eulerbooks/notebooks/problem0063.ipynb

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    "# [Powerful Digit Counts](https://projecteuler.net/problem=63)\n",
    "\n",
    "This problem can be solved with a calculator.\n",
    "\n",
    "For $x^n$ to have $n$ digits, the following inequality must hold:\n",
    "$$10^{n-1} \\leq x^n < 10^n$$\n",
    "If we solve for $x$, we get\n",
    "$$10^\\frac{n-1}{n} \\leq x < 10$$\n",
    "In other words, if $10^\\frac{n-1}{n} \\leq x < 10$ for some $x$, then $x^n$ will have $n$ digits. Therefore, the number of $n$ digit numbers that are also $n$th powers is simply the number of integers between $10^\\frac{n-1}{n}$ and 10. This can be computed as\n",
    "$$\\lfloor 10 - 10^\\frac{n-1}{n} \\rfloor$$\n",
    "At $n=22$, the lower bound is greater than 9, so we only need to consider values between $n=1$ and $n=21$.\n",
    "\n",
    "Evaluating this expression for $n=1,2,3,\\ldots,21$ and summing, we get\n",
    "$$9 + 6 + 5 + 4 + 3 + 3 + 2 + 2 + 2 + 2 + 1 \\times 11 = 49$$\n",
    "\n",
    "## Relevant sequences\n",
    "* Numbers with $k$ digits that are also $k$th powers: [A132722](https://oeis.org/A132722)\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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