Sum of Factors of a Number using Prime Factorization
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++
#include <bits/stdc++.h>
using namespace std;
void sieveOfEratosthenes( int N, int s[])
{
vector< bool > prime(N + 1, false );
for ( int i = 2; i <= N; i += 2)
s[i] = 2;
for ( int i = 3; i <= N; i += 2) {
if (prime[i] == false ) {
s[i] = i;
for ( int j = i; j * i <= N; j += 2) {
if (prime[i * j] == false ) {
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
int findSum( int N)
{
int s[N + 1];
int ans = 1;
sieveOfEratosthenes(N, s);
int currFactor = s[N];
int power = 1;
while (N > 1) {
N /= s[N];
if (currFactor == s[N]) {
power++;
continue ;
}
int sum = 0;
for ( int i=0; i<=power; i++)
sum += pow (currFactor,i);
ans *= sum;
currFactor = s[N];
power = 1;
}
return ans;
}
int main()
{
int n = 12;
cout << "Sum of the factors is : " ;
cout << findSum(n);
return 0;
}
|
Java
public class GFG {
static void sieveOfEratosthenes( int N, int s[])
{
boolean [] prime = new boolean [N + 1 ];
for ( int i = 0 ; i < N+ 1 ; i++)
prime[i] = false ;
for ( int i = 2 ; i <= N; i += 2 )
s[i] = 2 ;
for ( int i = 3 ; i <= N; i += 2 ) {
if (prime[i] == false ) {
s[i] = i;
for ( int j = i; j * i <= N; j += 2 ) {
if (prime[i * j] == false ) {
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
static int findSum( int N)
{
int [] s = new int [N + 1 ];
int ans = 1 ;
sieveOfEratosthenes(N, s);
int currFactor = s[N];
int power = 1 ;
while (N > 1 ) {
N /= s[N];
if (currFactor == s[N]) {
power++;
continue ;
}
int sum = 0 ;
for ( int i= 0 ; i<=power; i++)
sum += Math.pow(currFactor,i);
ans *= sum;
currFactor = s[N];
power = 1 ;
}
return ans;
}
public static void main(String[] args) {
int n = 12 ;
System.out.print( "Sum of the factors is : " );
System.out.print(findSum(n));
}
}
|
Python 3
def sieveOfEratosthenes(N, s) :
prime = [ False ] * (N + 1 )
for i in range ( 2 , N + 1 , 2 ) :
s[i] = 2
for i in range ( 3 , N + 1 , 2 ) :
if prime[i] = = False :
s[i] = i
for j in range (i, (N + 1 ) / / i, 2 ) :
if prime[i * j] = = False :
prime[i * j] = True
s[i * j] = i
def findSum(N) :
s = [ 0 ] * (N + 1 )
ans = 1
sieveOfEratosthenes(N, s)
currFactor = s[N]
power = 1
while N > 1 :
N / / = s[N]
if currFactor = = s[N] :
power + = 1
continue
sum = 0
for i in range (power + 1 ) :
sum + = pow (currFactor, i)
ans * = sum
currFactor = s[N]
power = 1
return ans
if __name__ = = "__main__" :
n = 12
print ( "Sum of the factors is :" , end = " " )
print (findSum(n))
|
C#
using System;
class GFG
{
static void sieveOfEratosthenes( int N, int []s)
{
bool[] prime = new bool[N + 1 ];
for ( int i = 0 ; i < N + 1 ; i++)
prime[i] = false ;
for ( int i = 2 ; i <= N; i += 2 )
s[i] = 2 ;
for ( int i = 3 ; i <= N; i += 2 )
{
if (prime[i] == false )
{
s[i] = i;
for ( int j = i; j * i <= N; j += 2 )
{
if (prime[i * j] == false )
{
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
static int findSum( int N)
{
int [] s = new int [N + 1 ];
int ans = 1 ;
sieveOfEratosthenes(N, s);
int currFactor = s[N];
int power = 1 ;
while (N > 1 )
{
N /= s[N];
if (currFactor == s[N])
{
power++;
continue ;
}
int sum = 0 ;
for ( int i = 0 ; i <= power; i++)
sum += ( int )Math.Pow(currFactor, i);
ans *= sum;
currFactor = s[N];
power = 1 ;
}
return ans;
}
public static void Main()
{
int n = 12 ;
Console.Write( "Sum of the factors is : " );
Console.WriteLine(findSum(n));
}
}
|
PHP
<?php
function sieveOfEratosthenes( $N , & $s )
{
$prime = array_fill (0, $N + 1, false);
for ( $i = 2; $i <= $N ; $i += 2)
$s [ $i ] = 2;
for ( $i = 3; $i <= $N ; $i += 2) {
if ( $prime [ $i ] == false) {
$s [ $i ] = $i ;
for ( $j = $i ; $j * $i <= $N ; $j += 2) {
if ( $prime [ $i * $j ] == false) {
$prime [ $i * $j ] = true;
$s [ $i * $j ] = $i ;
}
}
}
}
}
function findSum( $N )
{
$s = array_fill (0, $N + 1,0);
$ans = 1;
sieveOfEratosthenes( $N , $s );
$currFactor = $s [ $N ];
$power = 1;
while ( $N > 1) {
$N /= $s [ $N ];
if ( $currFactor == $s [ $N ]) {
$power ++;
continue ;
}
$sum = 0;
for ( $i =0; $i <= $power ; $i ++)
$sum += (int)pow( $currFactor , $i );
$ans *= $sum ;
$currFactor = $s [ $N ];
$power = 1;
}
return $ans ;
}
$n = 12;
echo "Sum of the factors is : " ;
echo findSum( $n );
?>
|
Javascript
<script>
function sieveOfEratosthenes(N,s)
{
let prime = new Array(N + 1);
for (let i = 0; i < N+1; i++)
prime[i] = false ;
for (let i = 2; i <= N; i += 2)
s[i] = 2;
for (let i = 3; i <= N; i += 2) {
if (prime[i] == false ) {
s[i] = i;
for (let j = i; j * i <= N; j += 2) {
if (prime[i * j] == false ) {
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
function findSum(N)
{
let s = new Array(N + 1);
let ans = 1;
sieveOfEratosthenes(N, s);
let currFactor = s[N];
let power = 1;
while (N > 1) {
N = Math.floor(N/s[N]);
if (currFactor == s[N]) {
power++;
continue ;
}
let sum = 0;
for (let i=0; i<=power; i++)
sum += Math.pow(currFactor,i);
ans *= sum;
currFactor = s[N];
power = 1;
}
return ans;
}
let n = 12;
document.write( "Sum of the factors is : " );
document.write(findSum(n));
</script>
|
Output:
Sum of the factors is : 28
Time Complexity: O(N)
Auxiliary Space: O(n) // an extra n sized array has been created.
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