# Sum of all the factors of a number

Given a number n, the task is to find the sum of all the divisors.

Examples :

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 +
10 + 15 + 30 = 72

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

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

A simple solution is to traverse through all divisors and add them.

 // Simple C++ program to  // find sum of all divisors  // of a natural number #include using namespace std;     // Function to calculate sum of all  //divisors of a given number int divSum(int n) {     if(n == 1)       return 1;        // Sum of divisors     int result = 0;         // find all divisors which divides 'num'     for (int i = 2; i <= sqrt(n); i++)     {         // if 'i' is divisor of 'n'         if (n % i == 0)         {             // if both divisors are same             // then add it once else add             // both             if (i == (n / i))                 result += i;             else                 result += (i + n/i);         }     }         // Add 1 and n to result as above loop     // considers proper divisors greater      // than 1.     return (result + n + 1); }     // Driver program to run the case int main() {     int n = 30;     cout << divSum(n);     return 0; }

 // Simple Java program to  // find sum of all divisors  // of a natural number import java.io.*;     class GFG {        // Function to calculate sum of all      //divisors of a given number     static int divSum(int n)     {          if(n == 1)            return 1;         // Final result of summation          // of divisors         int result = 0;                // find all divisors which divides 'num'         for (int i = 2; i <= Math.sqrt(n); i++)         {             // if 'i' is divisor of 'n'             if (n % i == 0)             {                 // if both divisors are same                 // then add it once else add                 // both                 if (i == (n / i))                     result += i;                 else                     result += (i + n / i);             }         }                // Add 1 and n to result as above loop         // considers proper divisors greater         // than 1.         return (result + n + 1);                }            // Driver program to run the case     public static void main(String[] args)     {         int n = 30;         System.out.println(divSum(n));     } }    // This code is contributed by Prerna Saini.

 # Simple Python 3 program to  # find sum of all divisors of # a natural number import math      # Function to calculate sum  # of all divisors of given #  natural number def divSum(n) :     if(n == 1):        return 1        # Final result of summation      # of divisors     result = 0          # find all divisors which     # divides 'num'     for i in range(2,(int)(math.sqrt(n))+1) :            # if 'i' is divisor of 'n'         if (n % i == 0) :                # if both divisors are same              # then add it only once             # else add both             if (i == (n/i)) :                 result = result + i             else :                 result = result + (i + n//i)                           # Add 1 and n to result as above      # loop considers proper divisors     # greater than 1.     return (result + n + 1)      # Driver program to run the case n = 30 print(divSum(n))    # This code is contributed by Nikita Tiwari.

 // Simple C# program to  // find sum of all divisors  // of a natural number using System;    class GFG {        // Function to calculate sum of all      //divisors of a given number     static int divSum(int n)     {         if(n == 1)            return 1;            // Final result of summation          // of divisors         int result = 0;                // find all divisors which divides 'num'         for (int i = 2; i <= Math.Sqrt(n); i++)         {             // if 'i' is divisor of 'n'             if (n % i == 0)             {                 // if both divisors are same                 // then add it once else add                 // both                 if (i == (n / i))                     result += i;                 else                     result += (i + n / i);             }         }                // Add 1 and n to result as above loop         // considers proper divisors greater         // than 1.         return (result + n + 1);     }            // Driver program to run the case     public static void Main()     {                    int n = 30;                    Console.WriteLine(divSum(n));     } }    // This code is contributed by vt_m.



Output :

72

An efficient solution is to use below formula.
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)

We can notice that individual terms of above
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) *
(p2a2+1 - 1)/(p2 -1) *
..................................
(pkak+1 - 1)/(pk -1)

How does above formula work?

Consider the number 18.

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
(21)(31) + (20)(3^2) + (2^1)(32)
= (20)(30) + (2^0)(31) + (2^0)(32) +
(21)(3^0) + (21)(31) + (21)(32)
= (20)(30 + 31 + 32) +
(21)(30 + 31 + 32)
= (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

So the task reduces to finding all prime factors and their powers.

 // 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) {     // Traversing through all prime factors.     int res = 1;     for (int i = 2; i <= sqrt(n); i++)     {                       int curr_sum = 1;         int curr_term = 1;         while (n % i == 0) {                // THE BELOW STATEMENT MAKES             // IT BETTER THAN ABOVE METHOD              //  AS WE REDUCE VALUE OF n.             n = n / i;                curr_term *= i;             curr_sum += curr_term;         }            res *= curr_sum;     }        // This condition is to handle      // the case when n is a prime     // number greater than 2.     if (n >= 2)         res *= (1 + n);        return res; }    // Driver code int main() {     int n = 30;     cout << sumofFactors(n);     return 0; }

 // Formula based Java program to  // find sum of all divisors of n.    import java.io.*; import java.math.*; public class GFG{            // Returns sum of all factors of n.     static int sumofFactors(int n)     {         // Traversing through all prime factors.         int res = 1;         for (int i = 2; i <= Math.sqrt(n); i++)         {                                   int  curr_sum = 1;             int curr_term = 1;                            while (n % i == 0)              {                        // THE BELOW STATEMENT MAKES                 // IT BETTER THAN ABOVE METHOD                  // AS WE REDUCE VALUE OF n.                 n = n / i;                        curr_term *= i;                 curr_sum += curr_term;             }                    res *= curr_sum;         }                // This condition is to handle          // the case when n is a prime          // number greater than 2         if (n > 2)             res *= (1 + n);                return res;     }            // Driver code     public static void main(String args[])     {         int n = 30;         System.out.println(sumofFactors(n));     } }    /*This code is contributed by Nikita Tiwari.*/

 # Formula based Python3 code to find  # sum of all divisors of n. import math as m    # Returns sum of all factors of n. def sumofFactors(n):            # Traversing through all     # prime factors     res = 1     for i in range(2, int(m.sqrt(n) + 1)):                    curr_sum = 1         curr_term = 1                    while n % i == 0:                            n = n / i;                curr_term = curr_term * i;             curr_sum += curr_term;                        res = res * curr_sum            # This condition is to handle the      # case when n is a prime number      # greater than 2     if n > 2:         res = res * (1 + n)        return res;    # driver code     sum = sumofFactors(30) print ("Sum of all divisors is: ",sum)    # This code is contributed by Saloni Gupta

 // Formula based Java program to  // find sum of all divisors of n. using System;    public class GFG {            // Returns sum of all factors of n.     static int sumofFactors(int n)     {                    // Traversing through all prime factors.         int res = 1;         for (int i = 2; i <= Math.Sqrt(n); i++)         {                                   int curr_sum = 1;             int curr_term = 1;                            while (n % i == 0)              {                        // THE BELOW STATEMENT MAKES                 // IT BETTER THAN ABOVE METHOD                  // AS WE REDUCE VALUE OF n.                 n = n / i;                        curr_term *= i;                 curr_sum += curr_term;             }                    res *= curr_sum;         }                // This condition is to handle          // the case when n is a prime          // number greater than 2         if (n > 2)             res *= (1 + n);                return res;     }            // Driver code     public static void Main()     {                    int n = 30;                    Console.WriteLine(sumofFactors(n));     } }    /*This code is contributed by vt_m.*/

 2)         \$res *= (1 + \$n);        return \$res; }    // Driver Code \$n = 30; echo sumofFactors(\$n);    // This code is contributed by Anuj_67. ?>

Output :
72

Further Optimization.
If there are multiple queries, we can use Sieve to find prime factors and their powers.

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