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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.

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

#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); }

Enter a number: 1092 Sum of all odd prime factors: 23

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