10.117 Find the largest positive integer with the following property: its last digit is not 0, and after removing one of its digits (but not the first digit), the resulting number is a divisor of the original number.
(57th Moscow Mathematical Olympiad, 1994)
$$
\begin{aligned}
& \overline{a_{k} \cdots a_{t+1}} \cdot 10^{t}+a_{t} \cdot 10^{t-1}+\overline{a_{t-1} \cdots a_{1}} \\
= & b\left(\overline{a_{k} \cdots a_{t+1}} \cdot 10^{t}+\overline{a_{t-1} \cdots a_{1}}\right) .
\end{aligned}
$$

It is easy to see that $110$, then $t=k-1$. Thus, from (1) we get
$$
\left[a_{k-1}-(b-10) a_{n}\right] \cdot 10^{k-2}=(b-1) \overline{a_{k-2} \cdots a_{1}}
$$
Both sides of (2) are positive integers, i.e., $a_{k-1}>(b-10) a_{k}$.

At this point, (2) becomes $(b-1) \overline{a_{k-2} \cdots a_{1}}=2^{k-2} \cdot 5^{k-2}\left[a_{k-1}-(b-10) a_{k}\right]$
Since $a_{1} \neq 0$, the number $a_{k-2} \cdots a_{1}$ cannot be both a multiple of 2 and a multiple of 5.
When it is a multiple of 2, $b-1$ should be a multiple of $5^{k-2}$, but because $b-17 a_{b}$
Since $a_{6} \geqslant 1$, then $a_{5}>7$, so $a_{5}=8, a_{5}=9$.
When $a_{5}=8$, from (3), $a_{6}=1, b-1=16$, substituting into (4) we get
$$
\overline{a_{4} a_{3} a_{2} a_{1}}=\frac{10000}{16}=625 .
$$

This indicates that the original number is $\overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1}}=180625$.
In fact, $180625=17 \times 10625$. This meets the problem's requirements.
When $a_{5}=9$, from (5), $a_{6}=1$, and $b-1=16$, substituting into (4), we get
$$
\overline{a_{4} a_{3} a_{2} a_{1}}=\frac{20000}{16}=1250 \text {. }
$$

At this point, $a_{1}=0$, which does not meet the requirements.
If $b1$. After similar discussions, when $b-1=8, t-1=3, \overline{a_{3} a_{2} a_{1}}=875$, the right side of (6) is maximized, with $\overline{a_{n} \cdots a_{5}}+a_{4}=7$, so $k=5, a_{5} \leqslant 7$. Therefore, the largest integer at this time is 70875, and $70875=9 \times 7875$, which also meets the conditions.
However, $70875<180625$, so the largest positive integer sought is 180625.