Four. (50 points) A decimal number of the form 0. $a_{1} a_{2}^{(k)} \cdots a_{n}^{(k)} \cdots$ (denoted as $A\left(a_{1}, k\right)$), where the digit $a_{1}$ is any natural number from 1 to 9; and when $a_{1}$ is given, $a_{2}^{(k)}$ is the unit digit of the product $k a_{1}$, and $a_{n}^{(k)}$ is the unit digit of the product $k a_{n-1}^{(k)}$ $(n=3,4, \cdots, k)$. Find the sum $\sum_{k=1}^{9} \sum_{a_{1}=1}^{9} A\left(a_{1}, k\right)$.
Four, Solution: We have $\sum_{k=1}^{9} \sum_{a_{1}=1}^{9} A\left(a_{1}, k\right)=\sum_{\substack{k=1 \\ k \neq 5}}^{9} A(5, k)+\sum_{a_{1}=1}^{9} A\left(a_{1}, 5\right)+\sum_{\substack{k=1 \\ k \neq 5}}^{9} \sum_{\substack{a_{1}=1 \\ a_{1} \neq 5}}^{9} A\left(a_{1}, k\right)(1)$
(1) It is clear that when $k=1,3,5,7,9$, $A(5, k)=0 . \dot{5}=\frac{5}{9}$; when $k=2,4,6,8$, $A(5, k)=0.5$. Therefore, $\sum_{\substack{k=1 \\ k \neq 5}}^{9} A(5, k)=\frac{5}{9} \times 4+0.5 \times 4=\frac{38}{9}$.
(2) Furthermore, when $a_{1}=1,3,5,7,9$, $A\left(a_{1}, 5\right)=0 . a_{1} 5$. When $a_{1}=2,4,6,8$, $A\left(a_{1}, 5\right)=0.2,0.4, 0.6,0.8$. Therefore, $\sum_{a_{1}=1}^{9} A\left(a_{1}, 5\right)=\frac{5}{9} \times 5+2=\frac{43}{9}$ (2);
At this point, we only need to study $\sum_{\substack{k=1 \\ k \neq 5}}^{9} \sum_{\substack{a_{1}=1 \\ a_{1} \neq 5}}^{9} A\left(a_{1}, k\right)=\sum_{\substack{k=1 \\ k \neq 5}}^{9} \sum_{\substack{ \\a_{1}=1}}^{4}\left[A\left(a_{1}, k\right)+A\left(10-a_{1}, k\right)\right]$. At this time, for a given $a_{1}, k$, there must exist a natural number $l$ such that $10(l-1)<a_{1} k<10 l,(3)$;
Thus, $10 k-10 l<10 k-a_{1} k<10 k-10(l-1)$,
which means $10(k-l)<\left(10-a_{1}\right) k<10(k-l+1)(4)$;
From equations (3) and (4), we know that the sum of the second digits of the decimals $A\left(a_{1}, k\right)$ and $A\left(10-a_{1}, k\right)$ is $\left[a_{1} k-10(l-1)\right]+ \left[\left(10-a_{1}\right) k-10(k-l)\right]=10$.
Similarly, it can be proven that the sum of the digits in any corresponding decimal place is always 10.
Therefore, at this time, $A\left(a_{1}, k\right)+A\left(10-a_{1}, k\right)=1 . i=\frac{10}{9}$ (5);
Thus, we have $\sum_{k=1}^{9} \sum_{a_{1}=1}^{9} A\left(a_{1}, k\right)=\frac{38}{9}+\frac{43}{9}+\sum_{\substack{k=1 \\ k \neq 5}}^{9} \sum_{a_{1}=1}^{1} \frac{10}{9}=\frac{38}{9}+\frac{43}{9}+\frac{10}{9} \times 32=\frac{401}{9}=44 \frac{5}{9}$.