Problem 4. The magician taught Kashtanka to bark as many times as he showed her secretly. When Kashtanka correctly answered how much two times two is in this way, he hid a delicious cake in a suitcase with a combination lock and said:

- The eight-digit code for the suitcase is the solution to the puzzle УЧУЙ $=\kappa \mathrm{E} \times$ КС. You need to replace the same letters with the same digits, and different letters with different digits, so that the equation is correct. Bark the necessary number of times for each of the eight letters, and you will get a treat.

But then a mishap occurred. Kashtanka, from excitement, barked one more time for each letter than she should have. Of course, the suitcase did not open. Suddenly, a child's voice exclaimed: "Unfair! The dog solved the puzzle correctly!" And indeed, if each digit of the solution the magician had in mind is increased by 1, another solution to the puzzle is obtained!

Can we restore: a) what exact solution the magician had in mind; b) what the number УЧУЙ was in this solution?
[7 points] (A. K. Kulikov, T. A. Korchemkina, I. V. Raskina)
Answer. a) No. b) Yes, УЧУЙ $=2021$.

Solution. a) Note that КЕ and КС represent different numbers, but swapping them does not change the product УЧУЙ. Therefore, for each solution to the puzzle, there is a paired solution where the digits corresponding to $\mathrm{E}$ and $\mathrm{C}$ are swapped. Thus, it is impossible to uniquely determine the solution the magician intended.

b) After increasing all digits by 1, we again get a solution to the puzzle, so,

$$
\begin{aligned}
& \text { УЧУЙ }+1111=(К \mathrm{~K}+11)(К \mathrm{~K}+11)= \\
& \quad=\mathrm{КE} \cdot К \mathrm{C}+11 \cdot \mathrm{KE}+11 \cdot \text { КС }+11 \cdot 11 .
\end{aligned}
$$

Since УЧУЙ $=$ КЕ $\cdot$ КС, we get that

$$
\begin{gathered}
1111=11 \cdot \kappa \mathrm{E}+11 \cdot \kappa \mathrm{C}+11 \cdot 11 \\
101=\kappa \mathrm{E}+\kappa \mathrm{C}+11 \\
\kappa \mathrm{E}+\kappa \mathrm{C}=90
\end{gathered}
$$

which means

$$
\begin{gathered}
10 \cdot \mathrm{K}+\mathrm{E}+10 \cdot \mathrm{K}+\mathrm{C}=90 \\
20 \cdot \mathrm{K}+\mathrm{E}+\mathrm{C}=90
\end{gathered}
$$

Note that $\mathrm{E}+\mathrm{C}$ is a number from 1 to 17, so $20 \cdot \mathrm{K}$ is a number from 73 to 89 that is divisible by 20. Therefore, $20 \cdot К=80$ and $К=4$, and $\mathrm{E}+\mathrm{C}=10$.

The digits $\mathrm{E}$ and $\mathrm{C}$ are different, neither is 4, and neither is 9 (otherwise, adding 1 to each digit would cause a carry-over, which would change exactly one of the digits in the positions of У or К, and the result would not be a solution to the puzzle). Therefore, for the equation $\mathrm{E}+\mathrm{C}=10$, only two options remain: $2+8$ and $3+7$.

Then, $К Е \cdot К С=42 \cdot 48=2016$ or $К \mathrm{E} \cdot К \mathrm{C}=43 \cdot 47=2021$. But in the word УЧУЙ, the first and third letters are the same, so only the variant УЧУЙ $=2021$ fits.