1987 AIME Problems/Problem 12
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 .
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, 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
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