Solved 16, 17, and 67 in ipython.

ipython/EulerProblem016.ipynb

@@ -0,0 +1,142 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 16\n",
+    "\n",
+    "$2^{15}$ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.\n",
+    "\n",
+    "What is the sum of the digits of the number $2^{1000}$?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Seriously? Okay, probably more difficult in C."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1366\n"
+     ]
+    }
+   ],
+   "source": [
+    "s = sum(map(int, str(2**1000)))\n",
+    "assert(s == 1366)\n",
+    "print(s)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Okay, let's do it without big ints."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def double_number_string(number_string):\n",
+    "    number_string = number_string[::-1] # reverse\n",
+    "    result = []\n",
+    "    carriage = 0\n",
+    "    for c in number_string:\n",
+    "        s = carriage + int(c) + int(c)\n",
+    "        carriage = s // 10\n",
+    "        next_digit = s % 10\n",
+    "        result.append(str(next_digit))\n",
+    "    if carriage > 0:\n",
+    "        result.append(str(carriage))\n",
+    "    result = \"\".join(result)[::-1]\n",
+    "    return result\n",
+    "\n",
+    "assert(double_number_string(\"132\") == \"264\")\n",
+    "assert(double_number_string(\"965\") == \"1930\")"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1366\n"
+     ]
+    }
+   ],
+   "source": [
+    "def powers_of_two(n):\n",
+    "    \"\"\" Retuns nth power of two as a string. \"\"\"\n",
+    "    assert(n >= 0)\n",
+    "    if n == 0:\n",
+    "        return \"1\"\n",
+    "    s = \"2\"\n",
+    "    for _ in range(n - 1):\n",
+    "        s = double_number_string(s)\n",
+    "    return s\n",
+    "\n",
+    "assert(powers_of_two(3) == \"8\")\n",
+    "assert(powers_of_two(10) == \"1024\")\n",
+    "\n",
+    "number_string = powers_of_two(1000)\n",
+    "print(sum(map(int, number_string)))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Here we go. The conversion to integer and the sum would be easy in C."
+   ]
+  }
+ ],
+ "metadata": {
+  "completion_date": "Sun, 31 Aug 2014, 20:49",
+  "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": [
+   "c",
+   "manual"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}