Final answer:
Only one positive integer less than 100 has the property that the square of the sum of its digits is greater than the number itself, and that integer is 81.
Step-by-step explanation:
The question asks for the number of positive integers less than 100 for which the square of the sum of its digits is greater than the number itself. To find this, we recognize that any number less than 100 can be written as 10a+b, where a is the tens digit and b is the ones digit. The sum of the digits squared would be (a+b)^2.
We need to test each number where 0 <= a <= 9 and 0 <= b <= 9. However, if we consider the upper limit for the sum of digits (9+9) we get 18, and its square is 324 which is greater than the maximum number we are testing (99). This means we only need to consider the cases where the squares of the lower sums are greater than the number represented as 10a+b.
When testing, we find that only one number, 81, satisfies this condition. This is because its digits sum to 9 (8+1), and 9 squared is 81, which is equal to the number itself. Thus, the answer is b) 1.