Difference between revisions of "1987 AIME Problems/Problem 12"
(→Solution 2) |
Hyprox1413 (talk | contribs) (added solution with calculus) |
||
Line 22: | Line 22: | ||
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> | 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 | 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 n for which LHS will be less than <math>\frac{1}{1000}</math>. ~ Hyprox1413 | ||
== See also == | == See also == |
Revision as of 14:38, 31 December 2021
Contents
[hide]Problem
Let be the smallest integer whose cube root is of the form , where is a positive integer and is a positive real number less than . Find .
Solution 1
In order to keep as small as possible, we need to make as small as possible.
. Since and is an integer, we must have that . This means that the smallest possible should be quite a bit smaller than 1000. In particular, should be less than 1, so and . , so we must have . Since we want to minimize , we take . Then for any positive value of , , so it is possible for to be less than . However, we still have to make sure a sufficiently small exists.
In light of the equation , we need to choose as small as possible to ensure a small enough . The smallest possible value for is 1, when . Then for this value of , , and we're set. The answer is .
Solution 2
To minimize , we should minimize . We have that . For a given value of , if , there exists an integer between and , and the cube root of this integer would be between and as desired. We seek the smallest such that .
Trying values of , we see that the smallest value of that works is .
Why is it and not greater than or equal to? - awesomediabrine
Solution 3 (Similar to Solution 2)
Since is less than , we have . Notice that since we want minimized, should also be minimized. Also, should be as close as possible, but not exceeding . This means should be set to . Substituting and simplifying, we get The last two terms in the right side can be ignored in the calculation because they are too small. This results in . The minimum positive integer that satisfies this is . ~ Hb10
Solution 4 (Calculus)
Note that the cube root is increasing for positive reals while its derivative is decreasing, so linear approximation gives and From this, it is clear that is the smallest n for which LHS will be less than . ~ Hyprox1413
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.