Daily Math Puzzle: 2026-10-23

Such a scenario is logically impossible. — In a round-robin tournament with 5 teams, each team plays 4 games. Since there are no draws, each game results in one win and one loss. The total number of games played is (5 teams * 4 games per team) / 2 = 10 games. A fundamental property of such a tournament is that the sum of all losses across all teams must equal the total number of games played. So, the sum of actual losses for all five teams must be 10. Let's list the stated number of losses for each team: * Team P: 1 loss * Team Q: 2 losses * Team R: 3 losses * Team S: 4 losses * Team T: 0 losses The sum of these stated losses is 1 + 2 + 3 + 4 + 0 = 10. The puzzle states that exactly one manager is lying, and the other four are telling the truth. Let 'L_X' be the stated number of losses for a team X, and 'l_X' be the true number of losses for that team X. Suppose manager X is the liar. This means L_X ≠ l_X. For all other managers Y (Y ≠ X), they are telling the truth, so L_Y = l_Y. We know that the sum of all true losses must be 10: l_P + l_Q + l_R + l_S + l_T = 10. If manager X is the liar, then the sum of true losses can be written as: l_X + (Sum of L_Y for all Y ≠ X) = 10. We also know that the sum of all stated losses is 10: L_P + L_Q + L_R + L_S + L_T = 10. So, (Sum of L_Y for all Y ≠ X) = 10 - L_X. Substituting this back into the equation for true losses: l_X + (10 - L_X) = 10. This simplifies to l_X = L_X. This means that if exactly one manager is lying, their actual number of losses must be identical to the number of losses they stated. However, if they are 'lying', their actual number of losses must be *different* from their stated number of losses. This creates a logical contradiction. Therefore, the premise that exactly one manager is lying is impossible under these conditions. No manager can be the single liar.