{"id":216678,"date":"2024-05-05T04:41:52","date_gmt":"2024-05-05T07:41:52","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=216678"},"modified":"2024-05-05T05:08:37","modified_gmt":"2024-05-05T08:08:37","slug":"problemao-soma-dos-divisores","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-soma-dos-divisores\/","title":{"rendered":".Problem\u00e3o: Soma dos divisores"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=217721\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#A0522D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1n\" style=\"line-height: 25px; 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 9\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;\" \/>\nSabendo-se que a soma de todos os divisores positivos [tex]d_1,\\, d_2,\\, \\cdots,\\, d_k[\/tex] de um n\u00famero natural [tex]n[\/tex] \u00e9 [tex]124[\/tex], com [tex]d_i \\ne d_j[\/tex] para [tex]i \\ne j[\/tex], calcule o valor de [tex]\\dfrac{1}{d_1}+\\dfrac{1}{d_2}+\\cdots + \\dfrac{1}{d_k}[\/tex] em fun\u00e7\u00e3o de [tex]n.[\/tex]\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"g1n\" style=\"line-height: 28px;\">\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;\" \/>\nObserve que entre os divisores  [tex]\\{d_1,\\, d_2,\\, \\cdots,\\, d_k\\}[\/tex] se encontra o pr\u00f3prio n\u00famero [tex]n[\/tex]. Assim, o m\u00ednimo m\u00faltiplo comum desses divisores deve ser o pr\u00f3prio [tex]n[\/tex], j\u00e1 que, por defini\u00e7\u00e3o de divisor, [tex]n[\/tex] \u00e9 m\u00faltiplo de cada um dos seus divisores. Logo, [tex]\\text{mmc}  (d_1, d_2, \\cdots, d_k)=n[\/tex].<br \/>\nDessa forma, segue que:<br \/>\n[tex]\\qquad \\dfrac{1}{d_1}+\\dfrac{1}{d_2}+\\cdots + \\dfrac{1}{d_k}=\\dfrac{\\dfrac{n}{d_1}+\\dfrac{n}{d_2}+\\cdots+ \\dfrac{n}{d_k}}{n}.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Agora, observe que para cada divisor [tex]d_i[\/tex] de [tex]n[\/tex] existe um \u00fanico [tex]d_j[\/tex] (tamb\u00e9m divisor de [tex]n[\/tex]), tal que [tex]n=d_i\\cdot d_j[\/tex], ou seja, [tex]\\dfrac{n}{d_i}=d_j[\/tex].<br \/>\nComo [tex]d_j[\/tex] existe e \u00e9 \u00fanico para cada [tex]d_i[\/tex], segue que os quocientes [tex]\\,\\dfrac{n}{d_1},\\,\\dfrac{n}{d_2},\\,\\cdots, \\, \\dfrac{n}{d_k}\\, [\/tex] esgotam os divisores de [tex]n[\/tex]; com isso temos que:<br \/>\n[tex]\\qquad \\dfrac{n}{d_1}+\\dfrac{n}{d_2}+\\cdots+ \\dfrac{n}{d_k}\\stackrel{\\color{red}{*}}{=}d_1+d_2+\\cdots+ d_k=124[\/tex].<br \/>\n[tex]\\qquad \\color{red}{*}[\/tex] n\u00e3o necessariamente nesta ordem.<br \/>\nFinalmente, conclu\u00edmos que<br \/>\n[tex]\\qquad \\dfrac{1}{d_1}+\\dfrac{1}{d_2}+\\cdots + \\dfrac{1}{d_k}=\\dfrac{\\dfrac{n}{d_1}+\\dfrac{n}{d_2}+\\cdots+ \\dfrac{n}{d_k}}{n}\\\\<br \/>\n\\quad \\boxed{\\dfrac{1}{d_1}+\\dfrac{1}{d_2}+\\cdots + \\dfrac{1}{d_k} =\\dfrac{124}{n}}.[\/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\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (Indicado a partir do 9\u00ba ano do E. F.) Sabendo-se que a soma de todos os divisores positivos [tex]d_1,\\, d_2,\\, \\cdots,\\, d_k[\/tex] de um n\u00famero natural [tex]n[\/tex] \u00e9 [tex]124[\/tex], com [tex]d_i \\ne d_j[\/tex] para [tex]i \\ne j[\/tex], calcule o valor de [tex]\\dfrac{1}{d_1}+\\dfrac{1}{d_2}+\\cdots + \\dfrac{1}{d_k}[\/tex] em fun\u00e7\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-soma-dos-divisores\/\">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-216678","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/216678","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=216678"}],"version-history":[{"count":4,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/216678\/revisions"}],"predecessor-version":[{"id":217724,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/216678\/revisions\/217724"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=216678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}