WMC 2018 Senior Problem 40

There are $\sqrt{1000000}=1000$ perfect squares in the given interval and $\sqrt[3]{1000000}=100$ perfect cubes. There are also $\sqrt[6]{1000000}=10$ perfect 6th powers (which are both squares and cubes). so there are $1000+100-10 = \boxed{1090}$ perfect squares and cubes.

$\sqrt[3]{10^6} = 10^2 =100; \sqrt{10^6} = 10^3 = 1000$ So there are $1000 + 100$ perfect squares and cubes. Wait! Aren't we missing the overlap? (Numbers that are both squares and cubes) Well, since, $\sqrt[3]{10^6} = 100$ There are $\sqrt{100} = 10$ squares and cubing each of them gives 10 numbers that are both squares and cubes. Our final step is $1100 - 10 = 1090$