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

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Felix Martin
2018-06-14 20:11:26 -04:00
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<h1 id="Euler-Problem-26">Euler Problem 26<a class="anchor-link" href="#Euler-Problem-26">&#182;</a></h1><p>A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:</p>
<h1 id="Euler-Problem-26">Euler Problem 26<a class="anchor-link" href="#Euler-Problem-26"></a></h1><p><a href="/euler">Back to overview.</a></p><p>A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:</p>
<pre><code>1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
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<p>Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.</p>
<p>Find the value of d &lt; 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.</p>
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<p>The trick here is to identify a cycle. The easiest way I see to do this is to memorize the remainders. If we have a remainder that we occurred previously there is a cycle.</p>
<p>Let's consider 1/3. The initial remainder is 10. For ten divided by three the new remainder is again ten (or one times ten). So we have a one-cycle of 3.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">get_cycle_count</span><span class="p">(</span><span class="n">nominator</span><span class="p">,</span> <span class="n">denominator</span><span class="p">):</span>
<span class="kn">from</span> <span class="nn">itertools</span> <span class="k">import</span> <span class="n">count</span>
<span class="k">assert</span><span class="p">(</span><span class="n">nominator</span> <span class="o">==</span> <span class="mi">1</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">get_cycle_count</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">10</span><span class="p">)</span> <span class="o">==</span> <span class="mi">0</span><span class="p">)</span>
<span class="k">assert</span><span class="p">(</span><span class="n">get_cycle_count</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">6</span><span class="p">)</span> <span class="o">==</span> <span class="mi">1</span><span class="p">)</span>
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<p>This is a simple divison algorithm. The only special thing is the remainder and that we remember when it occurs the first time. If a remainder occurrs for the second time we substract the position and thus have the lenght of the cycle. With this solution we should be efficient enough to brute force.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">s</span> <span class="o">=</span> <span class="nb">max</span><span class="p">([(</span><span class="n">get_cycle_count</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">i</span><span class="p">),</span> <span class="n">i</span><span class="p">)</span><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="mi">1</span><span class="p">,</span> <span class="mi">1000</span><span class="p">)])</span>
<span class="n">s</span> <span class="o">=</span> <span class="n">s</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span>
<span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">983</span><span class="p">)</span>
<span class="nb">print</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
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<pre>983
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