{"id":45437,"date":"2015-09-11T12:42:28","date_gmt":"2015-09-11T15:42:28","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=45437"},"modified":"2023-03-29T03:14:16","modified_gmt":"2023-03-29T06:14:16","slug":"probleminha-de-domingo-a-domingo","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/probleminha-de-domingo-a-domingo\/","title":{"rendered":".Problema: De Domingo a Domingo"},"content":{"rendered":"\n<p style=\"margin:20px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 7\u00ba ano do E. F.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n(<strong>OBM <\/strong>2005) Em um ano, no m\u00e1ximo quantos meses t\u00eam cinco domingos?<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nLembre-se de que um ano que n\u00e3o \u00e9 bissexto tem [tex]365[\/tex] dias e um ano bissexto tem [tex]366[\/tex] dias e observe as seguintes divis\u00f5es:<\/p>\n<table class=\"bf\" style=\"width:40%;\">\n<tr  style=\"width:50%;\">\n<td class=\"bf\">\n[tex]\\qquad \\qquad \\begin{array}{r}<br \/>\n365 \\, \\end{array}   \\begin{array}{|r}<br \/>\n \\,  \\, 7 \\, \\, \\,  \\,  \\,  \\\\  \\hline<br \/>\n\\end{array}[\/tex]<br \/>\n[tex]\\qquad \\qquad\\begin{array}{r}<br \/>\n \\,  \\,  \\,  \\,  \\,   1<br \/>\n\\end{array}\\begin{array}{r}<br \/>\n  \\,  \\,  \\, \\, \\, 52<br \/>\n\\end{array}\\qquad \\qquad[\/tex]<\/td>\n<td class=\"bf\">\n[tex]\\qquad \\qquad \\begin{array}{r}<br \/>\n366 \\, \\end{array}   \\begin{array}{|r}<br \/>\n \\,  \\, 7 \\, \\, \\,  \\,  \\,  \\\\  \\hline<br \/>\n\\end{array}[\/tex]<br \/>\n[tex]\\qquad \\qquad\\begin{array}{r}<br \/>\n \\,  \\,  \\,  \\,  \\,   2<br \/>\n\\end{array}\\begin{array}{r}<br \/>\n  \\,  \\,  \\, \\, \\, 52<br \/>\n\\end{array}\\qquad \\qquad[\/tex]<\/td>\n<\/tr>\n<\/table>\n<ul>\n<li>Como [tex]365[\/tex] dividido por [tex]7[\/tex] d\u00e1 quociente [tex]52[\/tex] e resto [tex]1[\/tex] e [tex]366[\/tex] dividido por [tex]7[\/tex] d\u00e1 o mesmo quociente e resto [tex]2[\/tex]; ent\u00e3o um ano n\u00e3o bissexto tem &#8220;[tex]52[\/tex] semanas completas e [tex]1[\/tex] dia&#8221; e, um bissexto, &#8220;[tex]52[\/tex] semanas completas e [tex]2[\/tex] dias&#8221;. Assim,\n<p style=\"margin:20px\">\n<ul>\n<li><strong>em um ano, bissexto ou n\u00e3o, h\u00e1 no m\u00e1ximo [tex]53[\/tex] domingos.<\/strong> <\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Agora, observe as seguintes divis\u00f5es:<\/p>\n<table class=\"bf\" style=\"width:40%;\">\n<tr  style=\"width:50%;\">\n<td class=\"bf\">\n[tex]\\qquad \\qquad \\begin{array}{r}<br \/>\n28 \\, \\end{array}   \\begin{array}{|r}<br \/>\n \\,  \\, 7 \\, \\, \\,  \\,    \\\\  \\hline<br \/>\n\\end{array}[\/tex]<br \/>\n[tex]\\qquad \\qquad\\begin{array}{r}<br \/>\n \\,  \\,  \\,     0<br \/>\n\\end{array}\\begin{array}{r}<br \/>\n  \\,  \\,  \\, \\, \\, 4<br \/>\n\\end{array}\\qquad \\qquad[\/tex]<\/td>\n<td class=\"bf\">\n[tex]\\qquad \\qquad \\begin{array}{r}<br \/>\n31 \\, \\end{array}   \\begin{array}{|r}<br \/>\n \\,  \\, 7 \\, \\, \\,  \\,  \\\\  \\hline<br \/>\n\\end{array}[\/tex]<br \/>\n[tex]\\qquad \\qquad\\begin{array}{r}<br \/>\n \\,  \\,  \\,     3<br \/>\n\\end{array}\\begin{array}{r}<br \/>\n  \\,  \\,  \\, \\, \\, 4<br \/>\n\\end{array}\\qquad \\qquad[\/tex]<\/td>\n<\/tr>\n<\/table>\n<ul>\n<li>Como um m\u00eas tem entre [tex]28 = 4 \\times 7\\,[\/tex] e [tex]\\,31 = 4 \\times 7+3[\/tex] dias, ent\u00e3o todo m\u00eas tem &#8220;[tex]4[\/tex] semanas completas&#8221; e &#8220;[tex]0[\/tex], [tex]1[\/tex], [tex]2[\/tex] ou [tex]3[\/tex] dias&#8221;. Com isso conclu\u00edmos que\n<p style=\"margin:20px\">\n<ul>\n<li><strong>todo m\u00eas tem [tex]4[\/tex] ou [tex]5[\/tex] domingos.<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Vamos, ent\u00e3o, distribuir os [tex]53[\/tex] domingos poss\u00edveis de um ano entre os seus [tex]12[\/tex] meses:<\/p>\n<table class=\"bf\" style=\"width:40%;\">\n<tr  style=\"width:50%;\">\n<td class=\"bf\">\n[tex]\\qquad \\qquad \\begin{array}{r}<br \/>\n53 \\, \\end{array}   \\begin{array}{|r}<br \/>\n \\,  \\, 12 \\, \\,  \\,    \\\\  \\hline<br \/>\n\\end{array}[\/tex]<br \/>\n[tex]\\qquad \\qquad\\begin{array}{r}<br \/>\n \\,  \\,  \\,     5<br \/>\n\\end{array}\\begin{array}{r}<br \/>\n  \\,  \\,  \\, \\, \\, 4<br \/>\n\\end{array}\\qquad \\qquad[\/tex]<\/td>\n<\/tr>\n<\/table>\n<p>Como [tex]53[\/tex] dividido por [tex]12[\/tex] d\u00e1 quociente [tex]4[\/tex] e resto [tex]5[\/tex],<span style=\"color: red\"><strong> h\u00e1 no m\u00e1ximo [tex]5[\/tex] meses com [tex]5[\/tex] domingos em um ano<\/strong><\/span>.<br \/>\nExemplos de anos com cinco meses com cinco domingos cada s\u00e3o aqueles iniciados no domingo, como os anos de [tex]2012[\/tex], [tex]2017[\/tex] e [tex]2022[\/tex], entre outros.<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong> Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g3\">\nParticipou da discuss\u00e3o o Clube <span style=\" color: #686461;\"><strong> Todos pela Matem\u00e1tica<\/strong><\/span>.\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 7\u00ba ano do E. F.) (OBM 2005) Em um ano, no m\u00e1ximo quantos meses t\u00eam cinco domingos? Solu\u00e7\u00e3o Lembre-se de que um ano que n\u00e3o \u00e9 bissexto tem [tex]365[\/tex] dias e um ano bissexto tem [tex]366[\/tex] dias e observe as seguintes divis\u00f5es: [tex]\\qquad \\qquad \\begin{array}{r} 365 \\, \\end{array} \\begin{array}{|r} \\, &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/probleminha-de-domingo-a-domingo\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-45437","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/45437","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=45437"}],"version-history":[{"count":16,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/45437\/revisions"}],"predecessor-version":[{"id":185703,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/45437\/revisions\/185703"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=45437"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}