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,
(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< 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 ?> |
Javascript
<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 als 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> |
Sum of the factors is : 28