(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 $2^{n +1}$.

As noted, the sequence goes like
$$ 0,\underset{2}{\underbrace{1,1}},\underset{4}{\underbrace{2,2,2,2}},\underset{8}{\underbrace{3,3,3,3,3,3,3,3}},4,4,\ldots$$
i.e, every natural number $k$ occurs $2^k$ times.
So desired is $$ \sum k\cdot 2^k =1994$$
It's quick enough to attack directly:
$$ 1\cdot2 + 2\cdot4 + 3\cdot8 + 4\cdot16 + 5\cdot 32 + 6\cdot 64 + 7\cdot 128 = 1538$$
Next up is $8$ repeating $x$ times till $1994$ $$1538 + 8\cdot x = 1994$$
$$\Rightarrow x=57 $$
Last term of our sequence can be found by counting the number of repeating units: $$n = (1+2+4+\ldots+128) + 57 = \boxed{312}$$
