Sum of Factors of a Number using Prime Factorization (original) (raw)

Last Updated : 20 Mar, 2023

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++ `

// 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 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

//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

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 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

JavaScript

`

Output:

Sum of the factors is : 28

Time Complexity: O(N)

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