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# C++ Program to 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 try your approach on {IDE} first, before moving on to the solution.

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.

## CPP

 `// 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;``}`

Output:

`26`

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

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