Sum of Factors of a Number using Prime Factorization

Given a number N. The task is to find the sum of all factors of the given number N.

Examples:

Input : N = 12
Output : 28
All factors of 12 are: 1,2,3,4,6,12

Input : 60
Output : 168

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Suppose N = 1100, the idea is to first find the prime factorization of the given number N.
Therefore, prime factorization of 1100 = 2² * 5² * 11.

So, the formulae to calculate sum of all factors can be given as,

(2⁰ + 2¹ + 2²) * (5⁰ + 5¹ + 5²) * (11⁰ + 11¹)
(upto the power of factor in factorization i.e. power of 2 and 5 is 2 and 11 is 1.)
= (1 + 2 + 2²) * (1 + 5 + 5²) * (1 + 11)
= 7 * 31 * 12
= 2604

So, sum of all factors of 1100 = 2604

Below is the implementation of the above approach:

C++

// C++ Program to find sum of all  
// factors of a given number 
  
#include <bits/stdc++.h> 
  
using namespace std; 
  
// Using SieveOfEratosthenes to find smallest prime 
// factor of all the numbers. 
// For example, if N is 10, 
// s[2] = s[4] = s[6] = s[10] = 2 
// s[3] = s[9] = 3 
// s[5] = 5 
// s[7] = 7 
void sieveOfEratosthenes(int N, int s[]) 
{ 
    // Create a boolean array "prime[0..n]" and 
    // initialize all entries in it as false. 
    vector<bool> prime(N + 1, false); 
  
    // Initializing smallest factor equal to 2 
    // for all the even numbers 
    for (int i = 2; i <= N; i += 2) 
        s[i] = 2; 
  
    // For odd numbers less then equal to n 
    for (int i = 3; i <= N; i += 2) { 
        if (prime[i] == false) { 
  
            // s(i) for a prime is the number itself 
            s[i] = i; 
  
            // For all multiples of current prime number 
            for (int j = i; j * i <= N; j += 2) { 
                if (prime[i * j] == false) { 
                    prime[i * j] = true; 
  
                    // i is the smallest prime factor for 
                    // number "i*j". 
                    s[i * j] = i; 
                } 
            } 
        } 
    } 
} 
  
// Function to find sum of all prime factors 
int findSum(int N) 
{ 
    // Declaring array to store smallest prime 
    // factor of i at i-th index 
    int s[N + 1]; 
  
    int ans = 1; 
  
    // Filling values in s[] using sieve 
    sieveOfEratosthenes(N, s); 
      
    int currFactor = s[N]; // Current prime factor of N 
    int power = 1; // Power of current prime factor 
  
    while (N > 1) { 
        N /= s[N]; 
  
        // N is now N/s[N]. If new N als has smallest 
        // prime factor as currFactor, increment power 
        if (currFactor == s[N]) { 
            power++; 
            continue; 
        } 
  
        int sum = 0; 
          
        for(int i=0; i<=power; i++) 
            sum += pow(currFactor,i); 
          
        ans *= sum; 
          
          
        // Update current prime factor as s[N] and 
        // initializing power of factor as 1. 
        currFactor = s[N]; 
        power = 1; 
    } 
  
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    int n = 12; 
  
    cout << "Sum of the factors is : "; 
  
    cout << findSum(n); 
      
    return 0; 
} 

Java

//Java Program to find sum of all  
//factors of a given number 
public class GFG { 
  
    //Using SieveOfEratosthenes to find smallest prime 
    //factor of all the numbers. 
    //For example, if N is 10, 
    //s[2] = s[4] = s[6] = s[10] = 2 
    //s[3] = s[9] = 3 
    //s[5] = 5 
    //s[7] = 7 
    static void sieveOfEratosthenes(int N, int s[]) 
    { 
     // Create a boolean array "prime[0..n]" and 
     // initialize all entries in it as false. 
     boolean[] prime = new boolean[N + 1]; 
       
     for(int i = 0; i < N+1; i++) 
         prime[i] = false; 
  
     // Initializing smallest factor equal to 2 
     // for all the even numbers 
     for (int i = 2; i <= N; i += 2) 
         s[i] = 2; 
  
     // For odd numbers less then equal to n 
     for (int i = 3; i <= N; i += 2) { 
         if (prime[i] == false) { 
  
             // s(i) for a prime is the number itself 
             s[i] = i; 
  
             // For all multiples of current prime number 
             for (int j = i; j * i <= N; j += 2) { 
                 if (prime[i * j] == false) { 
                     prime[i * j] = true; 
  
