102 lines
2.1 KiB
Plaintext
102 lines
2.1 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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"# Euler Problem 34\n",
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"\n",
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"145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.\n",
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"\n",
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"Find the sum of all numbers which are equal to the sum of the factorial of their digits.\n",
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"\n",
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"Note: as 1! = 1 and 2! = 2 are not sums they are not included."
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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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"The algorithm for checking if a number is curious should be efficient. The more difficult thing is to select the upper bound for the brute force. It can be seen that $9 999 999 < 9! * 7$. Hence we can select $10^7$ as our bound."
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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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"source": [
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"from math import factorial\n",
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"\n",
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"def is_curious(n):\n",
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" s = sum([factorial(int(d)) for d in str(n)])\n",
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" return n == s\n",
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"\n",
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"assert(is_curious(145) == True)"
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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": true
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},
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"outputs": [],
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"source": [
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"s = sum([n for n in range(3, 10**7) if is_curious(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": 3,
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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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"data": {
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"text/plain": [
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"40730"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"assert(s == 40730)\n",
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"s"
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]
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}
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],
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"metadata": {
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"completion_date": "Mon, 12 Feb 2018, 17:57",
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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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"tags": [
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"factorial",
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"brute force"
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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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