Fast factoring algorithms? [closed]

Answer 1

You probably wish to go for the modulus operator( % ).

E.g

import java.util.Scanner;

public class Factor {
     public static void main(String[] args) {
     System.out.println("Enter a number whose factors are to be calculated: ");
     Scanner scanNum = new Scanner(System.in);
     int numFac = 0;
     if(scanNum.hasNextInt())  {
        numFac = scanNum.nextInt();
       }

     System.out.println("The Factors of the entered number are:-");

     for(int i = 1; i <= numFac; i++)   {
        if(numFac%i == 0)   {
            System.out.print(i+" ");
          }
     }
   }
  }

Answer 2

public class FactorGenerator{      
  private int number;       
  private int i;           
  public FactorGenerator(int numberToFactor){      
    number = numberToFactor;   
  }

  public int nextFactor(){
     while(number % i == 0){
        System.out.print((Math.round(i)) + " "); 
        number =((number / i));
        return i;     
     }     
     return i;     
 }     

 public boolean hasMoreFactors(){      
   for ( i = 2; i <= number; i++){      
     nextFactor(); 
   }
  return false;      
 }      

}

Test Program:      
import java.util.Scanner;        
 public class FactorgeneratorTester{          
   public static void main (String [] args){           
       Scanner in= new Scanner(System.in);         
       System.out.println("input the value");       
       int number = in.nextInt();       
       FactorGenerator fg = new FactorGenerator(number);             
           if (fg.hasMoreFactors()){      
             System.out.println(fg.hasMoreFactors());      
           }            
   }               

}

the input
210
output
2 3 5 7

Answer 3

Go through a loop applying modulus to all of the intermediate numbers.

X=1;
WHILE(X<=20)
 IF 20%x == 0
 THEN FACTOR!
 X++;
END

Answer 4

List<Integer> factors = new ArrayList<Integer>();
for (int i = 1; i < NUMBER; i++) {
    if (NUMBER % i == 0) {
        factors.add(i);
    }
}

Answer 5

public static Integer[] findFactors(int d)
{
    List<Integer> list = new ArrayList<Integer>();
    for(int i = 1; i <= d/2; i++)
    {
        if(d % i == 0) list.add(new Integer(i));
    }
    list.add(new Integer(d));
    return (Integer[]) list.toArray(new Integer[0]);
}

public static void main(String[] args)
{
    Integer[] list = findFactors(20);
    for(Integer i : list) System.out.println(i);
}

Output:

1
2
4
5
10
20

Answer 6

If the number that you want to find the factors for happens to be odd, you only need to test odd numbers because it is impossible to have an even factor for an odd number. So with a pre-check up front, you can save yourself some processing.

private static List<Integer> findFactors(int num)
{
    int incrementer = 1;
    if (num % 2 != 0)
    {
        incrementer = 2; //only test the odd ones
    }
    List<Integer> list = new ArrayList<Integer>();
    for (int i = 1; i <= num / 2; i=i+incrementer)
    {
        if (num % i == 0)
        {
            list.add(i);
        }
    }
    list.add(num);
    return list;
}