Finished 33 and 34.

ipython/EulerProblem033.ipynb

@@ -0,0 +1,172 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 33\n",
+    "\n",
+    "The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.\n",
+    "\n",
+    "We shall consider fractions like, 30/50 = 3/5, to be trivial examples.\n",
+    "\n",
+    "There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.\n",
+    "\n",
+    "If the product of these four fractions is given in its lowest common terms, find the value of the denominator."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We start write a function which checks if a number is curios and then brute force."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def is_curious(n, d):\n",
+    "    assert(len(str(n)) == 2 and len(str(d)) == 2)\n",
+    "    if n == d:\n",
+    "        return False\n",
+    "    for i in range(1, 10):\n",
+    "        if str(i) in str(n) and str(i) in str(d):\n",
+    "            try:\n",
+    "                n_ = int(str(n).replace(str(i), \"\"))\n",
+    "                d_ = int(str(d).replace(str(i), \"\"))\n",
+    "            except ValueError:\n",
+    "                return False\n",
+    "            try:\n",
+    "                if n_ / d_ == n / d:\n",
+    "                    return True\n",
+    "            except ZeroDivisionError:\n",
+    "                return False\n",
+    "    return False\n",
+    "\n",
+    "assert(is_curious(49, 98) == True)\n",
+    "assert(is_curious(30, 50) == False)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[(16, 64), (19, 95), (26, 65), (49, 98)]"
+      ]
+     },
+     "execution_count": 25,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "fs = [(n, d) for n in range(10, 100) for d in range(n, 100) if is_curious(n, d)]\n",
+    "fs"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "387296/38729600\n"
+     ]
+    }
+   ],
+   "source": [
+    "n = 1\n",
+    "d = 1\n",
+    "for n_, d_ in fs:\n",
+    "    n *= n_\n",
+    "    d *= d_\n",
+    "\n",
+    "print(\"{}/{}\".format(n, d))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Now we can see that the solution is $100$. But actually it would be nice to calculate the GCD."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "100"
+      ]
+     },
+     "execution_count": 40,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def gcd_euclid(a, b):\n",
+    "    if a == b:\n",
+    "        return a\n",
+    "    elif a > b:\n",
+    "        return gcd_euclid(a - b, b)\n",
+    "    elif a < b:\n",
+    "        return gcd_euclid(a, b - a)\n",
+    "        \n",
+    "gcd_nd = gcd_euclid(n, d)\n",
+    "\n",
+    "s = d // gcd_nd\n",
+    "assert(s == 100)\n",
+    "s"
+   ]
+  }
+ ],
+ "metadata": {
+  "completion_date": "Mon, 12 Feb 2018, 17:29",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "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": "3.5.4"
+  },
+  "tags": [
+   "gcd",
+   "curious",
+   "faction"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}

ipython/EulerProblem034.ipynb

@@ -0,0 +1,101 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 34\n",
+    "\n",
+    "145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.\n",
+    "\n",
+    "Find the sum of all numbers which are equal to the sum of the factorial of their digits.\n",
+    "\n",
+    "Note: as 1! = 1 and 2! = 2 are not sums they are not included."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "The algorithm for checking if a number is curious should be efficient. The more difficult thing is to select the upper bound for the brute force. It can be seen that $9 999 999 < 9! * 7$. Hence we can select $10^7$ as our bound."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "from math import factorial\n",
+    "\n",
+    "def is_curious(n):\n",
+    "    s = sum([factorial(int(d)) for d in str(n)])\n",
+    "    return n == s\n",
+    "\n",
+    "assert(is_curious(145) == True)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "s = sum([n for n in range(3, 10**7) if is_curious(n)])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "40730"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "assert(s == 40730)\n",
+    "s"
+   ]
+  }
+ ],
+ "metadata": {
+  "completion_date": "Mon, 12 Feb 2018, 17:57",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "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": "3.5.4"
+  },
+  "tags": [
+   "factorial",
+   "brute force"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}

ipython/index.html

@@ -492,6 +492,34 @@
                   </td>
                 </tr>
                 
+                <tr>
+                  <td>Problem 033</td>
+                  <td>Mon, 12 Feb 2018, 17:29</td>
+                  <td><a href="EulerProblem033.html">Problem 033</a></td>
+                  <td>
+                  
+                    <kbd>gcd</kbd>
+                  
+                    <kbd>curious</kbd>
+                  
+                    <kbd>faction</kbd>
+                  
+                  </td>
+                </tr>
+                
+                <tr>
+                  <td>Problem 034</td>
+                  <td>Mon, 12 Feb 2018, 17:57</td>
+                  <td><a href="EulerProblem034.html">Problem 034</a></td>
+                  <td>
+                  
+                    <kbd>factorial</kbd>
+                  
+                    <kbd>brute force</kbd>
+                  
+                  </td>
+                </tr>
+                
                 <tr>
                   <td>Problem 067</td>
                   <td>Fri, 5 Sep 2014, 07:36</td>