Блог 529 - Project Eulerhttps://likemath.ru/2016-10-30T17:40:00+03:00Project Euler и остальноеМоё решение задачи 1342016-10-30T17:40:00+03:002016-10-30T17:40:00+03:00Алексей Лобановtag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/<p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></p>
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</script>Моё решение задачи 1462016-10-21T17:40:00+03:002016-10-21T17:40:00+03:00Алексей Лобановtag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/<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 числами.</p>
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</script>Моё решение задачи 602015-11-22T23:41:00+03:002015-11-22T23:41:00+03:00Алексей Лобановtag:likemath.ru,2015-07-17:posts/moio-reshenie-zadachi-60/<p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после “склеивания” в любом порядке любых двух чисел из него тоже будет простое число.</p>