Checking if a number is prime
Unfortunately, your isPrime function returns incorrect results for the squares of odd numbers.

This is a pretty common oversight though, so don't feel too bad.

A better implementation of isPrime looks more like this:

bool isPrime(int num) {
    if (num &lt;= 3) {
        return num &gt; 1;
    } else if (num % 2 == 0 || num % 3 == 0) {
        return false;
    } else {
        for (int i = 5; i * i &lt;= num; i += 6) {
            if (num % i == 0 || num % (i + 2) == 0) {
                return false;
            }
        }
        return true;
    }
}


In the first if, we handle the special cases of 0 through 3 as well as all of the negative numbers.

In the second, we eliminate all of the multiples of 2 and 3.

Finally, in the catch-all else, we're handling everything else.

By starting at 5 and incrementing by 6, we're able to skip all of our multiples of 2 and 3 which we already eliminated.  So we're checking 2/3rds of the numbers your original implementation checks.

Moreover, because we're dealing with integers, i * i &lt;= num is a bit better than i &lt; sqrt(num) (which actually needs to be &lt;=).