Find sum of even factors of a number

• Difficulty Level : Medium
• Last Updated : 13 Apr, 2021

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 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.
Below is the implementation of the above approach.

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

<?php
// Formula based php program to find sum
// of all divisors of n.

// Returns sum of all factors of n.
function 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 = 2; \$i <= sqrt(\$n); \$i++) {

// While i divides n, print i
// and divide n
\$count = 0;
\$curr_sum = 1;
\$curr_term = 1;
while (\$n % \$i == 0) {
\$count++;

\$n = floor(\$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
\$n = 18;
echo sumofFactors(\$n);

// This code is contributed by mits
?>

Javascript

<script>

// javascript program to
// find sum of all divisors of n.

// Returns sum of all factors of n.
function sumofFactors(n)
{
// If n is odd, then there
// are no even factors.
if (n % 2 != 0)
return 0;

// Traversing through all prime
// factors.
let res = 1;
for (let i = 2; i <= Math.sqrt(n); i++)
{
let count = 0, curr_sum = 1;
let 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

let n = 18;
document.write(sumofFactors(n));

// This code is contributed by susmitakundugoaldanga.
</script>

Output:

26

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