4. There are four-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It is known that the number $m$ is divisible by 100, and its decimal representation consists of four pairs of identical adjacent digits. Find $m$ and $n$.
Answer: $6325 \cdot 5236=33117700$.

Solution. Let's write

$$
m=\overline{a b c d}, \quad n=\overline{d c b a}, \quad m n=\overline{x x y y z z 00}
$$

It is clear that one of the numbers $m$ and $n$ ends in 5 (let it be $m$). Then $d=5$, and the digit $a$ is even. Obviously, $m n$ is divisible by 11. Since

$$
m \bmod 11=(b+d-a-c) \bmod 11=(-n) \bmod 11
$$

the numbers $m$ and $n$ are divisible by 11, and their product is divisible by 121. Since $\overline{x x y y z z}=11 \cdot \overline{x 0 y 0 z}$, we get $x+y+z \vdots 11$, so $x+y+z$ is equal to 11 or 22. Note that

$$
(a+b+c+d)^{2} \bmod 9=m n \bmod 9=\overline{x x y y z z 00} \bmod 9=2(x+y+z) \bmod 9
$$

The relationship between the numbers $p$ and $p^{2} \bmod 9$ is given in the table:

| $p$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p^{2} \bmod 9$ | 1 | 4 | 0 | 7 | 7 | 0 | 4 | 1 | 0 |

(remainders will cyclically repeat thereafter). Since $44 \bmod 9=8$, the case $x+y+z=22$ is impossible. If $x+y+z=11$, taking into account $d=5$ we get $(a+b+c+5)^{2} \bmod 9=22 \bmod 9=4$. In addition, $b+5-a-c$ is divisible by 11. Let's consider two cases.

1) Suppose $a+c=b+5 \pm 11$. Then

$4=(a+b+c+5)^{2} \bmod 9=(2(b+5) \pm 11)^{2} \bmod 9=4(b+5 \pm 1)^{2} \bmod 9 \Longleftrightarrow(b+5 \pm 1)^{2}=1 \bmod 9$,

and by the table $b+5 \pm 1$ is 8 or 10. In the case of a plus, $a+c$ will be 18 or 20, which is impossible. In the case of a minus, $a+c$ will be non-positive, which is also impossible due to $a>0$.

2) Suppose $a+c=b+5$. Then by the table $b+5$ is 8 or 10, that is, $b$ is 3 or 5. Calculating the second-to-last digit of $m n$, we get

$$
0=\left(5 b+a c+\frac{a}{2}\right) \bmod 10=\left(5+a c+\frac{a}{2}\right) \bmod 10
$$

Since $a$ is even, the permissible pairs $(a, c)$ for $b=3$ are $(2,6),(4,4),(6,2),(8,0)$, and for $b=5$ they are $(2,8),(4,6),(6,4),(8,2)$. Substituting these pairs into the right side of $(*)$, we get

$$
8,3,0,9,2,1,2,5
$$

Thus, only $m=6325$ satisfies $(*)$. Since $6325 \cdot 5263=33117700$, this number gives the answer.