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