Problem: For each real number $x$, let $\lfloor x \rfloor$ denote the greatest integer that does not exceed $x$. For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer? Solution: Playing around, we see $n = 2^1$ or $2^1+1$ fails and $2^2+1, 2^2+2, 2^2+3$ work. Then, $2^3, 2^3+1, 2^3+2, \ldots, 2^3+7$ fail. The pattern is that if $k$ is odd, then $2^k+r$ for $0 \leq r \leq 2^k-1$ is not such an integer, and everything else is. The $k$ we concern ourselves with are $k = 1,3,5,7,9$. For each, we see there are $2^1$ numbers that fail when $k=1$, $2^3$ numbers that fail for $k=3$, and so on. We must be careful, though, since for $k=9$, we only concern ourselves with numbers lt; 1000$ and $2^9 = 512$ meaning there are $1000-512 = 488$ $(+1)$ numbers that fail for $k=9$ (as opposed to the usual $2^9$). We add $1$ to account for the edge case. Thus, we get $1000-(2^1+2^3+2^5+2^7+489) = 341$. But this counts $1$ which does not yield a positive even integer when taking logs so we get $340$. In this case, it is better to directly count instead of do so complementary.