{"id":111692,"date":"2018-09-23T19:36:00","date_gmt":"2018-09-23T22:36:00","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=111692"},"modified":"2024-06-02T09:03:44","modified_gmt":"2024-06-02T12:03:44","slug":"problema-para-ajudar-na-escola-qual-e-o-produto","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-qual-e-o-produto\/","title":{"rendered":".Problema para ajudar na escola: Qual \u00e9 o produto?"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=219694\" 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:24px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#00BF00\"> (A partir do 8\u00ba ano do E. F. &#8211; <span style=\"font-weight: normal;\"> N\u00edvel de dificuldade: M\u00e9dio<\/span>)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nO esquema abaixo mostra a multiplica\u00e7\u00e3o entre um n\u00famero de tr\u00eas algarismos e um n\u00famero de dois algarismos.<br \/>\nSe cada quadradinho representa um algarismo, calcule o resultado da multiplica\u00e7\u00e3o.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_220.png\" style=\"border:none\" alt=\"\" width=\"180\" height=\"153\" class=\"aligncenter size-full wp-image-105956\" \/><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\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\"\/>\nSeja [tex]a[\/tex] o algarismo das unidades do n\u00famero que aparece na primeira linha do esquema da nossa multiplica\u00e7\u00e3o.<\/p>\n<ul>\n<li> Observe que o primeiro produto obtido no esquema apresentado termina em [tex]5[\/tex].<\/li>\n<p style=\"text-align:center;\">\n[tex]\\begin{array}{c c c c c}<br \/>\n&#038; &#038; \\Box &#038; \\Box &#038; \\textcolor{red}{a}\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; \\textcolor{red}{3}\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; \\Box &#038; 0 &#038;\\textcolor{red}{5}\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038; \\textcolor{red}{5}<br \/>\n\\end{array}\\qquad [\/tex]\t<\/p>\n<p>Dessa forma, o produto  [tex] \\textcolor{red}{3 \\times a}[\/tex] termina em [tex] \\textcolor{red}{5}[\/tex] e [tex] \\textcolor{red}{ a}[\/tex] \u00e9 um algarismo; ent\u00e3o, [tex] \\textcolor{red}{a=5}.[\/tex]<br \/>\nFicamos, ent\u00e3o,  com um esquema melhor:<\/p>\n<p style=\"text-align:center;\">\n [tex]\\qquad \\qquad \\begin{array}{c c c c c}<br \/>\n&#038; &#038;  &#038; \\underline{1} &#038;  \\\\<br \/>\n&#038; &#038; \\Box &#038; \\Box &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; \\Box &#038; 0 &#038; 5\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]<\/p>\n<\/ul>\n<p>Seja [tex]b[\/tex] o algarismo das dezenas do n\u00famero que aparece na primeira linha do esquema da nossa multiplica\u00e7\u00e3o.<\/p>\n<ul>\n<li>No segundo passo do algoritmo da multiplica\u00e7\u00e3o fazemos um produto e uma soma e o n\u00famero que obtemos termina em [tex]0[\/tex].<\/li>\n<p style=\"text-align:center;\">\n[tex] \\begin{array}{c c c c c}<br \/>\n&#038; &#038;  &#038; \\underline{\\textcolor{red}{1}} &#038;  \\\\<br \/>\n&#038; &#038; \\Box &#038; \\textcolor{red}{b} &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; \\textcolor{red}{3}\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; \\Box &#038; \\textcolor{red}{0} &#038; 5\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]\t<\/p>\n<p>Dessa forma, o n\u00famero [tex] \\textcolor{red}{3 \\times b+1}[\/tex] termina em [tex] \\textcolor{red}{0}[\/tex] e sendo [tex] \\textcolor{red}{ b}[\/tex]  um algarismo; ent\u00e3o, [tex] \\textcolor{red}{b=3}.