2. Find all eight-digit numbers with this property: if we strike out the first two and the last two digits of the number, we get a four-digit number that is 2019 times smaller than the original number.

(Pavel Calábek)
Solution. In any suitable eight-digit number $N$, let $A$ be its first two-digit number, $B$ the following four-digit number, and $C$ the last two-digit number. If we interpret $A, B, C$ as numbers written in the decimal system, then $10 \leq A \leq 99$, $1000 \leq B \leq 9999$ (according to the problem, the number $B$ is four-digit, so its representation does not start with zero), and $0 \leq C \leq 99$. The original number $N$ then has the representation $N=10^{6} A+10^{2} B+C$, which, according to the problem, must satisfy the equation

$$
10^{6} A+10^{2} B+C=2019 B \text { or } 10^{6} A+C=1919 B \text {. }
$$

Let's write it considering the factorization $1919=19 \cdot 101$ as the equation

$$
(101-1)^{3} A+C=19 \cdot 101 \cdot B
$$

from which it follows that the number 101 is a divisor of the integer $C-A$, because

$$
(101-1)^{3} A+C=\left(101^{3} A-3 \cdot 101^{2} A+3 \cdot 101 A-A\right)+C=101 k+(C-A),
$$

where $k$ is a suitable integer. However, from the given estimates for the numbers $A$ and $C$, the inequalities $-99 \leq C-A \leq 89$ follow, and in the interval $[-99,89]$ there is only one multiple of the number 101, namely the number 0, so it must be $C-A=0$. Substituting $C=A$ into the simplified equation, we get

$$
\left(10^{6}+1\right) A=19 \cdot 101 \cdot B
$$

whose both sides we can now divide by the number 101, since $10^{6}+1=101 \cdot 9901$, and thus arrive at the final equation

$$
9901 A=19 B \text {. }
$$

From this, due to the coprimality of the numbers 9901 and 19 (since $9901=521 \cdot 19+2$), it follows that the four-digit number $B$ is a multiple of the number 9901, and therefore the last equation is satisfied in our situation only if $B=9901$ and $A=19$ (and thus also $C=19$).

Answer. The problem is satisfied by the unique eight-digit number 19990119.

For a complete solution, award 6 points. Award 1 point for appropriate notation of the numbers $A, B, C$ and setting up the equation, 2 points for the observation that 101 must divide $A \cdot 10^{6}+1$, 1 point for deriving $101 \mid C-A$, 1 point for deriving the equality $C=A$, and 1 point for solving the equation $9901=19 B$ (the coprimality of the numbers 9901 and 19 must be at least mentioned, but its proof is not required).

If the solver does not apply divisibility by 101 and only reaches the conclusion $1919 \mid A \cdot 10^{6}+1$, award 1 point.