Find sum of even factors of a number (original) (raw)

Last Updated : 25 Oct, 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

Prerequisite : Sum of factors
As discussed in above mentioned previous post, sum of factors of a number is
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 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.
Below is the implementation of the above approach.

Try It Yourself

C++ `

// Formula based CPP program to find sum of all // divisors of n. #include <bits/stdc++.h> using namespace std;

// Returns sum of all factors of n. 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 <= sqrt(n); i++) {

    // While i divides n, print i and divide n
    int count = 0, curr_sum = 1, curr_term = 1;
    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 code int main() { int n = 18; cout << sumofFactors(n); return 0; }

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 */

Python3

Formula based Python3

program to find sum

of alldivisors of n.

import math

Returns sum of all

factors of n.

def sumofFactors(n) :

# If n is odd, then
# there are no even
# factors.
if (n % 2 != 0) :
    return 0 

# Traversing through
# all prime factors.
res = 1
for i in range(2, (int)(math.sqrt(n)) + 1) :
    
    # While i divides n
    # print i and divide n
    count = 0
    curr_sum = 1
    curr_term = 1
    while (n % i == 0) :
        count= count + 1

        n = n // i

        # here we remove the
        # 2^0 that is 1. All
        # other factors
        if (i == 2 and count == 1) :
            curr_sum = 0

        curr_term = curr_term * i
        curr_sum = curr_sum + curr_term
    
    res = res * curr_sum
    

# This condition is to
# handle the case when
# n is a prime number.
if (n >= 2) :
    res = res * (1 + n)

return res

Driver code

n = 18 print(sumofFactors(n))

This code is contributed by Nikita Tiwari.

C#

// Formula based C# program to // find sum of all divisors of n. using System;

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 Code
public static void Main() {
    int n = 18;
    Console.WriteLine(sumofFactors(n));
}

}

// This code is contributed by vt_m

PHP

JavaScript

`

Output:

26

Time Complexity: O(?n log n)
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.