Implemented solutions 5-7 in ipython.

ipython/EulerProblem005.ipynb

@@ -0,0 +1,130 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem\n",
+    "\n",
+    "2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.\n",
+    "\n",
+    "What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "My easiest guess is to multiply all prime numbers till the number."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "def get_primes_smaller(number):\n",
+    "    primes = []\n",
+    "    prospects = [n for n in range(2, number)]\n",
+    "    while prospects:\n",
+    "        p = prospects[0]\n",
+    "        prospects = [x for x in prospects if x % p != 0]\n",
+    "        primes.append(p)\n",
+    "    return primes"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "210\n"
+     ]
+    }
+   ],
+   "source": [
+    "from operator import mul\n",
+    "from functools import reduce\n",
+    "\n",
+    "def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n",
+    "    ps= get_primes_smaller(n + 1)\n",
+    "    return reduce(mul, ps, 1)\n",
+    "\n",
+    "print(get_number_which_is_divisible_by_all_numbers_from_one_to(10))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "That obviously didn't work. The reason is that the same prime can occur multiple times in the factorization of a divisor. For example $2^{3} = 8$. We can always brute force of course. We do a smart brute force and only check multiples from the product of primes because this factor must be part of the solution."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "232792560\n"
+     ]
+    }
+   ],
+   "source": [
+    "def is_divisible_by_numbers_smaller_or_equal(number, maximum_number):\n",
+    "    for n in range(2, maximum_number + 1):\n",
+    "        if number % n != 0:\n",
+    "            return False\n",
+    "    return True\n",
+    "\n",
+    "def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n",
+    "    ps = get_primes_smaller(n + 1)\n",
+    "    factor = reduce(mul, ps, 1)\n",
+    "    multiples_of_factor = factor\n",
+    "    while True:\n",
+    "        if is_divisible_by_numbers_smaller_or_equal(multiples_of_factor, n):\n",
+    "            return multiples_of_factor\n",
+    "        multiples_of_factor += factor\n",
+    "\n",
+    "assert(get_number_which_is_divisible_by_all_numbers_from_one_to(10) == 2520)\n",
+    "print(get_number_which_is_divisible_by_all_numbers_from_one_to(20))"
+   ]
+  }
+ ],
+ "metadata": {
+  "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"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}

ipython/EulerProblem006.ipynb

@@ -0,0 +1,94 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 6\n",
+    "\n",
+    "The sum of the squares of the first ten natural numbers is,\n",
+    "\n",
+    "$1^2 + 2^2 + ... + 10^2 = 385$\n",
+    "\n",
+    "The square of the sum of the first ten natural numbers is,\n",
+    "\n",
+    "$(1 + 2 + ... + 10)^2 = 55^2 = 3025$\n",
+    "\n",
+    "Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 − 385 = 2640$.\n",
+    "\n",
+    "Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Okay, this is as straightforward as it can get."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "25164150\n"
+     ]
+    }
+   ],
+   "source": [
+    "s = sum([x for x in range(1, 101)])**2  - sum([x**2 for x in range(1, 101)])\n",
+    "assert(s == 25164150)\n",
+    "print(s)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "General solution."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def diff_between_sum_of_squares_and_square_sum(n):\n",
+    "    return sum([x for x in range(1, n + 1)])**2  - sum([x**2 for x in range(1, n + 1)])\n",
+    "\n",
+    "assert(diff_between_sum_of_squares_and_square_sum(100) == 25164150)\n",
+    "assert(diff_between_sum_of_squares_and_square_sum(10) == 2640)"
+   ]
+  }
+ ],
+ "metadata": {
+  "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"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}

ipython/EulerProblem007.ipynb

@@ -0,0 +1,87 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 7\n",
+    "\n",
+    "By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.\n",
+    "\n",
+    "What is the 10 001st prime number?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We reuse our prime functions and use a trick from [stackoverflow](https://stackoverflow.com/questions/2300756/get-the-nth-item-of-a-generator-in-python) to get the nth element."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "104743\n"
+     ]
+    }
+   ],
+   "source": [
+    "import itertools\n",
+    "\n",
+    "def is_prime(n, smaller_primes):\n",
+    "    for s in smaller_primes:\n",
+    "        if n % s == 0:\n",
+    "            return False\n",
+    "        if s * s > n:\n",
+    "            return True\n",
+    "    return True\n",
+    "\n",
+    "def prime_generator_function():\n",
+    "    primes = [2, 3, 5, 7]\n",
+    "    for p in primes:\n",
+    "        yield p\n",
+    "    while True:\n",
+    "        p += 2\n",
+    "        if is_prime(p, primes):\n",
+    "            primes.append(p)\n",
+    "            yield p\n",
+    "\n",
+    "def get_nth_prime(n):\n",
+    "    ps = prime_generator_function()\n",
+    "    return next(itertools.islice(ps, n - 1, n))\n",
+    "\n",
+    "assert(get_nth_prime(6) == 13)\n",
+    "print(get_nth_prime(10001))"
+   ]
+  }
+ ],
+ "metadata": {
+  "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"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}