China Mathematical Competition (Jiangxi)

Let $[a_1{a}_2\cdots a_k{]}_p$ be a number base $p$ with $k$ digits. Multiply each number in $M$ by $7^4$, and we get $$M^{\prime }=\{a_1{7}^3+a_2{7}^2+a_{3}7+a_4;a_i\in T,i=1,2,3,4\}=\{[a_1{a}_2{a}_3{a}_4{]}_7^{\prime }\mid a_i\in T,i=1,2,3,4\}.$$ The maximum number in $M^{\prime }$ is $[6666{]}_7=[2400{]}_{10}$. In the decimal system, starting from 2 400 in the descending order, the 2 005-th number is $2400-2004=396$. But $[396{]}_{10}=[1104{]}_7$. Divide this number by $7^4$, we get a number in $M$, that is $\frac{1}{7}+\frac{1}{7^2}+\frac{0}{7^3}+\frac{4}{7^4}$. Answer: C.