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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
parent 894160d1a0
commit c34ebd6181
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<title>EulerProblem023</title><script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script>
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<h1 id="Euler-Problem-23">Euler Problem 23<a class="anchor-link" href="#Euler-Problem-23">&#182;</a></h1><p>A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.</p>
<h1 id="Euler-Problem-23">Euler Problem 23<a class="anchor-link" href="#Euler-Problem-23"></a></h1><p><a href="/euler">Back to overview.</a></p><p>A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.</p>
<p>A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.</p>
<p>As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.</p>
<p>Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.</p>
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<p>We reuse the sum of proper divisors function and use it to tell whether a number is abundant or not.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">sum_of_proper_divisors</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="kn">from</span> <span class="nn">math</span> <span class="k">import</span> <span class="n">sqrt</span>
<span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
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<span class="n">abundant_numbers_smaller_30000</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">1</span><span class="p">,</span> <span class="mi">30001</span><span class="p">)</span> <span class="k">if</span> <span class="n">is_abundant</span><span class="p">(</span><span class="n">n</span><span class="p">)]</span>
<span class="n">abundant_numbers_smaller_30000_set</span> <span class="o">=</span> <span class="nb">set</span><span class="p">(</span><span class="n">abundant_numbers_smaller_30000</span><span class="p">)</span>
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<p>Now that we have all abundant numbers in sorted order we can write a function which tests whether a number can be written as a sum of two abundant numbers.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_sum_of_two_abundants</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="k">for</span> <span class="n">a</span> <span class="ow">in</span> <span class="n">abundant_numbers_smaller_30000</span><span class="p">:</span>
<span class="k">if</span> <span class="n">a</span> <span class="o">&gt;</span> <span class="p">(</span><span class="n">n</span> <span class="o">//</span> <span class="mi">2</span><span class="p">):</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="mi">23</span><span class="p">)</span> <span class="o">==</span> <span class="kc">False</span><span class="p">)</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="mi">28123</span><span class="p">)</span> <span class="o">==</span> <span class="kc">True</span><span class="p">)</span>
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<p>Now we try 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">sum</span><span class="p">([</span><span class="n">i</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">30000</span><span class="p">)</span> <span class="k">if</span> <span class="ow">not</span> <span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="n">i</span><span class="p">)])</span>
<span class="nb">print</span><span class="p">(</span><span class="n">s</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">4179871</span><span class="p">)</span>
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<pre>4179871
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