Finished problem 39 and added some new ones.

ipython/EulerProblem039.ipynb

@@ -13,21 +13,116 @@
     "For which value of p ≤ 1000, is the number of solutions maximised?"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a <= b <= c$ is true for every solution. Let's start with a function that tests the given examples."
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 13,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def is_right_angle_triangle(a, b, c, p):\n",
+    "    if a + b + c != p:\n",
+    "        return False\n",
+    "    if a**2 + b**2 != c**2:\n",
+    "        return False\n",
+    "    return True\n",
+    "\n",
+    "given = [(20, 48, 52, 120), (24, 45, 51, 120), (30, 40, 50, 120)]\n",
+    "for g in given:\n",
+    "    assert(is_right_angle_triangle(*g))\n",
+    "assert(is_right_angle_triangle(29, 41, 50, 120) == False)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "This seems bruteforceable. Let's try to find the solutions for p = 120."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[(30, 40, 50), (24, 45, 51), (20, 48, 52)]\n"
+     ]
+    }
+   ],
+   "source": [
+    "def brute_force(p):\n",
+    "    solutions = [(a,b, p - a - b)\n",
+    "        for b in range(1, p // 2 + 1)\n",
+    "        for a in range(1, b)\n",
+    "        if a*a + b*b == (p - a - b) * (p - a - b)\n",
+    "    ]\n",
+    "    return solutions\n",
+    "\n",
+    "print(brute_force(120))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Looks good. Let's try for all $p <= 1000$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
    "metadata": {
     "collapsed": true
    },
    "outputs": [],
-   "source": []
+   "source": [
+    "solutions = [(len(brute_force(p)), p) for p in range(1, 1001)]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Well, not too fast, but also not too slow. Let's get the max and we have our solution."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "840\n"
+     ]
+    }
+   ],
+   "source": [
+    "s, p = max(solutions)\n",
+    "print(p)\n",
+    "assert(p == 840)"
+   ]
   }
  ],
  "metadata": {
-  "completion_date": "",
+  "completion_date": "Sun, 20 May 2018, 21:16",
   "kernelspec": {
    "display_name": "Python 3",
-   "language": "python",
+   "language": "python3.6",
    "name": "python3"
   },
   "language_info": {
@@ -40,7 +135,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.5.5"
+   "version": "3.6.5"
   },
   "tags": [
    "triangle",
@@ -49,5 +144,5 @@
   ]
  },
  "nbformat": 4,
- "nbformat_minor": 0
+ "nbformat_minor": 1
 }

ipython/EulerProblem041.ipynb

@@ -0,0 +1,57 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Pandigital prime (Euler Problem 41)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.\n",
+    "\n",
+    "What is the largest n-digit pandigital prime that exists?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "completion_date": "",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python3.6",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.5"
+  },
+  "tags": [
+   "pandigital",
+   "prime"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

ipython/EulerProblem042.ipynb

@@ -0,0 +1,59 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Coded triangle numbers (Euler Problem 42)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "The nth term of the sequence of triangle numbers is given by, $t_n = \\frac{1}{2}n(n+1)$; so the first ten triangle numbers are:\n",
+    "\n",
+    "    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...\n",
+    "\n",
+    "By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = $t_{10}$. If the word value is a triangle number then we shall call the word a triangle word.\n",
+    "\n",
+    "Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words? (The file is saved in the same directory as this notebook file as EulerProblem042.txt)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "completion_date": "",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python3.6",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.5"
+  },
+  "tags": [
+   "triangle",
+   "words"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

ipython/EulerProblem043.ipynb

@@ -0,0 +1,73 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Sub-string divisibility (Euler Problem 43)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.\n",
+    "\n",
+    "Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:\n",
+    "\n",
+    "$d_2d_3d_4$=406 is divisible by 2\n",
+    "\n",
+    "$d_3d_4d_5$=063 is divisible by 3\n",
+    "\n",
+    "$d_4d_5d_6$=635 is divisible by 5\n",
+    "\n",
+    "$d_5d_6d_7=357$ is divisible by 7\n",
+    "\n",
+    "$d_6d_7d_8=572$ is divisible by 11\n",
+    "\n",
+    "$d_7d_8d_9=728$ is divisible by 13\n",
+    "\n",
+    "$d_8d_9d_{10} =289$ is divisible by 17\n",
+    "\n",
+    "Find the sum of all 0 to 9 pandigital numbers with this property."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "completion_date": "",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python3.6",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.5"
+  },
+  "tags": [
+   "pandigital",
+   "divisibility"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

ipython/html/EulerProblem039.html

@@ -11779,21 +11779,152 @@ div#notebook {
 <p>$\{20,48,52\}, \{24,45,51\}, \{30,40,50\}$</p>
 <p>For which value of p ≤ 1000, is the number of solutions maximised?</p>
 
