AIME 1990 · 第 1 题

Solution 1

Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$. This happens to be $23^2=529$. Notice that there are $23$ squares and $8$ cubes less than or equal to $529$, but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numbers in our sequence less than $529$. Magically, we want the $500th$ term, so our answer is the biggest non-square and non-cube less than $529$, which is $\boxed{528}$.

Solution 2

This solution is similar as Solution 1, but to get the intuition why we chose to consider $23^2 = 529$, consider this:

We need $n - T = 500$, where $n$ is an integer greater than 500 and $T$ is the set of numbers which contains all $k^2,k^3\le 500$.

Firstly, we clearly need $n > 500$, so we substitute n for the smallest square or cube greater than $500$. However, if we use $n=8^3=512$, the number of terms in $T$ will exceed $n-500$. Therefore, $n=23^2=529$, and the number of terms in $T$ is $23+8-2=29$ by the Principle of Inclusion-Exclusion, fulfilling our original requirement of $n-T=500$. As a result, our answer is $529-1 = \boxed{528}$.