                     // i is the smallest prime factor for 
                     // number "i*j". 
                     s[i * j] = i; 
                 } 
             } 
         } 
     } 
    } 
  
    //Function to find sum of all prime factors 
    static int findSum(int N) 
    { 
     // Declaring array to store smallest prime 
     // factor of i at i-th index 
     int[] s = new int[N + 1]; 
  
     int ans = 1; 
  
     // Filling values in s[] using sieve 
     sieveOfEratosthenes(N, s); 
        
     int currFactor = s[N]; // Current prime factor of N 
     int power = 1; // Power of current prime factor 
  
     while (N > 1) { 
         N /= s[N]; 
  
         // N is now N/s[N]. If new N als has smallest 
         // prime factor as currFactor, increment power 
         if (currFactor == s[N]) { 
             power++; 
             continue; 
         } 
  
         int sum = 0; 
            
         for(int i=0; i<=power; i++) 
             sum += Math.pow(currFactor,i); 
            
         ans *= sum; 
            
            
         // Update current prime factor as s[N] and 
         // initializing power of factor as 1. 
         currFactor = s[N]; 
         power = 1; 
     } 
  
     return ans; 
    } 
  
    //Driver code 
    public static void main(String[] args) { 
          
        int n = 12; 
  
         System.out.print("Sum of the factors is : "); 
  
         System.out.print(findSum(n)); 
    } 
}  

Python 3

# Python 3 Program to find  
# sum of all factors of a 
# given number  
  
# Using SieveOfEratosthenes  
# to find smallest prime  
# factor of all the numbers.  
# For example, if N is 10,  
# s[2] = s[4] = s[6] = s[10] = 2  
# s[3] = s[9] = 3  
# s[5] = 5  
# s[7] = 7  
def sieveOfEratosthenes(N, s) : 
  
    # Create a boolean list "prime[0..n]"  
    # and initialize all entries in it  
    # as false. 
    prime = [False] * (N + 1) 
  
    # Initializing smallest  
    # factor equal to 2 for  
    # all the even numbers  
    for i in range(2, N + 1, 2) : 
        s[i] = 2
  
    # For odd numbers less 
    # then equal to n  
    for i in range(3, N + 1, 2) : 
  
        if prime[i] == False : 
  
            # s[i] for a prime is 
            # the number itself  
            s[i] = i 
  
            # For all multiples of  
            # current prime number  
            for j in range(i, (N + 1) // i, 2) : 
                  
                if prime[i * j] == False : 
                    prime[i * j] = True
                      
                    # i is the smallest 
                    # prime factor for  
                    # number "i*j".  
                    s[i * j] = i 
  
                #J += 2 
  
# Function to find sum 
# of all prime factors 
def findSum(N) : 
  
    # Declaring list to store  
    # smallest prime factor of 
    # i at i-th index  
    s = [0] * (N + 1) 
    ans = 1
  
    # Filling values in s[] using  
    # sieve function calling 
    sieveOfEratosthenes(N, s) 
  
    # Current prime factor of N 
    currFactor = s[N] 
  
    # Power of current prime factor 
    power = 1
  
    while N > 1 : 
        N //= s[N] 
  
        # N is now N//s[N]. If new N  
        # also has smallest prime  
        # factor as currFactor,  
        # increment power 
        if currFactor == s[N] : 
            power += 1
            continue
  
        sum = 0
  
        for i in range(power + 1) : 
            sum += pow(currFactor, i) 
  
        ans *= sum
  
        # Update current prime factor  
        # as s[N] and initializing  
        # power of factor as 1.  
        currFactor = s[N] 
        power = 1
          
    return ans  
  
# Driver Code 
if __name__ == "__main__" : 
  
    n = 12
    print("Sum of the factors is :", end = " ") 
    print(findSum(n)) 
      
# This code is contributed by ANKITRAI1 

C#

// C# Program to find sum of all  
// factors of a given number 
using System; 
  
class GFG  
{ 
  
// Using SieveOfEratosthenes to find smallest  
// prime factor of all the numbers. 
// For example, if N is 10, 
// s[2] = s[4] = s[6] = s[10] = 2 
// s[3] = s[9] = 3 
// s[5] = 5 
// s[7] = 7 
static void sieveOfEratosthenes(int N, int []s) 
{ 
      
// Create a boolean array "prime[0..n]" and 
// initialize all entries in it as false. 
bool[] prime = new bool[N + 1]; 
  
for(int i = 0; i < N + 1; i++) 
    prime[i] = false; 
  
