$6 \cdot 47 \lg M$ has the first digit $a$, then the first digit of $\lg \frac{1000}{M}$ is
(A) $2-a$.
(B) $3-a$.
(C) $2-a$ or $3-a$.
(D) None of the above.
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1989)
［Solution］From $\lg \frac{1000}{M}=3-\lg M=3-a($ if $\lg M=a)$, or $(2-a)+[1-(\lg M-a)]$ (if $a<\lg M<a+1)$, so the first digit of $\lg \frac{1000}{M}$ is
$3-a$ (if $\lg M=a$) or $2-a$ (if $a<\lg M<a+1$. Therefore, the answer is $(C)$.