{"id":6336,"date":"2021-05-05T06:00:14","date_gmt":"2021-05-05T04:00:14","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=6336"},"modified":"2021-05-12T08:58:30","modified_gmt":"2021-05-12T06:58:30","slug":"multiplos-repitunos-ome1993-p4","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/multiplos-repitunos-ome1993-p4\/","title":{"rendered":"M\u00faltiplos repitunos (OME1993 P4)"},"content":{"rendered":"<p>El enunciado del <a href=\"https:\/\/bit.ly\/3evNIvr\">problema 4 de la OME (Olimpiada Matem\u00e1tica Espa\u00f1ola) del 1993<\/a> es<\/p>\n<blockquote><p>\n  Demostrar que para todo n\u00famero primo p distinto de 2 y de 5, existen infinitos m\u00faltiplos de p de la forma 1111&#8230;&#8230;1 (escrito s\u00f3lo con unos).\n<\/p><\/blockquote>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   multiplosRepitunos :: Integer -> [Integer]\n<\/pre>\n<p>tal que (multiplosRepitunos p n) es la lista de los m\u00faltiplos repitunos de p (es decir, de la forma 1111&#8230;1 escrito s\u00f3lo con unos), donde p es un n\u00famero primo distinto de 2 y 5. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   take 2 (multiplosRepitunos 7) == [111111,111111111111]\n   head (multiplosRepitunos 19)  == 111111111111111111\n   length (show (head (multiplosRepitunos (primes !! (10^5))))) == 43324\n<\/pre>\n<p>Comprobar con QuickCheck que para todo primo p mayor que 5 y todo n\u00famero entero positivo n, existe un m\u00fatiplo repituno de p mayor que n.<\/p>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nimport Data.Numbers.Primes (primes)\nimport Test.QuickCheck (Property, (==>), quickCheck)\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nmultiplosRepitunos :: Integer -> [Integer]\nmultiplosRepitunos p =\n  [x | x <- repitunos\n     , mod x p == 0]\n\n-- repitunos es la lista de los n\u00fameros de la forma 111...1 (escrito s\u00f3lo con\n-- unos). Por ejemplo,\n--    take 5 repitunos  ==  [1,11,111,1111,11111]\nrepitunos :: [Integer]\nrepitunos = 1 : [10*x+1 | x <- repitunos]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\nmultiplosRepitunos2 :: Integer -> [Integer]\nmultiplosRepitunos2 p =\n  [x | x <- repitunos2\n     , mod x p == 0]\n\nrepitunos2 :: [Integer]\nrepitunos2 = [div (10^n-1) 9 | n <- [1..]]\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> length (show (head (multiplosRepitunos (primes !! (10^5)))))\n--    43324\n--    (0.58 secs, 1,272,561,448 bytes)\n--    \u03bb> length (show (head (multiplosRepitunos2 (primes !! (10^5)))))\n--    43324\n--    (5.50 secs, 2,563,458,656 bytes)\n\n-- Comprobaci\u00f3n\n-- ============\n\n-- La propiedad es\nprop_multiplosRepitunos :: Int -> Property\nprop_multiplosRepitunos k =\n  k > 2  ==>\n  not (null (multiplosRepitunos (primes !! k)))\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_multiplosRepitunos\n--    +++ OK, passed 100 tests.\n<\/pre>\n<h4>Nuevas soluciones<\/h4>\n<ul>\n<li>En los comentarios se pueden escribir nuevas soluciones.\n<li>El c\u00f3digo se debe escribir entre una l\u00ednea con &#60;pre lang=&quot;haskell&quot;&#62; y otra con &#60;\/pre&#62;\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>El enunciado del problema 4 de la OME (Olimpiada Matem\u00e1tica Espa\u00f1ola) del 1993 es Demostrar que para todo n\u00famero primo p distinto de 2 y de 5, existen infinitos m\u00faltiplos de p de la forma 1111&#8230;&#8230;1 (escrito s\u00f3lo con unos). Definir la funci\u00f3n multiplosRepitunos :: Integer -> [Integer] tal que (multiplosRepitunos p n) es la&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[2],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6336"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=6336"}],"version-history":[{"count":2,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6336\/revisions"}],"predecessor-version":[{"id":6451,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6336\/revisions\/6451"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=6336"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=6336"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=6336"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}