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Đề thi Olympic sinh viên thế giới năm 2008

Tham khảo tài liệu 'đề thi olympic sinh viên thế giới năm 2008', khoa học tự nhiên, toán học phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | n Problem 6. For a permutation a ---jin of 1 2 . n define -D ff fc . Let Q n d be k l the number of permutations a of f 2 . n with d D 7 . Prove that Q n d is even for d 2n. Solution. Consider the n x n determinant 1 A a 1 where the 7-th entry is x A From the definition of the determinant we get A x 52 where Sn is the set of all permutations of 1 2 n and inv i1 . denotes the number of inversions in the sequence i1 in . So Q n d has the same parity as the coefficient of xd in A x . It remains to evaluate A x . In order to eliminate the entries below the diagonal subtract the n l -th row multiplied by x from the n-th row. Then subtract the n 2 -th row multiplied by x from the n l -th and so on. Finally subtract the first row multiplied by x from the second row. 1 x 0 1 x2 For d in the coefficient of xd is 0 so Q n d is even. 4 IMC2008 Blagoevgrad Bulgaria Day 1 July 27 2008 Problem 1. Find all continuous functions R R such that x f y is rational for all reals x and y such that x y is rational. Solution. We prove that f x ax b where a E Q and b G R. These functions obviously satify the conditions. Suppose that a function f x fulfills the required properties. For an arbitrary rational q consider the function gq x f x q f x . This is a continuous function which attains only rational values therefore gq is constant. Set a 1 0 and b 0 . Let n be an arbitrary positive integer and let r f i n 0 . Since f x 1 n f x f i n 0 r for all x we have and - 0 - 0 - -1 n - -1 n - -2 n - . - - fc - l n - f -k n -kr for k 1. In the case k n we get a 1 0 nr so r a n. Hence fc n 0 kr ak n and then fc n a k n b for all integers k and n 0. So we have f x ax b for all rational x. Since the function is continous and the rational numbers form a dense subset of R the same holds for all real x. Problem 2. Denote by V the real vector space of all real polynomials in one variable and let P V -n R be a linear map. Suppose that for all g G V with P fg 0 we have P f 0 or P g 0. Prove that there exist