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
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 | |
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