{"id":1071,"date":"2019-03-02T17:19:13","date_gmt":"2019-03-02T16:19:13","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=1071"},"modified":"2024-02-08T14:06:07","modified_gmt":"2024-02-08T13:06:07","slug":"nombre-de-chiffres-dun-nombre","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2019\/03\/02\/nombre-de-chiffres-dun-nombre\/","title":{"rendered":"Nombre de chiffres d&rsquo;un nombre"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Il arrive parfois qu&rsquo;un nombre s&rsquo;\u00e9crive de mani\u00e8re tr\u00e8s condens\u00e9e mais que le nombre de chiffres qui le compose soit tr\u00e8s grand. <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Par exemple, le nombre \\(9^{8^7}\\) ne s&rsquo;affiche m\u00eame pas avec Xcas&#8230; tellement le nombre de chiffres qui le composent est grand. Mais comment savoir ce nombre de chiffres ?<\/p>\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_85 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/03\/02\/nombre-de-chiffres-dun-nombre\/#En_base_decimale\" >En base d\u00e9cimale<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/03\/02\/nombre-de-chiffres-dun-nombre\/#En_binaire\" >En binaire<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/03\/02\/nombre-de-chiffres-dun-nombre\/#Generalites\" >G\u00e9n\u00e9ralit\u00e9s<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"En_base_decimale\"><\/span>En base d\u00e9cimale<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Notons:$$N=\\sum_{k=0}^{n-1}a_k\\times10^k.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ce nombre est compos\u00e9 de \\(n\\) chiffres : \\(a_0,\\ a_1,\\ a_2,\\ \\ldots,\\ a_{n-1}\\). On l&rsquo;\u00e9crit:$$N=\\overline{a_{n-1}a_{n-2}\\cdots a_2a_1a_0}^{10}.$$Un nombre \u00e0 \\(n\\) chiffres est n\u00e9cessairement compris entre \\(10^{n-1}\\) et \\(10^n\\), donc:$$10^{n-1}\\leq N &lt; 10^n.$$En composant par le logarithme d\u00e9cimal, on obtient l&rsquo;encadrement:$$\\log(10^{n-1}) \\leq \\log(N) &lt; \\log(10^n),$$soit:$$(n-1)\\log(10)\\leq\\log(N)&lt;n\\log(10).$$Or, par d\u00e9finition, \\(\\log(10)=1\\) d&rsquo;o\u00f9 finalement:$$n-1\\leq\\log(N)&lt;n.$$On en d\u00e9duit alors que \\(n\\) est l&rsquo;entier imm\u00e9diatement sup\u00e9rieur (ou \u00e9gal) \u00e0 \\(\\log(N)\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Par exemple, $$\\log\\left(9^{8^7}\\right)=8^7\\log(9)\\approx4607913,91681$$ donc le nombre de chiffres de \\(9^{8^7}\\) est 4607914.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"En_binaire\"><\/span>En binaire<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Le principe est le m\u00eame. On consid\u00e8re un nombre:$$N=\\overline{a_{n-1}\\cdots a_1a_0}^{2}=\\sum_{k=0}^{n-1}a_k\\times2^k\\,,\\,\\ a_{n-1}\\neq0.$$Alors, pour \\(N\\) exprim\u00e9 en d\u00e9cimal, $$2^{n-1} \\leq N&lt; 2^n$$ soit: $$\\frac{\\ln(2^{n-1})}{\\ln2} \\leq \\frac{\\ln(N)}{\\ln2} &lt; \\frac{\\ln(2^n)}{\\ln2},$$ d&rsquo;o\u00f9: $$n-1 \\leq \\frac{\\ln(N)}{\\ln2} &lt; n.$$ Ainsi, le nombre \\(n\\) de chiffres (en binaire) du nombre \\(N\\) est-il \u00e9gal \u00e0 l&rsquo;entier imm\u00e9diatement sup\u00e9rieur (ou \u00e9gal) \u00e0 \\( \\frac{\\ln(N)}{\\ln2}  \\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Par exemple, \\( \\overline{1101}^{2}=\\overline{13}^{10}\\), et \\( ENT\\left( \\frac{\\ln(13)}{\\ln2}\\right)+1  =4\\). Il y a bien 4 chiffres dans le nombre binaire correspondant \u00e0 13 (en base d\u00e9cimale).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Generalites\"><\/span>G\u00e9n\u00e9ralit\u00e9s<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">On peut bien entendu g\u00e9n\u00e9raliser cette formule en disant que si \\(N\\) est un nombre d\u00e9cimal alors le nombre de chiffres du nombre en base \\(a\\) correspondant est \u00e9gal \u00e0:$$n=ENT\\left(\\frac{\\ln(N)}{\\ln(a)}\\right)+1.$$<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Il arrive parfois qu&rsquo;un nombre s&rsquo;\u00e9crive de mani\u00e8re tr\u00e8s condens\u00e9e mais que le nombre de chiffres qui le compose soit tr\u00e8s grand. Par exemple, le nombre \\(9^{8^7}\\) ne s&rsquo;affiche m\u00eame pas avec Xcas&#8230; tellement le nombre de chiffres qui le composent est grand. Mais comment savoir ce nombre de chiffres [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4,6],"tags":[89],"class_list":["post-1071","post","type-post","status-publish","format-standard","hentry","category-informatique","category-mathematiques","tag-logarithme"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.8 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Nombre de chiffres d&#039;un nombre - Mathweb.fr<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/03\/02\/nombre-de-chiffres-dun-nombre\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Nombre de chiffres d&#039;un nombre - Mathweb.fr\" \/>\n<meta property=\"og:description\" content=\"Il arrive parfois qu&rsquo;un nombre s&rsquo;\u00e9crive de mani\u00e8re tr\u00e8s condens\u00e9e mais que le nombre de chiffres qui le compose soit tr\u00e8s grand. Par exemple, le nombre (9^{8^7}) ne s&rsquo;affiche m\u00eame pas avec Xcas&#8230; tellement le nombre de chiffres qui le composent est grand. 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