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<!--[if lt IE 7]> <html class="no-js lt-ie9 lt-ie8 lt-ie7"> <![endif]--><!--[if IE 7]> <html class="no-js lt-ie9 lt-ie8"> <![endif]--><!--[if IE 8]> <html class="no-js lt-ie9"> <![endif]--><!--[if gt IE 8]><!--><html class="no-js"> <head><meta charset="utf-8"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><title>Блог 529</title><meta name="description" content><meta name="viewport" content="width=device-width"><link rel="stylesheet" href="./theme/css/normalize.css"><link href="http://fonts.googleapis.com/css?family=Philosopher&subset=latin,cyrillic" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Forum&subset=cyrillic" rel="stylesheet" type="text/css"><link href="//fonts.googleapis.com/css?family=Oswald" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Ubuntu+Mono" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=PT+Sans" rel="stylesheet" type="text/css"><link rel="stylesheet" href="./theme/css/font-awesome.min.css"><link rel="stylesheet" href="./theme/css/main.css"><link href="http://likemath.ru/feeds/all.atom.xml" type="application/atom+xml" rel="alternate" title="Блог 529 Atom Feed"><link href="http://likemath.ru/feeds/all.rss.xml" type="application/rss+xml" rel="alternate" title="Блог 529 RSS Feed"><script src="./theme/js/vendor/modernizr-2.6.2.min.js"></script></head><body><!--[if lt IE 7]> <!--[if lt IE 7]> <html class="no-js lt-ie9 lt-ie8 lt-ie7"> <![endif]--><!--[if IE 7]> <html class="no-js lt-ie9 lt-ie8"> <![endif]--><!--[if IE 8]> <html class="no-js lt-ie9"> <![endif]--><!--[if gt IE 8]><!--><html class="no-js"> <head><meta charset="utf-8"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><title>Блог 529</title><meta name="description" content><meta name="viewport" content="width=device-width"><link rel="stylesheet" href="./theme/css/normalize.css"><link href="http://fonts.googleapis.com/css?family=Philosopher&subset=latin,cyrillic" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Forum&subset=cyrillic" rel="stylesheet" type="text/css"><link href="//fonts.googleapis.com/css?family=Oswald" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Ubuntu+Mono" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=PT+Sans" rel="stylesheet" type="text/css"><link rel="stylesheet" href="./theme/css/font-awesome.min.css"><link rel="stylesheet" href="./theme/css/main.css"><link href="http://likemath.ru/feeds/all.atom.xml" type="application/atom+xml" rel="alternate" title="Блог 529 Atom Feed"><link href="http://likemath.ru/feeds/all.rss.xml" type="application/rss+xml" rel="alternate" title="Блог 529 RSS Feed"><script src="./theme/js/vendor/modernizr-2.6.2.min.js"></script></head><body><!--[if lt IE 7]>
<p class="chromeframe">You are using an <strong>outdated</strong> browser. Please <a href="http://browsehappy.com/">upgrade your browser</a> or <a href="http://www.google.com/chromeframe/?redirect=true">activate Google Chrome Frame</a> to improve your experience.</p> <p class="chromeframe">You are using an <strong>outdated</strong> browser. Please <a href="http://browsehappy.com/">upgrade your browser</a> or <a href="http://www.google.com/chromeframe/?redirect=true">activate Google Chrome Frame</a> to improve your experience.</p>
<![endif]--><div id="wrapper"><h1>Archives for Блог 529</h1><dl><dt>Пт 21 Октябрь 2016</dt><dd><a href="./posts/moio-reshenie-zadachi-146/">Моё решение задачи&nbsp;146</a></dd><dt>Пт 22 Июль 2016</dt><dd><a href="./posts/nakhozhdenie-summy-k-ykh-stepenei/">Нахождение суммы k-ых&nbsp;степеней</a></dd><dt>Чт 17 Март 2016</dt><dd><a href="./posts/wallabag-i-realnaia-zhizn/">Wallabag и реальная&nbsp;жизнь</a></dd><dt>Вс 10 Январь 2016</dt><dd><a href="./posts/kak-ia-shakhmatnogo-bota-pisal/">Как я шахматного бота&nbsp;писал</a></dd><dt>Вс 02 Август 2015</dt><dd><a href="./posts/crossgen-v10/">CrossGen&nbsp;v1.0</a></dd><dt>Пт 17 Июль 2015</dt><dd><a href="./posts/moio-reshenie-zadachi-60/">Моё решение задачи&nbsp;60</a></dd><dt>Пт 03 Июль 2015</dt><dd><a href="./posts/eshchio-odno-vychislenie-vyrazhenii/">Ещё одно вычисление&nbsp;выражений</a></dd><dt>Пт 17 Апрель 2015</dt><dd><a