On the card, a four-digit natural number is written, from which we can strike out any two digits and always get a two-digit natural number that is divisible by 5 without a remainder. How many such four-digit natural numbers exist? (Note, for example, the number 06 is not a two-digit number.)

(L. Šimünek)
We denote the digits of the suitable four-digit number as follows: $v$ is in the thousands place, $x$ in the hundreds place, $y$ in the tens place, and $z$ in the units place. A natural number is divisible by five if and only if it has the digit 0 or 5 in the units place. After crossing out two digits from the original number, the units place can be occupied by the digits $x, y$, or $z$, so these digits can only be 0 or 5. According to the problem, after crossing out any two digits, a two-digit number is formed. The tens place of such a number can be occupied by the digits $v, x$, or $y$, so these digits cannot be 0.

Summarizing the above observations, the digit $v$ can be any digit from 1 to 9, the digits $x$ and $y$ can only be 5, and the digit $z$ can be 0 or 5. Therefore, there are $9 \cdot 2 = 18$ four-digit numbers that satisfy the conditions.

Scoring: 1 point for recognizing that after crossing out, the digits $x, y, z$ can end up in the units place; 1 point for the condition that $v, x, y \neq 0$; 1 point for determining that $x$ and $y$ are 5; 1 point for recognizing that $z$ is 0 or 5; 1 point for the number of possibilities for the digit $v$; 1 point for the result of 18.

Award the full score if the student correctly identifies the conditions and arrives at the number 18 by listing all the valid numbers. If the solution lists all possibilities without any commentary, award a maximum of 4 points.