Difference between revisions of "1987 AIME Problems/Problem 12"
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− | We have that <math>(n + 1 | + | 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 + 1 | + | <cmath>(n + \frac{1}{1000})^3 - n^3 > 1</cmath> |
− | <cmath>3 | + | <cmath>\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1</cmath> |
− | <cmath>3n^2 + 3 | + | <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
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 .
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.