Find sum of even factors of a number

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 = (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.
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.

// 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:

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.

Find sum of even factors of a number

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Find sum of even factors of a number

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