(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $ 
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor  + \ldots + \lfloor \log_2 n \rfloor  = 1994 $$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.

Denote the sum by $S_{n}$. Note that for any $k\in\mathbb{N}$ there are $2^k$ positive integers $x$ for which $[\log_{2}(x)]=k$, and those are $x=2^{k},2^{k}+1,\ldots,2^{k+1}-1$. Thus we have $$S_{2^{k}-1}=0 + (1+1) + (2+2+2+2+) + \cdots + \bigl((k-1)+(k-1)+\cdots + (k-1)\bigr)$$ where there number of $(k-1)$ terms is $2^{k-1}$.  It follows that $$S_{2^{k}-1} = (k-2)2^{k}+2$$ Putting $k=8$ we see that $S_{255}=1538<1994$ and putting $k=9$ we see that $S_{511}=3586>1994$. Thus it's clear that our $n$ should satisfy $2^{8}-1<n<2^{9}-1$. Now we have $$1994=S_{n}=S_{255}+(n-255)8=8n-502$$ which gives $n=312$.
