tag:blogger.com,1999:blog-2400513859305780710.post7182891396648822022..comments2023-04-04T07:53:53.789-07:00Comments on Mathematical Food For Thought: Pigeon, I Choose You! Topic: Pigeonhole Principle. Level: AIME.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-2400513859305780710.post-34883489004671475512005-12-02T20:14:47.000-08:002005-12-02T20:14:47.000-08:00I've got a better solution: n | 2n. :) More...I've got a better solution: n | 2n. :) More specifically, your solution stipulates that the element with the smaller power of 2 divides the element with the larger power of 2, but the only two elements of the form 2^a * n and 2^{a+1} * n are n and 2n, since, for all k > n, 2k does not appear in the set, and for all kQCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-45996035651237060462005-12-02T20:15:10.000-08:002005-12-02T20:15:10.000-08:00And for all kAnd for all kQCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-47031208442959515662005-12-02T20:15:46.000-08:002005-12-02T20:15:46.000-08:00And for all k less than 2n, k over two does not ap...And for all k less than 2n, k over two does not appear in the set. Stupid cutoff. x.xQCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-39449895769479408772005-12-02T22:46:51.000-08:002005-12-02T22:46:51.000-08:00Bah. Copied the problem wrong... it's not supp...Bah. Copied the problem wrong... it's not supposed to be that easy -___-. Edited now.paladin8noreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-69613299362599119162005-12-03T22:30:55.000-08:002005-12-03T22:30:55.000-08:00#1s easy. Its almost like that example that AoPS b...#1s easy. Its almost like that example that AoPS book gives in the beginning, only with 2x2. Although by generalizing, i guess it means that for n*n square of unit x, if there are n^2+1 points then at least two of them have to be within x\sqrt{2} unit?yjeongnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-83917327414885011832005-12-03T22:35:13.000-08:002005-12-03T22:35:13.000-08:00For #2, we denote such set {a_n}. They must be all...For #2, we denote such set {a_n}. They must be all different mod 100, or else the difference or the sum will be divisible by 100 (easily proven.) <br><br>So there are 51 supposedly different mod 100, but since there are 50 groups of mod 100 that add up to 100 mod 100... i.e. {0}, {1,99}, ..., {49,51}, the last one must be in the same group as one of the other previous 50 integers, making those a and b such that 100|a+b.<br><br>generalize: For integer set S of x terms, there are two, a and b, such that 2(x-1)|<br>(a+b) or 2(x-1)|(a-b)yjeongnoreply@blogger.com