065-Solution.ipynb

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   "cell_type": "markdown",
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   "source": [
    "# The square root of 2 can be written as an infinite continued fraction.\n",
    "\n",
    "![](images/65-1.PNG)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)]."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![](images/65-2.PNG)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Hence the sequence of the first ten convergents for √2 are: 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ..."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# What is most surprising is that the important mathematical constant, e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...]."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# The first ten terms in the sequence of convergents for e are:\n",
    "\n",
    "# 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ..."
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "____"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {},
   "outputs": [],
   "source": [
    "from fractions import Fraction"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [],
   "source": [
    "list_100 = [2,1,2]+[item for sublist in [[1,1]+[2*k] for k in range(2,35)] for item in sublist][:97]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {},
   "outputs": [],
   "source": [
    "def test(n):\n",
    "    list_temp = list_100[:n]\n",
    "    total = Fraction(0,1)\n",
    "    while len(list_temp)>1:\n",
    "        last_val = list_temp[-1]\n",
    "        list_temp = list_temp[:-1]\n",
    "        total = Fraction(1,(last_val + total))\n",
    "    last_val = list_temp[-1]\n",
    "    return Fraction(last_val,1) + total"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 60,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1: 2\n",
      "2: 3\n",
      "3: 8/3\n",
      "4: 11/4\n",
      "5: 19/7\n",
      "6: 87/32\n",
      "7: 106/39\n",
      "8: 193/71\n",
      "9: 1264/465\n",
      "10: 1457/536\n"
     ]
    }
   ],
   "source": [
    "for i in range(1,11):\n",
    "    print('{}: {}'.format(i, test(i)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "- It looks like they're tying out"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Fraction(6963524437876961749120273824619538346438023188214475670667, 2561737478789858711161539537921323010415623148113041714756)"
      ]
     },
     "execution_count": 58,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "test(100)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "- The numerator of the 100th convergent is 6963524437876961749120273824619538346438023188214475670667"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 61,
   "metadata": {},
   "outputs": [],
   "source": [
    "i_100 = 6963524437876961749120273824619538346438023188214475670667"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "272"
      ]
     },
     "execution_count": 64,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sum([int(x) for x in list(str(i_100))])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Answer: 272"
   ]
  }
 ],
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