Solved 64.
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@ -13,11 +13,61 @@
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"collapsed": true
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},
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"source": [
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"The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.\n",
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"The 5-digit number, 16807=$7^5$, is also a fifth power. Similarly, the 9-digit number, 134217728=$8^9$, is a ninth power.\n",
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"\n",
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"How many n-digit positive integers exist which are also an nth power?"
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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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"def get_digit_count(n):\n",
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" if n < 10:\n",
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" return 1\n",
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" c = 0\n",
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" while n:\n",
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" n //= 10\n",
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" c += 1\n",
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" return c\n",
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"\n",
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"assert(get_digit_count(0) == 1)\n",
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"assert(get_digit_count(1) == 1)\n",
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"assert(get_digit_count(33) == 2)\n",
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"assert(get_digit_count(100) == 3)"
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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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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"49\n"
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]
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}
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],
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"source": [
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"def get_n_digit_positive_integers(n):\n",
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" r = []\n",
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" i = 1\n",
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" while True:\n",
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" if get_digit_count(i ** n) == n:\n",
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" r.append(i ** n)\n",
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" if get_digit_count(i ** n) > n:\n",
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" return r\n",
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" i += 1\n",
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"\n",
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"s = sum([len(get_n_digit_positive_integers(n)) for n in range(1, 1000)])\n",
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"print(s)\n",
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"assert(s == 49)"
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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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@ -29,7 +79,7 @@
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}
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],
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"metadata": {
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"completion_date": "",
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"completion_date": "Sun, 6 Jan 2019, 06:17",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python3.6",
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@ -47,7 +97,9 @@
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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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"tags": [
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"powers"
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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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@ -0,0 +1,213 @@
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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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"# Odd period square roots (Euler Problem 64)"
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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=64](https://projecteuler.net/problem=64)\n",
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"\n",
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"The first ten continued fraction representations of (irrational) square roots are:\n",
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"\n",
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"√2=[1;(2)], period=1\n",
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"\n",
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"√3=[1;(1,2)], period=2\n",
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"\n",
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"√5=[2;(4)], period=1\n",
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"\n",
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"√6=[2;(2,4)], period=2\n",
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"\n",
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"√7=[2;(1,1,1,4)], period=4\n",
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"\n",
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"√8=[2;(1,4)], period=2\n",
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"\n",
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"√10=[3;(6)], period=1\n",
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"\n",
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"√11=[3;(3,6)], period=2\n",
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"\n",
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"√12= [3;(2,6)], period=2\n",
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"\n",
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"√13=[3;(1,1,1,1,6)], period=5\n",
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"\n",
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"Exactly four continued fractions, for N ≤ 13, have an odd period.\n",
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"\n",
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"How many continued fractions for N ≤ 10000 have an odd period?"
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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 math\n",
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"\n",
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"def get_floor_sqrt(n):\n",
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" return math.floor(math.sqrt(n))\n",
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"\n",
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"assert(get_floor_sqrt(5) == 2)\n",
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"assert(get_floor_sqrt(23) == 4)"
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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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{
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"data": {
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"text/plain": [
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"(3, 3, 2)"
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]
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},
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"execution_count": 2,
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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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"def next_expansion(current_a, current_nominator, current_denominator, original_number):\n",
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" # Less typing\n",
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" cn = current_nominator\n",
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" cd = current_denominator\n",
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"\n",
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" # Step 1: Multiply the fraction so that we can use the third binomial formula.\n",
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" # Make sure we can reduce the nominator.\n",
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" assert((original_number - cd * cd) % cn == 0)\n",
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" # The new nominator is the denominator since we multiply with (x + cd) and then\n",
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" # reduce the previous nominator.\n",
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" # The new denominator is calculated by applying the third binomial formula and\n",
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" # then by divided by the previous nominator.\n",
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" cn, cd = cd, (original_number - cd * cd) // cn\n",
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" \n",
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" # Step 2: Calculate the next a by finding the next floor square root.\n",
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" next_a = math.floor((math.sqrt(original_number) + cn) // cd)\n",
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" \n",
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" # Step 3: Remove next a from the fraction by substracting it.\n",
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" cn = cn - next_a * cd\n",
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" cn *= -1\n",
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" \n",
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" return next_a, cn, cd\n",
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"\n",
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"next_expansion(1, 7, 3, 23)"
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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 get_continued_fraction_sequence(n):\n",
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" \n",
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" # If number is a square number we return it.\n",
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" floor_sqrt = get_floor_sqrt(n)\n",
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" if n == floor_sqrt * floor_sqrt:\n",
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" return ((floor_sqrt), [])\n",
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" \n",
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" # Otherwise, we calculate the next expansion till we\n",
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" # encounter a step a second time. \n",
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" a = floor_sqrt\n",
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" cn = a\n",
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" cd = 1\n",
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" sequence = []\n",
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" previous_steps = []\n",
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" \n",
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" while not (a, cn, cd) in previous_steps :\n",