[\/tex]<br \/>\nFicamos, ent\u00e3o, com o seguinte esquema:<\/p>\n<p style=\"text-align:center;\">\n [tex]\\qquad \\qquad \\begin{array}{c c c c c}<br \/>\n&#038; &#038;  \\underline{1} &#038; \\underline{1} &#038;  \\\\<br \/>\n&#038; &#038; \\Box &#038; 3 &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; \\Box &#038; 0 &#038; 5\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]\n<\/p>\n<\/ul>\n<p>Agora, seja [tex]c[\/tex] o algarismo das centenas do n\u00famero que aparece na terceira linha do esquema da nossa multiplica\u00e7\u00e3o.<\/p>\n<ul>\n<li>Observe que o algarismo [tex]c[\/tex] somado com [tex]4[\/tex] produz um n\u00famero que termina em [tex]1.[\/tex]<\/li>\n<p style=\"text-align:center;\">\n [tex] \\begin{array}{c c c c c}<br \/>\n&#038; &#038;  \\underline{1} &#038; \\underline{1} &#038;  \\\\<br \/>\n&#038; &#038; \\Box &#038; 3 &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; \\textcolor{red}{c} &#038; 0 &#038; 5\\\\<br \/>\n&#038; \\Box &#038; \\textcolor{red}{4} &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; \\textcolor{red}{1} &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]<\/p>\n<p>Ent\u00e3o, a \u00fanica alternativa que temos \u00e9  [tex] \\textcolor{red}{c=7}[\/tex]. Com isso o n\u00famero que aparece na terceira linha est\u00e1 completamente determinado e ficamos com este esquema:<\/p>\n<p style=\"text-align:center;\">\n [tex]\\qquad \\qquad \\begin{array}{c c c c c}<br \/>\n&#038; &#038;  \\underline{1} &#038; \\underline{1} &#038;  \\\\<br \/>\n&#038; &#038; \\Box &#038; 3 &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; 7 &#038; 0 &#038; 5\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]\n<\/p>\n<\/ul>\n<p>Denotaremos por [tex]d[\/tex] o algarismo das centenas do n\u00famero que aparece na primeira linha.<\/p>\n<ul>\n<li> Perceba que [tex]\\textcolor{red}{3 \\times d35=705}[\/tex].<\/li>\n<p style=\"text-align:center;\">\n  [tex]\\begin{array}{c c c c c}<br \/>\n&#038; &#038; \\textcolor{red}{d} &#038; \\textcolor{red}{3} &#038; \\textcolor{red}{5}\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; \\textcolor{red}{3}\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; \\textcolor{red}{7} &#038; \\textcolor{red}{0} &#038; \\textcolor{red}{5}\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]<\/p>\n<p>Logo, [tex]d35=705 \\div3=235[\/tex] e, portanto, [tex]\\textcolor{red}{d=2}.[\/tex]<br \/>\nCom mais essa informa\u00e7\u00e3o, nossa multiplica\u00e7\u00e3o fica assim:<\/p>\n<p style=\"text-align:center;\">\n [tex]\\qquad \\qquad \\begin{array}{c c c c c}<br \/>\n&#038; &#038; 2 &#038; 3 &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; \\Box &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; 7 &#038; 0 &#038; 5\\\\<br \/>\n&#038; \\Box &#038; 4 &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]\n<\/p>\n<\/ul>\n<p>Vamos denotar por [tex]e[\/tex] o algarismo das dezenas do n\u00famero da segunda linha.<\/p>\n<ul>\n<li> Praticamente finalizaremos o problema.<\/li>\n<p style=\"text-align:center;\">\n[tex] \\begin{array}{c c c c c}<br \/>\n&#038; &#038; \\textcolor{red}{2} &#038; \\textcolor{red}{3} &#038;\\textcolor{red}{ 5}\\\\<br \/>\n&#038; \\times&#038;  &#038; \\textcolor{red}{e} &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; 7 &#038; 0 &#038; 5\\\\<br \/>\n&#038; \\Box &#038; \\textcolor{red}{4} &#038; \\Box &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n\\Box &#038; \\Box &#038; 1 &#038; \\Box &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]<\/p>\n<p>Observe que  [tex]e \\times 235[\/tex] resulta em um n\u00famero de tr\u00eas algarismos, cujo algarismo central \u00e9 [tex]\\textcolor{red}{4}.