6) Let the set $T=\{0,1,2,3,4,5,6\}, M=\left\{\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\right.$ $\left.\left.\frac{a_{4}}{7^{4}} \right\rvert\, a_{i} \in T, i=1,2,3,4\right\}$, arrange the elements of $M$ in descending order, then the 2005th number is ( ).
(A) $\frac{5}{7}+\frac{5}{7^{2}}+\frac{6}{7^{3}}+\frac{3}{7^{4}}$
(B) $\frac{5}{7}+\frac{5}{7^{2}}+\frac{6}{7^{3}}+\frac{2}{7^{4}}$
(C) $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{4}{7^{4}}$
(D) $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{3}{7^{4}}$
Let $\left[a_{1} a_{2} \cdots a_{k}\right]_{p}$ represent a $k$-digit number in base $p$. Multiplying each number in the set $M$ by $7^{4}$, we get
$$
\begin{aligned}
M^{\prime} & =\left\{a_{1} \cdot 7^{3}+a_{2} \cdot 7^{2}+a_{3} \cdot 7+a_{4} \mid a_{i} \in T, i=1,2,3,4\right\} \\
& =\left\{\left[a_{1} a_{2} a_{3} a_{4}\right]_{7} \mid a_{i} \in T, i=1,2,3,4\right\} .
\end{aligned}
$$
The largest number in $M^{\prime}$ is $[6666]_{7}=[2400]_{10}$.
In decimal numbers, the 2005th number from the largest starting from 2400 is $2400-2004=396$. And $[396]_{10}=[1104]_{7}$. Dividing this number by $7^{4}$, we get the number in $M$ as $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{4}{7^{4}}$. Therefore, the answer is C.