173 lines
3.9 KiB
Plaintext
173 lines
3.9 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 33\n",
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"\n",
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"The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.\n",
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"\n",
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"We shall consider fractions like, 30/50 = 3/5, to be trivial examples.\n",
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"\n",
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"There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.\n",
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"\n",
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"If the product of these four fractions is given in its lowest common terms, find the value of the denominator."
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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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"We start write a function which checks if a number is curios and then brute force."
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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": 21,
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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 is_curious(n, d):\n",
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" assert(len(str(n)) == 2 and len(str(d)) == 2)\n",
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" if n == d:\n",
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" return False\n",
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" for i in range(1, 10):\n",
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" if str(i) in str(n) and str(i) in str(d):\n",
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" try:\n",
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" n_ = int(str(n).replace(str(i), \"\"))\n",
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" d_ = int(str(d).replace(str(i), \"\"))\n",
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" except ValueError:\n",
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" return False\n",
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" try:\n",
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" if n_ / d_ == n / d:\n",
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" return True\n",
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" except ZeroDivisionError:\n",
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" return False\n",
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" return False\n",
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"\n",
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"assert(is_curious(49, 98) == True)\n",
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"assert(is_curious(30, 50) == False)"
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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": 25,
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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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"[(16, 64), (19, 95), (26, 65), (49, 98)]"
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]
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},
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"execution_count": 25,
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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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"fs = [(n, d) for n in range(10, 100) for d in range(n, 100) if is_curious(n, d)]\n",
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"fs"
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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": 32,
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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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"387296/38729600\n"
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]
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}
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],
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"source": [
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"n = 1\n",
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"d = 1\n",
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"for n_, d_ in fs:\n",
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" n *= n_\n",
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" d *= d_\n",
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"\n",
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"print(\"{}/{}\".format(n, d))"
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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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"Now we can see that the solution is $100$. But actually it would be nice to calculate the GCD."
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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": 40,
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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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"100"
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]
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},
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"execution_count": 40,
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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 gcd_euclid(a, b):\n",
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" if a == b:\n",
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" return a\n",
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" elif a > b:\n",
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" return gcd_euclid(a - b, b)\n",
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" elif a < b:\n",
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" return gcd_euclid(a, b - a)\n",
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" \n",
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"gcd_nd = gcd_euclid(n, d)\n",
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"\n",
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"s = d // gcd_nd\n",
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"assert(s == 100)\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, 17:29",
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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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"gcd",
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"curious",
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"faction"
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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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