4. On the board, the product of the numbers $\overline{\text{IKS}}$ and $\overline{\text{KSI}}$ is written, where the letters correspond to different non-zero decimal digits. This product contains three digits $C$ and one each of $I$, $K$, and 0, and its leading digit is $C$. What is written on the board?
Answer: $112015=521 \cdot 215$.

Solution. Let $p=\overline{\text { UKC }} \cdot \overline{\mathrm{KCV}}$. The numbers $(\mathrm{U}+\mathrm{K}+\mathrm{C})^{2}$ and $(\mathrm{U}+\mathrm{K}+3 \mathrm{C})$ give the same remainders when divided by 9, from which $\mathrm{C} \bmod 9=\frac{(\text { U }+\mathrm{K}+\mathrm{C})(\text { U }+\mathrm{K}+\mathrm{C}-1)}{2} \bmod 9$. The relationship between the numbers $n$ and $\frac{n(n-1)}{2} \bmod 9$ is given in the table:

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

(thereafter the remainders will cyclically repeat). Thus, $\mathrm{C} \in\{1,3,6,9\}$. The case $\mathrm{C}=9$ is impossible, since $\mathrm{C}$ is the leading digit of $p$ and $p<(U+1) \cdot(\mathrm{K}+1) \cdot 10000 \leqslant 900000$. The inequalities hold:

$$
\mathrm{C} \geqslant\left[\frac{\square \cdot \mathrm{K}}{10}\right] \quad \text { and } \quad \mathrm{C} \leqslant \frac{(\mathrm{U}+1)(\mathrm{K}+1)}{10}
$$

Consider three cases.

1) $\mathrm{C}=1$. From the table, we find that $\mathrm{U}+\mathrm{K}+1 \in\{8,11,14,17\}$, that is, $\mathrm{U}+\mathrm{K} \in\{7,10,13,16\}$. Let Ц be the second-to-last digit of $p$. Then $Ц=(1+\mathrm{U} \cdot \mathrm{K}) \bmod 10$. Note that $Ц \neq \mathrm{U}$, since $\mathrm{U}$ is the last digit of $p$. Moreover, Ц $\neq 0$, because the product of different digits $\mathrm{U}$ and $\mathrm{K}$ cannot end in 9. If $Ц=\mathrm{K}$, then $1+(\mathrm{U}-1) \cdot \mathrm{K} \vdots 10$, so $(\mathrm{U}-1) \cdot \mathrm{K}$ ends in 9. The permissible pairs $(\mathrm{U},\mathrm{K})$ are $(4,3),(8,7),(2,9)$. A suitable value of $\mathrm{U}+\mathrm{K}$ will only be obtained for the pair $(4,3)$, but it does not suit us, since $431 \cdot 314=135334$. Thus, $Ц=\mathrm{C}=1$, from which $\mathrm{U} \cdot \mathrm{K} \vdots 10$. The first of the inequalities $(*)$ gives $\mathrm{U} \cdot \mathrm{K} \leqslant 19$. Therefore, $\mathrm{U} \cdot \mathrm{K}=10$, that is, the pair $(\mathrm{U}, \mathrm{K})$ is $(2,5)$ or $(5,2)$. Since $251 \cdot 512=128512$ and $521 \cdot 215=112015$, only the second case suits us.
2) $\mathrm{C}=3$. From the table, we find that $\mathrm{U}+\mathrm{K}+3 \in\{7,12,16\}$, that is, $\mathrm{U}+\mathrm{K} \in\{9,13\}$, since the digits $\mathrm{U}$, $\mathrm{K}$, $\mathrm{C}$ are different. Let $\mathrm{U}+\mathrm{K}=9$. The second of the inequalities $(*)$ gives $(\mathrm{U}+1)(\mathrm{K}+1) \geqslant 30$, from which the pair $(\mathrm{U}, \mathrm{K})$ is $(4,5)$ or $(5,4)$. The first case is impossible, since $p$ will end in 2, and in the second we get $543 \cdot 435=236205$, which also does not suit us. Let $\mathrm{U}+\mathrm{K}=13$. From the first inequality $(*)$ it follows that $\mathrm{U} \cdot \mathrm{K} \leqslant 39$, so the pair $(\mathrm{U}, \mathrm{K})$ is $(4,9)$ or $(9,4)$. Then $p$ will end in 7 or 2, which is impossible.
3) $\mathrm{C}=6$. From the table, we find that $\mathrm{U}+\mathrm{K}+6 \in\{13,15,22\}$, from which $\mathrm{U}+\mathrm{K} \in\{7,9,16\}$. In the first two cases $(\mathrm{U}+1)(\mathrm{K}+1) \leqslant 30$, which is impossible due to $(*)$. Therefore, $\mathrm{U}+\mathrm{K}=16$, that is, the pair $(\mathrm{U},\mathrm{K})$ is $(7,9)$ or $(9,7)$. Then $p$ will end in 2 or 4, which is impossible.