16. Find the number of integers between 1 and 2013 with the property that the sum of its digits equals 9 .
16. Answer: 101.
Case 1: $n<1000$. Write $n=\overline{a b c}$. Then
$$
a+b+c=9, \quad a, b, c \in\{0,1, \ldots, 9\} .
$$

Case 2: $1000 \leq n<2000$. Write $n=\overline{1 a b c}$. Then
$$
a+b+c=8, \quad a, b, c \in\{0,1, \ldots, 8\} .
$$

Case 3: $2000 \leq n \leq 2013$. Then $n=2007$.
Therefore, there are $\binom{9+3-1}{9}+\binom{8+3-1}{8}+1=55+45+1=101$ such numbers