{"id":136581,"date":"2019-10-01T11:31:17","date_gmt":"2019-10-01T14:31:17","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=136581"},"modified":"2024-06-19T22:57:40","modified_gmt":"2024-06-20T01:57:40","slug":"problema-para-ajudar-na-escola-obtendo-quadrados-perfeitos","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-obtendo-quadrados-perfeitos\/","title":{"rendered":".Problema para ajudar na escola: Obtendo quadrados perfeitos"},"content":{"rendered":"

✏ Link do problema para dispositivos da Apple<\/em>.<\/span><\/a><\/strong><\/p>\n

\n

Problema
\n (A partir do 8\u00ba ano do E. F. – N\u00edvel de dificuldade: Dif\u00edcil<\/span>)<\/span><\/p>\n

\n

Um n\u00famero natural [tex]A[\/tex] \u00e9 dito um quadrado perfeito<\/em><\/strong> se existir um n\u00famero natural [tex]n[\/tex] tal que [tex]A=n^2.[\/tex] Assim, um quadrado perfeito \u00e9 simplesmente o produto de um n\u00famero natural por si pr\u00f3prio: [tex]\\boxed{A=n^2=n\\cdot n}.[\/tex]<\/p>\n

\n
\n

Preencha a tabela abaixo utilizando os [tex]16[\/tex] primeiros n\u00fameros naturais n\u00e3o nulos, sem repeti\u00e7\u00e3o, de modo que a soma de dois \u201dn\u00fameros vizinhos\u201d seja um quadrado perfeito.<\/p>\n

\"\"<\/p>\n

\n<\/div>\n

\n

\n

Solu\u00e7\u00e3o <\/p>\n

\n

\n

Observe inicialmente que o menor e o maior n\u00famero resultante da soma de dois n\u00fameros naturais distintos escolhidos dentre os [tex]16[\/tex] primeiros n\u00fameros naturais n\u00e3o nulos s\u00e3o [tex]1+2=3[\/tex] e [tex]15+16=31[\/tex], respectivamente.
\nPor outro lado, os quadrados perfeitos entre [tex]3[\/tex] e [tex]31[\/tex] s\u00e3o [tex]4=2^2[\/tex], [tex]9=3^2[\/tex], [tex]16=4^2[\/tex] e [tex]25=5^2[\/tex], e esses quatro quadrados perfeitos podem ser escritos como soma de n\u00fameros naturais distintos entre [tex]1[\/tex] e [tex]16[\/tex] das seguintes formas:<\/p>\n

    \n
  • [tex]4=\\textcolor{red}{1}+\\textcolor{#FF00FF}{3}[\/tex];<\/li>\n
  • [tex]9=\\textcolor{red}{1}+\\textcolor{#4CC417}{8}[\/tex]; [tex]9=\\textcolor{blue}{2}+\\textcolor{#FFA62F}{7}[\/tex]; [tex]9=\\textcolor{#FF00FF}{3}+\\textcolor{#A52A2A}{6}[\/tex]; [tex]9=\\textcolor{#00FFFF}{4}+\\textcolor{#00FF00}{5}[\/tex];<\/li>\n
  • [tex]16=\\textcolor{red}{1}+\\textcolor{#FFB86B}{15}[\/tex]; [tex]16=\\textcolor{blue}{2}+\\textcolor{#60D4FF}{14}[\/tex]; [tex]16=\\textcolor{#FF00FF}{3}+\\textcolor{#934CFF}{13}[\/tex]; [tex]16=\\textcolor{#00FFFF}{4}+\\textcolor{#D98BFF}{12}[\/tex]; [tex]16=\\textcolor{#00FF00}{5}+\\textcolor{#43BFC7}{11}[\/tex]; [tex]16=\\textcolor{#A52A2A}{6}+\\textcolor{#F535AA}{10}[\/tex]; [tex]16=\\textcolor{#FFA62F}{7}+\\textcolor{#800080}{9}[\/tex];<\/li>\n
  • [tex]25=\\textcolor{#800080}{9}+\\textcolor{# 008000}{16}[\/tex]; [tex]25=\\textcolor{#F535AA}{10}+\\textcolor{#FFB86B}{15}[\/tex]; [tex]25=\\textcolor{#43BFC7}{11}+\\textcolor{#60D4FF}{14}[\/tex]; [tex]25=\\textcolor{#D98BFF}{12}+\\textcolor{#934CFF}{13}[\/tex].<\/li>\n<\/ul>\n

    Para melhor visualiza\u00e7\u00e3o dos vizinhos poss\u00edveis para cada n\u00famero, vamos construir uma tabela com todas as combina\u00e7\u00f5es poss\u00edveis.<\/p>\n

    \"\"<\/p>\n

    Note que os n\u00fameros [tex]8[\/tex] e [tex]16[\/tex] s\u00f3 podem ter um vizinho, j\u00e1 que s\u00e3o parcelas de apenas uma soma cada um; esses vizinhos s\u00e3o [tex]1[\/tex] e [tex]9[\/tex], respectivamente. Dessa forma, [tex]8[\/tex] e [tex]16[\/tex] ficar\u00e3o nos extremos da tabela.<\/p>\n

