$6 \cdot 49$ Given $10^{-221}<M<10^{-220}, 10^{-2}<n<10^{-1},$ the mantissa of the common logarithm of $M$ is $n$, then the mantissa of the common logarithm of $\frac{1}{M^{9}}$ is
(A) $\frac{1}{n^{9}}$.
(B) $-9 n$.
(C) $9 n-1$.
(D) $1-9 n$.
(China Guangzhou, Wuhan, Fuzhou and other five cities junior high school mathematics league, 1988)
[Solution] From $\lg \frac{1}{M^{9}}=-9 \lg M=-9(-221+n)$
$$
=1988+(1-9 n) \text {. }
$$

It is also known that $10^{-2}-9 n>-0.9 \text {. }
$$

Then $0.1<1-9 n<0.91$, so $0<1-9 n<1$, meaning $1-9 n$ is the mantissa of the logarithm. Therefore, the answer is $(D)$.