Suppose S<\/em> is a nonempty set of positive integers with the property that if n<\/em> is in S,<\/em> then every positive divisor of 2025n<\/em><\/sup>\u221215n<\/em><\/sup> is in S. <\/em>Must S<\/em> contain all positive integers?<\/p>\n\n\n\n Yes.<\/p>\n\n\n\n First note that 2025\/15 = 135 = 33<\/sup>\u00b75, so that 2025n<\/sup><\/sub><\/em> \u2212 15n<\/sup><\/em> = 15n<\/sup><\/em>(135n <\/em><\/sup>\u2212 1). If n<\/em> is in S,<\/em> then S <\/em>contains all divisors of 2025n<\/sup><\/sub><\/em> \u2212 15n<\/sup><\/em>, which include 1, 2, and 3. We show by induction that S<\/em> contains all positive integers. For the inductive step, suppose all integers less than n<\/em> are in S,<\/em> and let n = 3a<\/sup><\/em>5b<\/sup>c,<\/em> where c<\/em> is relatively prime to 15. By Euler\u2019s theorem (or by the pigeonhole principle), there exists k <\/em>with 1 \u2264 k <\/em>\u2264 c<\/em>\u22121 such that c <\/em>divides 135k <\/sup><\/em>\u2212 1. If a = b =<\/em> 0, we are done. If a<\/em> + b<\/em> > 0, then by, say, an easy induction or the tangent line approximation to f(x) = yx<\/sup><\/em> at x<\/em> = 0, one sees that a <\/em>< 3a<\/sup><\/em> and b<\/em> < 5b<\/sup><\/em>, implying<\/p>\n\n\n\n a<\/em> + b<\/em> < 3a<\/sup><\/em> + 5b<\/em><\/sup> \u2212 1 = 3a<\/sup><\/em>5b<\/sup><\/em> + (3a<\/em><\/sup> \u2212 1) (5b<\/sup><\/em> \u2212 1) \u2264 3a<\/sup><\/em>5b<\/sup>.<\/p>\n\n\n\n Hence<\/p>\n\n\n\n (a <\/em>+ b<\/em>)k<\/em> \u2264 3a<\/sup><\/em>5b<\/sup><\/em>(c <\/em>\u2212 1) < n,<\/em><\/p>\n\n\n\n c <\/em>divides 135(a+b<\/em>)k<\/em><\/sup> \u2212 1, and 3a<\/sup><\/em>5b<\/em><\/sup> divides 15(a+b<\/em>)k<\/em><\/sup>, so that n<\/em> divides 2025(a<\/em>+b<\/em>)k<\/em><\/sup> \u221215(a<\/em>+b<\/em>)k<\/em><\/sup>. <\/p>\n\n\n\n Therefore, n<\/em> is in S.<\/em><\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Putnam Mathematical Competition Problem B3 Suppose S is a nonempty set of positive integers with the property that if n is in S, then every positive divisor of 2025n\u221215n is in S. Must S contain all positive integers? Answer: Yes. Solution: First note that 2025\/15 = 135 = 33\u00b75, so that 2025n \u2212 15n = […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-89","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":[],"_links":{"self":[{"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/posts\/89","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/comments?post=89"}],"version-history":[{"count":5,"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/posts\/89\/revisions"}],"predecessor-version":[{"id":259,"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/posts\/89\/revisions\/259"}],"wp:attachment":[{"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/media?parent=89"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/categories?post=89"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/magazine.hmc.edu\/spring-summer-2026\/wp-json\/wp\/v2\/tags?post=89"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Answer:<\/h3>\n\n\n\n
Solution:<\/h3>\n\n\n\n