answer:1. **Representation of the Problem Using Graph Theory:** - We represent the 21 people as 21 vertices in a graph G. - If two people talk on the phone, we draw an edge between the corresponding vertices. 2. **Key Information and Assumptions:** - In graph G, we know there are a total of 102 edges. - We are also given that there exists a cycle of odd length m in G. - Our goal is to prove that there exists a triangle (a cycle of length 3) within G, indicating that there are three people who each talked to the other two. 3. **Analysis of the Shortest Odd Cycle:** - Let C be the shortest odd cycle in G, with length 2k+1 where k geq 1. - If k=1, then C is a triangle (a cycle of length 3), and our proof is complete. - Suppose k geq 2. Represent C as: [ v_1, v_2, ldots, v_{2k+1}, v_1 ] - Since C is the shortest odd cycle, any pair of vertices v_i and v_j (where i neq j, and |i-j| neq 1, 2k) must not be directly connected by an edge. Otherwise, we would form a shorter odd cycle. 4. **Using Turán's Theorem:** - Consider the graph formed by removing the vertices of C. There are 21-(2k+1) = 20-2k other vertices in G. - According to Turán's Theorem, the maximum number of edges in a triangle-free graph with n vertices is leftlfloor frac{n^2}{4} rightrfloor. For n = 20 - 2k, this gives: [ leftlfloor frac{(20-2k)^2}{4} rightrfloor = (10-k)^2 ] 5. **Summing the Edges:** - Any additional vertex not in C can be connected to at most k vertices within C to avoid creating a triangle. - Total edges in the graph thus are: [ e leq (edges , in , C) + (edges , from , vertices , not , in , C) + (edges , between , remaining , vertices) ] [ = 2k+1 + k(20-2k) + (10-k)^2 ] [ = 2k+1 + 20k - 2k^2 + 100 - 20k + k^2 ] [ = 102 - (k-1)^2 ] 6. **Contradiction:** - Since the number of edges is fixed at 102: [ k = 2 implies 102 - 1 = 101 ] - As this counting yields fewer than 102 edges, there's a contradiction unless there is a triangle in the graph G. # Conclusion: [ boxed{text{There exists at least one triangle in graph } G text{, which means there are three people who mutually talked to each other.}} ]

answer:1. **Problem description**: We have to decide which player wins given two initial matchstick piles and the rule of reducing the number of matchsticks in one pile by a divisor of the number of matchsticks in the other pile. We need to analyze the positions to determine winning strategies for each scenario. 2. **Key observation**: Winning positions are states where each pile has an odd number of matchsticks. Therefore, to determine the winner, we must check the parity (odd or even) of the amount of matchsticks in the piles. 3. **Analysis of position (a)**: - Initial piles: 101 matchsticks and 201 matchsticks. - Check parity: Both 101 and 201 are odd numbers. - Thus, (101, 201) is a winning position for the player who faces it. Since (101, 201) is a winning position, if it's the beginning configuration, the second player (the one who faces the moves of the first player) will always have a winning strategy provided they play optimally. 4. **Conclusion for (a)**: - The second player wins. - The reason: If the position at the start is winning and it is the second player's turn to play, they can always make a move to ensure the opponent is left with a losing position. 5. **Analysis of position (b)**: - Initial piles: 100 matchsticks and 201 matchsticks. - Check parity: 100 is even, 201 is odd. - Thus, (100, 201) is not a winning position for the player who faces it. Since (100, 201) is not a winning position, it means the first player who faces this position can guarantee a win by forcing the opponent into a losing position after their turn. 6. **Conclusion for (b)**: - The first player wins. - The reason: The position at the start is losing if you face it and advantageous if you are the one to move next. # Conclusion: - Point (a): boxed{text{The second player wins}} - Point (b): boxed{text{The first player wins}}

answer:For the function to be meaningful, it is only necessary that x-1>0, which solves to x>1. Therefore, the domain of the function is (1, +infty). Hence, the correct choice is boxed{D}.

answer:Let's denote the number of eggs with double yolks as ( x ). Each of these eggs contributes 2 yolks to the total count. The remaining eggs in the carton, which would be ( 12 - x ), each contribute 1 yolk to the total count. The total number of yolks in the carton is given as 17. So we can set up the following equation: [ 2x + (12 - x) = 17 ] Solving for ( x ): [ 2x + 12 - x = 17 ] [ x + 12 = 17 ] [ x = 17 - 12 ] [ x = 5 ] So, there were boxed{5} eggs with double yolks in the carton.