Miguel chose a three-digit number and another two-digit number. What is the sum of these numbers if their difference is 989?
(a) 1000
(b) 1001
(c) 1009
(d) 1010
(e) 2005
The correct option is (c).

Since the difference is 989 and the smaller number has two digits (thus being greater than 9), the three-digit number must be greater than $989+9=998$, so the only option is 999. Therefore, the two-digit number is 10 and the sum of the two is $999+10=1009$.