Can a 300-digit number whose digits are 100 0s, 100 1s and 100 2s (in some order) be a perfect square?

The question:

Can a 300-digit number whose digits are 100 0s, 100 1s and 100 2s (in some order) be a

I'm new to digital roots in divisibility and I am trying to figure this out.

Am I right in saying that the digital roots for this would be: $$100(0) + 100(1) + 100(2) = 300\\ 100(0+1+2) = 300 \to 3+0+0 = 3$$

And the digital root of a square number is always 1, 4, 7 or 9. (It is never 2, 3, 5, 6 or 8.) This means that this number can never be a perfect square regardless of the order of the digits.

Is this correct?

EDIT:

I found this in some of my notes, however this goes against what I can deduce, the reasoning seems correct, but it should be false and not true? • Can you define the term "digital root" Apr 5 '17 at 14:52
• @StellaBiderman (decimal) digit sum of digit sum of digit sum of ... Apr 5 '17 at 14:56
• @Stella: Also it is the origin of the number on your computer. In this case, the SE servers. :P Apr 5 '17 at 14:56
• Digital root is what you get if you sum the digits, then sum the digits of the answer, and so on, until you get to a single digit. If you do this operation in base $b$ the answer will tell you about $n$ mod $b-1$. Apr 5 '17 at 14:57
• @Monkleys It would be interesting to see your attempt on the proof of digital roots of squares. Apr 5 '17 at 15:07

Yes, that's correct. A simpler way of putting it in this particular case is that the number is divisible by $3$, but not by $9$, so can't be square.