{"id":57882,"date":"2016-03-16T18:18:40","date_gmt":"2016-03-16T21:18:40","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=57882"},"modified":"2023-03-31T20:42:53","modified_gmt":"2023-03-31T23:42:53","slug":"problema-de-olimpiada-fracoes-de-domino","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-olimpiada-fracoes-de-domino\/","title":{"rendered":".Problema de Olimp\u00edada:  Fra\u00e7\u00f5es de Domin\u00f3"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSe ignorarmos o &#8220;duplo zero&#8221;, as restantes [tex]27[\/tex] pe\u00e7as de um jogo de domin\u00f3 podem ser vistas como fra\u00e7\u00f5es inferiores ou iguais a 1. Observe o exemplo abaixo.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2016\/03\/olimpico_27_11.png\" style=\"border:0px;\" alt=\"olimpico_27_11\" width=\"300\" height=\"140\" class=\"aligncenter size-full wp-image-57885\" \/><\/p>\n<p>Qual a soma dessas [tex]27[\/tex] fra\u00e7\u00f5es?<\/p>\n<p style=\"font-size: 12px;text-align:right;line-height:10px;\">(<strong>OLIMP\u00cdADA DE MATEM\u00c1TICA DO ESTADO DO RIO DE JANEIRO<\/strong> &#8211; 2011.)<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:45px\">\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\"\/>\nPara obter essa soma, basta que somemos as [tex]21[\/tex] fra\u00e7\u00f5es correspondentes \u00e0s pe\u00e7as que n\u00e3o cont\u00e9m nenhum zero, j\u00e1 que as fra\u00e7\u00f5es [tex]\\dfrac{0}{1}, \\,  \\dfrac{0}{2}, \\,  \\dfrac{0}{3}, \\,  \\dfrac{0}{4}, \\,  \\dfrac{0}{5}, \\,  \\dfrac{0}{6}[\/tex] correspondem ao n\u00famero zero que, como parcela, n\u00e3o altera uma soma.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2016\/03\/dominopronto.png\" style=\"border:0px;\" alt=\"dominopronto\" width=\"387\" height=\"165\" class=\"aligncenter size-full wp-image-57904\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2016\/03\/dominopronto.png 387w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2016\/03\/dominopronto-300x128.png 300w\" sizes=\"auto, (max-width: 387px) 100vw, 387px\" \/>Vamos, ent\u00e3o, \u00e0 soma:<\/p>\n<p>[tex]\\dfrac{1}{1}+\\left(\\dfrac{1}{2}+\\dfrac{2}{2}\\right)+ \\left(\\dfrac{1}{3}+\\dfrac{2}{3}+\\dfrac{3}{3}\\right)+ \\left(\\dfrac{1}{4}+\\dfrac{2}{4}+\\dfrac{3}{4}+\\dfrac{4}{4}\\right)+\\\\<br \/>\n\\qquad + \\left(\\dfrac{1}{5}+\\dfrac{2}{5}+\\dfrac{3}{5}+\\dfrac{4}{5}+\\dfrac{5}{5}\\right)+ \\left(\\dfrac{1}{6}+\\dfrac{2}{6}+\\dfrac{3}{6}+\\dfrac{4}{6}+\\dfrac{5}{6}+\\dfrac{6}{6}\\right)=\\\\<br \/>\n\\qquad = 1+\\dfrac{3}{2}+2+\\dfrac{5}{2}+3+\\dfrac{7}{2}=\\\\<br \/>\n\\qquad  =\\dfrac{27}{2}[\/tex].<br \/>\nAssim, a soma das [tex]27[\/tex] fra\u00e7\u00f5es de domin\u00f3 \u00e9 [tex]\\dfrac{27}{2}[\/tex].<\/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=\"gg\">\nOlimp\u00edada: <strong>XII ONM<\/strong> &#8211; Portugal<br \/>\n Categoria A &#8211; 8\u00ba\/ 9\u00ba ano <\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema Se ignorarmos o &#8220;duplo zero&#8221;, as restantes [tex]27[\/tex] pe\u00e7as de um jogo de domin\u00f3 podem ser vistas como fra\u00e7\u00f5es inferiores ou iguais a 1. Observe o exemplo abaixo. Qual a soma dessas [tex]27[\/tex] fra\u00e7\u00f5es? (OLIMP\u00cdADA DE MATEM\u00c1TICA DO ESTADO DO RIO DE JANEIRO &#8211; 2011.) Solu\u00e7\u00e3o Para obter essa soma, basta que somemos as &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-de-olimpiada-fracoes-de-domino\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-57882","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/57882","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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=57882"}],"version-history":[{"count":35,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/57882\/revisions"}],"predecessor-version":[{"id":185974,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/57882\/revisions\/185974"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=57882"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}