Finished 26 and added 27 to ipython.

ipython/EulerProblem026.ipynb

@@ -0,0 +1,126 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 26\n",
+    "\n",
+    "A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:\n",
+    "\n",
+    "~~~\n",
+    "1/2\t = \t0.5\n",
+    "1/3\t = \t0.(3)\n",
+    "1/4\t = \t0.25\n",
+    "1/5\t = \t0.2\n",
+    "1/6\t = \t0.1(6)\n",
+    "1/7\t = \t0.(142857)\n",
+    "1/8\t = \t0.125\n",
+    "1/9\t = \t0.(1)\n",
+    "1/10    = \t0.1\n",
+    "~~~\n",
+    "\n",
+    "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.\n",
+    "\n",
+    "Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "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.\n",
+    "\n",
+    "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."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def get_cycle_count(nominator, denominator):\n",
+    "    from itertools import count\n",
+    "    assert(nominator == 1)\n",
+    "    remainders = {}\n",
+    "    remainder = nominator\n",
+    "    results = []\n",
+    "    for i in count():\n",
+    "        result = remainder // denominator\n",
+    "        remainder = remainder % denominator\n",
+    "        results.append(result)\n",
+    "        if remainder in remainders:\n",
+    "            return i - remainders[remainder]\n",
+    "        else:\n",
+    "            remainders[remainder] = i\n",
+    "        if remainder == 0:\n",
+    "            return 0\n",
+    "        remainder *= 10\n",
+    "\n",
+    "assert(get_cycle_count(1, 7) == 6)\n",
+    "assert(get_cycle_count(1, 10) == 0)\n",
+    "assert(get_cycle_count(1, 6) == 1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "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."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "983\n"
+     ]
+    }
+   ],
+   "source": [
+    "s = max([(get_cycle_count(1, i), i)for i in range(1, 1000)])\n",
+    "s = s[1]\n",
+    "assert(s == 983)\n",
+    "print(s)"
+   ]
+  }
+ ],
+ "metadata": {
+  "completion_date": "Mon, 9 Nov 2015, 22:11",
+  "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": [
+   "reciprocal",
+   "division",
+   "nominator",
+   "denominator"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}

ipython/EulerProblem027.ipynb

@@ -0,0 +1,61 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 27\n",
+    "\n",
+    "Euler discovered the remarkable quadratic formula:\n",
+    "\n",
+    "$n^2 + n + 41$\n",
+    "\n",
+    "It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when $n=40$, $40^2 + 40 + 41 = 40(40+1)+41$ is divisible by $41$, and certainly when $n=41,41^2+41+41$ is clearly divisible by 41.\n",
+    "\n",
+    "The incredible formula $n^2−79n+1601$ was discovered, which produces 80 primes for the consecutive values 0≤n≤79. The product of the coefficients, $−79$ and $1601$, is $−126479$.\n",
+    "\n",
+    "Considering quadratics of the form:\n",
+    "\n",
+    "$n^2 + an +b$, where |a|<1000 and |b|≤1000\n",
+    "\n",
+    "where |n| is the modulus/absolute value of n e.g. |11|=11 and |−4|=4.\n",
+    "\n",
+    "Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "completion_date": "Mon, 21 Aug 2017, 21:11",
+  "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": [
+   "quadratic primes"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}