answer:1. **Assumption**: Suppose that the desired situation does not occur. That is, there do not exist teams ( A ), ( B ), and ( C ) such that ( A ) defeated ( B ), ( B ) defeated ( C ), and ( C ) defeated ( A ). 2. **Implication**: This implies a linear hierarchical order of wins and losses. Specifically, if ( A ) defeated ( B ) and ( B ) defeated ( C ), then ( A ) must also defeat ( C ). This results in a transitive property of wins, where victory chains do not form cycles. 3. **Tournament Hierarchy**: Evaluate the implications of such a hierarchy: - The team with the highest number of points, say ( X ), must have defeated all other teams (since no team scored 7 points and no cyclic relationships exist). - The team with the second-highest number of points, say ( Y ), must have defeated all teams except ( X ). - This pattern continues down to the lowest-ranked team. 4. **Maximum Points Calculation**: Now consider the number of wins: - Since there are 12 teams in the tournament, each team can play against 11 other teams. - The highest-ranked team ( X ) wins all of its 11 matches. - The second-ranked team ( Y ) wins against 10 other teams (losing only to ( X )). - The same reduction pattern applies to the nth-ranked team: it only defeats ( (12 - n - 1) ) other teams. 5. **Contradiction with Problem Constraints**: - According to our assumptions, consider the team ranked 5th. This team must defeat ( (12 - 5 - 1) = 6 ) teams (since they must only lose to 6 higher-ranked teams). - However, this directly contradicts the problem's assertion that no team has exactly 7 points. - Hence, in this hierarchy, the 5th placed team having exactly 7 points is inevitable. 6. **Conclusion**: Since our assumption led us to a contradiction, there must be at least one occurrence of a cyclic relationship among the teams forming the trio ( A ), ( B ), and ( C ) where ( A ) defeated ( B ), ( B ) defeated ( C ), and ( C ) defeated ( A ). Therefore, the final conclusion: (blacksquare)

answer:To demonstrate that for any number ( n ) greater than 2, there exists another number such that the sum of the squares of these two numbers forms a perfect square, we consider the cases of ( n ) being even or odd separately. Case 1: ( n ) is even 1. Let ( n = 2r ), where ( r ) is an integer greater than 1. 2. Calculate ( n^2 ): [ n^2 = (2r)^2 = 4r^2 ] 3. Consider the number ( m = r^2 - 1 ). 4. Calculate the square of ( m ): [ m^2 = (r^2 - 1)^2 = r^4 - 2r^2 + 1 ] 5. Sum the squares of ( n ) and ( m ): [ n^2 + m^2 = 4r^2 + (r^4 - 2r^2 + 1) ] 6. Simplify the expression: [ 4r^2 + r^4 - 2r^2 + 1 = r^4 + 2r^2 + 1 ] 7. Note that: [ r^4 + 2r^2 + 1 = (r^2 + 1)^2 ] 8. Hence, [ n^2 + m^2 = (r^2 + 1)^2 ] Thus, for even ( n ), there exists a number ( m = r^2 - 1 ) such that ( n^2 + m^2 ) is a perfect square. Case 2: ( n ) is odd 1. Let ( n ) be an odd integer. 2. Consider the number ( m = frac{n^2 - 1}{2} ). 3. Calculate ( n^2 ): [ n^2 = n^2 ] 4. Calculate the square of ( m ): [ m^2 = left(frac{n^2 - 1}{2}right)^2 = frac{(n^2 - 1)^2}{4} = frac{n^4 - 2n^2 + 1}{4} ] 5. Sum the squares of ( n ) and ( m ): [ n^2 + m^2 = n^2 + frac{n^4 - 2n^2 + 1}{4} ] 6. Combine the fractions: [ n^2 + frac{n^4 - 2n^2 + 1}{4} = frac{4n^2}{4} + frac{n^4 - 2n^2 + 1}{4} = frac{n^4 + 2n^2 + 1}{4} ] 7. Note that: [ frac{n^4 + 2n^2 + 1}{4} = left( frac{n^2 + 1}{2} right)^2 ] 8. Hence, [ n^2 + m^2 = left( frac{n^2 + 1}{2} right)^2 ] Thus, for odd ( n ), there exists a number ( m = frac{n^2 - 1}{2} ) such that ( n^2 + m^2 ) is a perfect square. # Conclusion: In both cases, ( n ) being either even or odd, we have shown that there exists ( m ) such that ( n^2 + m^2 ) is a perfect square. [ boxed{} ]

answer:According to the rules of addition and subtraction of integers, we set up the equation 15 + 37 = 52. **Analysis:** When calculating vertically, align the same digits before performing the calculation. The result is boxed{52}.

answer:Let's denote the weight of the person who was replaced as W. The total weight increase for the group of 8 persons is 8 persons * 2.5 kg/person = 20 kg. This increase is due to the new person who weighs 87 kg. Therefore, the weight of the person who was replaced is the weight of the new person minus the total weight increase: W = 87 kg - 20 kg W = 67 kg The person who was replaced weighed boxed{67} kg.