1. Does there exist a natural number for which the sum of the digits of its square is:
a) 80;
b) 81?
Solution. a) It is easily proven that $a \equiv S(a)(\bmod 3)$ for every natural number $a$ (where $S(a)$ denotes the sum of the digits of the number $a$). Since the square of a natural number can only give a remainder of 0 or 1 modulo 3, we get that there does not exist a perfect square with a digit sum of 80.

## b) It exists. For example, 111111111.