China Mathematical Competition (Jiangxi)

Since the number of non-negative integer solutions of the equation $x_1+x_2+\cdots +x_k=m$ is $C_{m+k-1}^k$, the number of integer solutions, when $x_1\ge 1$ and $x_i\ge 0(i\ge 2)$, is $C_{m+k-2}^{m-1}$. Let $m=7$, the number of lucky numbers with $k$ digits is $p(k)=C_{k+5}^6$.

Since $2005$ is the minimum lucky number of the type $\overline{2abc}$ and $p(1)=C_6^6=1$, $p(2)=C_7^6=7$, $p(3)=C_8^6=28$. Note that the number of four-digit lucky numbers of the type $\overline{1abc}$ is the number of non-negative integer solutions of $a+b+c=6$, i.e. $C_{6+3-1}^6=28$. Thus $1+7+28+28+1=65$ and $2005$ is the $65$-th lucky number, i.e. $a_{65}=2005$, so $n=65$, $5n=325$.

Furthermore $p(4)=C_9^9=84$, $p(5)=C_{10}^9=210$ and ${\sum }_{k=1}^{5}p(k)=330$.

Therefore the last six lucky numbers with $5$ digits, from the largest to the smallest, are $70000$, $61000$, $60100$, $60010$, $60001$, $52000$. So the $325$-th lucky number is $52000$, i.e. $a_{325}=52000$.