6.46 Given that the mantissa of the common logarithm of a positive number $M$ is $\frac{1}{a}$, then the mantissa of the common logarithm of $\frac{1}{M}$ is
(A) $\frac{1}{a}$.
(B) $\frac{a-1}{a}$.
(C) $\frac{-1}{a}$.
(D) $a$.
(China Guangzhou, Wuhan, Fuzhou, and other five cities Junior High School Mathematics League, 1984)
[Solution] $\lg \frac{1}{M}=\lg 1-\lg M=-\lg M$.
Given that $\frac{1}{a}$ is the mantissa of $\lg M$, then $0<\frac{1}{a}<1$.
Let its characteristic be $b$, then
$$
\lg M=b+\frac{1}{a} .
$$

And $-\lg M=-b-\frac{1}{a}=(-b-1)+\left(1-\frac{1}{a}\right)$. Therefore, from $0<\frac{1}{a}<1$ we know $0<1-\frac{1}{a}<1$, thus the mantissa of $\lg \frac{1}{M}$ is $1-\frac{1}{a}$, i.e., $\frac{a-1}{a}$. Hence, the answer is (B).