    \\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#4CC417}{8}&\\textcolor{red}{1}&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\textcolor{#800080}{9}&\\textcolor{#008000}{16}\\\\
    \n\\hline \\end{array}<\/p>\n

    \\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#008000}{16}&\\textcolor{#800080}{9}&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\textcolor{red}{1}&\\textcolor{#4CC417}{8}\\\\
    \n\\hline
    \n\\end{array}
    \nVeja que o [tex]1[\/tex] e o [tex]3[\/tex] t\u00eam tr\u00eas poss\u00edveis vizinhos, mas os n\u00fameros restantes t\u00eam apenas dois poss\u00edveis vizinhos. Particularmente, o segundo vizinho do [tex]9[\/tex] \u00e9 o [tex]7[\/tex] e com isso completamos mais um espa\u00e7o das duas poss\u00edveis tabelas.<\/p>\n

    [tex]\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#4CC417}{8}&\\textcolor{red}{1}&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\textcolor{#FFA62F}{7}&\\textcolor{#800080}{9}&\\textcolor{#008000}{16}\\\\
    \n\\hline
    \n\\end{array}\\\\
    \n\\,\\\\
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#008000}{16}&\\textcolor{#800080}{9}&\\textcolor{#FFA62F}{7}&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\,\\,&\\textcolor{red}{1}&\\textcolor{#4CC417}{8}\\\\
    \n\\hline
    \n\\end{array}[\/tex]<\/p>\n

    O segundo vizinho do [tex]7[\/tex] \u00e9 o [tex]2[\/tex]. O segundo vizinho do [tex]2[\/tex] \u00e9 o [tex]14[\/tex]. O segundo vizinho do [tex]14[\/tex] \u00e9 o [tex]11[\/tex]. O segundo vizinho do [tex]11[\/tex] \u00e9 o [tex]5[\/tex]. O segundo vizinho do [tex]5[\/tex] \u00e9 o [tex]4[\/tex]. O segundo vizinho do [tex]4[\/tex] \u00e9 o [tex]12[\/tex]. O segundo vizinho do [tex]12[\/tex] \u00e9 o [tex]13[\/tex]. O segundo vizinho do [tex]13[\/tex] \u00e9 o [tex]3[\/tex]. Paramos por aqui, por enquanto.
    \nVejam nossas duas tabelas:<\/p>\n

    [tex]\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#4CC417}{8}&\\textcolor{red}{1}&\\,\\,&\\,\\,&\\,\\,&\\textcolor{#FF00FF}{3}&\\textcolor{#934CFF}{13}&\\textcolor{#D98BFF}{12}&\\textcolor{#00FFFF}{4}&\\textcolor{#00FF00}{5}&\\textcolor{#43BFC7}{11}&\\textcolor{#60D4FF}{14}&\\textcolor{blue}{2}&\\textcolor{#FFA62F}{7}&\\textcolor{#800080}{9}&\\textcolor{#008000}{16}\\\\
    \n\\hline
    \n\\end{array}\\\\
    \n\\,\\\\
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#008000}{16}&\\textcolor{#800080}{9}&\\textcolor{#FFA62F}{7}&\\textcolor{blue}{2}&\\textcolor{#60D4FF}{14}&\\textcolor{#43BFC7}{11}&\\textcolor{#00FF00}{5}&\\textcolor{#00FFFF}{4}&\\textcolor{#D98BFF}{12}&\\textcolor{#934CFF}{13}&\\textcolor{#FF00FF}{3}&\\,\\,&\\,\\,&\\,\\,&\\textcolor{red}{1}&\\textcolor{#4CC417}{8}\\\\
    \n\\hline
    \n\\end{array}.[\/tex]<\/p>\n

    Os poss\u00edveis vizinhos do [tex]3[\/tex] seriam [tex]1[\/tex], [tex]6[\/tex] e [tex]13[\/tex]; mas, como [tex]1[\/tex] e [tex]13[\/tex] j\u00e1 foram utilizados, resta apenas o [tex]6.[\/tex]<\/p>\n

    [tex]
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#4CC417}{8}&\\textcolor{red}{1}&\\,\\,&\\,\\,&\\textcolor{#A52A2A}{6}&\\textcolor{#FF00FF}{3}&\\textcolor{#934CFF}{13}&\\textcolor{#D98BFF}{12}&\\textcolor{#00FFFF}{4}&\\textcolor{#00FF00}{5}&\\textcolor{#43BFC7}{11}&\\textcolor{#60D4FF}{14}&\\textcolor{blue}{2}&\\textcolor{#FFA62F}{7}&\\textcolor{#800080}{9}&\\textcolor{#008000}{16}\\\\
    \n\\hline
    \n\\end{array}\\\\
    \n\\,\\\\
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#008000}{16}&\\textcolor{#800080}{9}&\\textcolor{#FFA62F}{7}&\\textcolor{blue}{2}&\\textcolor{#60D4FF}{14}&\\textcolor{#43BFC7}{11}&\\textcolor{#00FF00}{5}&\\textcolor{#00FFFF}{4}&\\textcolor{#D98BFF}{12}&\\textcolor{#934CFF}{13}&\\textcolor{#FF00FF}{3}&\\textcolor{#A52A2A}{6}&\\,\\,&\\,\\,&\\textcolor{red}{1}&\\textcolor{#4CC417}{8}\\\\
    \n\\hline
    \n\\end{array}[\/tex]<\/p>\n

