Question

Possible Duplicate:
Project Euler Problem 12 – C++

I am trying to get the first triangle number with more than 400 divisors (Triangle Number eg: 1,3,6,10). For an example, triangle number 6 has four divisors 1,2,3,6. The following is my attempt to get the triangle number with 400 divisors

import java.math.BigInteger;

public class IQ3
{

        static int num1 = 1;    
        static int devideResult = 0;      

    public static void main(String[]args)
    {


        while(true)
        {
            int triangle = num1*(num1+1)/2;

            if(devide(triangle))
            {
                break;
            }

            num1++;
        }

    }

    static boolean devide(int num)
    {
        boolean result = false;
        int devideCounter = 2;       


        for(int i=1;i<=num/2;i++)
            {
                if(num%i == 0)
                {
                    devideCounter++;
                    System.out.println("Devide Counter: "+devideCounter);
                    //System.out.println("i number: "+i);
                    //System.out.println("input number: "+num);

                    if(devideCounter>400)
                    {
                        System.out.println("Number: "+num);
                        result = true;
                        break;
                    }
                }
            }

        return result;
    }
}

But this takes a huge time, and some times it crashes.

However, since the answer could be really big, I thought of using BigInteger.

import java.math.BigInteger;

public class IQ2P2
{

        static BigInteger num1 = new BigInteger("1");
        static BigInteger two = new BigInteger("2");
        static BigInteger one = new BigInteger("1");
        static BigInteger i = new BigInteger("1");
        static BigInteger zero = new BigInteger("0");


        static int devideResult = 0;        
    //    static int devideCounter = 0;

    public static void main(String[]args)
    {


        while(true)
        {
            BigInteger triangle = num1.multiply(num1.add(one)).divide(two);

            if(devide(triangle))
            {
                break;
            }

            num1.add(one);
        }

    }

    static boolean devide(BigInteger num)
    {
        boolean result = false;
        int devideCounter = 2;       


        while((i.compareTo(num))<(num.divide(two).intValue()))
            {
                if(num.remainder(i) == zero)
                {
                    devideCounter++;
                    System.out.println("Devide Counter: "+devideCounter);
                    //System.out.println("i number: "+i);
                    //System.out.println("input number: "+num);

                    if(devideCounter>400)
                    {
                        System.out.println("Number: "+num);
                        result = true;
                        break;
                    }
                }
                i.add(one);
            }

        return result;
    }
}

But the biginteger one never returned anything.

Please help me to get first Triangle number with more than 400 divisors.

Note: This is not a homework. I am not a student.

The following is a response to an answer

import java.math.BigInteger;

public class IQ2
{

        static long num1 = 1;
        static long numberToAdd = 0;
        static long devideResult = 0;  


       static   long triangleNum = 1;
    static long incrementer = 2;
    //    static int devideCounter = 0;

    public static void main(String[]args)
    {


        while(true)
        {
            triangleNum += incrementer++;

            if(devide(triangleNum))
            {
                break;
            }

            num1++;
        }

    }

    static boolean devide(long num)
    {
        boolean result = false;
        int devideCounter = 2;       


        for(long i=1;i<=num/2;i++)
            {
                if(num%i == 0)
                {
                    devideCounter++;
                    System.out.println("Devide Counter: "+devideCounter);
                    //System.out.println("i number: "+i);
                    //System.out.println("input number: "+num);

                    if(devideCounter>400)
                    {
                        System.out.println("Number: "+num);
                        result = true;
                        break;
                    }
                }
            }

        return result;
    }
}

Answer 1

you need to optimize the way you find out the amount of divisors for a given number. First, for every d <= sqrt(n) such that n%d==0, there is m=n/d such that n%m==0 and m >= sqrt(n). That means you can count both of them at once, stopping at sqrt(n).

But the real optimization is to calculate the prime factorization of a number instead, and find out the amount of divisors from there.