In the decimal system, the 18-digit number

$x 3640548981270644 y$

is divisible by 99. Determine the digits $x$ and $y$.
Solution. It is known that if $g \geqq 2$, then a number A written in the $g$-base number system is divisible by $(g-1)$ if and only if the sum of its "digits" - of course, we are talking about the digits in the $g$-base number system - is divisible by $(g-1)$.

In our case, we can easily get the representation of the given number in the $g=100$-base number system: the new digits can be obtained by reading the digits of the decimal representation in pairs from the back; that is,

\[
\begin{gathered}
A=a_{0}+a_{1} \cdot 100+a_{2} \cdot 100^{2}+a_{3} \cdot 100^{3}+a_{4} \cdot 100^{4}+a_{5} \cdot 100^{5}+a_{6} \cdot 100^{6}+a_{7} \cdot 100^{7}+a_{8} \cdot 100^{8} \\
\quad \text{where } a_{0}=40+y ; a_{1}=64 ; a_{2}=70 ; a_{3}=12 ; a_{4}=98 ; a_{5}=48 ; a_{6}=5 ; a_{7}= \\
=64 ; a_{8}=10 x+3.
\end{gathered}
\]

The sum of the "digits" is $404+10 x+y=4 \cdot 99+8+10 x+y$, so the given number is divisible by 99 if and only if

\[
r=8+10 x+y
\]

is also divisible by 99.

Since $0<x \leqq 9$ and $0 \leqq y \leqq 9$, it follows that $0<10 x+y \leqq 99$, that is, $8<r \leqq 107$. Among the found bounds, there is only one number divisible by 99, which is 99 itself. Thus, $r=99, 10 x+y=91$, so - due to the uniqueness of the decimal representation - $x=9$ and $y=1$.

Since our steps were reversible, the above choice indeed results in a number divisible by 99.

Based on the work of Gábor Simon (Szolnok, Varga Katalin Gymnasium, II. grade)

Remark. Many solvers split the condition into two parts, separately using the known conditions for divisibility by 9 and 11 in the decimal system: a number written in the decimal system is divisible by 9 if and only if the sum of its digits is divisible, and it is divisible by 11 if and only if the sum of the digits in even and odd positions gives the same remainder when divided by 11, or the difference of these two sums is divisible by 11. Based on this, they obtained two conditions for $x$ and $y$, and - correctly - reasoned further that a positive integer is divisible by 99 if and only if it is divisible by 9 and 11.