{"id":105157,"date":"2018-06-20T18:11:05","date_gmt":"2018-06-20T21:11:05","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=105157"},"modified":"2023-05-02T04:15:59","modified_gmt":"2023-05-02T07:15:59","slug":"problemao-uma-soma-de-produtos","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-uma-soma-de-produtos\/","title":{"rendered":".Problem\u00e3o: Uma soma de produtos"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:27px;\">\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\"\/>\nSeja [tex]p(n)[\/tex] o produto dos d\u00edgitos de um n\u00famero natural [tex]n[\/tex]. Por exemplo,  <\/p>\n<ul>\n<li>[tex]p(8)=8[\/tex],<\/li>\n<li> [tex]p(32)=6[\/tex],<\/li>\n<li> [tex]p(241)=8[\/tex].<\/li>\n<\/ul>\n<p>Determine o valor da soma [tex]\\boxed{p(1)+p(2)+ \\dots + p(99)+p(100)} \\, .[\/tex]\n<\/p><\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\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 a soma [tex]\\boxed{p(1)+p(2)+ \\dots + p(99)+p(100)} \\, [\/tex], vamos agrupar de maneira conveniente as parcelas. Mas, antes,  observe que os m\u00faltiplos de [tex]10[\/tex] em nada contribuem para a soma solicitada no problema, j\u00e1 que o produto dos d\u00edgitos desses n\u00fameros ser\u00e1 zero. Portanto, vamos ignor\u00e1-los nas somas parciais que faremos.<br \/>\nVeja que:<\/p>\n<ul>\n<li>[tex]p(1)+p(2)+ \\dots + p(8)+p(9)=1+2+\\cdots+8+9 \\, [\/tex];<\/li>\n<li>[tex]\\begin{align*} p(11)+p(12)+ \\dots + p(18)+p(19)&#038;=1\\cdot 1+1\\cdot 2+\\cdots+1\\cdot 8+1\\cdot 9\\\\<br \/>\n&#038;=1\\cdot \\left(1+2+\\cdots+8+9 \\right) \\, ; \\end{align*}[\/tex]<\/li>\n<li>[tex]\\begin{align*} p(21)+p(22)+ \\dots + p(28)+p(29)&#038;=2\\cdot 1+2\\cdot 2+\\cdots+2\\cdot 8+2\\cdot 9\\\\<br \/>\n&#038;=2\\cdot \\left(1+2+\\cdots+8+9 \\right) \\, ; \\end{align*}[\/tex]<\/li>\n<li>[tex]\\begin{align*} p(31)+p(32)+ \\dots + p(38)+p(39)&#038;=3\\cdot 1+3\\cdot 2+\\cdots+3\\cdot 8+3\\cdot 9\\\\<br \/>\n&#038;=3\\cdot \\left(1+2+\\cdots+8+9 \\right) \\, ; \\end{align*}[\/tex]<\/li>\n<\/ul>\n<p>e assim sucessivamente, at\u00e9 obtermos a \u00faltima soma parcial:<\/p>\n<ul>\n<li>[tex]\\begin{align*} p(91)+p(92)+ \\dots + p(98)+p(99)&#038;=9\\cdot 1+9\\cdot 2+\\cdots+9\\cdot 8+9\\cdot 9\\\\<br \/>\n&#038;=9\\cdot \\left(1+2+\\cdots+8+9 \\right) \\, ; \\end{align*}[\/tex]<\/li>\n<\/ul>\n<p>Portanto,<br \/>\n[tex]\\qquad \\begin{align*}p(1)+p(2)+ \\dots + p(99)+p(100)=&#038;\\left(p(1)+p(2)+ \\dots + p(8)+p(9)\\right)+\\\\<br \/>\n&#038;+\\left( p(11)+p(12)+ \\dots + p(18)+p(19)\\right)+\\\\<br \/>\n&#038;+\\left( p(21)+p(22)+ \\dots + p(28)+p(29)\\right)+\\\\<br \/>\n&#038;+\\left(p(31)+p(32)+ \\dots + p(38)+p(39)\\right)+\\\\<br \/>\n&#038;+ \\cdots+\\\\<br \/>\n&#038;+\\left(p(91)+p(92)+ \\dots + p(98)+p(99)\\right) \\, ,<br \/>\n\\end{align*}[\/tex]<br \/>\nou seja,<br \/>\n[tex]\\qquad \\begin{align*}p(1)+p(2)+ \\dots + p(99)+p(100)=&#038;\\left(1+2+\\cdots+8+9\\right)+\\\\<br \/>\n&#038;+ 1\\cdot \\left(1+2+\\cdots+8+9 \\right)+\\\\<br \/>\n&#038;+ 2\\cdot \\left(1+2+\\cdots+8+9 \\right)+\\\\<br \/>\n&#038;+3\\cdot \\left(1+2+\\cdots+8+9 \\right)+\\\\<br \/>\n&#038;+ \\cdots+\\\\<br \/>\n&#038;+9\\cdot \\left(1+2+\\cdots+8+9 \\right),<br \/>\n\\end{align*}[\/tex]<br \/>\nou, ainda,<br \/>\n[tex]\\qquad \\begin{align*}p(1)+p(2)+ \\dots + p(99)+p(100)=&#038;\\left(1+2+\\cdots+8+9\\right)+\\\\<br \/>\n&#038;+\\left[\\left(1+2+\\cdots+8+9\\right)\\cdot\\left(1+2+\\cdots+8+9\\right)\\right]\\\\<br \/>\n=&#038;\\left(1+2+\\cdots+8+9\\right)+\\left(1+2+\\cdots+8+9\\right)^2 \\, .<br \/>\n\\end{align*}[\/tex]<br \/>\nFinalmente, como [tex]1+2+\\cdots+8+9=45 \\, [\/tex], segue que:<br \/>\n[tex]\\qquad \\fcolorbox{black}{#eee0e5}{$p(1)+p(2)+ \\dots + p(99)+p(100)=45+45^2=45+2025=2070$} \\, .[\/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","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 7\u00ba ano do E. F.) Seja [tex]p(n)[\/tex] o produto dos d\u00edgitos de um n\u00famero natural [tex]n[\/tex]. Por exemplo, [tex]p(8)=8[\/tex], [tex]p(32)=6[\/tex], [tex]p(241)=8[\/tex]. Determine o valor da soma [tex]\\boxed{p(1)+p(2)+ \\dots + p(99)+p(100)} \\, .[\/tex] Solu\u00e7\u00e3o Para obter a soma [tex]\\boxed{p(1)+p(2)+ \\dots + p(99)+p(100)} \\, [\/tex], vamos agrupar de maneira conveniente as parcelas. &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-uma-soma-de-produtos\/\">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-105157","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/105157","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=105157"}],"version-history":[{"count":39,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/105157\/revisions"}],"predecessor-version":[{"id":188611,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/105157\/revisions\/188611"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=105157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}