$6 \cdot 30$ If $\lg m=b-\lg n$, then $m$ equals
(A) $\frac{b}{n}$.
(B) $b n$.
(C) $10^{b} n$.
(D) 5 .
(E) $\frac{10^{b}}{n}$.
(1st American High School Mathematics Examination, 1950)
[Solution] Notice that $\lg m=b-\lg n=\lg 10^{b}-\lg n=\lg \frac{10^{b}}{n}$, then $m=\frac{10^{b}}{n}$.
Therefore, the answer is $(E)$.