href="./posts/moi-pervyi-post/">Мой первый&nbsp;пост</a></dd></dl></div><script> <![endif]--><div id="wrapper"><h1>Archives for Блог 529</h1><dl><dt>Вс 30 Октябрь 2016</dt><dd><a href="./posts/moio-reshenie-zadachi-134/">Моё решение задачи&nbsp;134</a></dd><dt>Пт 21 Октябрь 2016</dt><dd><a href="./posts/moio-reshenie-zadachi-146/">Моё решение задачи&nbsp;146</a></dd><dt>Пт 22 Июль 2016</dt><dd><a href="./posts/nakhozhdenie-summy-k-ykh-stepenei/">Нахождение суммы k-ых&nbsp;степеней</a></dd><dt>Чт 17 Март 2016</dt><dd><a href="./posts/wallabag-i-realnaia-zhizn/">Wallabag и реальная&nbsp;жизнь</a></dd><dt>Вс 10 Январь 2016</dt><dd><a href="./posts/kak-ia-shakhmatnogo-bota-pisal/">Как я шахматного бота&nbsp;писал</a></dd><dt>Вс 02 Август 2015</dt><dd><a href="./posts/crossgen-v10/">CrossGen&nbsp;v1.0</a></dd><dt>Пт 17 Июль 2015</dt><dd><a href="./posts/moio-reshenie-zadachi-60/">Моё решение задачи&nbsp;60</a></dd><dt>Пт 03 Июль 2015</dt><dd><a href="./posts/eshchio-odno-vychislenie-vyrazhenii/">Ещё одно вычисление&nbsp;выражений</a></dd><dt>Пт 17 Апрель 2015</dt><dd><a href="./posts/moi-pervyi-post/">Мой первый&nbsp;пост</a></dd></dl></div><script>
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<!--[if lt IE 7]> <html class="no-js lt-ie9 lt-ie8 lt-ie7"> <![endif]--><!--[if IE 7]> <html class="no-js lt-ie9 lt-ie8"> <![endif]--><!--[if IE 8]> <html class="no-js lt-ie9"> <![endif]--><!--[if gt IE 8]><!--><html class="no-js"> <head><meta charset="utf-8"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><title>Блог 529 - Алексей Лобанов</title><meta name="description" content><meta name="viewport" content="width=device-width"><link rel="stylesheet" href="../theme/css/normalize.css"><link href="http://fonts.googleapis.com/css?family=Philosopher&subset=latin,cyrillic" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Forum&subset=cyrillic" rel="stylesheet" type="text/css"><link href="//fonts.googleapis.com/css?family=Oswald" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Ubuntu+Mono" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=PT+Sans" rel="stylesheet" type="text/css"><link rel="stylesheet" href="../theme/css/font-awesome.min.css"><link rel="stylesheet" href="../theme/css/main.css"><link rel="stylesheet" href="../theme/css/blog.css"><link rel="stylesheet" href="../theme/css/github.css"><link href="http://likemath.ru/feeds/all.atom.xml" type="application/atom+xml" rel="alternate" title="Блог 529 Atom Feed"><link href="http://likemath.ru/feeds/all.rss.xml" type="application/rss+xml" rel="alternate" title="Блог 529 RSS Feed"><script src="../theme/js/vendor/modernizr-2.6.2.min.js"></script></head><body><!--[if lt IE 7]> <!--[if lt IE 7]> <html class="no-js lt-ie9 lt-ie8 lt-ie7"> <![endif]--><!--[if IE 7]> <html class="no-js lt-ie9 lt-ie8"> <![endif]--><!--[if IE 8]> <html class="no-js lt-ie9"> <![endif]--><!--[if gt IE 8]><!--><html class="no-js"> <head><meta charset="utf-8"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><title>Блог 529 - Алексей Лобанов</title><meta name="description" content><meta name="viewport" content="width=device-width"><link rel="stylesheet" href="../theme/css/normalize.css"><link href="http://fonts.googleapis.com/css?family=Philosopher&subset=latin,cyrillic" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Forum&subset=cyrillic" rel="stylesheet" type="text/css"><link href="//fonts.googleapis.com/css?family=Oswald" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Ubuntu+Mono" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=PT+Sans" rel="stylesheet" type="text/css"><link rel="stylesheet" href="../theme/css/font-awesome.min.css"><link rel="stylesheet" href="../theme/css/main.css"><link rel="stylesheet" href="../theme/css/blog.css"><link rel="stylesheet" href="../theme/css/github.css"><link href="http://likemath.ru/feeds/all.atom.xml" type="application/atom+xml" rel="alternate" title="Блог 529 Atom Feed"><link href="http://likemath.ru/feeds/all.rss.xml" type="application/rss+xml" rel="alternate" title="Блог 529 RSS Feed"><script src="../theme/js/vendor/modernizr-2.6.2.min.js"></script></head><body><!--[if lt IE 7]>
