4. On the board, the product of the numbers $\overline{\text { IKS }}$ and $\overline{\text { ISK }}$ is written, where the letters correspond to different non-zero decimal digits, one of which is even. This product is a five-digit number and reads the same from left to right and from right to left. What is written on the board?
Answer: $29392=167 \cdot 176$.

Solution. Since the problem does not change when swapping $\mathrm{K}$ and $\mathrm{C}$, we can assume that $\mathrm{K}<\mathrm{C}$. Let $p=\overline{\text { IKS }} \cdot \overline{\text { ISK }}$. The number $p$ is a five-digit number and $p \geqslant 10000 \cdot \text{I}^2$, from which it follows that $\text{I} \leqslant 3$. We consider three cases.

1) $\text{I} = 1$. Since $p<40000$, the leading digit of $p$ does not exceed 3. It coincides with the last digit of $p$, which is $\mathrm{K} \cdot \mathrm{C} \bmod 10$. By the condition, $\mathrm{K} \cdot \mathrm{C}$ is even, so $\mathrm{K} \cdot \mathrm{C} \bmod 10=2$. Since the digits $\mathrm{K}$ and $\mathrm{C}$ are different from $\text{I}$ and have different parity, the pair $(\mathrm{K}, \mathrm{C})$ can be $(3,4)$, $(6,7)$, $(8,9)$. Since

$$
134 \cdot 143=19162, \quad 167 \cdot 176=29392, \quad 189 \cdot 198=37422,
$$

only the pair $(6,7)$ fits.

2) $\text{I} = 2$. The leading digit of $p$ lies between 4 and 8. It coincides with the last digit of $p$, which is $\mathrm{K} \cdot \mathrm{C} \bmod 10$ and, by the condition, is odd. Therefore, $\mathrm{K} \cdot \mathrm{C} \bmod 10$ is 7 or 5. In the first case, the pair $(\mathrm{K}, \mathrm{C})$ can be $(1,7)$ or $(3,9)$. Since $217 \cdot 271=58807$ and $237 \cdot 273=64701$, neither option fits. Let $\mathrm{K} \cdot \mathrm{C} \bmod 10=5$. Then one of the digits $\mathrm{K}$ and $\mathrm{C}$ is 5. Note that $p \geqslant 40000+2000(\mathrm{K}+\mathrm{C})$. From this, $\mathrm{K}+\mathrm{C} \leqslant 9$, otherwise the leading digit of $p$ will be greater than 5. Therefore, the pair $(\mathrm{K}, \mathrm{C})$ is $(1,5)$ or $(3,5)$. Since $215 \cdot 251=53965$ and $235 \cdot 253=59455$, neither case fits.
3) $\text{I}=3$. Then $p \geqslant 90000+3000(\mathrm{K}+\mathrm{C})$, from which $\mathrm{K}+\mathrm{C} \leqslant 3$ and $(\mathrm{K}, \mathrm{C})=(1,2)$. But the product $312 \cdot 321$ is a six-digit number, which does not fit.