{"id":6302,"date":"2021-04-23T06:00:43","date_gmt":"2021-04-23T04:00:43","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=6302"},"modified":"2021-04-30T09:36:19","modified_gmt":"2021-04-30T07:36:19","slug":"mayores-potencias-de-5-que-dividen-a-la-sucesion","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/mayores-potencias-de-5-que-dividen-a-la-sucesion\/","title":{"rendered":"Mayores potencias de 5 que dividen a la sucesi\u00f3n (OME2014 P4)"},"content":{"rendered":"<p>El enunciado del <a href=\"https:\/\/bit.ly\/2QrAck5\">problema 4 de la OME (Olimpiada Matem\u00e1tica Espa\u00f1ola) del 2014<\/a> es<\/p>\n<blockquote><p>\n  Sea {x(n)} la sucesi\u00f3n de enteros positivos definida por x(1) = 2 y x(n+1) = 2*x(n)^3+x(n) para todo n >= 1.  Determinar la mayor potencia de 5 que divide al n\u00famero x(2014)^2+1.\n<\/p><\/blockquote>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   mayorExponente :: Integer -> Integer\n<\/pre>\n<p>tal que (mayorExponente n) es el mayor m tal que 5^m divide al n\u00famero x(n)^2+1.<\/p>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nmayorExponente :: Integer -> Integer\nmayorExponente n =\n  mayorExponenteDe5QueDivide ((termino n)^2 + 1)\n\n-- (termino n) es el t\u00e9rmino n-\u00e9simo de la sucesi\u00f3n. Por ejemplo,\n--    termino 2  ==  18\n--    termino 3  ==  11682\n--    termino 4  ==  3188464624818\ntermino :: Integer -> Integer\ntermino 1 = 2\ntermino n = 2*x^3+x\n  where x = termino (n-1)\n\n-- (mayorExponenteDe5QueDivide n) es el mayor exponente de 5 que divide a n. Por\n-- ejemplo,\n--    mayorExponenteDe5QueDivide 50  ==  2\nmayorExponenteDe5QueDivide :: Integer -> Integer\nmayorExponenteDe5QueDivide n\n  | n `mod` 5 \/= 0 = 0\n  | otherwise      = 1 + mayorExponenteDe5QueDivide (n `div` 5)\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\n-- Con el siguente c\u00e1lculo\n--    \u03bb> [mayorExponente n | n <- [1..5]]\n--    [1,2,3,4,5]\n-- se observa que el mayor exponente de 5 que divide a x(n)^2+1 es n.\n\nmayorExponente2 :: Integer -> Integer\nmayorExponente2 = id\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 2014 es Sea {x(n)} la sucesi\u00f3n de enteros positivos definida por x(1) = 2 y x(n+1) = 2*x(n)^3+x(n) para todo n >= 1. Determinar la mayor potencia de 5 que divide al n\u00famero x(2014)^2+1. Definir la funci\u00f3n mayorExponente :: Integer -> Integer tal&#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\/6302"}],"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=6302"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6302\/revisions"}],"predecessor-version":[{"id":6378,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6302\/revisions\/6378"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=6302"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=6302"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=6302"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}