tailieunhanh - Đề thi Olympic sinh viên thế giới năm 2002 ngày 2

" Đề thi Olympic sinh viên thế giới năm 2002 ngày 2 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 đại học trên thế giới. Tài liệu hay giúp ích cho việc tham khảo | Solutions for problems in the 9th International Mathematics Competition for University Students Warsaw July 19 - July 25 2002 Second Day Problem 1. Compute the determinant of the n x n matrix A aij _i -1 i-j if i j 2 if i j. Solution. Adding the second row to the first one then adding the third row to the second one . adding the nth row to the n 1 th the determinant does not change and we have 2 -1 1 . 1 T1 1 1 0 0 . 0 0 -1 2 -1 . T1 1 0 1 1 0 . 0 0 1 -1 2 . 1 T1 0 0 1 1 . 0 0 det A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T1 1 T1 . 2 -1 0 0 0 0 . 1 1 1 T1 1 . -1 2 1 T1 1 T1 . -1 2 Now subtract the first column from the second then subtract the resulting second column from the third . and at last subtract the n 1 th column from the nth column. This way we have 10 0 . 0 0 0 10 . 0 0 det A . . . . . . . . . . n 1 0 0 0 . 1 0 0 0 0 . 0 n 1 Problem 2. Two hundred students participated in a mathematical contest. They had 6 problems to solve. It is known that each problem was correctly solved by at least 120 participants. Prove that there must be two participants such that every problem was solved by at least one of these two students. Solution. For each pair of students consider the set of those problems which was not solved by them. There exist 22 19900 sets we have to prove that at least one set is empty. 1 For each problem there are at most 80 students who did not solve it. From these students at most 820 3160 pairs can be selected so the problem can belong to at most 3160 sets. The 6 problems together can belong to at most 6 3160 18960 sets. Hence at least 19900 18960 940 sets must be empty. Problem 3. For each n 1 let . kn a L k k 0 OO -j n X kr k 0 Show that an bn is an integer. Solution. We prove by induction on n that an e and bne are integers we prove this for n 0 as well. For n 0 the term 00 in the definition of the sequences must be replaced by 1. From the power series of ex an e1 e and bn e-1 1 e. Suppose that for some n 0 a0 a1 . an