Monday 27 April 2009

PROJECT EULER #71

Link to Project Euler problem 71

Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d 8 in ascending order of size, we get:


1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that 2/5 is the fraction immediately to the left of 3/7.
By listing the set of reduced proper fractions for d 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.


This turned out to be too easy, althought the wording of the problem does say

By listing the set of reduced proper fractions......

and I didn't do that...


using System;

namespace ProjectEuler
{
class Program
{
static void Main()
{
//Problem 71
DateTime start = DateTime.Now;
int a=2,b=5,c=3,d=7;
while (b+d<=1000000)
{
a = a + c;
b = b + d;
}
Console.WriteLine(a + "/"+b);
TimeSpan time = DateTime.Now - start;
Console.WriteLine("This took {0}", time);
Console.ReadKey();
}
}
}

No comments: