How many five-digit integers exist with the following conditions,
$1.$ be divisible by $125$
$2.$ Sum of their digits be divisible by $4$.
$1) 176 \qquad\qquad 2)178 \qquad\qquad 3)180 \qquad\qquad 4)198\qquad\qquad 5)200$
To solve this problem I noted that the last three digits of these numbers should be either one of the following, $000, 125, 250, 375, 500, 625, 750, 875$. Now if I divide the sum of each of these three digits by $4$, the remainder will be, $0, 0, 3,3,1,1,0,0$ respectively. Now for four cases, sum of the first two digits should be divisible by four, and for two cases it should be in the form $4k+1$ and finally for the last two cases it should be $4k'-1$.
Now after counting all these cases, which was relatively time-consuming, I got $178$ as the answer.
I'm wondering if it is possible to solve this problem with alternative approaches. Since sum of the digits of a number is related to the remainder of dividing a number by $9$ I think this idea would be helpful, but not sure how to implement it.