From 679d66e661e03f656409b5616508601155e113c2 Mon Sep 17 00:00:00 2001 From: filifa Date: Mon, 14 Apr 2025 20:32:50 -0400 Subject: [PATCH] add problem 26 --- problem0026.ipynb | 129 ++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 129 insertions(+) create mode 100644 problem0026.ipynb diff --git a/problem0026.ipynb b/problem0026.ipynb new file mode 100644 index 0000000..e9f4cee --- /dev/null +++ b/problem0026.ipynb @@ -0,0 +1,129 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "id": "b7b7784e", + "metadata": {}, + "source": [ + "# [Reciprocal Cycles](https://projecteuler.net/problem=26)\n", + "\n", + "Another easy problem with SageMath." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "id": "20a3987a", + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "983" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "max(range(2, 1000), key=lambda d: (1/d).period())" + ] + }, + { + "cell_type": "markdown", + "id": "6dc68724", + "metadata": {}, + "source": [ + "But let's talk about how we would write our own algorithm for calculating the period.\n", + "\n", + "If $d = 2^a 5^b n$ where $n$ is coprime to 2 and 5, then the [period](https://mathworld.wolfram.com/DecimalPeriod.html) of $\\frac{1}{d}$ is the [multiplicative order](https://en.wikipedia.org/wiki/Multiplicative_order) of 10 modulo $n$. This is the same as finding the smallest positive $k$ such that\n", + "$$10^k \\equiv 1 \\pmod{n}$$\n", + "(Wondering why this is called multiplicative order? It has to do with the mathematical concept of [groups](https://en.wikipedia.org/wiki/Group_(mathematics)), but you don't need to be familiar with them to apply the formula.)\n", + "\n", + "Based on the definition, we can easily write a function that computes multiplicative order that is efficient enough for this problem, but it's not very efficient in general. The computation is a special case of the [discrete logarithm](https://en.wikipedia.org/wiki/Discrete_logarithm), which has no known efficient algorithm in general." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "id": "d967be38", + "metadata": {}, + "outputs": [], + "source": [ + "def multiplicative_order(a, n):\n", + " if n == 1:\n", + " return 1\n", + " \n", + " g = 1\n", + " for k in range(1, n):\n", + " g *= a\n", + " g %= n\n", + " if g == 1:\n", + " return k" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "id": "dc51eb0e", + "metadata": {}, + "outputs": [], + "source": [ + "def period(d):\n", + " while d % 2 == 0:\n", + " d //= 2\n", + " while d % 5 == 0:\n", + " d //= 5\n", + " \n", + " return multiplicative_order(10, d)" + ] + }, + { + "cell_type": "markdown", + "id": "14f53b42", + "metadata": {}, + "source": [ + "Personally, I don't feel it's immediately obvious *why* multiplicative order can be used to calculate these periods. Here's a deeper dive into why this works, if you're curious.\n", + "\n", + "## Explanation\n", + "As above, let $d = 2^a 5^b n$, where $n$ is coprime to 2 and 5, and consider the unit fraction $u=\\frac{1}{d}$. The decimal representation of this fraction (as with any fraction) may have a [fractional part](https://en.wikipedia.org/wiki/Fractional_part) with $q$ leading digits, followed by a repetend of $r$ digits.\n", + "\n", + "Let's make note of a trick you probably already know: multiplying a number by 10 gives the same result as if you just moved the original number's decimal point to the right one digit (e.g. $1.25 \\times 10 = 12.5$). This means that $10^q u$ has a decimal representation with only the repetend in its fractional part. If you were to then multiply it by $10^r$, the integer part would increase, but the fractional part would stay the same since it repeats every $r$ digits.\n", + "\n", + "The above is important for understanding the following: $10^q 10^r u - 10^q u$ *is an integer*. How do we know? Because by the logic above, the fractional part of both $10^q 10^r u$ and $10^q u$ only have the repetend, so they cancel out when we subtract, leaving us with an integer. It immediately follows that $d(10^q 10^r u - 10^q u) = 10^q 10^r - 10^q$ is also an integer that has $d$ as a factor. We can express this as\n", + "$$10^{q+r} \\equiv 10^q \\pmod{d}$$\n", + "Through the properties of [modular arithmetic](https://en.wikipedia.org/wiki/Modular_arithmetic), we can cancel common factors of 2 and 5 in $10^{q+r}$, $10^q$, and $d$ until we end up with\n", + "$$10^r \\equiv 1 \\pmod{n}$$\n", + "By definition, $r$ is the multiplicative order of 10 modulo $n$.\n", + "\n", + "As a concrete example of the above, consider $d = 2^4 \\times 5 \\times 63 = 5040$. The decimal representation of $u = \\frac{1}{d}$ is $0.0001(984126)$, where 984126 is the repetend. Therefore $q=4$ and $r=6$. Sure enough, $10^q 10^r u - 10^q u = 1984125$ is an integer; therefore, $10^{10} \\equiv 10^4 \\pmod{5040}$, and $10^6 \\equiv 1 \\pmod{63}$.\n", + "\n", + "## Relevant sequences\n", + "* Periods of reciprocals: [A007732](https://oeis.org/A007732)" + ] + } + ], + "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 +}