8. Let the set $T=\{0,1,2,3,4,5,6\}, M=\left\{\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\frac{a_{4}}{7^{4}}\right\}, a_{i} \in \mathbf{T}, i=\{1,2,3,4\}$, arrange the numbers in $M$ in descending order, then the 2005th number is $\qquad$.
8. Use $\left[a_{1} a_{2} \cdots a_{k}\right]_{p}$ to represent a $k$-digit $p$-ary number. Multiplying each number in set $M$ by $\mathcal{T}^{4}$ gives $M^{\prime}=\left\{a_{1} \cdot 7^{3}+a_{2} \cdot 7^{2}+a_{3} \cdot 7+a_{4}\left|a_{i} \in T, i=1,2,3,4\right|=\left\{\left[a_{1} a_{2} a_{3} a_{4}\right]_{7}\left|a_{i} \in T, i=1,2,3,4\right|, M^{\prime}\right.\right\}$, where the largest number in $M^{\prime}$ is $[6666]_{7}=[2400]_{10}$. Starting from 2400 and listing numbers in descending order, the 2005th number is $2400-2004=396$, and $[396]_{10}=[1104]_{7}$. Dividing this number by $\tau^{4}$ gives the required number as $\frac{1}{7}+\frac{1}{7^{2}}+\frac{0}{7^{3}}+\frac{4}{7^{4}}$.