Euler has design of a homepage and there is a link back to the overview for each solution.

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-<h1 id="Euler-Problem">Euler Problem<a class="anchor-link" href="#Euler-Problem">&#182;</a></h1><p>2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.</p>
+<h1 id="Euler-Problem">Euler Problem<a class="anchor-link" href="#Euler-Problem">¶</a></h1><p><a href="/euler">Back to overview.</a></p><p>2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.</p>
 <p>What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?</p>
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 <p>My easiest guess is to multiply all prime numbers till the number.</p>
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 <div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">get_primes_smaller</span><span class="p">(</span><span class="n">number</span><span class="p">):</span>
    <span class="n">primes</span> <span class="o">=</span> <span class="p">[]</span>
    <span class="n">prospects</span> <span class="o">=</span> <span class="p">[</span><span class="n">n</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">number</span><span class="p">)]</span>
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        <span class="n">primes</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">p</span><span class="p">)</span>
    <span class="k">return</span> <span class="n">primes</span>
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 <div class=" highlight hl-ipython3"><pre><span></span><span class="kn">from</span> <span class="nn">operator</span> <span class="k">import</span> <span class="n">mul</span>
 <span class="kn">from</span> <span class="nn">functools</span> <span class="k">import</span> <span class="n">reduce</span>

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 <span class="nb">print</span><span class="p">(</span><span class="n">get_number_which_is_divisible_by_all_numbers_from_one_to</span><span class="p">(</span><span class="mi">10</span><span class="p">))</span>
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 <pre>210
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 <p>That obviously didn't work. The reason is that the same prime can occur multiple times in the factorization of a divisor. For example $2^{3} = 8$. We can always brute force of course. We do a smart brute force and only check multiples from the product of primes because this factor must be part of the solution.</p>
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 <div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_divisible_by_numbers_smaller_or_equal</span><span class="p">(</span><span class="n">number</span><span class="p">,</span> <span class="n">maximum_number</span><span class="p">):</span>
    <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">maximum_number</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
        <span class="k">if</span> <span class="n">number</span> <span class="o">%</span> <span class="n">n</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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 <span class="k">assert</span><span class="p">(</span><span class="n">get_number_which_is_divisible_by_all_numbers_from_one_to</span><span class="p">(</span><span class="mi">10</span><span class="p">)</span> <span class="o">==</span> <span class="mi">2520</span><span class="p">)</span>
 <span class="nb">print</span><span class="p">(</span><span class="n">get_number_which_is_divisible_by_all_numbers_from_one_to</span><span class="p">(</span><span class="mi">20</span><span class="p">))</span>
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 <pre>232792560
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