Let’s analyze the problem step by step: **Restatement:** - There are \( n \) teams. - Each team plays every other team **once**: \(\binom{n}{2}\) matches. - Win: 3 points, Draw: 1 point, Loss: 0 points. - After the tournament, **some team has:** - **the most points** - **but has won the fewest games** (among all teams). We are to find the **smallest \( n \)** for which this is possible. --- ### 1. **Understanding the Requirement** - "Most points, fewest wins": the team with the highest points must have the lowest number of wins of all teams. If a team has the most points but the fewest wins, it must have earned many points from draws (since a win is worth more, but they have fewer wins). --- ### 2. **Let’s Try Small Values of \( n \)** #### **Case \( n = 2 \):** - Only one match. - Impossible: the winner of the match has most points and most wins. #### **Case \( n = 3 \):** - 3 teams: A, B, C. - Matches: AB, AC, BC. Suppose one team has fewest wins but most points. But a win is 3 points; if you have fewest wins, you must get most points from draws. Try all draws: all teams have 2 points. - No possibility. #### **Case \( n = 4 \):** - Teams: A, B, C, D. - Each plays 3 matches. Suppose team A has 0 wins, but the most points. All draws: each team earns 3 points (not unique). Suppose A has 1 win: then others must have at least 1 win as well (since there are 6 matches, 6 wins in total). Try to distribute more draws to A, but the other teams will have at least as many wins. **Conclusion:** For \( n = 2, 3, 4 \), **not possible.** --- ### 3. **Try \( n = 5 \)** - 5 teams: A, B, C, D, E. - Each plays 4 matches. Suppose team A has **0 wins** and gets only draws: 4 points. If any team wins at least one match, it gets at least 3 points (for one win, more if it draws others). But other teams will also have at least as many points (since A only draws, the points are distributed). Let’s try to see if there's a way for a team with fewer wins (maybe 1) to have more points than all others. Let's try to **construct a scenario:** Suppose Team A: 1 win, 3 draws: \(1 \cdot 3 + 3 \cdot 1 = 6\) points. Other teams: Let's see if they can have at least 2 wins (since A has the fewest, and only 1), but get less than 6 points. But 2 wins alone is 6 points already, and possibly more with draws. So for \(n=5\), it *still seems impossible*. --- ### 4. **Try \( n = 6 \)** - 6 teams: Each plays 5 matches. Suppose Team A has **all draws:** 5 points. Suppose Team A has 1 win, 4 draws: 3 + 4 = 7 points. Can we arrange for other teams to have at least 2 wins, but fewer points than Team A? Suppose Team B: 2 wins (6 points), 3 losses (0 points): 6 points. So now, Team A (with 1 win) has 7 points, Team B (with 2 wins) has 6 points. Is it possible for every other team to have at least 2 wins, but fewer points than A? Let’s check the *total number of wins* in the tournament. Total number of matches: \( \binom{6}{2} = 15 \) matches. Each match is either a win/loss or draw. Let’s say Team A: 1 win, 4 draws = 7 points. The other 5 teams: Each plays 5 matches (one against A, 4 against others). For A's results: - 1 win (vs B) - 4 draws (vs C, D, E, F) So: - B loses to A (0 points), plays C, D, E, F. - C, D, E, F all drew with A. Let’s try to assign two wins to each of the other teams, but arrange so that their points are less than 7. But as before, 2 wins = 6 points, even with 3 losses, that's 6 points. But if a team has 3 wins, it's 9 points. So **Yes, it is possible for a team with 1 win (A) to have more points than teams with more wins (if those wins are not accompanied by draws).** But the key condition is **A has the most points and the fewest wins.** Let's try to construct an explicit example for 6 teams. --- ### 5. **Explicit Construction for \( n = 7 \)** Let's generalize: Suppose: - Team A draws all its matches: 6 draws, 6 points. - Other teams beat each other in a cycle (e.g., B beats C, C beats D, ..., G beats B), and lose/draw with A. Let’s try: - A draws all matches: 6 points, 0 wins. - B: loses to A (draw), beats C, loses to D, loses to E, loses to F, loses to G. (1 win, 1 draw, 4 losses): 4 points. - C: loses to A (draw), loses to B, beats D, loses to E, loses to F, loses to G. (1 win, 1 draw, 4 losses): 4 points. - D: loses to A (draw), loses to B, loses to C, beats E, loses to F, loses to G. (1 win, 1 draw, 4 losses): 4 points. - E: loses to A (draw), loses to B, loses to C, loses to D, beats F, loses to G. (1 win, 1 draw, 4 losses): 4 points. - F: loses to A (draw), loses to B, loses to C, loses to D, loses to E, beats G. (1 win, 1 draw, 4 losses): 4 points. - G: loses to A (draw), beats B, loses to C, loses to D, loses to E, loses to F. (1 win, 1 draw, 4 losses): 4 points. But the sums do not add up: with 7 teams, each playing 6 matches, total matches is 21; total points is at most 3 * 21 = 63, but with many draws, less. But the key is with more teams, the more draws A can have, and the more likely A can be the unique highest scorer with only draws. The **classic construction** is for \( n = 7 \): Let’s try: - Team A draws all matches: 6 points, 0 wins. - All other matches are played as wins/losses (no draws), with each of B, C, D, E, F, G winning exactly one game among themselves: - B beats C - C beats D - D beats E - E beats F - F beats G - G beats B Each of B–G: 1 win (3 points), 0 draws, 5 losses (0 points): 3 points. Each also drew with A: 1 point extra, so each has **4 points**. So A: 0 wins, 6 draws = **6 points** Everyone else: 1 win, 1 draw, 4 losses = **4 points** A has the **most points (6)** and **fewest wins (0)**. #### **Check for \( n = 6 \):** Try the same construction: - A draws all matches: 5 points, 0 wins. - B beats C, C beats D, D beats E, E beats B. F? Can't form a cycle. Actually, with 5 teams (aside from A), you can't form a perfect cycle. So, **n = 7 is the smallest number for which this construction works.** --- ## **Final Answer** \[ \boxed{7} \] is the **smallest number of teams** in the tournament for which it is possible for a team to have the most points and the fewest wins. **Construction:** - Team A draws every match: 0 wins, 6 draws, 6 points. - Every other team beats one and loses to 4, with 1 draw (against A): 1 win, 1 draw, 4 losses = 4 points. - Thus, A has the most points and the fewest wins.