Difference between revisions of "1987 AIME Problems/Problem 12"
m (→See also: c) |
m (→Solution) |
||
Line 6: | Line 6: | ||
<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 | + | 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 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
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, so 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 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 insure a small enough . The smallest possible value for is 1, when . Then for this value of , , and we're set. The answer is .
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 |