1-19 Below is a division problem of an eight-digit number by a three-digit number. Try to find the quotient and explain your reasoning.



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[Sol] The original expression is labeled as follows:
All characters represent decimal digits, and characters with subscript 1 are not equal to 0.
(1) Since after the first trial division by 7, two digits $\overline{a_{5} a_{6}}$ are simultaneously moved down, it indicates that $\overline{c_{1} c_{2} a_{5}} < \overline{x_{1} x_{2} x_{3}}$; thus, $y_{2}=0$, and similarly, $y_{4}=0$.
(2) The difference between the four-digit number $\overline{a_{1} a_{2} a_{3} a_{4}}$ and the three-digit number $\overline{b_{1} b_{2} b_{3}}$ is the two-digit number $\overline{c_{1} c_{2}}$. This is only possible when $a_{1}=1, a_{2}=0, b_{1}=9$, and $\overline{a_{3} a_{4}}<\overline{b_{2} b_{3}}$; otherwise, the difference would be a three-digit number.
Similarly, $C_{1}=1, C_{2}=0, d_{1}=9, \overline{a_{5} a_{6}}<\overline{d_{2} d_{3}}$.
(3) Since $C_{2}=0$, it follows that $a_{4}=b_{3}$. Also, because $C_{1}=1$ and $\overline{a_{3} a_{4}}<\overline{b_{2} b_{3}}$, it follows that $a_{3}<b_{2}$, so $a_{3}=0, b_{2}=9$.
(4) Since $7 \cdot \overline{x_{1} x_{2} x_{3}}=\overline{b_{1} b_{2} b_{3}}=\overline{99 b_{3}}$, it follows that $x_{1}=1, x_{2}=4$; otherwise, if $x_{1} \geqslant 2$ or $x_{1}=1, x_{2} \geqslant 5$, then $7 \cdot \overline{x_{1} x_{2} x_{3}} \geqslant 1050$; and if $x_{1}=1, x_{2} \leqslant 3$, then $7 \cdot \overline{x_{1} x_{2} x_{3}}<980$.

From $7 \cdot \overline{14 x_{3}}=\overline{99 b_{3}}$, i.e., $7 \cdot (140 + x_{3}) = 990 + b_{3}$, it follows that $7 \cdot x_{3} = 10 + b_{3}$, so $x_{3}=2, b_{3}=4$, thus $a_{4}=b_{3}=4$.
(5) Since $y_{5} \cdot 142$ is a four-digit number, it follows that $y_{5} \geqslant 8$, i.e., $y_{5}=8$ or 9; and since the last digit of $y_{5} \cdot 142$ must be 8, it follows that $y_{5}=9$.
(6) Since $\overline{e_{1} e_{2} a_{7}} < \overline{x_{1} x_{2} x_{3}}$, it follows that $e_{1}=1$. From $1 a_{5} - d_{2} = e_{1} = 1$, regardless of what $a_{5}$ is, it must be that $d_{2}=9$.

From $y_{3} \cdot \overline{x_{1} x_{2} x_{3}} = \overline{d_{1} d_{2} d_{3}}$, i.e., $y_{3} \cdot 142 = \overline{99 d_{3}}$, it follows that $y_{3}=7, d_{3}=4$ (for the same reason as (4)).
(7) From $\overline{e_{1} e_{2} a_{7} 8} = 9 \times 142 = 1278$, it follows that $e_{2}=2, a_{7}=7$.
Thus, $\overline{c_{1} c_{2} a_{5} a_{6}} = \overline{d_{1} d_{2} d_{3}} + \overline{e_{1} e_{2}} = 994 + 12 = 1006$.
Therefore, $a_{5}=0, a_{6}=6$.
So, the quotient is 70709. The original expression is: