4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible.
Answer: 220000001.

Solution. Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that

$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a_{7}+a_{k} a_{8} \quad \text { for any } \quad k \in\{0, \ldots, 8\}
$$

Clearly, $b_{0}=a_{0} a_{8} \bmod 10$. Since $10^{17}>m n \geqslant a_{0} a_{8} \cdot 10^{16}$, we get $a_{0} a_{8} \leqslant 9$, which means $b_{0}=a_{0} a_{8}$. Suppose that for some $k0$, we get $a_{0}=1$ and $a_{k}=0$ for $k=1, \ldots, 6$. The number $m=220000001$ clearly satisfies the condition of the problem.