102 Let $1991=2^{\alpha_{1}}+2^{a_{2}}+\cdots+2^{\alpha_{n}}$, where $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are distinct non-negative integers, then $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=$ $\qquad$
102 43. Since $1991=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{2}+2^{1}+2^{0}$, we have $\alpha_{1}+\alpha_{2}+\cdots$ $+\alpha_{n}=10+9+\cdots+0=43$.