国际大学生数学竞赛试题及解答（2004-2011）.pdfVIP

11th International Mathematical Competition for University Students Skopje, 25–26 July 2004 Solutions for problems on Day 1 Problem 1. Let S be an infinite set of real numbers such that |s1 + s2 + · · · + sk | 1 for every finite subset {s , s , . . . , s } ⊂ S . Show that S is countable. [20 points] 1 2 k Solution. Let Sn = S ∩ ( 1 , ∞) for any integer n 0. It follows from the inequality that |Sn | n. Similarly, if we n define S−n = S ∩ (−∞, − 1 ), then |S−n | n. Any nonzero x ∈ S is an element of some Sn or S−n , because there n exists an n such that x 1 , or x − 1 . Then S ⊂ {0} ∪ (Sn ∪ S−n ), S is a countable union of finite sets, and n n n∈N hence countable. Problem 2. Let P (x) = x2 − 1. How many distinct real solutions does the following equation have: P (P (. . . (P (x)) . . . )) = 0 ? [20 points] 2004 Solution. Put P (x) = P (P (...(P (x))...)). As P (x) ≥ −1, for each x ∈ R, it must be that P (x) = P (P (x)) ≥ n 1 n+1 1 n n −1, for each n ∈ N and each x ∈ R. Therefore the equation Pn (x) = a, where a −1 has no real solutions. Let us prove that the equation Pn (x) = a, where a 0, has exactly two distinct real solutions. To this end we use mathematical induction by n. If n = 1 the assertion foll