11-29 Let $1990=2^{\alpha_{1}}+2^{\alpha_{2}}+\cdots+2^{\alpha_{n}}$, where $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are distinct non-negative integers. Find $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$.
(China Beijing High School Grade 1 Mathematics Competition, 1990)
$$
\text { [Solution] } \begin{aligned}
1990 & =1024+512+256+128+64+4+2 \\
& =2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{2}+2 .
\end{aligned}
$$

Therefore, $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=10+9+8+7+6+2+1=43$.