Let's focus on the digits....

As previously stated:
Beginning digit is (d+n)
Recurring digit is the number following the beginning digit.
5*n = d0,d5,(d+1)0,(d+1)5,(d+2)0,(d+2)5.....(d+n)0,d(d+n)5
5*n = 0,5,(1)0,(1)5,(2)0,(2)5........

2*n = (d)2,(d)4,(d)6,(d)8,(d)0,(d+1)2,(d+1)4,(d+1)6,(d+1)8,(d+1)0...........(d+n)2,(d+n)4,(d+n)6,(d+n)8,(d+n)0
2*n = 2,4,6,8,10,12,14,16,18

They intersect and are therefore both divisible by (d+n)0. This leaves one other pattern for 5*n and that's (d+n)5.

1000/5 = 200
The above quotient is actually saying "There are 200 numbers which follow the pattern (d+n)0 or (d+n)5." Since (d+n)0 is a pattern shared by 2*n and 5*n you can discard all numbers of this form. Which mines you must divide that quotient 200 by 2. 200/2 = 100.

Now our set is restricted to 100 numbers of the form (d+n)5: 5,15,25,35.......965,975,985,995.
You may have noticed you add the first number in our set to the last number in our set and get a 1000. This is consistent for all additions of the form set[d] + set[d+n] if
(D+ (D+n))=99.
set[0] = 5
set[1] = 15
......
set[98] = 985
set[99] = 995

Of that 100 number set you can have 50 pairs of (D+(D+n)) = 99.
50 pairs all equaling 1000 = 50000
😎 #NJIT