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) { <![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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<![endif]--><div id="wrapper"><h1>Authors on Блог 529</h1><ul> <li><a href="./author/aleksei-lobanov.html">Алексей Лобанов</a> (8)</li></ul></div><script> <![endif]--><div id="wrapper"><h1>Authors on Блог 529</h1><ul> <li><a href="./author/aleksei-lobanov.html">Алексей Лобанов</a> (9)</li></ul></div><script>
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) { <![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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<feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529</title><link href="http://likemath.ru/" rel="alternate"></link><link href="http://likemath.ru/feeds/all.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-21T17:40:00+03:00</updated><entry><title>Моё решение задачи 146</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-146/" rel="alternate"></link><published>2016-10-21T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/</id><summary type="html">&lt;p&gt;Краткое условие: необходимо найти сумму всех натуральных &lt;span class="math"&gt;\(n\)&lt;/span&gt;, что &lt;span class="math"&gt;\(n^2+1\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+3\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+7\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+9\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+13\)&lt;/span&gt;, и &lt;span class="math"&gt;\(n^2+27\)&lt;/span&gt; будут последовательными простыми&amp;nbsp;числами.&lt;/p&gt; <feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529</title><link href="http://likemath.ru/" rel="alternate"></link><link href="http://likemath.ru/feeds/all.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html">&lt;p&gt;Краткое условие: назовём &lt;em&gt;порождающим&lt;/em&gt; для двух последовательных простых &lt;span class="math"&gt;\(p_1 &amp;lt; p_2\)&lt;/span&gt; наименьшее натуральное число, что оно закачивается на &lt;span class="math"&gt;\(p_1\)&lt;/span&gt; и при этом делится на &lt;span class="math"&gt;\(p_2\)&lt;/span&gt;. Необходимо найти сумму порождающих для всех &lt;span class="math"&gt;\(p_1 \in \left[ 5; 10^6&amp;nbsp;\right]\)&lt;/span&gt;&lt;/p&gt;
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&lt;/script&gt;</summary><category term="Project Euler"></category><category term="Python"></category><category term="sympy"></category></entry><entry><title>Моё решение задачи 146</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-146/" rel="alternate"></link><published>2016-10-21T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/</id><summary type="html">&lt;p&gt;Краткое условие: необходимо найти сумму всех натуральных &lt;span class="math"&gt;\(n\)&lt;/span&gt;, что &lt;span class="math"&gt;\(n^2+1\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+3\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+7\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+9\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+13\)&lt;/span&gt;, и &lt;span class="math"&gt;\(n^2+27\)&lt;/span&gt; будут последовательными простыми&amp;nbsp;числами.&lt;/p&gt;
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&lt;/script&gt;</description><dc:creator xmlns:dc="http://purl.org/dc/elements/1.1/">Алексей Лобанов</dc:creator><pubDate>Sun, 30 Oct 2016 17:40:00 +0300</pubDate><guid isPermaLink="false">tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</guid><category>Project Euler</category><category>Python</category><category>sympy</category></item><item><title>Моё решение задачи 146</title><link>http://likemath.ru/posts/moio-reshenie-zadachi-146/</link><description>&lt;p&gt;Краткое условие: необходимо найти сумму всех натуральных &lt;span class="math"&gt;\(n\)&lt;/span&gt;, что &lt;span class="math"&gt;\(n^2+1\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+3\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+7\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+9\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+13\)&lt;/span&gt;, и &lt;span class="math"&gt;\(n^2+27\)&lt;/span&gt; будут последовательными простыми&amp;nbsp;числами.&lt;/p&gt;