+</div>
+</div>
+</div>
+<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
+</div>
+<div class="inner_cell">
+<div class="text_cell_render border-box-sizing rendered_html">
+<p>We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a &lt;= b &lt;= c$ is true for every solution. Let's start with a function that tests the given examples.</p>
+
 </div>
 </div>
 </div>
 <div class="cell border-box-sizing code_cell rendered">
 <div class="input">
-<div class="prompt input_prompt">In&nbsp;[&nbsp;]:</div>
+<div class="prompt input_prompt">In&nbsp;[13]:</div>
 <div class="inner_cell">
     <div class="input_area">
-<div class=" highlight hl-ipython3"><pre><span></span> 
+<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_right_angle_triangle</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="n">b</span><span class="p">,</span> <span class="n">c</span><span class="p">,</span> <span class="n">p</span><span class="p">):</span>
+    <span class="k">if</span> <span class="n">a</span> <span class="o">+</span> <span class="n">b</span> <span class="o">+</span> <span class="n">c</span> <span class="o">!=</span> <span class="n">p</span><span class="p">:</span>
+        <span class="k">return</span> <span class="kc">False</span>
+    <span class="k">if</span> <span class="n">a</span><span class="o">**</span><span class="mi">2</span> <span class="o">+</span> <span class="n">b</span><span class="o">**</span><span class="mi">2</span> <span class="o">!=</span> <span class="n">c</span><span class="o">**</span><span class="mi">2</span><span class="p">:</span>
+        <span class="k">return</span> <span class="kc">False</span>
+    <span class="k">return</span> <span class="kc">True</span>
+
+<span class="n">given</span> <span class="o">=</span> <span class="p">[(</span><span class="mi">20</span><span class="p">,</span> <span class="mi">48</span><span class="p">,</span> <span class="mi">52</span><span class="p">,</span> <span class="mi">120</span><span class="p">),</span> <span class="p">(</span><span class="mi">24</span><span class="p">,</span> <span class="mi">45</span><span class="p">,</span> <span class="mi">51</span><span class="p">,</span> <span class="mi">120</span><span class="p">),</span> <span class="p">(</span><span class="mi">30</span><span class="p">,</span> <span class="mi">40</span><span class="p">,</span> <span class="mi">50</span><span class="p">,</span> <span class="mi">120</span><span class="p">)]</span>
+<span class="k">for</span> <span class="n">g</span> <span class="ow">in</span> <span class="n">given</span><span class="p">:</span>
+    <span class="k">assert</span><span class="p">(</span><span class="n">is_right_angle_triangle</span><span class="p">(</span><span class="o">*</span><span class="n">g</span><span class="p">))</span>
+<span class="k">assert</span><span class="p">(</span><span class="n">is_right_angle_triangle</span><span class="p">(</span><span class="mi">29</span><span class="p">,</span> <span class="mi">41</span><span class="p">,</span> <span class="mi">50</span><span class="p">,</span> <span class="mi">120</span><span class="p">)</span> <span class="o">==</span> <span class="kc">False</span><span class="p">)</span>
 </pre></div>
 
