euler/ipython/EulerProblem032.ipynb

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   "source": [
    "# Euler Problem 32\n",
    "\n",
    "We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.\n",
    "\n",
    "The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.\n",
    "\n",
    "Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.\n",
    "\n",
    "HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum."
   ]
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   "source": [
    "We start by thinking about how many digits are used for multiplicand/multiplier/product. Obviously, the product must have more then three digits. Can it have more then 4 digits? Obviously it can because then there are only four digits for multiplicand and multiplier which is always smaller than ten thousand. So our product must be in the form $12 \\cdot 345 = 5789$. This makes the rest very easy in Python."
   ]
  },
  {
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   "execution_count": 1,
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "45228"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from itertools import permutations\n",
    "\n",
    "def is_solution(s):\n",
    "    a, b, c = int(s[0:2]), int(s[2:5]), int(s[5:])\n",
    "    if a * b == c:\n",
    "        return c\n",
    "    a, b, c = int(s[0:1]), int(s[1:5]), int(s[5:])  \n",
    "    if a * b == c:\n",
    "        return c\n",
    "    return 0\n",
    "\n",
    "assert(is_solution(\"391867254\") == 7254)\n",
    "assert(is_solution(\"391867245\") == 0)\n",
    "\n",
    "\n",
    "s = sum(set([is_solution(\"\".join(p)) for p in permutations(\"123456789\")]))\n",
    "assert(s == 45228)\n",
    "s"
   ]
  }
 ],
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  "completion_date": "Wed, 30 Aug 2017, 14:25",
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