    Dos poss\u00edveis vizinhos do [tex]1[\/tex] resta apenas o [tex]15.[\/tex]<\/p>\n

    [tex]
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#4CC417}{8}&\\textcolor{red}{1}&\\textcolor{#FFB86B}{15}&\\,\\,\\,&\\textcolor{#A52A2A}{6}&\\textcolor{#FF00FF}{3}&\\textcolor{#934CFF}{13}&\\textcolor{#D98BFF}{12}&\\textcolor{#00FFFF}{4}&\\textcolor{#00FF00}{5}&\\textcolor{#43BFC7}{11}&\\textcolor{#60D4FF}{14}&\\textcolor{blue}{2}&\\textcolor{#FFA62F}{7}&\\textcolor{#800080}{9}&\\textcolor{#008000}{16}\\\\
    \n\\hline
    \n\\end{array}\\\\
    \n\\,\\\\
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#008000}{16}&\\textcolor{#800080}{9}&\\textcolor{#FFA62F}{7}&\\textcolor{blue}{2}&\\textcolor{#60D4FF}{14}&\\textcolor{#43BFC7}{11}&\\textcolor{#00FF00}{5}&\\textcolor{#00FFFF}{4}&\\textcolor{#D98BFF}{12}&\\textcolor{#934CFF}{13}&\\textcolor{#FF00FF}{3}&\\textcolor{#A52A2A}{6}&\\,\\,\\,&\\textcolor{#FFB86B}{15}&\\textcolor{red}{1}&\\textcolor{#4CC417}{8}\\\\
    \n\\hline
    \n\\end{array}[\/tex]<\/p>\n

    Finalizando e conferindo, o n\u00famero [tex]10[\/tex], que sobrou, \u00e9 de fato um poss\u00edvel vizinho do [tex]6[\/tex] e do [tex]15.[\/tex] Portanto, s\u00e3o estas as poss\u00edveis tabelas que atendem \u00e0s hip\u00f3teses do problema:<\/p>\n

    [tex]\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#4CC417}{8}&\\textcolor{red}{1}&\\textcolor{#FFB86B}{15}&\\textcolor{#F535AA}{10}&\\textcolor{#A52A2A}{6}&\\textcolor{#FF00FF}{3}&\\textcolor{#934CFF}{13}&\\textcolor{#D98BFF}{12}&\\textcolor{#00FFFF}{4}&\\textcolor{#00FF00}{5}&\\textcolor{#43BFC7}{11}&\\textcolor{#60D4FF}{14}&\\textcolor{blue}{2}&\\textcolor{#FFA62F}{7}&\\textcolor{#800080}{9}&\\textcolor{#008000}{16}\\\\
    \n\\hline
    \n\\end{array}\\\\
    \n\\,\\\\
    \n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
    \n\\hline
    \n\\textcolor{#008000}{16}&\\textcolor{#800080}{9}&\\textcolor{#FFA62F}{7}&\\textcolor{blue}{2}&\\textcolor{#60D4FF}{14}&\\textcolor{#43BFC7}{11}&\\textcolor{#00FF00}{5}&\\textcolor{#00FFFF}{4}&\\textcolor{#D98BFF}{12}&\\textcolor{#934CFF}{13}&\\textcolor{#FF00FF}{3}&\\textcolor{#A52A2A}{6}&\\textcolor{#F535AA}{10}&\\textcolor{#FFB86B}{15}&\\textcolor{red}{1}&\\textcolor{#4CC417}{8}\\\\
    \n\\hline
    \n\\end{array}.[\/tex]<\/p>\n

    \n

    Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

    \n\n\n
    \"\"<\/td>\n\n

    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.
    \nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (Chrome<\/em>, Edge<\/em>, Firefox<\/em>, Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito Adobe Acrobat Reader<\/strong><\/em>.
    \nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n

    \n

    \nArquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    ✏ Link do problema para dispositivos da Apple. Problema (A partir do 8\u00ba ano do E. F. – N\u00edvel de dificuldade: Dif\u00edcil) Um n\u00famero natural [tex]A[\/tex] \u00e9 dito um quadrado perfeito se existir um n\u00famero natural [tex]n[\/tex] tal que [tex]A=n^2.[\/tex] Assim, um quadrado perfeito \u00e9 simplesmente o produto de um n\u00famero natural por si pr\u00f3prio: … <\/p>\n

    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-136581","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/136581","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=136581"}],"version-history":[{"count":15,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/136581\/revisions"}],"predecessor-version":[{"id":220557,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/136581\/revisions\/220557"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=136581"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}