1987 AIME Problems/Problem 12
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 .
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 to . This means should be set to . Substituting and simplifying, we get Note that 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
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.