Project Euler Problem 3: Finding the Largest Prime Factor

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SUMMARY

The largest prime factor of the number 600851475143 can be efficiently found using the C++ programming language. The initial code provided by the user failed due to the use of 32-bit integers, which cannot accommodate the large value. The recommended solution involves using the "unsigned long long" data type and optimizing the factor search by starting from the square root of the number. The final implementation correctly identifies the largest prime factor and outputs it using the C++ standard library.

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  • Basic algorithm design for prime factorization
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Programmers, algorithm enthusiasts, and anyone interested in solving mathematical problems using C++. This discussion is particularly beneficial for those tackling Project Euler challenges or similar computational problems.

SELFMADE
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What is the largest prime factor of the number 600851475143 ?

My attempt:
Code:
#include <stdafx.h>

int main() 
{
	int num1, num2, num3, num4, lrgprime, sum;
	lrgprime=num2=num4=sum=0;
	for (num1=2; num1<300425737572; num1++) {
		num2=600851475143%num1; 

		if (num2==0) {
			for (num3=2; num3<=num1; num3++) {
				num4=num1%num3;
				if (num4==0 && num1==num3 && num1>=lrgprime)
					lrgprime=num1; 
			}		
		}	
	}
	printf("This is the largest prime factor: %d", lrgprime);
	return 0;
}

The machine is just not running this code

What did I do wrong?
 
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check with some smaller numbers first. see whether you get anything
 
Two problems. First, 600851475143 is too large to store in an int (which is usually 32 bits and only goes up to about 2 billion). Try using long longs instead of ints. Second,
Code:
		if (num2==0) {
			for (num3=2; num3<=num1; num3++) {
				num4=num1%num3;
				if (num4==0 && num1==num3 && num1>=lrgprime)
					lrgprime=num1; 
			}		
		}
doesn't do what you want. It eventually always sets lrgprime to num1, even when num1 has divisors. You need to use a break statement to break out of the loop if num4 == 0 and num1 != num3.
 
SELFMADE said:
What is the largest prime factor of the number 600851475143 ?

My attempt:
Code:
#include <stdafx.h>

int main() 
{
	int num1, num2, num3, num4, lrgprime, sum;
	lrgprime=num2=num4=sum=0;
	for (num1=2; num1<300425737572; num1++) {
		num2=600851475143%num1; 

		if (num2==0) {
			for (num3=2; num3<=num1; num3++) {
				num4=num1%num3;
				if (num4==0 && num1==num3 && num1>=lrgprime)
					lrgprime=num1; 
			}		
		}	
	}
	printf("This is the largest prime factor: %d", lrgprime);
	return 0;
}

The machine is just not running this code

What did I do wrong?

1) stdafx.h is not a standard header
2) The maximum value of a 32-bit signed int is 2147483647. Of course, there is no reason to use integers for this task. The largest basic type you could use would be "unsigned long long", which has a maximum value of 18446744073709551615
3) Your code is a mess...you have tons of useless auxiliary variables and have split things into multiple lines just making it more confusing than it should be.
4) This is not an efficient approach, and will take too long on a big number.

Code:
#include <iostream>
#include <limits>

typedef unsigned long long BigInt;
using namespace std;

BigInt largest_prime_factor( const BigInt &x )
{
	//find factors
	for(BigInt y = x/2; y>2; --y)
	{
		//check factors
		if( x%y == 0 )
		{
			//check if factor is prime
			for(BigInt z=2; z<y; ++z)
				if( y%z == 0 )
					goto not_prime;
			return y;
		}
not_prime: ;
	}
	return x;
}

int main() 
{

	cout << "Max value = " << numeric_limits<BigInt>::max() << "\n";
	cout << largest_prime_factor( 600851475143 );
}
 
Last edited:
Code:
for(BigInt y = x/2; y>2; --y)
Just a suggestion: since the largest prime factor of x cannot not exceed x0=(int)sqrt(x), it pays to start at x0, especially for large numbers.
 
Thank you so much, I am going to save this page up, go home and try it. I have been struggling on this problem for 3 days.
 

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