From 45a14137f3c786299d2fa156d0d02b0eea4f491c Mon Sep 17 00:00:00 2001 From: filifa Date: Thu, 10 Apr 2025 23:53:32 -0400 Subject: [PATCH] add problem 25 --- problem0025.ipynb | 57 +++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 57 insertions(+) create mode 100644 problem0025.ipynb diff --git a/problem0025.ipynb b/problem0025.ipynb new file mode 100644 index 0000000..d9f2629 --- /dev/null +++ b/problem0025.ipynb @@ -0,0 +1,57 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "id": "b5da29ec", + "metadata": {}, + "source": [ + "# [1000-digit Fibonacci Number](https://projecteuler.net/problem=25)\n", + "\n", + "This problem can be solved with a scientific calculator!\n", + "\n", + "We're looking at [Fibonacci numbers](https://en.wikipedia.org/wiki/Fibonacci_sequence) again, after last seeing them in [problem 2](https://projecteuler.net/problem=2). Our goal is to find the first 1000 digit number in the sequence. Mathematically, we can state this is as the lowest $n$ such that\n", + "$$1+\\log{F_n} \\geq 1000$$\n", + "(Here, $\\log$ is the base 10 logarithm).\n", + "\n", + "To assist in finding $n$, we can employ Binet's formula:\n", + "$$F_n = \\frac{\\phi^n - (-\\phi)^{-n}}{\\sqrt{5}}$$\n", + "Here, $\\phi$ is the [golden ratio](https://en.wikipedia.org/wiki/Golden_ratio).\n", + "\n", + "Binet's formula is exact, but we can make our work a little easier by approximating.\n", + "$$F_n \\approx \\frac{\\phi^n}{\\sqrt{5}}$$\n", + "This approximation works since $(-\\phi)^{-n}$ approaches 0 as $n$ increases. This also means this approximation gets more accurate as $n$ increases, which is especially convenient since we're looking for a large $n$. Substituting this approximation into the above inequality, we have\n", + "$$1 + \\log{\\frac{\\phi^n}{\\sqrt{5}}} \\geq 1000$$\n", + "\n", + "Now with a little bit of algebra, we can solve for $n$.\n", + "$$n \\geq 999\\log_{\\phi}10 + \\log_{\\phi}\\sqrt{5}$$\n", + "If you're trying to solve this with a scientific calculator, you probably don't have a $\\log_{\\phi}$ button (*please let me know if you do*), but we can just use the [logarithmic change of base formula](https://en.wikipedia.org/wiki/List_of_logarithmic_identities). This ends up getting us\n", + "$$n \\geq 4781.85927\\ldots$$\n", + "Therefore, we want $n=4782$.\n", + "\n", + "## Relevant sequences\n", + "* Fibonacci numbers: [A000045](https://oeis.org/A000045)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "SageMath 9.5", + "language": "sage", + "name": "sagemath" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.11.2" + } + }, + "nbformat": 4, + "nbformat_minor": 5 +}