4. On the board, the product of the three-digit numbers $\overline{\text{IKS}}$ and $\overline{\text{KSI}}$ is written, where the letters correspond to different non-zero decimal digits. This product reads the same from left to right and from right to left, two of its digits are 4, and the other four digits match I. What is written on the board?
Answer: $477774=762 \cdot 627$ or $554455=593 \cdot 935$.

Solution. Let $p=\overline{\text { ИКС }} \cdot \overline{\mathrm{KCИ}}$. From the condition on $p$, it follows that $p \vdots 11$ and, therefore, one of the factors of $p$ is divisible by 11. Let's first assume that $\overline{\text { ИКС }} \vdots 11$. This means that И $+\mathrm{C}-\mathrm{K} \vdots 11$, from which either И $+\mathrm{C}=\mathrm{K}$ or $И+\mathrm{C}=\mathrm{K}+11$. Also note that the numbers $(И+\mathrm{K}+\mathrm{C})^{2}$ and 4 И +8 give the same remainders when divided by 9. Let's consider two cases.

1) $И+\mathrm{C}=\mathrm{K}$. Then $(2 \mathrm{~K})^{2} \bmod 9=(4 И+8) \bmod 9$, that is, $\mathrm{K}^{2} \bmod 9=(И+2) \bmod 9$. The dependence 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). Therefore, $(И+2) \bmod 9 \in\{0,1,4,7\}$ and И $\in\{7,8,2,5\}$. Substituting $(И+2) \bmod 9$ into the bottom row of the table, we will get $К$ in the top row. If $И=5$, then $\mathrm{K}$ is 4 or 5, which is impossible. For И $=8$ we get $\mathrm{K}=8=$ И, which does not fit the condition. The values И $=7$ and И $=2$ correspond to the pairs $(\mathrm{K}, \mathrm{C})=(9,2)$ and $(\mathrm{K}, \mathrm{C})=(7,5)$. Since

$$
792 \cdot 927=734184, \quad 275 \cdot 752=206800
$$

these cases do not fit us either.
2) $И+\mathrm{C}=\mathrm{K}+11$. Then

$$
(4 \mathrm{U}+8) \bmod 9=(2 \mathrm{~K}+11)^{2} \bmod 9=4(\mathrm{~K}+1)^{2} \bmod 9 \Longleftrightarrow(\mathrm{K}+1)^{2} \bmod 9=(\mathrm{U}+2) \bmod 9
$$

and again И $\in\{7,8,2,5\}$. The cases И $=2$ and И $=8$ do not fit: for permissible values of $\mathrm{K}$ from the equality $\mathrm{C}=\mathrm{K}-$ И +11 we get $\mathrm{C} \geqslant 10$. For the remaining И, the following triples (И,К,С) are found according to the table: $(5,3,9),(7,2,6),(7,5,9)$. Since

$$
539 \cdot 395=212905, \quad 726 \cdot 267=193842, \quad 759 \cdot 597=453123
$$

they do not fit us either.

Now let $\overline{\mathrm{KCИ}} \vdots$ and 11. This means that either И $+\mathrm{K}=\mathrm{C}$ or $\mathrm{U}+\mathrm{K}=\mathrm{C}+11$. By swapping $\mathrm{K}$ and С in the previous reasoning, we get that the triple (И, К, С) takes one of the following values: $(2,5,7),(7,2,9),(7,9,5),(7,6,2),(5,9,3)$. Since

$$
257 \cdot 572=147004, \quad 729 \cdot 297=216513, \quad 795 \cdot 957=760815, \quad 762 \cdot 627=477774, \quad 593 \cdot 935=554455
$$

only the last two triples fit us.