There is 1 integer whose digits sum to $1: 10$. There are 4 integers whose digits sum to $4: 13,22,31$, and 40 . There are 9 integers whose digits sum to $9: 18,27,36,45,54,63,72,81$, and 90 . There are 3 integers whose digits sum to $16: 79,88$, and 97 . Two digits cannot sum to 25 or any greater square since the greatest sum of digits of a twodigit number is $9+9=18$ Thus, the answer is $1+4+9+3=$ (C)17. 
How many two-digit numbers have digits whose sum is a perfect square? (2006 AMC 8 Problem, Question \#11) 
Options:
A. $$13$$ 
B. $$16$$ 
C. $$17$$ 
D. $$18$$ 
E. $$19$$ 