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<feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529</title><link href="http://likemath.ru/" rel="alternate"></link><link href="http://likemath.ru/feeds/tag-sympy.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html">&lt;p&gt;Краткое условие: назовём &lt;em&gt;порождающим&lt;/em&gt; для двух последовательных простых &lt;span class="math"&gt;\(p_1 &amp;lt; p_2\)&lt;/span&gt; наименьшее натуральное число, что оно закачивается на &lt;span class="math"&gt;\(p_1\)&lt;/span&gt; и при этом делится на &lt;span class="math"&gt;\(p_2\)&lt;/span&gt;. Необходимо найти сумму порождающих для всех &lt;span class="math"&gt;\(p_1 \in \left[ 5; 10^6&amp;nbsp;\right]\)&lt;/span&gt;&lt;/p&gt;
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href="."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="./">Главная</a></li><li><a href="./pages/projects.html">Мои проекты</a></li><li><a href="./pages/about.html">Об авторе</a></li><li><a href="./feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="./posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) { <![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href="."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="./">Главная</a></li><li><a href="./pages/projects.html">Мои проекты</a></li><li><a href="./pages/about.html">Об авторе</a></li><li><a href="./feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="./posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="./posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href="../.."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../../">Главная</a></li><li><a href="../../pages/projects.html">Мои проекты</a></li><li><a href="../../pages/about.html">Об авторе</a></li><li><a href="../../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="../../posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><p>Например, если <span class="math">\(p_1 = 19\)</span>, то следующее простое <span class="math">\(p_2 = 23\)</span>. Тогда порождающим будет число <span class="math">\(1219\)</span>, при этом <span class="math">\(1219 \: \vdots \: 23\)</span>.</p><p>Полное условие можно найти <a href="https://projecteuler.net/problem=134">тут</a></p><p>Несмотря на то, что сложность задачи 45%, для её решения достаточно выписать&nbsp;условие.</p><p>Пусть <span class="math">\(p_1\)</span> содержит в себе <span class="math">\(k\)</span> цифр, т.е. <span class="math">\(n = r \cdot 10^k + p_1\)</span>, где <span class="math">\(r\)</span> &#8212; какое-то натуральное число с отрезка <span class="math">\(\left[ 1; p_2-1&nbsp;\right]\)</span></p><p>Давайте посчитаем остатки по модулю <span class="math">\(p_2\)</span>: <span class="math">\(n \equiv r \cdot 10^k + p_1 \equiv 0\)</span>. Отсюда получим явную формулу для <span class="math">\(r\)</span>: <div class="math">$$ r \equiv -p_1 \cdot 10^{-k} \equiv -p_1 \cdot 10^{p_2 -1-k} $$</div></p><p>Комментарии:</p><ol><li>Так как <span class="math">\(a^p \equiv a \mod p\)</span>, то верно что <span class="math">\(a^{-k} \equiv a^{p -1-k} \mod&nbsp;p\)</span></li><li>Это всё бессмысленно, если не знать про <a href="https://ru.wikipedia.org/wiki/Алгоритмы_быстрого_возведения_в_степень">алгоритм быстрого возведения в степень</a>, который делает асимптотическую сложность возведения в степень&nbsp;логарифмической.</li></ol><p>У нас есть явная формула для порождающего, и мы знаем как её быстро посчитать. Ниже приведён код на Python с использованием <a href="http://www.sympy.org/ru/">sympy</a>.</p><div class="highlight"><pre><span class="code-line"><span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">primerange</span> <span class="c1"># для получения простых чисел</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="c1"># быстрое возведение в степень по модулю</span></span>
<span class="code-line"><span class="k">def</span> <span class="nf">fast_pow</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">y</span><span class="p">,</span> <span class="n">modulo</span><span class="p">):</span></span>
