Difference between revisions of "1987 AIME Problems/Problem 12"
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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. 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. | <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. 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>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 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 == | == Solution 2 == | ||
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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>. | ||
+ | |||
+ | 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) == | == Solution 3 (Similar to Solution 2) == | ||
− | Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < | + | 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 == | == See also == |
Latest revision as of 16:41, 6 October 2022
Contents
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
Because if its equal to, then there is no integer in between the two values. - resources
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 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 |
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