Question 178
Let $T = \{0, 1, 2, 3, 4, 5, 6\}$ and $M = \{\frac{a_1}{7} + \frac{a_2}{7^2} + \frac{a_3}{7^3} + \frac{a_4}{7^4}; a_i \in T, i = 1, 2, 3, 4\}$. Arrange the numbers in $M$ in the descending order. Then the 2 005-th 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) $\frac{1}{7} + \frac{1}{7^2} + \frac{0}{7^3} + \frac{3}{7^4}$
(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) $\frac{1}{7} + \frac{1}{7^2} + \frac{0}{7^3} + \frac{3}{7^4}$