177 lines
4.0 KiB
Plaintext
177 lines
4.0 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Problem 6:\n",
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"\n",
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"[Euler Project #6](https://projecteuler.net/problem=6)\n",
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"\n",
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"> *The sum of the squares of the first ten natural numbers is,\n",
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"1^2+2^2+...+10^2=385\n",
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"\n",
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">The square of the sum of the first ten natural numbers is,\n",
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"(1+2+...+10)^2=55^2=3025\n",
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"\n",
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">Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025−385=2640.\n",
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"\n",
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">Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.*\n",
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"\n",
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"---"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Reserved Space For Imports\n",
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"---"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"import os\n",
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"import pprint\n",
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"import time # Typically imported for sleep function, to slow down execution in terminal.\n",
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"import typing\n",
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"import decorators # Typically imported to compute execution duration of functions.\n",
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"import numpy"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Reserved Space For Method Definition\n",
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"---"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"def sum_of_squares(i: int,j: int):\n",
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" '''\n",
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" Function that computes the sum of a series of squared integers.\n",
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" '''\n",
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" series = [k**2 for k in range(i,j+1)]\n",
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" #pprint.pprint(series)\n",
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" summ = sum(series)\n",
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" return summ"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"def square_of_sum(m: int,n: int):\n",
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" '''\n",
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" Function that computes the square of a summation of a series of integers.\n",
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" '''\n",
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" series = [p for p in range(m,n+1)]\n",
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" summ = sum(series)**2\n",
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" #print(summ)\n",
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" return summ"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 1. Begin with testing the problem statement's example case.\n",
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"---"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"*Mathematically, the trick is to recognize that this a LCM (Least Common Multiple) problem. The solution is list all the prime factors of each number in the factor list, then compute their collective product.*\n",
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"\n",
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" - [ ] Create a method that performs the first computation.\n",
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" - [ ] Create a method that performs the second computation.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"25164150\n"
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]
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}
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],
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"source": [
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"# Receive problem inputs...\n",
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"smallest_factor = 1\n",
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"largest_factor = 100\n",
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"\n",
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"a = square_of_sum(smallest_factor,largest_factor) - sum_of_squares(smallest_factor,largest_factor)\n",
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"print(a)\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}
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