Find sum of odd factors of a number
Last Updated :
14 Oct, 2023
Given a number n, the task is to find the odd factor sum.
Examples :
Input : n = 30
Output : 24
Odd dividers sum 1 + 3 + 5 + 15 = 24
Input : 18
Output : 13
Odd dividers sum 1 + 3 + 9 = 13
Approach :- Naive approach in o(n) time complexity
The first approach involves iterating over all factors of the given number and checking if each factor is odd. If a factor is odd, it is added to the running sum of odd factors.
Begin by including the necessary header files, including iostream, which will allow us to perform input and output operations.
Define a function named sumOfOddFactors that takes an integer n as input and returns an integer as output.
Initialize an integer variable sum to zero, which will be used to accumulate the sum of odd factors of the input integer.
Use a for loop to iterate through all integers from 1 to n, incrementing by 1 at each iteration.
Check if the current integer i is a factor of n by using the modulo operator (%). If the remainder of n divided by i is zero, then i is a factor of n.
Check if the current factor i is odd by using the modulo operator (%). If the remainder of i divided by 2 is 1, then i is odd.
If i is both a factor of n and odd, then add i to the sum variable.
After the loop has completed, return the final value of sum.
In the main function, initialize an integer variable n to the value 30.
Call the sumOfOddFactors function with n as input and output the result to the console.
Return 0 to indicate successful program execution.
C++
#include <iostream>
using namespace std;
int sumOfOddFactors( int n) {
int sum = 0;
for ( int i = 1; i <= n; i++) {
if (n % i == 0 && i % 2 == 1) {
sum += i;
}
}
return sum;
}
int main() {
int n = 30;
cout << sumOfOddFactors(n) << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static int sumOfOddFactors( int n) {
int sum = 0 ;
for ( int i = 1 ; i <= n; i++) {
if (n % i == 0 && i % 2 == 1 ) {
sum += i;
}
}
return sum;
}
public static void main(String[] args) {
int n = 30 ;
System.out.println(sumOfOddFactors(n));
}
}
|
Python3
def sumOfOddFactors(n):
sum = 0
for i in range ( 1 , n + 1 ):
if n % i = = 0 and i % 2 = = 1 :
sum + = i
return sum
if __name__ = = '__main__' :
n = 30
print (sumOfOddFactors(n))
|
C#
using System;
class Program
{
static int SumOfOddFactors( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++)
{
if (n % i == 0 && i % 2 == 1)
{
sum += i;
}
}
return sum;
}
static void Main()
{
int n = 30;
Console.WriteLine(SumOfOddFactors(n));
}
}
|
Javascript
function sumOfOddFactors(n) {
let sum = 0;
for (let i = 1; i <= n; i++) {
if (n % i === 0 && i % 2 === 1) {
sum += i;
}
}
return sum;
}
const n = 30;
console.log(sumOfOddFactors(n));
|
Time complexity – O(N)
Auxiliary Space – O(1)
Prerequisite : Sum of all the factors of a number
As discussed in above mentioned previous post, sum of factors of a number is
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).
Sum of divisors = (1 + p1 + p12 ... p1a1) *
(1 + p2 + p22 ... p2a2) *
.............................................
(1 + pk + pk2 ... pkak)
To find sum of odd factors, we simply need to ignore even factors and their powers. For example, consider n = 18. It can be written as 2132 and sum of all factors is (1)*(1 + 2)*(1 + 3 + 32). Sum of odd factors (1)*(1+3+32) = 13.
To remove all even factors, we repeatedly divide n while it is divisible by 2. After this step, we only get odd factors. Note that 2 is the only even prime.
C++
#include <bits/stdc++.h>
using namespace std;
int sumofoddFactors( int n)
{
int res = 1;
while (n % 2 == 0)
n = n / 2;
for ( int i = 3; i <= sqrt (n); i++)
{
int count = 0, curr_sum = 1;
int curr_term = 1;
while (n % i == 0) {
count++;
n = n / i;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
int main()
{
int n = 30;
cout << sumofoddFactors(n);
return 0;
}
|
Java
import java.io.*;
import java.math.*;
class GFG {
static int sumofoddFactors( int n)
{
int res = 1 ;
while (n % 2 == 0 )
n = n / 2 ;
for ( int i = 3 ; i <= Math.sqrt(n); i++)
{
int count = 0 , curr_sum = 1 ;
int curr_term = 1 ;
while (n % i == 0 )
{
count++;
n = n / i;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2 )
res *= ( 1 + n);
return res;
}
public static void main(String args[])
throws IOException
{
int n = 30 ;
System.out.println(sumofoddFactors(n));
}
}
|
Python3
import math
def sumofoddFactors( n ):
res = 1
while n % 2 = = 0 :
n = n / / 2
for i in range ( 3 , int (math.sqrt(n) + 1 )):
count = 0
curr_sum = 1
curr_term = 1
while n % i = = 0 :
count + = 1
n = n / / i
curr_term * = i
curr_sum + = curr_term
res * = curr_sum
if n > = 2 :
res * = ( 1 + n)
return res
n = 30
print (sumofoddFactors(n))
|
C#
using System;
class GFG {
static int sumofoddFactors( int n)
{
int res = 1;
while (n % 2 == 0)
n = n / 2;
for ( int i = 3; i <= Math.Sqrt(n); i++)
{
int count = 0, curr_sum = 1;
int curr_term = 1;
while (n % i == 0)
{
count++;
n = n / i;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
public static void Main(String[] argc)
{
int n = 30;
Console.Write(sumofoddFactors(n));
}
}
|
Javascript
<script>
function sumofoddFactors(n)
{
let res = 1;
while (n % 2 == 0)
n = n / 2;
for (let i = 3; i <= Math.sqrt(n); i++)
{
let count = 0;
let curr_sum = 1;
let curr_term = 1;
while (n % i == 0)
{
count++;
n = n / i;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
let n = 30;
document.write(sumofoddFactors(n));
</script>
|
PHP
<?php
function sumofoddFactors( $n )
{
$res = 1;
while ( $n % 2 == 0)
$n = $n / 2;
for ( $i = 3; $i <= sqrt( $n ); $i ++)
{
$count = 0; $curr_sum = 1;
$curr_term = 1;
while ( $n % $i == 0)
{
$count ++;
$n = $n / $i ;
$curr_term *= $i ;
$curr_sum += $curr_term ;
}
$res *= $curr_sum ;
}
if ( $n >= 2)
$res *= (1 + $n );
return $res ;
}
$n = 30;
echo sumofoddFactors( $n );
?>
|
Time complexity: O(sqrt(n))
Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...