Project Euler Question 3 Help

Question

I\'m trying to work through Project Euler and I\'m hitting a barrier on problem 03. I have an algorithm that works for smaller numbers, but problem 3 uses a very, very large

Answer 1

Easy peasy in C++:

#include <iostream>

using namespace std;


int main()
{
    unsigned long long int largefactor = 600851475143;
    for(int i = 2;;)
    {
        if (largefactor <= i)
            break;
        if (largefactor % i == 0)
        {
            largefactor = largefactor / i;
        }
        else
            i++;
    }

    cout << largefactor << endl;

    cin.get();
    return 0;
}

Answer 2

you might want to see this: Is there a simple algorithm that can determine if X is prime, and not confuse a mere mortal programmer?

and I like lill mud's solution:

require "mathn.rb"
puts 600851475143.prime_division.last.first

I checked it here

Answer 3

Once you find the answer, enter the following in your browser ;)

http://www.wolframalpha.com/input/?i=FactorInteger(600851475143)

Wofram Alpha is your friend

Answer 4

Although the question asks for the largest prime factor, it doesn't necessarily mean you have to find that one first...

Answer 5

long n = 600851475143L; //not even, so 2 wont be a factor
int factor = 3; 
while( n > 1)
{
    if(n % factor == 0)
    {
        n/=factor;
    }else
        factor += 2; //skip even numbrs
}
        print factor;

This should be quick enough...Notice, there's no need to check for prime...

Answer 6

The way I did it was to search for primes (p), starting at 2 using the Sieve of Eratosthenes. This algorithm can find all the primes under 10 million in <2s on a decently fast machine.

For every prime you find, test divide it into the number you are testing against untill you can't do integer division anymore. (ie. check n % p == 0 and if true, then divide.)

Once n = 1, you're done. The last value of n that successfully divided is your answer. On a sidenote, you've also found all the prime factors of n on the way.

PS: As been noted before, you only need to search for primes between 2 <= n <= sqrt(p). This makes the Sieve of Eratosthenes a very fast and easy to implement algorithm for our purposes.