[ Rebus $]$

Authors: Kumygein A.K., Raskina I.V.

The magician taught Kastanka to bark as many times as he showed her secretly from the audience. When Kastanka 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 rebus UCHUI $=$ KE $\times$ KS. 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. Kastanka, from excitement, barked one more time than necessary for each letter. Of course, the suitcase did not open. Suddenly, a child's voice said, "Unfair! The dog solved the rebus correctly!" And indeed, if each digit of the solution the magician had in mind is increased by 1, it results in another solution to the rebus!

Can we determine: a) what exact solution the magician had in mind; b) what the number UCHUI was in this solution?

#
a) Note that KE and KS replace different numbers, but swapping them does not change the product UCHU. Therefore, for each solution to the puzzle, there is a paired solution where the digits corresponding to E and S are swapped. Thus, it is impossible to uniquely reconstruct the solution the magician intended.

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

UCHU $+1111=(KE+11)(KS+11)=$ KE $\cdot$ KS + $11 \cdot \text{KE}+11 \cdot$ KS $+11 \cdot 11$.

Since UCHU $=$ KE $\cdot$ KS, we get that

$1111=11 \cdot \text{KE}+11 \cdot \text{KS}+11 \cdot 11$

$101=\text{KE}+\text{KS}+11$

$\text{KE}+\text{KS}=90$,

which means

$10 \cdot \text{K}+\text{E}+10 \cdot \text{K}+\text{C}=90$

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

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

Then $KE \cdot KS=42 \cdot 48=2016$ or $\text{KE} \cdot \text{KS}=43 \cdot 47=2021$. But in the word UCHU, the first and third letters are the same, so only the variant UCHU $=2021$ fits.

## Answer

a) No. b) Yes, UCHU = 2021.