If the integer is a perfect square, what is the sum of the digits of its square root?
x^2 = 152AB1
sqrt(152001) = 389.873056263
So the number x is slightly greater than 389.873056263.
Since 152001 ends with a 1, the number x has to end with a 1 or 9.
391^2 = 152881
=^._.^=
