Wrote a fast solution for 3 in ipython.

ipython/EulerProblem003.ipynb

@@ -21,9 +21,7 @@
   {
    "cell_type": "code",
    "execution_count": 1,
-   "metadata": {
+   "metadata": {},
-    "collapsed": false
-   },
    "outputs": [],
    "source": [
     "def get_primes_smaller(number):\n",
@@ -49,9 +47,7 @@
   {
    "cell_type": "code",
    "execution_count": 2,
-   "metadata": {
+   "metadata": {},
-    "collapsed": false
-   },
    "outputs": [],
    "source": [
     "import math\n",
@@ -85,10 +81,8 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 5,
+   "execution_count": 3,
-   "metadata": {
+   "metadata": {},
-    "collapsed": false
-   },
    "outputs": [
     {
      "name": "stdout",
@@ -103,7 +97,8 @@
     "    return get_prime_factors(number)[-1]\n",
     "\n",
     "assert(get_largest_prime(13195) == 29)\n",
-    "print(get_largest_prime(600851475143))"
+    "#print(get_largest_prime(600851475143))\n",
+    "print(6857) # computed the previously but remove it so that we can reexecute the complete kernel"
    ]
   },
   {
@@ -117,12 +112,59 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 4,
-   "metadata": {
+   "metadata": {},
-    "collapsed": true
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6857\n"
+     ]
+    }
+   ],
+   "source": [
+    "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_prime_factors(number):\n",
+    "    prime_generator = prime_generator_function()\n",
+    "    remainder = number\n",
+    "    factors = []\n",
+    "    for p in prime_generator:\n",
+    "        while remainder % p == 0:\n",
+    "            remainder /= p\n",
+    "            factors.append(p)\n",
+    "        if remainder == 1 or p * p > number:\n",
+    "            break\n",
+    "    if remainder != 1:\n",
+    "        factors.append(remainder)\n",
+    "    return factors\n",
+    "\n",
+    "print(get_largest_prime(600851475143))\n"
+   ]
   },
-   "outputs": [],
+  {
-   "source": []
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Here we go. Obviously much better than precalculation primes that we never need."
+   ]
   }
  ],
  "metadata": {
@@ -141,9 +183,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.5.4"
+   "version": "3.6.3"
   }
  },
  "nbformat": 4,
- "nbformat_minor": 0
+ "nbformat_minor": 1
 }