[\/tex]<br \/>\nMas s\u00f3 existem tr\u00eas m\u00faltiplos de [tex] 235[\/tex] com tr\u00eas algarismos:<br \/>\n[tex] \\quad 235=235 \\times 1\\; ;\\quad 470=235 \\times 2\\; ;\\; 705=235 \\times 3\\; ;\\; \\boxed{940=235 \\times \\textcolor{red}{4}}\\,.[\/tex]<br \/>\nComo podemos ver, desses, apenas o n\u00famero [tex]940[\/tex] satisfaz a condi\u00e7\u00e3o de o algarismo central ser [tex]4[\/tex]. Assim, [tex]\\textcolor{red}{e=4}[\/tex] e fechamos a nossa multiplica\u00e7\u00e3o, j\u00e1 que, com essas informa\u00e7\u00f5es, conclu\u00edmos que o esquema apresentado no problema \u00e9 o algoritmo da multiplica\u00e7\u00e3o [tex]235 \\times 43.[\/tex]<\/ul>\n<p style=\"text-align:center;\">\n[tex]\\qquad \\qquad \\begin{array}{c c c c c}<br \/>\n&#038; &#038; 2 &#038; 3 &#038; 5\\\\<br \/>\n&#038; \\times&#038;  &#038; 4 &#038; 3\\\\<br \/>\n\\hline<br \/>\n&#038; &#038; 7 &#038; 0 &#038; 5\\\\<br \/>\n&#038; 9 &#038; 4 &#038; 0 &#038; &#038; \\\\<br \/>\n\\hline<br \/>\n1 &#038; 0 &#038; 1 &#038; 0 &#038;  5<br \/>\n\\end{array}\\qquad [\/tex]<\/p>\n<p>Dessa forma, o resultado da multiplica\u00e7\u00e3o \u00e9 [tex]\\, \\fcolorbox{black}{#eee0e5}{$10105$}[\/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=\"g3\">\nParticipou da discuss\u00e3o o Clube <strong>OCTETO MATEM\u00c1TICO<\/strong>.<\/div>\n<p style=\"margin:20px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala\" style=\"width:520px;background-color: #FFF5EE; padding:10px; font-family: calibri; font-size: 13px;color: #1C1C1C; line-height: 20px; \"> Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.<br \/>\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (<em>Chrome<\/em>, <em>Edge<\/em>, <em>Firefox<\/em>, <em>Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito <em><strong>Adobe Acrobat Reader<\/strong><\/em>.<br \/>\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o <strong><em>Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar <a href=\"http:\/\/get.adobe.com\/br\/reader\/\" rel=\"noopener\" target=\"_blank\"><strong>AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\" float:right;  width: 100px;\">\n<p style=\"text-align: center; border-style: dotted; border-color :#FFFAFA; border-width :thin; width:90px;  height:40px;  padding: 5px; background-color: #800000;\">\n<a href=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/12\/Qual-\u00e9-o-produto.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Qual \u00e9 o produto.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (A partir do 8\u00ba ano do E. F. &#8211; N\u00edvel de dificuldade: M\u00e9dio) O esquema abaixo mostra a multiplica\u00e7\u00e3o entre um n\u00famero de tr\u00eas algarismos e um n\u00famero de dois algarismos. Se cada quadradinho representa um algarismo, calcule o resultado da multiplica\u00e7\u00e3o. Solu\u00e7\u00e3o Seja [tex]a[\/tex] o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-qual-e-o-produto\/\">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-111692","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/111692","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=111692"}],"version-history":[{"count":113,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/111692\/revisions"}],"predecessor-version":[{"id":219696,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/111692\/revisions\/219696"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=111692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}