Question

why doesn´t ((num / i) % 1 == 0) work in C++ when num is a double? and how would I instead write this code, that checks for factorials by checking if it leaves a remainder (etc 0.3333).

int getFactorials (double num)
{
    int total = 0;      // if (total / 2) is equal too 'num' it is a perfect number.

    for (int i = 1; i < num; i++)
    {
        if ((num / i) % 1 == 0)
        {
            cout << i << endl;
        }
    }
    return 0;
}

Answer 1

Actually what you want to do is check if n is divisible by i so all you have to change is

if ((num / i) % 1 == 0)

into

if (num % i == 0) 

You should know that this is error-prone because you are using double as a type for num. You should use an int instead.