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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

Approach: 

Suppose N = 1100, the idea is to first find the prime factorization of the given number N. 
Therefore, the prime factorization of 1100 = 22 * 52 * 11.
So, the formula to calculate the sum of all factors can be given as, 



A dry run is as shown below as follows: 

(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, the sum of all factors of 1100 = 2604 

Below is the implementation of the above approach: 




// 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 also 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 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 also 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 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# 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 also 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 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 also 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


?>


















                        






                        























                                    



                                    




                                    




                                    


                                



















<script>


//Javascript 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.




         let prime = new Array(N + 1);




        



         for(let i = 0; i < N+1; i++)




            prime[i] = false;




   



         // Initializing smallest factor equal to 2




         // for all the even numbers




         for (let i = 2; i <= N; i += 2)




            s[i] = 2;




   



         // For odd numbers less then equal to n




         for (let 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 (let 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




         let s = new Array(N + 1);




   



         let ans = 1;




   



         // Filling values in s[] using sieve




         sieveOfEratosthenes(N, s);




         



         let currFactor = s[N]; // Current prime factor of N




         let power = 1; // Power of current prime factor




   



         while (N > 1) {




            N = Math.floor(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++;




                 continue;




             }




   



             let sum = 0;




             



             for(let 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




    let n = 12;




    document.write("Sum of the factors is : ");




    document.write(findSum(n));




         


 


 


// This code is contributed by rag2127


</script>


















                        






                        








Output: 

Sum of the factors is : 28


 




Time Complexity: O(N)


Auxiliary Space: O(n) // an extra n sized array has been created.



	

        Article Tags : 
        

            DSA
Mathematical        


		
	

	


      

      

          
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