[b]p1.[/b] How many five-digit numbers are there which will produce the number $2012$ if one digit is crossed out?


[b]p2.[/b] A classroom floor is colored using two colors. Prove that there are two identically colored points exactly $1$ foot apart.


[b]p3.[/b] Can you draw a closed
a) $6$-segment
b) $7$-segment
c) $8$-segment
polygonal line which self-crosses each of its segments exactly once? Example on the picture shows a $5$-segment closed polygonal line which self-crosses each of its segments twice.
[img]https://cdn.artofproblemsolving.com/attachments/0/d/3926c84ced20315c9776c111b6661fc226fde4.png[/img]


[b]p4.[/b] Jack wants to enter a wonder cave. In front of the entrance there is a round table with $4$ identical hats lying symmetrically along the circle. There are $4$ identical coins, one under each hat. Jack can lift any two hats, examine the two coins, turn them as he likes and put the hats back. After this Jack closes his eyes and the table starts spinning, and when it stops Jack cannot tell by how much the table rotated. Then again he can choose two hats and so on. The wonder cave will open if and only if the coins are either all HEADS or all TAILS up. How must Jack act to enter the cave?

Remark: Lifting hats randomly and turning all coins, say, HEADS up is not a winning strategy. Jack may be so unlucky that he never lifts a hat which covers TAIL.


PS. You should use hide for answers.
1. To solve the problem of finding how many five-digit numbers will produce the number $2012$ if one digit is crossed out, we need to consider the following cases:

    - Case 1: The digit to be crossed out is at the beginning of the number.
    - Case 2: The digit to be crossed out is not at the beginning of the number.

    Let's analyze each case in detail:

    **Case 1: The digit to be crossed out is at the beginning of the number.**

    If the digit to be crossed out is at the beginning, then the number must be of the form $x2012$, where $x$ is any digit from $1$ to $9$ (since it is a five-digit number and cannot start with $0$). Therefore, there are $9$ possible numbers in this case.

    **Case 2: The digit to be crossed out is not at the beginning of the number.**

    If the digit to be crossed out is not at the beginning, then the number must be of the form $2x012$, $20x12$, $201x2$, or $2012x$, where $x$ can be any digit from $0$ to $9$. Each of these forms provides $10$ possible numbers (one for each digit from $0$ to $9$).

    Since there are $4$ positions where the digit can be crossed out (excluding the first position), we have $4 \times 10 = 40$ possible numbers in this case.

    Adding the numbers from both cases, we get:
    \[
    9 + 40 = 49
    \]

    Therefore, the total number of five-digit numbers that will produce the number $2012$ if one digit is crossed out is $\boxed{49}$.