Counting numbers having three different simple factors

Question

There are 1000 numbers in number.txt, from 2 to 9 digits, each on a separate line. The exercise is to count the number of numbers that fill the condition: when factorizing, this number has exactly 3 different simple factors, they can occur several times, and all of them are even numbers.

for example   105 - factors: 3, 5, 7 - YES, 1287 - factors: 3, 3, 11, 13 - YES,

1157625 - factors: 3,3,3,5,5,5,7,7,7 - YES,

55 - factors: 5, 11 - NO.

#include <iostream>
#include <fstream>
using namespace std;
int number, threefnumbers=0;
int main()
{
  ifstream file("numbers.txt");
  ofstream outputf("results.txt");
  int count_factors;
  while (file >> number)
  {
    count_factors=0;
    int factor=3;
    if (number%2!=0)
    {
       while (number>1)
       {
        if (number%factor==0)
          count_factors++;
        while (number%factor==0)
        {
          number=number/factor;
        }
        factor+=2;
       }
       if (count_factors==3) threefnumbers++;
     }
  }

  outputf << "59.1) " << endl << threefnumbers;

  file.close();
  outputf.close();
  return 0;
}

I know from number.txt that there are many numbers that fulfill the condition, but the program returns only 1. Why is this?

Answer 1

, 2 - . , , , 2. -, :

while(read number)
{
    int factor_count = 0;
    // check 2 by itself
    if (number % 2 == 0)
    {
        factor_count++;
        while(number % 2 == 0)
            number /= 2;
    }
    for (int factor = 3; factor < number; factor += 2)
    {
        if (number % factor == 0)
        {
            factor_count++;
            while(number % factor == 0)
                number /= factor;
        }
    }
    if(factor_count == 3)
        do something
}

, , , , 999,999,999. , .

std::vector<int> primes = get_prime_list(999999999);  
// returns a list of all prime numbers less than the number passed in.
// leaving it to you to implement but a Sieve of Eratosthenes should work well
while(read number)
{
    int factor_count = 0;
    for(auto e : primes)
    {
        if (number % e == 0)
        {
            factor_count++;
            while(number % e == 0)
                number /= e;
        }
        if (number == 1) // number is fully factorized
            break;
    }
    if(factor_count == 3)
        do something
}