GAIC Math 2024 Problem 26

Let $T=\{0,1,2,3,4,5,6\}$ and $M=\left\{\left.\frac{a_1}{7}+\frac{a_2}{7^2}+\frac{a_3}{7^3}+\frac{a_4}{7^4}\right|a_i\in T,i=1,2,3,4\right\}$. If the elements of $M$ are arranged in descending order, then the 2005th number is \underline{\hspace{2cm}}. \begin{align*} \text{A)}\ & \frac{5}{7}+\frac{5}{7^2}+\frac{6}{7^3}+\frac{3}{7^4}& \text{B)}\ & \frac{5}{7}+\frac{5}{7^2}+\frac{6}{7^3}+\frac{2}{7^4}\ \text{C)}\ & \frac{1}{7}+\frac{1}{7^2}+\frac{0}{7^3}+\frac{4}{7^4}& \text{D)}\ & \frac{1}{7}+\frac{1}{7^2}+\frac{0}{7^3}+\frac{3}{7^4} \ \end{align*}