611. Can a natural number be a perfect square if the sum of its digits is: a) $20 ;$ b) $10 ?$
$\triangle$ a) Let's divide the number 20 by 3 with a remainder: $20=3 \cdot 6+2$. The remainder is 2. Then the number itself, when divided by 3, also gives a remainder of 2 (see the statement of support problem 120 from § 5). However, the square of a natural number, when divided by 3, cannot give a remainder of 2 (see the statement of problem 163 from § 6). Therefore, a natural number with a sum of digits equal to 20 cannot be a perfect square.

b) The sum of the digits of a perfect square can be equal to 10. For example, the number 64. Answer: a) cannot; b) can.