4. For a three-digit number $\overline{a b c}$, the number of three-digit numbers that satisfy $\overline{a b c}=37(a+b+c)$ is $\qquad$
A. 13
B. 14
C. 15
D. 16
Solve:
$$
100 a+10 b+c=37(a+b+c) \Rightarrow 7 a=3 b+4 c \text {, }
$$

Thus, when $a=1$, $(b, c)=(1,1)$; when $a=2$, $(b, c)=(2,2)$; when $a=3$, $(b, c)=(3,3),(7,0)$; when $a=4$, $(b, c)=(0,7),(4,4),(8,1)$; when $a=5$, $(b, c)=(1,8),(5,5),(9,2)$; when $a=6$, $(b, c)=(2,9),(6,6)$; when $a=7$, $(b, c)=(7,7)$; when $a=8$, $(b, c)=(8,8)$; when $a=9$, $(b, c)=(9,9)$. There are 15 solutions in total, so the answer is $C$.