1. If the seven-digit integer $2020 a b c$ is divisible by 45, then the number of all possible values of $a+b+c$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
-1. C.
Since $45=9 \times 5$, the seven-digit number can be divisible by 5 and 9. Therefore, $c=0$ or 5.
Notice that,
$$
2+0+2+0+a+b+c=a+b+c+4
$$

is divisible by 9.
When $c=0$,
9| $(a+b+4) \Rightarrow a+b=5$ or 14
$\Rightarrow a+b+c=5$ or 14.
When $c=5$,
9| $(a+b+9) \Rightarrow a+b=0,9$ or 18
$\Rightarrow a+b+c=5,14$ or 23.
In summary, all possible values of $a+b+c$ are 5, 14, or 23, i.e., 3.