3-ча 1. In a chess tournament, 12 people participated. After the tournament, each participant made 12 lists. The first list includes only himself, the second list includes him and those he defeated, the third list includes all people from the second list and those they defeated, and so on. The 12th list includes all people from the 11th list and those they defeated. It is known that for any participant in the tournament, the 12th list includes a person who was not in the 11th list. How many draws were played in the tournament?
Solution 1. Answer: 54. If the \((k+1)\)-th list is the same as the \(k\)-th, then the lists numbered \(k+2, \ldots, 11, 12\) will also be exactly the same. However, according to the condition, the 11th list and the 12th list are different. Therefore, each participant's \(k\)-th list contains exactly \(k\) people. In particular, the 2nd list contains exactly two people. This means that each participant won exactly one game. Therefore, the number of draws is \(\frac{12 \cdot 11}{2} - 12 = 54\).

Part 2. Given the numbers: \(4, 14, 24, \ldots, 94, 104\). Prove that it is impossible to first cross out one number, then two more from the remaining, then three more, and finally four more numbers such that after each crossing out, the sum of the remaining numbers is divisible by eleven.

Solution 2. There are 11 numbers in total, and 10 numbers need to be crossed out. Therefore, in the end, one number that is divisible by 11 should remain, i.e., the number 44. On the other hand, the number 44 must be crossed out first. Indeed, the sum of all the given numbers is \(\frac{11 \cdot 108}{2}\), so it is divisible by 11. Consequently, if the sum of the remaining numbers is divisible by 11 after crossing out one number, then the crossed-out number must also be divisible by 11.