n + \frac{1}{1000})^3 - n^3 > 1</cmath>
+<cmath>\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1</cmath>
+<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>.
+Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not greater than or equal to? - awesomediabrine
+Because if its equal to, then there is no integer in between the two values. - resources
+== Solution 3 (Similar to Solution 2) ==
+Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath>
+The last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10
+== Solution 4 (Calculus) ==
+Note that the cube root is increasing for positive reals while its derivative is decreasing, so linear approximation gives
+<cmath>\sqrt[3]{n^3+1} - n < \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3} = \frac{1}{3n^2}</cmath>
+and
+<cmath>\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}}</cmath>
+From this, it is clear that <math>n = \boxed{019}</math> is the smallest <math>n</math> for which LHS will be less than <math>\frac{1}{1000}</math>. ~ Hyprox1413
 == See also ==
-* [[1987 AIME Problems]]
+{{AIME box|year=1987|num-b=11|num-a=13}}
+[[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

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