Skip to content
Related Articles

Related Articles

Java Program to Find sum of even factors of a number
  • Last Updated : 05 Dec, 2018

Given a number n, the task is to find the even factor sum of a number.

Examples:

Input : 30
Output : 48
Even dividers sum 2 + 6 + 10 + 30 = 48

Input : 18
Output : 26
Even dividers sum 2 + 6 + 18 = 26

Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  ...........................
                  (1 + pk + pk2 ... pkak) 

If number is odd, then there are no even factors, so we simply return 0.

If number is even, we use above formula. We only need to ignore 20. All other terms multiply to produce even factor sum. For example, consider n = 18. It can be written as 2132 and sun of all factors is (20 + 21)*(30 + 31 + 32). if we remove 20 then we get the
Sum of even factors (2)*(1+3+32) = 26.



To remove odd number in even factor, we ignore then 20 whaich is 1. After this step, we only get even factors. Note that 2 is the only even prime.




// Formula based Java program to
// find sum of all divisors of n.
import java.util.*;
import java.lang.*;
  
public class GfG {
  
    // Returns sum of all factors of n.
    public static int sumofFactors(int n)
    {
        // If n is odd, then there
        // are no even factors.
        if (n % 2 != 0)
            return 0;
  
        // Traversing through all prime
        // factors.
        int res = 1;
        for (int i = 2; i <= Math.sqrt(n); i++) {
            int count = 0, curr_sum = 1;
            int curr_term = 1;
  
            // While i divides n, print i and
            // divide n
            while (n % i == 0) {
                count++;
  
                n = n / i;
  
                // here we remove the 2^0 that
                // is 1. All other factors
                if (i == 2 && count == 1)
                    curr_sum = 0;
  
                curr_term *= i;
                curr_sum += curr_term;
            }
  
            res *= curr_sum;
        }
  
        // This condition is to handle the
        // case when n is a prime number.
        if (n >= 2)
            res *= (1 + n);
  
        return res;
    }
  
    // Driver function
    public static void main(String argc[])
    {
        int n = 18;
        System.out.println(sumofFactors(n));
    }
}
  
/* This code is contributed by Sagar Shukla */
Output:
26

Please refer complete article on Find sum of even factors of a number for more details!

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :