5*. Find the largest natural number, not ending in zero, which, when one (not the first) digit is erased, decreases by an integer factor.

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5. Let $x$ be the deleted digit, $a$ be the part of the number to the left of $x$, and $c$ be the part of the number to the right of $x$. Then the number has the form $\overline{a x c}$, see fact 11. Suppose the digit $x$ is in the $(n+1)$-th place (counting from the right). Then

$$
\overline{a x c}=a \cdot 10^{n+1}+x \cdot 10^{n}+c .
$$

After deleting the digit $x$, the resulting number is $\overline{a c}=a \cdot 10^{n}+c$. Consider the ratio of the original number to the resulting number

$$
r=\frac{a \cdot 10^{n+1}+x \cdot 10^{n}+c}{a \cdot 10^{n}+c}, \quad \text { where } c0$. Therefore,

$$
-8 \leqslant l \leqslant 9
$$

It is also clear that $l \neq 0$ (otherwise, the decimal representation of the number $c$ ends in zero).

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-132.jpg?height=40&width=886&top_left_y=444&top_left_x=179)

Proof. Consider two cases: $l>0$ and $l0$. From equation (3), it follows that $x-l a>0$, so,

$$
a18$ ). Therefore, $n \leqslant 4$. Let $n=4$, then $l+9=16$, and equation (3) can be rewritten as

$$
(x-7 a) \cdot 5^{4}=c
$$

Since $x$ is a digit, $a=1, x=8$ or $x=9$. For $x=9$, the number $c$ ends in zero, and thus is not suitable. For $x=8$, we get $c=625$ and the answer

$$
\overline{a x c}=180625
$$

$2^{\circ}$. Suppose the number $c$ does not contain fives. The number $l+9$ is divisible by a power of five no higher than the first, so $n \leqslant 1$, and the number is definitely not the maximum.