String manipulation

Yes, you need to convert the string to an int first:

try {
  int number = Integer.parseInt(theString);
  boolean isOdd = (number % 2) != 0;
  boolean isEven = (number % 2) == 0;
  boolean isZero = (number == 0);
} catch (NumberFormatException e) {
  System.out.println("Invalid number!");
}

A 30 digit number is very large. Not sure if it fits into integer or long.

You can do it this way if you want without converting the input into an integer or long.
1. Read in the number from the console. It will be a string.
2. Check the last digit of the number for whether it is even or odd. The last digit determines eveness or oddness.
3. Loop through the string from beginning to end, counting evens, odds, and zeros.

Sorry, I didn't notice about the 30 digits max. If it's that long, you should use a BigInteger object instead:

BigInteger number = new BigInteger("999999999999999999999999999999999999999999999999999");
BigInteger remainder = number.mod(new BigInteger("2"));
System.out.println("Odd: " + (remainder.intValue() != 0));
System.out.println("Even: " + (remainder.intValue() == 0));
System.out.println("isZero: " + (remainder.signum() == 0));

The last line should be:

System.out.println("Is zero: " + (number.signum() == 0));

If you wish to discover the properties of the individual digits, then it looks like you've got it pretty well taped. No, you don't need to convert it to a numerical value.

I appreciate everyone's help in this matter. I decided to leave it as a string then compare each digit in a for loop.
It is probably more coding than necessary but it worked.
Code below:

public void process(){
        String integerString = getIntegerString();
        int evenDigits = 0;
        int oddDigits = 0;
        int zeroDigits = 0;
        for (int i = 0; i < integerString.length(); i++){
            char c = integerString.charAt(i);
            if (c == '0') zeroDigits++;
            if (c == '0') evenDigits++;
            if (c == '1') oddDigits++;
            if (c == '2') evenDigits++;
            if (c == '3') oddDigits++;
            if (c == '4') evenDigits++;
            if (c == '5') oddDigits++;
            if (c == '6') evenDigits++;
            if (c == '7') oddDigits++;
            if (c == '8') evenDigits++;
            if (c == '9') oddDigits++;
        }
        System.out.println(integerString + " contains "
            + evenDigits + " even digits, "
            + oddDigits + " odd digits, and "
            + zeroDigits + " zero digits.");
    }

Yes, you could do:


for (int i = 0; i < integerString.length(); i++){
           char c = integerString.charAt(i);
           if ((c % 2) == 0) {
             evenDigits++;
           }
           else {
             oddDigits++
           }
           if (c == '0') zeroDigits++;
}