30.9. Find the largest three-digit number such that the number minus the sum of its digits is a perfect square.
30.9. 919
Let $a b c$ be a three-digit number such that difference between the number and the sum of its digits is a perfect square; that is,
$$
(100 a+10 b+c)-(a+b+c)=99 a+9 b=9(11 a+b)
$$
is a perfect square. To maximize the number $100 a+10 b+c$, we set $a=9, b=1$, and $c=9$.