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    "# 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,
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   "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,
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   "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))"
   ]
  }
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