SOLUTION: can there be a perfect square whose digits consist of exactly 4 ones, 4 twos and 4 zeros in any order?

Let N =  be such a number.


Since the sum of digits of the number  " N "   4*1 + 4*2 +4*0 = 4 + 8 + 0 = 12  is divisible by 3,

it implies that the number N itself is divisible by 3  (the "divisibility by 3 rule").



In turn, it implies that the number " n "  itself  is divisible by 3.



Then the number    is divisible by 3^2 = 9;  hence, the number N  is divisible by 9.



But the sum of the digits of the number N, which was calculated above as 12, is not divisible by 9.


It contradicts to the "divisibility by 9 rule". 


Hence, such a number N with assigned properties DOES NOT EXIST.