Java Program to Find sum of even factors of a number

Last Updated : 03 Nov, 2022

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 = (p1^a1)*(p2^a2)* … (pk^ak).

Sum of divisors = (1 + p1 + p1² ... p1^a1) * 
                  (1 + p2 + p2² ... p2^a2) *
                  ...........................
                  (1 + pk + pk² ... pk^ak)

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 sum 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 which is 1. After this step, we only get even factors. Note that 2 is the only even prime.

java

// 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 */