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In this section, we will see how we can get the sum of all even prime factors of a number in an efficient way. There is a number say n = 480, we have to get all factors of this. The prime factors of 480 are 2, 2, 2, 2, 2, 3, 5. The sum of all even factors is 2+2+2+2+2 = 10. To solve this problem, we have to follow this rule −

- When the number is divisible by 2, add them into the sum, and divide the number by 2 repeatedly.
- Now the number must be odd. So we will not find any factors which are even. Then simply ignore those factors.

Let us see the algorithm to get a better idea.

printPrimeFactors(n): begin sum := 0 while n is divisible by 2, do sum := sum + 2 n := n / 2 done end

#include<iostream> using namespace std; int sumEvenFactors(int n){ int i, sum = 0; while(n % 2 == 0){ sum += 2; n = n/2; //reduce n by dividing this by 2 } return sum; } main() { int n; cout << "Enter a number: "; cin >> n; cout << "Sum of all even prime factors: "<< sumEvenFactors(n); }

Enter a number: 480 Sum of all even prime factors: 10

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