Solved 47 and 48.
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"cells": [
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Distinct primes factors (Euler Problem 47)"
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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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"collapsed": true
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},
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"source": [
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"[https://projecteuler.net/problem=47](https://projecteuler.net/problem=47)\n",
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"\n",
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"The first two consecutive numbers to have two distinct prime factors are:\n",
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"\n",
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"14 = 2 × 7\n",
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"\n",
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"15 = 3 × 5\n",
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"\n",
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"The first three consecutive numbers to have three distinct prime factors are:\n",
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"\n",
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"644 = 2² × 7 × 23\n",
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"\n",
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"645 = 3 × 5 × 43\n",
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"\n",
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"646 = 2 × 17 × 19.\n",
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"\n",
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"Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?"
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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": 46,
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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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"[2, 3, 7]\n"
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]
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}
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],
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"source": [
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"def sieve_of_eratosthenes(number):\n",
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" primes = []\n",
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" prospects = [n for n in range(2, number + 1)]\n",
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" while prospects:\n",
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" p = prospects[0]\n",
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" prospects = [x for x in prospects if x % p != 0]\n",
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" primes.append(p)\n",
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" if p * p > number:\n",
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" break\n",
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" return primes + prospects\n",
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"\n",
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"import math\n",
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"\n",
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"def get_prime_factors(n):\n",
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" ps = sieve_of_eratosthenes(n)\n",
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" fs = []\n",
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" for p in ps:\n",
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" if n % p == 0:\n",
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" fs.append(p)\n",
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" while n % p == 0:\n",
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" n = n // p\n",
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" return fs\n",
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"\n",
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"def trial_division(n):\n",
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" a = [] \n",
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" if n % 2 == 0:\n",
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" a.append(2)\n",
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" while n % 2 == 0:\n",
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" n //= 2\n",
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" f = 3\n",
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" while f * f <= n:\n",
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" if n % f == 0:\n",
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" a.append(f)\n",
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" while n % f == 0:\n",
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" n //= f\n",
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" else:\n",
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" f += 2 \n",
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" if n != 1:\n",
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" a.append(n)\n",
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" return a\n",
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" \n",
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"assert(get_prime_factors(14) == [2, 7])\n",
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"assert(get_prime_factors(644) == [2, 7, 23])\n",
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"\n",
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"print(trial_division(126))"
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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": 54,
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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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"134043\n"
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]
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}
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],
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"source": [
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"s = []\n",
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"for n in range(2, 1000000):\n",
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" if len(trial_division(n)) == 4:\n",
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" s.append(n)\n",
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" else:\n",
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" s = []\n",
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" if len(s) == 4:\n",
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" s = s[0]\n",
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" break\n",
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"\n",
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"print(s)\n",
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"assert(s == 134043)"
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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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"collapsed": true
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},
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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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"completion_date": "Sun, 23 Dec 2018, 00:24",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python3.6",
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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": "3.6.5"
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},
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"tags": [
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"trial division",
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"prime",
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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": 2
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}
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@ -0,0 +1,85 @@
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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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"# Self powers (Euler Problem 48)"
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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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"collapsed": true
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},
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"source": [
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"[https://projecteuler.net/problem=48](https://projecteuler.net/problem=48)\n",
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"\n",
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"The series, $1^1 + 2^2 + 3^3 + ... + 10^{10} = 10405071317$.\n",
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"\n",
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"Find the last ten digits of the series, $1^1 + 2^2 + 3^3 + ... + 1000^{1000}$."
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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 would be way harder in C/C++. In every language with long int support it is easy. See the number of people who have solved it."
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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": 7,
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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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"9110846700\n"
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]
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}
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],
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"source": [
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"s = int(str(sum([i**i for i in range(1, 1001)]))[-10:])\n",
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"assert(s == 9110846700)\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": "code",
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"execution_count": null,
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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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}
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],
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"metadata": {
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"completion_date": "Sun, 23 Dec 2018, 00:32",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python3.6",
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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": "3.6.5"
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},
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"tags": [
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"brute force",
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"self power"
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]
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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