1990 AIME Problems/Problem 1
Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than . This happens to be . Notice that there are squares and cubes less than or equal to , but and are both squares and cubes. Thus, there are numbers in our sequence less than . Magically, we want the term, so our answer is the biggest non-square and non-cube less than , which is .
This solution is similar as Solution 1, but to get the intuition why we chose to consider , consider this:
We need , where is an integer greater than 500 and is the set of numbers which contains all .
Firstly, we clearly need , so we substitute n for the smallest square or cube greater than . However, if we use , the number of terms in will exceed . Therefore, , and the number of terms in is by the Principle of Inclusion-Exclusion, fulfilling our original requirement of . As a result, our answer is .
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