Added 24 and 25 to ipython.

ipython/EulerProblem023.ipynb

@@ -15,14 +15,100 @@
     "Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We reuse the sum of proper divisors function and use it to tell whether a number is abundant or not."
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 20,
    "metadata": {
-    "collapsed": true
+    "collapsed": false
    },
    "outputs": [],
-   "source": []
+   "source": [
+    "def sum_of_proper_divisors(n):\n",
+    "    from math import sqrt\n",
+    "    if n == 1:\n",
+    "        return 0\n",
+    "    s = 1\n",
+    "    for i in range(2, int(sqrt(n)) + 1):\n",
+    "        if n % i == 0:\n",
+    "            s += i\n",
+    "            x = (n // i)\n",
+    "            if x != i:\n",
+    "                s += x\n",
+    "    return s\n",
+    "\n",
+    "def is_abundant(n):\n",
+    "    return sum_of_proper_divisors(n) > n\n",
+    "\n",
+    "assert(is_abundant(7) == False)\n",
+    "assert(is_abundant(12) == True)\n",
+    "\n",
+    "abundant_numbers_smaller_30000 = [n for n in range(1, 30001) if is_abundant(n)]\n",
+    "abundant_numbers_smaller_30000_set = set(abundant_numbers_smaller_30000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Now that we have all abundant numbers in sorted order we can write a function which tests whether a number can be written as a sum of two abundant numbers."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def is_sum_of_two_abundants(n):\n",
+    "    for a in abundant_numbers_smaller_30000:\n",
+    "        if a > (n // 2):\n",
+    "            return False\n",
+    "        r = n - a\n",
+    "        if r in abundant_numbers_smaller_30000_set:\n",
+    "            return True\n",
+    "    return False\n",
+    "\n",
+    "assert(is_sum_of_two_abundants(24) == True)\n",
+    "assert(is_sum_of_two_abundants(23) == False)\n",
+    "assert(is_sum_of_two_abundants(28123) == True)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Now we try to brute force."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 34,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4179871\n"
+     ]
+    }
+   ],
+   "source": [
+    "s = sum([i for i in range(1, 30000) if not is_sum_of_two_abundants(i)])\n",
+    "print(s)\n",
+    "assert(s == 4179871)"
+   ]
   }
  ],
  "metadata": {
@@ -46,7 +132,8 @@
   },
   "tags": [
    "perfect number",
-   "abundant"
+   "abundant",
+   "factors"
   ]
  },
  "nbformat": 4,