{"id":171555,"date":"2020-11-19T03:54:14","date_gmt":"2020-11-19T05:54:14","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=171555"},"modified":"2023-07-24T21:51:50","modified_gmt":"2023-07-25T00:51:50","slug":"probleminha-uma-fracao-irredutivel","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/probleminha-uma-fracao-irredutivel\/","title":{"rendered":".Probleminha: Uma fra\u00e7\u00e3o irredut\u00edvel"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:28px;padding-bottom:15px\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 8\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\"\/>\nSeja [tex]\\dfrac{p}{q}[\/tex] a fra\u00e7\u00e3o irredut\u00edvel equivalente a [tex]\\dfrac{0,272727&#8230;}{3,131313&#8230;}[\/tex], com [tex]p[\/tex] e [tex]q[\/tex] inteiros positivos.<br \/>\n<strong>Qual o valor de [tex]p+q[\/tex]?<\/strong><\/p>\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:32px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 20px;border: 0 none\"\/>\n<p>A ideia \u00e9 encontrar as fra\u00e7\u00f5es geratrizes de [tex]0,272727&#8230;\\,[\/tex] e [tex]\\,3,131313&#8230;[\/tex] para depois encontrarmos a fra\u00e7\u00e3o irredut\u00edvel [tex]\\dfrac{p}{q}.[\/tex] <\/p>\n<ul>\n<li>Sendo [tex]\\boxed{x = 0,272727&#8230;}[\/tex], temos [tex]\\boxed{100x = 27,272727&#8230;}\\,[\/tex] e, ent\u00e3o: <\/li>\n<p>[tex]\\quad 100x-x = 27,272727&#8230; &#8211; 0,272727&#8230;\\\\<br \/>\n\\quad 99x = 27\\\\<br \/>\n\\quad  x=\\dfrac{27}{99}\\\\<br \/>\n\\quad  x = \\dfrac{3}{11}.[\/tex]<br \/>\nPortanto, a fra\u00e7\u00e3o geratriz de [tex]0,272727&#8230;[\/tex] \u00e9 [tex]\\,\\dfrac{3}{11}.[\/tex]<\/p>\n<li>Agora, sendo [tex]\\boxed{y = 3,131313&#8230;}[\/tex], temos [tex]\\boxed{100y = 313,131313&#8230;}\\,[\/tex] e, ent\u00e3o: <\/li>\n<p>[tex]\\quad 100y-y= 313,131313&#8230; &#8211; 3,131313&#8230;\\\\<br \/>\n\\quad 99y = 310\\\\<br \/>\n\\quad  y =\\dfrac{310}{99}.[\/tex]<br \/>\nOu seja, a fra\u00e7\u00e3o geratriz de [tex]3,131313&#8230;[\/tex] \u00e9 [tex]\\dfrac{310}{99}[\/tex]. <\/ul>\n<p>Logo,<br \/>\n[tex]\\quad \\dfrac{0,272727&#8230;}{3,131313&#8230;} = \\dfrac{x}{y} = \\dfrac{\\frac{3}{11}}{\\frac{310}{99}} = \\dfrac{3}{11}\\cdot\\dfrac{99}{310} = \\dfrac{3}{1}\\cdot\\dfrac{9}{310} = \\dfrac{27}{310}[\/tex].<\/p>\n<p>Como [tex]mdc(27,310) = 1[\/tex], ent\u00e3o[tex]\\dfrac{27}{310}[\/tex] j\u00e1 \u00e9 irredut\u00edvel.<br \/>\nAssim, conclu\u00edmos que [tex]\\boxed{p = 27}\\,[\/tex] e [tex]\\,\\boxed{q = 310}\\,[\/tex] e ent\u00e3o [tex]\\fcolorbox{black}{#eee0e5}{$p+q = 27+310 = 337$}\\,.[\/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:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 8\u00ba ano do E. F.) Seja [tex]\\dfrac{p}{q}[\/tex] a fra\u00e7\u00e3o irredut\u00edvel equivalente a [tex]\\dfrac{0,272727&#8230;}{3,131313&#8230;}[\/tex], com [tex]p[\/tex] e [tex]q[\/tex] inteiros positivos. Qual o valor de [tex]p+q[\/tex]? Solu\u00e7\u00e3o A ideia \u00e9 encontrar as fra\u00e7\u00f5es geratrizes de [tex]0,272727&#8230;\\,[\/tex] e [tex]\\,3,131313&#8230;[\/tex] para depois encontrarmos a fra\u00e7\u00e3o irredut\u00edvel [tex]\\dfrac{p}{q}.[\/tex] Sendo [tex]\\boxed{x = 0,272727&#8230;}[\/tex], temos [tex]\\boxed{100x = &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/probleminha-uma-fracao-irredutivel\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":254,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-171555","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171555","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\/254"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=171555"}],"version-history":[{"count":20,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171555\/revisions"}],"predecessor-version":[{"id":197859,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171555\/revisions\/197859"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=171555"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}