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Đề thi Olympic sinh viên thế giới năm 2006

" Đề thi Olympic sinh viên thế giới năm 2006 " . 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. | 13th International Mathematics Competition for University Students Odessa July 20-26 2006 First Day Problem 1. Let f R R be a real function. Prove or disprove each of the following statements. a If f is continuous and range f R then f is monotonic. b If f is monotonic and range f R then f is continuous. c If f is monotonic and f is continuous then range f R. 20 points Solution. a False. Consider function f x x3 x. It is continuous range f R but for example f 0 0 f 2 8 and f 1 0 therefore f 0 f 2 f 2 f 1 and f is not monotonic. b True. Assume first that f is non-decreasing. For an arbitrary number a the limits lim f and a lim f exist and lim f lim f. If the two limits are equal the function is continuous at a. Otherwise a a a if lim f b lim f c we have f x b for all x a and f x c for all x a therefore a a range f C 1 b U c 1 U f a g cannot be the complete R. For non-increasing f the same can be applied writing reverse relations or g x f x . c False. The function g x arctan x is monotonic and continuous but range g 2 2 R. Problem 2. Find the number of positive integers x satisfying the following two conditions 1. x 102006 2. x2 x is divisible by 102006. 20 points Solution 1. Let Sk 0 x 10k x2 x is divisible by 10k and s k Sk k 1. Let x ak 1ak . a1 be the decimal writing of an integer x 2 Sk 1 k 1. Then obviously y ak . a1 2 Sk. Now let y ak . a1 2 Sk be fixed. Considering ak 1 as a variable digit we have x2 x ak 110k y ak 110k y y2 y ak 110k 2y 1 ak 1102k. Since y2 y 10kz for an iteger z it follows that x2 x is divisible by 10k 1 if and only if z ak 1 2y 1 0 mod 10 . Since y 3 mod 10 is obviously impossible the congruence has exactly one solution. Hence we obtain a one-to-one correspondence between the sets Sk 1 and Sk for every k 1. Therefore s 2006 s 1 3 because S1 1 5 6g. Solution 2. Since x2 x x x 1 and the numbers x and x 1 are relatively prime one of them must be divisible by 22006 and one of them may be the same must be divisible by 52006. Therefore x must .