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    "# [Square Root Digital Expansion](https://projecteuler.net/problem=80)\n",
    "\n",
    "Unfortunately, just calling Python's `sqrt` won't cut it - floats don't have nearly enough digits of precision.\n",
    "\n",
    "The easiest approach here is to use SageMath's arbitrary precision routines, or Python's [decimal](https://docs.python.org/3/library/decimal.html) package. We need to watch for rounding issues, so we'll calculate slightly more than 100 digits and truncate."
   ]
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       "40886"
      ]
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   "source": [
    "def digital_sum(n):\n",
    "    s = sqrt(n).n(digits=110)\n",
    "    digits = (int(d) for d in str(s)[:101] if d != '.')\n",
    "    return sum(digits)\n",
    "    \n",
    "\n",
    "sum(digital_sum(n) for n in range(1, 101) if not is_square(n))"
   ]
  },
  {
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   "id": "ad2b9aee",
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   "source": [
    "If you'd rather not use those tools for some reason, another simple approach is to take advantage of Python's arbitrary precision integers and calculate the [integer square root](https://en.wikipedia.org/wiki/Integer_square_root) of $n \\times 10^k$ for a sufficiently large $k$ ($k \\geq 200$ for this problem)."
   ]
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    {
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       "40886"
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     "execution_count": 2,
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   ],
   "source": [
    "def digital_sum(n):\n",
    "    s = isqrt(n * 10^200)\n",
    "    digits = (int(d) for d in str(s)[:100])\n",
    "    return sum(digits)\n",
    "    \n",
    "\n",
    "sum(digital_sum(n) for n in range(1, 101) if not is_square(n))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c193ba51",
   "metadata": {},
   "source": [
    "If you want to go the extra mile, there are several [algorithms for computing square roots](https://en.wikipedia.org/wiki/Square_root_algorithms) that you can implement yourself, such as Heron's method, which is a special case of [Newton's method](https://en.wikipedia.org/wiki/Newton%27s_method) for solving $x^2 - n = 0$. The method works by starting with an initial estimate $x_0$ (such as $\\frac{n}{2}$), then repeatedly calculating\n",
    "$$x_{k+1} = \\frac{1}{2}\\left(x_k + \\frac{n}{x_k}\\right)$$\n",
    "until $|x_{k+1} - x_k|$ is sufficiently small. For computing the integer square root, this can be when $|x_{k+1} - x_k| < 1$.\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
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