1. Find all eight-digit numbers such that by erasing some quartet of consecutive digits, we obtain a four-digit number that is 2019 times smaller.
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(Pavel Calábek)
SOLUTION. In the first part of the solution, we will assume that in the sought eight-digit number $N$, we have crossed out the first four digits. These form a four-digit number, which we will denote as $A$. The remaining four digits of the number $N$ form a four-digit number $B$, which is mentioned in the problem, and it seems that $N=10^{4} A+B$. The required property is therefore expressed by the equation $10^{4} A+B=2019 B$ or $5000 A=1009 B$. Since the numbers 5000 and 1009 are coprime, the four-digit number $B$ must be a multiple (also a four-digit number) of 5000, whose double is already a five-digit number. The condition of divisibility is thus satisfied by only one number $B=5000$. For it, from the equation $5000 A=1009 B$ we get $A=1009$, so in total $N=10095000(=2019 \times 5000)$.

In the second part of the solution, we will consider all other cases of erasing four digits of the eight-digit number $N$. Then among them, there is no first digit of the number $N$, which we will denote as $a$. We will show that after such an erasure, the number $N$ decreases by more than $5000 k$ times (so it cannot be a solution to the problem).

Indeed, the number $N$ with the decimal representation $N=10^{7} a+\ldots$ will turn (after the considered erasure of 4 digits) into a number with the representation $10^{3} a+\ldots$, which is certainly less than $10^{3}(a+1)$, for the announced reduction, it is enough to prove the inequality $10^{7} a \geqq 5000 \cdot 10^{3}(a+1)$. However, this inequality holds, as after dividing both sides by the number $5 \cdot 10^{6}$, it turns into the inequality $2 a \geqq a+1$ or $a \geqq 1$. It is enough to add that the first digit $a$ of the representation of the number $N$ is always different from zero.

Answer. The sought eight-digit number is unique, and that is 10095000.

SUPPLEMENTARY AND ADDITIONAL PROBLEMS:

N1. Determine all four-digit numbers from which, after erasing the first two-digit number, we get a two-digit number that is 69 times smaller. [1725, 3450, and 5175. The sought number is of the form $10^{2} A+B$, where the two-digit numbers $A, B$ satisfy the equation $10^{2} A+B=69 B$ or $25 A=17 B$. Due to the coprimality of the numbers 17 and 25, it follows that $B$ is a multiple of the number 25.]

N2. Prove that if we erase two of the last three digits of a four-digit number, we get a number more than 50 times smaller. [The original number is of the form $\overline{c \star \star \star}$, the reduced one is of the form $\overline{c \star}$, so the first of them is at least $10^{3} c$, while the second number is less than $10(c+1)$. Therefore, it is enough to verify the inequality $10^{3} c \geqq 50 \cdot 10(c+1)$, which is satisfied for every $c \geqq 1$.]

D1. Find all four-digit numbers $\overline{a b c d}$ for which $\overline{a b c d}=20 \cdot \overline{a b}+16 \cdot \overline{c d}$. $[65-\mathrm{C}-\mathrm{S}-1]$

D2. Determine the largest two-digit number $k$ with the following property: there exists a natural number $N$, from which, after erasing the first digit from the left, we get a number $k$ times smaller. (After erasing the digit, the representation of the number may start with one or several zeros.) For the determined number $k$, then find the smallest suitable number $N$. [56-C-II-4]

D3. For which rational number $r>1$ does there exist the most of those four-digit numbers, from which, by erasing the first two-digit number, we get a two-digit number that is $r$ times smaller? [For $r=101$. According to the notation from problem N1, we are looking for such $r$ for which there exists the largest number of pairs $A$, $B$ of two-digit numbers satisfying the equation $10^{2} A+B=r B$ or $10^{2} A=(r-1) B$. The number $B$ is uniquely determined by the number $A$ for a given $r$, and $A \in\{10,11, \ldots, 99\}$. From this, it follows that for each $r$ there are at most 90 suitable numbers, and we get this number only when for each two-digit $A$, $B=10^{2} A /(r-1)$ is also a two-digit integer. This is satisfied only for $r=101$, when $B=A$, so all 90 suitable four-digit numbers are of the form $\overline{a b a b}$.]