Find the positive integer $n\,$ for which
\[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\]
(For real $x\,$, $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$)
Note that if $2^x \le a1994$. Thus, $k=8$. 
So, $\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}$.
Alternatively, one could notice this is an [arithmetico-geometric series](https://artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series) and avoid a lot of computation.