6.48 Given that the characteristic of the common logarithm of the reciprocal of a positive number $M$, $\frac{1}{M}$, is $a$, and the mantissa is $b(b \neq 0)$, then the characteristic and mantissa of the common logarithm of $M$ should be
(A) Characteristic $-a$, Mantissa $-b$.
(B) Characteristic $1-a$, Mantissa $-1-b$.
(C) Characteristic $\frac{1}{a}$, Mantissa $\frac{1}{b}$.
(D) Characteristic $-a-1$, Mantissa $-b+1$.
(China Fujian Province Junior High School Mathematics Competition, 1983)
[Solution] Let $\lg \frac{1}{M}=a+b$, i.e., $-\lg M=a+b$, then
$$
\lg M=-a-b,
$$

and since $1>b>0$, then $0<1-b<1$.
$$
\therefore \quad \lg M=-(a+1)+(1-b) \text {. }
$$

Thus, $-(a+1)$ is the characteristic of $\lg M$, and $1-b$ is its mantissa. Therefore, the answer is $(D)$.