95 lines
2.1 KiB
Plaintext
95 lines
2.1 KiB
Plaintext
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{
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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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"# Euler 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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"\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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"\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."
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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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"Okay, this is as straightforward as it can get."
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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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"collapsed": false
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},
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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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"s = sum([x for x in range(1, 101)])**2 - sum([x**2 for x in range(1, 101)])\n",
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"assert(s == 25164150)\n",
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"print(s)"
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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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"General solution."
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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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"collapsed": false
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},
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"outputs": [],
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"source": [
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"def diff_between_sum_of_squares_and_square_sum(n):\n",
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" return sum([x for x in range(1, n + 1)])**2 - sum([x**2 for x in range(1, n + 1)])\n",
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"\n",
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"assert(diff_between_sum_of_squares_and_square_sum(100) == 25164150)\n",
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"assert(diff_between_sum_of_squares_and_square_sum(10) == 2640)"
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]
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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": "3.5.4"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 0
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}
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