# Find sum of even factors of a number

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 ` `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

 `= 2) ` `        ``\$res` `*= (1 + ``\$n``); ` ` `  `    ``return` `\$res``; ` `} ` ` `  `// Driver code ` `    ``\$n` `= 18; ` `    ``echo` `sumofFactors(``\$n``); ` ` `  `// This code is contributed by mits ` `?> `

Output:

```26
```

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