 </div>
 </div>
 </div>
 
+</div>
+<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
+</div>
+<div class="inner_cell">
+<div class="text_cell_render border-box-sizing rendered_html">
+<p>This seems bruteforceable. Let's try to find the solutions for p = 120.</p>
+
+</div>
+</div>
+</div>
+<div class="cell border-box-sizing code_cell rendered">
+<div class="input">
+<div class="prompt input_prompt">In&nbsp;[16]:</div>
+<div class="inner_cell">
+    <div class="input_area">
+<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">brute_force</span><span class="p">(</span><span class="n">p</span><span class="p">):</span>
+    <span class="n">solutions</span> <span class="o">=</span> <span class="p">[(</span><span class="n">a</span><span class="p">,</span><span class="n">b</span><span class="p">,</span> <span class="n">p</span> <span class="o">-</span> <span class="n">a</span> <span class="o">-</span> <span class="n">b</span><span class="p">)</span>
+        <span class="k">for</span> <span class="n">b</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="n">p</span> <span class="o">//</span> <span class="mi">2</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
+        <span class="k">for</span> <span class="n">a</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="n">b</span><span class="p">)</span>
+        <span class="k">if</span> <span class="n">a</span><span class="o">*</span><span class="n">a</span> <span class="o">+</span> <span class="n">b</span><span class="o">*</span><span class="n">b</span> <span class="o">==</span> <span class="p">(</span><span class="n">p</span> <span class="o">-</span> <span class="n">a</span> <span class="o">-</span> <span class="n">b</span><span class="p">)</span> <span class="o">*</span> <span class="p">(</span><span class="n">p</span> <span class="o">-</span> <span class="n">a</span> <span class="o">-</span> <span class="n">b</span><span class="p">)</span>
+    <span class="p">]</span>
+    <span class="k">return</span> <span class="n">solutions</span>
+
+<span class="nb">print</span><span class="p">(</span><span class="n">brute_force</span><span class="p">(</span><span class="mi">120</span><span class="p">))</span>
+</pre></div>
+
+</div>
+</div>
+</div>
+
+<div class="output_wrapper">
+<div class="output">
+
+
+<div class="output_area">
+
+<div class="prompt"></div>
+
+
+<div class="output_subarea output_stream output_stdout output_text">
+<pre>[(30, 40, 50), (24, 45, 51), (20, 48, 52)]
+</pre>
+</div>
+</div>
+
+</div>
+</div>
+
+</div>
+<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
+</div>
+<div class="inner_cell">
+<div class="text_cell_render border-box-sizing rendered_html">
+<p>Looks good. Let's try for all $p &lt;= 1000$.</p>
+
+</div>
+</div>
+</div>
+<div class="cell border-box-sizing code_cell rendered">
+<div class="input">
+<div class="prompt input_prompt">In&nbsp;[18]:</div>
+<div class="inner_cell">
+    <div class="input_area">
+<div class=" highlight hl-ipython3"><pre><span></span><span class="n">solutions</span> <span class="o">=</span> <span class="p">[(</span><span class="nb">len</span><span class="p">(</span><span class="n">brute_force</span><span class="p">(</span><span class="n">p</span><span class="p">)),</span> <span class="n">p</span><span class="p">)</span> <span class="k">for</span> <span class="n">p</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">1001</span><span class="p">)]</span>
+</pre></div>
+
+</div>
+</div>
+</div>
+
+</div>
+<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
+</div>
+<div class="inner_cell">
+<div class="text_cell_render border-box-sizing rendered_html">
+<p>Well, not too fast, but also not too slow. Let's get the max and we have our solution.</p>
+
+</div>
+</div>
+</div>
+<div class="cell border-box-sizing code_cell rendered">
+<div class="input">
+<div class="prompt input_prompt">In&nbsp;[24]:</div>
+<div class="inner_cell">
+    <div class="input_area">
+<div class=" highlight hl-ipython3"><pre><span></span><span class="n">s</span><span class="p">,</span> <span class="n">p</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">solutions</span><span class="p">)</span>
+<span class="nb">print</span><span class="p">(</span><span class="n">p</span><span class="p">)</span>
+<span class="k">assert</span><span class="p">(</span><span class="n">p</span> <span class="o">==</span> <span class="mi">840</span><span class="p">)</span>
+</pre></div>
+
+</div>
+</div>
+</div>
+
+<div class="output_wrapper">
+<div class="output">
+
+
+<div class="output_area">
+
+<div class="prompt"></div>
+
+
+<div class="output_subarea output_stream output_stdout output_text">
+<pre>840
+</pre>
+</div>
+</div>
+
+</div>
+</div>
+
 </div>
     </div>
   </div>

ipython/html/index.html

@@ -576,7 +576,7 @@
                 
                 <tr>
                   <td>Problem 039</td>
-                  <td></td>
+                  <td>Sun, 20 May 2018, 21:16</td>
                   <td><a href="EulerProblem039.html">Problem 039</a></td>
                   <td>
                   
@@ -604,6 +604,45 @@
                   </td>
                 </tr>
                 
+                <tr>
+                  <td>Problem 041</td>
+                  <td></td>
+                  <td><a href="EulerProblem041.html">Problem 041</a></td>
+                  <td>
+                  
+                    <kbd>pandigital</kbd>
+                  
+                    <kbd>prime</kbd>
+                  
+                  </td>
+                </tr>
+                
+                <tr>
+                  <td>Problem 042</td>
+                  <td></td>
+                  <td><a href="EulerProblem042.html">Problem 042</a></td>
+                  <td>
+                  
+                    <kbd>triangle</kbd>
+                  
+                    <kbd>words</kbd>
+                  
+                  </td>
+                </tr>
+                
+                <tr>
+                  <td>Problem 043</td>
+                  <td></td>
+                  <td><a href="EulerProblem043.html">Problem 043</a></td>
+                  <td>
+                  
+                    <kbd>pandigital</kbd>
+                  
+                    <kbd>divisibility</kbd>
+                  
+                  </td>
+                </tr>
+                
                 <tr>
                   <td>Problem 067</td>
                   <td>Fri, 5 Sep 2014, 07:36</td>

ipython/publish.py

@@ -1,3 +1,5 @@
+#!/usr/bin/env python
+
 import jinja2
 import os
 import subprocess