6. Let the set $T=\{0,1,2,3,4,5,6\}, M=\left\{\left.\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\frac{a_{4}}{7^{4}} \right\rvert\, a_{i} \in T, i=1,2,3,4\right\}$. 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. $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{3}{7^{4}}$
Notice that $\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\frac{a_{4}}{7^{4}}=\frac{1}{7^{4}}\left(a_{1} \cdot 7^{3}+a_{2} \cdot 7^{2}+a_{3} \cdot 7+a_{4}\right)$, so when $a_{1}=6,5,4,3,2$, the number of elements is $5 \cdot 7^{3}=1715$; when $a_{1}=1$ and $a_{2}=6,5,4,3,2$, the number of elements is $5 \cdot 7^{2}=245$; when $a_{2}=1$ and $a_{3}=6,5,4,3,2,1$, the number of elements is $6 \cdot 7=42$; in total, $1715+245+42=2002$. The next three numbers are $1106,1105,1104$, so the 2005th number is $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{4}{7^{4}}$. Therefore, the answer is $C$.