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" #print(\"a: {} cn: {} cd: {}\".format(a, cn, cd))\n",
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" previous_steps.append((a, cn, cd))\n",
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" a, cn, cd = next_expansion(a, cd, cn, n)\n",
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" sequence.append(a)\n",
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" sequence.pop()\n",
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" return ((floor_sqrt), sequence)\n",
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"\n",
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"assert(get_continued_fraction_sequence(1) == ((1), []))\n",
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"assert(get_continued_fraction_sequence(4) == ((2), []))\n",
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"assert(get_continued_fraction_sequence(25) == ((5), []))\n",
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"assert(get_continued_fraction_sequence(2) == ((1), [2]))\n",
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"assert(get_continued_fraction_sequence(3) == ((1), [1,2]))\n",
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"assert(get_continued_fraction_sequence(5) == ((2), [4]))\n",
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"assert(get_continued_fraction_sequence(13) == ((3), [1,1,1,1,6]))"
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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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"collapsed": true
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},
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"outputs": [],
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"source": [
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"def get_period(n):\n",
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" _, sequence = get_continued_fraction_sequence(n)\n",
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" return len(sequence)\n",
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"\n",
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"assert(get_period(23) == 4)"
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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": 5,
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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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"1322\n"
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]
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}
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],
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"source": [
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"s = len([n for n in range(1, 10001) if get_period(n) % 2 != 0])\n",
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"print(s)\n",
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"assert(s == 1322)"
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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": "Tue, 22 Jan 2019, 05:10",
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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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"fractions",
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"square roots",
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"sequence",
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"periodic"
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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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@ -11778,13 +11778,70 @@ div#notebook {
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</div>
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<div class="inner_cell">
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<div class="text_cell_render border-box-sizing rendered_html">
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<p>The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.</p>
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<p>The 5-digit number, 16807=$7^5$, is also a fifth power. Similarly, the 9-digit number, 134217728=$8^9$, is a ninth power.</p>
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<p>How many n-digit positive integers exist which are also an nth power?</p>
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="prompt input_prompt">In [1]:</div>
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<div class="inner_cell">
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<div class="input_area">
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">get_digit_count</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
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<span class="k">if</span> <span class="n">n</span> <span class="o"><</span> <span class="mi">10</span><span class="p">:</span>
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<span class="k">return</span> <span class="mi">1</span>
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<span class="n">c</span> <span class="o">=</span> <span class="mi">0</span>
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<span class="k">while</span> <span class="n">n</span><span class="p">:</span>
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<span class="n">n</span> <span class="o">//=</span> <span class="mi">10</span>
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<span class="n">c</span> <span class="o">+=</span> <span class="mi">1</span>
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<span class="k">return</span> <span class="n">c</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">get_digit_count</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span> <span class="o">==</span> <span class="mi">1</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">get_digit_count</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span> <span class="o">==</span> <span class="mi">1</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">get_digit_count</span><span class="p">(</span><span class="mi">33</span><span class="p">)</span> <span class="o">==</span> <span class="mi">2</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">get_digit_count</span><span class="p">(</span><span class="mi">100</span><span class="p">)</span> <span class="o">==</span> <span class="mi">3</span><span class="p">)</span>
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</pre></div>
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</div>
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="prompt input_prompt">In [2]:</div>
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<div class="inner_cell">
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<div class="input_area">
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">get_n_digit_positive_integers</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
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<span class="n">r</span> <span class="o">=</span> <span class="p">[]</span>
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<span class="n">i</span> <span class="o">=</span> <span class="mi">1</span>
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<span class="k">while</span> <span class="kc">True</span><span class="p">:</span>
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<span class="k">if</span> <span class="n">get_digit_count</span><span class="p">(</span><span class="n">i</span> <span class="o">**</span> <span class="n">n</span><span class="p">)</span> <span class="o">==</span> <span class="n">n</span><span class="p">:</span>
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<span class="n">r</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">i</span> <span class="o">**</span> <span class="n">n</span><span class="p">)</span>
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<span class="k">if</span> <span class="n">get_digit_count</span><span class="p">(</span><span class="n">i</span> <span class="o">**</span> <span class="n">n</span><span class="p">)</span> <span class="o">></span> <span class="n">n</span><span class="p">:</span>
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<span class="k">return</span> <span class="n">r</span>
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<span class="n">i</span> <span class="o">+=</span> <span class="mi">1</span>
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<span class="n">s</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="nb">len</span><span class="p">(</span><span class="n">get_n_digit_positive_integers</span><span class="p">(</span><span class="n">n</span><span class="p">))</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1000</span><span class="p">)])</span>
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<span class="nb">print</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">49</span><span class="p">)</span>
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</pre></div>
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</div>
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</div>
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</div>
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<div class="output_wrapper">
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<div class="output">
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<div class="output_area">
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<div class="prompt"></div>
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<div class="output_subarea output_stream output_stdout output_text">
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<pre>49
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</pre>
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</div>
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</div>
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="prompt input_prompt">In [ ]:</div>
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<div class="inner_cell">
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@ -993,12 +993,32 @@
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</tr>
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<tr class="table-warning">
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<tr>
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<td><a href="EulerProblem063.html">Problem 063</a></td>
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<td></td>
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<td>Sun, 6 Jan 2019, 06:17</td>
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<td>
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<kbd>powers</kbd>
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</td>
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</tr>
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<tr>
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<td><a href="EulerProblem064.html">Problem 064</a></td>
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<td>Tue, 22 Jan 2019, 05:10</td>
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<td>
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<kbd>fractions</kbd>
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<kbd>square roots</kbd>
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<kbd>sequence</kbd>
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<kbd>periodic</kbd>
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</td>
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</tr>
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Reference in New Issue