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Đề thi Olympic sinh viên thế giới năm 2007

" Đề thi Olympic sinh viên thế giới năm 2007 " . Đây là một sân chơi lớn để sinh viên thế giới có dịp gặp gỡ, trao đổi, giao lưu và thể hiện khả năng học toán, làm toán của mình. Từ đó đến nay, các kỳ thi Olympic sinh viênthế giới đã liên tục được mở rộng quy mô rất lớn. Kỳ thi này là một sự kiện quan trọng đối với phong trào học toán của sinh viên thế giới trong trường. | IMC2007 Blagoevgrad Bulgaria Day 1 August 5 2007 Problem 1. Let f be a polynomial of degree 2 with integer coefficients. Suppose that f k is divisible by 5 for every integer k. Prove that all coefficients of f are divisible by 5. Solution 1. Let f x ax2 bx c. Substituting x 0 x 1 and x -1 we obtain that 5 f 0 c 5 f 1 a b c and 5 f 1 a b c . Then 5 f 1 f 1 2f 0 2a and 5 f 1 f 1 2b. Therefore 5 divides 2a 2b and c and the statement follows. Solution 2. Consider f x as a polynomial over the 5-element field i.e. modulo 5 . The polynomial has 5 roots while its degree is at most 2. Therefore f 0 mod 5 and all of its coefficients are divisible by 5. Problem 2. Let n 2 be an integer. What is the minimal and maximal possible rank of an n x n matrix whose n2 entries are precisely the numbers 1 2 . n2 Solution. The minimal rank is 2 and the maximal rank is n. To prove this we have to show that the rank can be 2 and n but it cannot be 1. i The rank is at least 2. Consider an arbitrary matrix A aij- with entries 1 2 . n2 in some order. Since permuting rows or columns of a matrix does not change its rank we can assume that 1 an 21 ni and n i2 in. Hence ni n and in n and at least one of these inequalities is strict. Then det ii The rank can be 2. Let 1 n2 n n 0so rk A rk aii a i in nn 1 2 . n n 1 n 2 . 2n T . . . . . . . n2 n 1 n2 n 2 . . . n2 aii ani in nn 2. The ith row is 1 2 . n n i 1 1 1 . 1 so each row is in the two-dimensional subspace generated by the vectors 1 2 . n and 1 1 . 1 . We already proved that the rank is at least 2 so rk T 2. iii The rank can be n i.e. the matrix can be nonsingular. Put odd numbers into the diagonal only even numbers above the diagonal and arrange the entries under the diagonal arbitrarily. Then the determinant of the matrix is odd so the rank is complete. Problem 3. Call a polynomial P x1 . xk good if there exist 2 x 2 real matrices A1 . Ak such that P xi . xfc det XiA . Find all values of k for which all homogeneous polynomials with k .