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?
I need to find the values of $n$ that satisfy two conditions: $n<1000$ and $\lfloor \log_{2} n \rfloor$ is a positive even integer. I can start by listing some examples of $n$ that meet the second condition. If $\lfloor \log_{2} n \rfloor = 2$, then $n$ must be between $4$ and $7$, inclusive. That's four values of $n$. If $\lfloor \log_{2} n \rfloor = 4$, then $n$ must be between $16$ and $31$, inclusive. That's sixteen values of $n$. If $\lfloor \log_{2} n \rfloor = 6$, then $n$ must be between $64$ and $127$, inclusive. That's sixty-four values of $n$. If $\lfloor \log_{2} n \rfloor = 8$, then $n$ must be between $256$ and $511$, inclusive. That's two hundred fifty-six values of $n$. If $\lfloor \log_{2} n \rfloor = 10$, then $n$ must be between $1024$ and $2047$, inclusive. But wait, that's too big for the first condition. So I can stop here. The total number of values of $n$ that meet both conditions is $4 + 16 + 64 + 256 = 340$. 