(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 $2^{n +1}$.

We'll find a maximal $m$ for which $$\sum_{k=1}^mk2^k\leq1994.$$
Indeed, $$\sum_{k=1}^mk2^k=2\sum_{k=1}^mk2^{k-1}=2\left(\sum_{k=1}^mx^{k}\right)'_{x=2}=2\left(\frac{x(x^m-1)}{x-1}\right)'_{x=2}=$$
$$=2\cdot\frac{(m+1)2^m-1-2^{m+1}+2)}{(1-1)^2}=(m+1)2^{m+1}-2^{m+2}+2.$$
Id est, $$(m+1)2^{m+1}-2^{m+2}+2\leq1994,$$ which gives $m=7$.
Now, $$\frac{1994-((7+1)2^{7+1}-2^{7+2}+2)}{8}=57$$ and we obtain:
$$n=1+2^1+...+2^7+57=2^8-1+57=312.$$