<span class="code-line"> <span class="k">if</span> <span class="n">y</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span></span>
<span class="code-line"> <span class="k">return</span> <span class="mi">1</span></span>
<span class="code-line"> <span class="n">p</span> <span class="o">=</span> <span class="n">fast_pow</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">y</span> <span class="o">//</span> <span class="mi">2</span><span class="p">,</span> <span class="n">modulo</span><span class="p">)</span></span>
<span class="code-line"> <span class="n">p</span> <span class="o">=</span> <span class="p">(</span><span class="n">p</span> <span class="o">*</span> <span class="n">p</span><span class="p">)</span> <span class="o">%</span> <span class="n">modulo</span></span>
<span class="code-line"> <span class="k">if</span> <span class="n">y</span> <span class="o">%</span> <span class="mi">2</span><span class="p">:</span></span>
<span class="code-line"> <span class="n">p</span> <span class="o">=</span> <span class="p">(</span><span class="n">p</span> <span class="o">*</span> <span class="n">x</span><span class="p">)</span> <span class="o">%</span> <span class="n">modulo</span></span>
<span class="code-line"> <span class="k">return</span> <span class="n">p</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="c1"># нам нужно первое простое, которое больше 10^6 -- 10^6+3</span></span>
<span class="code-line"><span class="n">primes</span> <span class="o">=</span> <span class="nb">list</span><span class="p">(</span><span class="n">primerange</span><span class="p">(</span><span class="mi">5</span><span class="p">,</span><span class="mi">10</span><span class="o">**</span><span class="mi">6</span><span class="o">+</span><span class="mi">4</span><span class="p">))</span> </span>
<span class="code-line"></span>
<span class="code-line"><span class="n">sm</span> <span class="o">=</span> <span class="mi">0</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">primes</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span></span>
<span class="code-line"> <span class="n">digs</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="nb">str</span><span class="p">(</span><span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]))</span> <span class="c1"># количество цифр</span></span>
<span class="code-line"> <span class="n">r</span> <span class="o">=</span> <span class="p">(</span><span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span><span class="o">**</span><span class="mi">2</span> <span class="o">-</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">*</span> <span class="n">fast_pow</span><span class="p">(</span><span class="mi">10</span><span class="p">,</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span> <span class="o">-</span> <span class="mi">1</span> <span class="o">-</span> <span class="n">digs</span><span class="p">,</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]))</span> <span class="o">%</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span></span>
<span class="code-line"> <span class="n">sm</span> <span class="o">+=</span> <span class="n">r</span> <span class="o">*</span> <span class="mi">10</span><span class="o">**</span><span class="n">digs</span> <span class="o">+</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="k">print</span><span class="p">(</span><span class="s1">&#39;Result is {}&#39;</span><span class="o">.</span><span class="n">format</span><span class="p">(</span><span class="n">sm</span><span class="p">))</span></span>
</pre></div><p>Ответ: <strong>18613426663617118</strong></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи&nbsp;146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) { <![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href=".."><h2><i class="icon-coffee"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../">Главная</a></li><li><a href="../pages/projects.html">Мои проекты</a></li><li><a href="../pages/about.html">Об авторе</a></li><li><a href="../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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