// Initializing smallest factor equal  
// to 2 for all the even numbers 
for (int i = 2; i <= N; i += 2) 
    s[i] = 2; 
  
// For odd numbers less then equal to n 
for (int i = 3; i <= N; i += 2) 
{ 
    if (prime[i] == false)  
    { 
  
        // s(i) for a prime is the 
        // number itself 
        s[i] = i; 
  
        // For all multiples of current 
        // prime number 
        for (int j = i; j * i <= N; j += 2) 
        { 
            if (prime[i * j] == false)  
            { 
                prime[i * j] = true; 
  
                // i is the smallest prime factor  
                // for number "i*j". 
                s[i * j] = i; 
            } 
        } 
    } 
} 
} 
  
// Function to find sum of all  
// prime factors 
static int findSum(int N) 
{ 
// Declaring array to store smallest  
// prime factor of i at i-th index 
int[] s = new int[N + 1]; 
  
int ans = 1; 
  
// Filling values in s[] using sieve 
sieveOfEratosthenes(N, s); 
  
int currFactor = s[N]; // Current prime factor of N 
int power = 1; // Power of current prime factor 
  
while (N > 1)  
{ 
    N /= s[N]; 
  
    // N is now N/s[N]. If new N als has smallest 
    // prime factor as currFactor, increment power 
    if (currFactor == s[N])  
    { 
        power++; 
        continue; 
    } 
  
    int sum = 0; 
      
    for(int i = 0; i <= power; i++) 
        sum += (int)Math.Pow(currFactor, i); 
      
    ans *= sum; 
      
    // Update current prime factor as s[N]  
    // and initializing power of factor as 1. 
    currFactor = s[N]; 
    power = 1; 
} 
  
return ans; 
} 
  
// Driver code 
public static void Main()  
{ 
    int n = 12; 
  
    Console.Write("Sum of the factors is : "); 
  
    Console.WriteLine(findSum(n)); 
} 
} 
  
// This code is contributed by Shashank 

PHP

<?php 
// PHP Program to find sum of all  
// factors of a given number 
  
// Using SieveOfEratosthenes to find smallest prime 
// factor of all the numbers. 
// For example, if N is 10, 
// s[2] = s[4] = s[6] = s[10] = 2 
// s[3] = s[9] = 3 
// s[5] = 5 
// s[7] = 7 
function sieveOfEratosthenes($N, &$s) 
{ 
    // Create a boolean array "prime[0..n]" and 
    // initialize all entries in it as false. 
    $prime=array_fill(0,$N + 1, false); 
  
    // Initializing smallest factor equal to 2 
    // for all the even numbers 
    for ($i = 2; $i <= $N; $i += 2) 
        $s[$i] = 2; 
  
    // For odd numbers less then equal to n 
    for ($i = 3; $i <= $N; $i += 2) { 
        if ($prime[$i] == false) { 
  
            // s(i) for a prime is the number itself 
            $s[$i] = $i; 
  
            // For all multiples of current prime number 
            for ($j = $i; $j * $i <= $N; $j += 2) { 
                if ($prime[$i * $j] == false) { 
                    $prime[$i * $j] = true; 
  
                    // i is the smallest prime factor for 
                    // number "i*j". 
                    $s[$i * $j] = $i; 
                } 
            } 
        } 
    } 
} 
  
// Function to find sum of all prime factors 
function findSum($N) 
{ 
    // Declaring array to store smallest prime 
    // factor of i at i-th index 
    $s=array_fill(0,$N + 1,0); 
  
    $ans = 1; 
  
    // Filling values in s[] using sieve 
    sieveOfEratosthenes($N, $s); 
      
    $currFactor = $s[$N]; // Current prime factor of N 
    $power = 1; // Power of current prime factor 
  
    while ($N > 1) { 
        $N /= $s[$N]; 
  
        // N is now N/s[N]. If new N als has smallest 
        // prime factor as currFactor, increment power 
        if ($currFactor == $s[$N]) { 
            $power++; 
            continue; 
        } 
        $sum = 0; 
          
        for($i=0; $i<=$power; $i++) 
            $sum += (int)pow($currFactor,$i); 
          
        $ans *= $sum; 
          
          
        // Update current prime factor as s[N] and 
        // initializing power of factor as 1. 
        $currFactor = $s[$N]; 
        $power = 1; 
    } 
  
    return $ans; 
} 
  
// Driver code 
  
    $n = 12; 
  
    echo "Sum of the factors is : "; 
  
    echo findSum($n); 
      
// This code is contributed by mits 
?> 

Output:

Sum of the factors is : 28

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

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