The 200-digit number 89252525... 2525 is multiplied by the number 444x18y27 (x and y are unknown digits). It turned out that the 53rd digit of the resulting number (counting from the right) is 1, and the 54th is 0. Find x and y.

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Answer: $x=4, y=6$. Let $A=89252525 \ldots 2525, B=444 x 18 y 27$. Represent the number $B$ as a sum

$B=444 x 00 y 00+18027$. The product $A \cdot 18027$ has the form $3 . .272727272727268175$, in particular, its 53rd digit is 7, and the 54th digit is 2. Let the number $C \cdot(B-18027) / 4$ be written with the digits $111 a b c d e f$. Then $A B=A \cdot 18027+4 A (B-18027)/4= 3..2727272727268175+3..1010101010100 \cdot C$. We will perform the multiplication and addition 'in column'. Then the 53rd digit in the number $A B$ will be equal to $7+1+1+b+d+f+x$, where $x$ is the number of tens that we carry from the 52nd place. The digit with number 54 is equal to $2+1+a+c+e+y$, where $y$ is the number of tens that we carry from the 53rd place. Clearly, $x \leq 4$. Note that $a, d \in 0 ; 1 ; 2, b, e \in 0 ; 2 ; 5 ; 7, c, f \in 0 ; 5$, and $b$ and $c$ cannot be equal to 5 simultaneously. From this we get the estimate $7+1+1+b+d+f+x \leq 7+1+1+7+2+5+4=27$, hence $y \leq 2$. But by the condition, the 53rd digit of the sum is 1, so $y \geq 1$. Further, $2+1+a+c+e+y \leq 2+1+2+5+7+2=19$. But the last digit of this number is zero, so $2+1+a+c+e+y=10$. In particular, from the 54th place to the 55th we carry exactly 1 tens. But from the form of our addends it follows that from the 54th place to the 55th we carry as many tens as from the 52nd to the 53rd. Therefore, $x=1$. We get that the following system of equations must be satisfied: either $2+1+a+c+e+2=10,7+1+1+b+d+f+1=21(1)$, or $2+1+a+c+e+1=10,7+1+1+b+d+f+1=11$ (2). Since each of the digits $a, b, c, d, e, f$ can take a small number of values, a simple enumeration gives the unique solution $a b c d e f=100150$ (considering that the digits $b c, e f \neq 05,55)$.

## Answer

$x=4, y=6$.