# Find sum of odd factors of a number using C++.

C++Server Side ProgrammingProgramming

In this section, we will see how we can get the sum of all odd prime factors of a number in an efficient way. There is a number say n = 1092, we have to get all factors of this. The prime factors of 1092 are 2, 2, 3, 7, 13. The sum of all odd factors is 3+7+13 = 23. To solve this problem, we have to follow this rule −

• When the number is divisible by 2, ignore that factor, and divide the number by 2 repeatedly.
• Now the number must be odd. Now starting from 3 to square root of the number, if the number is divisible by current value, then add the factor with the sum, and change the number by divide it with the current number then continue.
• Finally, the remaining number will also be added if the remaining number is odd

Let us see the algorithm to get a better idea.

## Algorithm

printPrimeFactors(n):
begin
sum := 0
while n is divisible by 2, do
n := n / 2
done
for i := 3 to , increase i by 2, do
while n is divisible by i, do
sum := sum + i
n := n / i
done
done
if n > 2, then
if n is odd, then
sum := sum + n
end if
end if
end

## Example

#include<iostream>
#include<cmath>
using namespace std;
int sumOddFactors(int n){
int i, sum = 0;
while(n % 2 == 0){
n = n/2; //reduce n by dividing this by 2
}
//as the number is not divisible by 2 anymore, all factors are odd
for(i = 3; i <= sqrt(n); i=i+2){ //i will increase by 2, to get only odd numbers
while(n % i == 0){
sum += i;
n = n/i;
}
}
if(n > 2){
if(n%2 == 1)
sum += n;
}
return sum;
}
main() {
int n;
cout << "Enter a number: ";
cin >> n;
cout <<"Sum of all odd prime factors: "<< sumOddFactors(n);
}

## Output

Enter a number: 1092
Sum of all odd prime factors: 23
Published on 30-Oct-2019 06:27:31