4. On the board, the product of the three-digit numbers $\overline{K C I}$ and $\overline{\text { ICK }}$ is written, where the letters correspond to different decimal digits. The record of this product consists of three pairs of identical adjacent digits. What is written on the board?
Answer: $224455=385 \cdot 583$.

Solution. Since the problem does not change when I and K are swapped, we will assume I $<\mathrm{K}$. Let $\overline{\mathrm{KCИ}} \cdot \overline{\text { ИСK }}=\overline{x x y y z z}$. It is obvious that the right side is divisible by 11. Since

$$
\overline{\mathrm{KCU}} \bmod 11=(\mathrm{U}+\mathrm{K}-\mathrm{C}) \bmod 11=\overline{\mathrm{UCK}} \bmod 11,
$$

the numbers $\overline{\mathrm{KCИ}}$ and $\overline{\text { ИСK }}$ 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 11 or 22. Note that

$$
(\mathrm{U}+\mathrm{K}+\mathrm{C})^{2} \bmod 9=\overline{\mathrm{KCU}} \cdot \overline{\mathrm{UCK}} \bmod 9=\overline{x x y y z z} \bmod 9=2(x+y+z) \bmod 9
$$

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

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

(thereafter the remainders will cyclically repeat). Since $44 \bmod 9=8$, the case $x+y+z=22$ is impossible. If $x+y+z=11$, we get $(И+\mathrm{K}+\mathrm{C})^{2} \bmod 9=22 \bmod 9=4$, and from the table $(И+\mathrm{K}+\mathrm{C}) \bmod 9 \in\{2,7\}$. In addition, И $+\mathrm{K}-\mathrm{C}$ is divisible by 11, so either $\mathrm{C}=\mathrm{U}+\mathrm{K}$ or $\mathrm{C}=\mathrm{U}+\mathrm{K}-11$. Let's consider these cases.

1) Suppose $\mathrm{C}=\mathrm{U}+\mathrm{K}$. Then $2 \mathrm{C} \bmod 9 \in\{2,7\}$. If $2 \mathrm{C} \bmod 9=2$, then $\mathrm{C}=1$, which is impossible since $И+\mathrm{K} \geqslant 2$. Let $2 \mathrm{C} \bmod 9=7$. Then $\mathrm{C}=8$, and the pair $(И, \mathrm{~K})$ can take the values $(1,7),(2,6),(3,5)$. Since $781 \cdot 187=146047,682 \cdot 286=195052,583 \cdot 385=224455$, only the third case fits.
2) Suppose $\mathrm{C}=И+\mathrm{K}-11$. Then $\mathrm{C} \leqslant 9+8-11=6$. Since

$$
2(C+1) \bmod 9=(2 C+11) \bmod 9=(И+K+C) \bmod 9 \in\{2,7\}
$$

we get $\mathrm{C}=9$ or $\mathrm{C}=7$, which is impossible.