$6 \cdot 54$ Using only the following simplified table: $10^{3}=1000,10^{4}=10000,2^{10}=$ 1024, $2^{11}=2048,2^{12}=4096,2^{13}=8192$, determine the best choice for the interval in which $\lg 2$ lies:
(A) $\frac{3}{10}$ and $\frac{4}{11}$.
(B) $\frac{3}{10}$ and $\frac{4}{12}$.
(C) $\frac{3}{10}$ and $\frac{4}{13}$.
(D) $\frac{3}{10}$ and $\frac{40}{132}$.
(E) $\frac{3}{11}$ and $\frac{40}{132}$.
(18th American High School Mathematics Examination, 1967)
［Solution］From the given information, we have
$$
10^{3}=1000<1024=2^{10}, \quad 2^{13}=8192<10^{4},
$$

Therefore,
$$
3<10 \lg 2, \quad 13 \lg 2<4 \text {. }
$$

That is,
$$
\frac{3}{10}<\lg 2<\frac{4}{13} \text {. }
$$

Note that $\lg 2<\frac{40}{132}$, i.e., $2^{132}<10^{40}$, cannot be determined from the given conditions, and (A), (B) are weaker conclusions compared to (C).
Therefore, the answer is (C).