Problem Solving

What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8

20!=2432902008176640000.
So the rightmost non-zero digit is 4.
OA is..?

err... I got 4.

I first eliminated all sets of 2*5 or 10 in the list of numbers between 1 and 20. So 2,5,10,15,20 mean that there are 4 zeroes in 20! (2*5,10, and 20*15=300 provides two zeroes and leaves a factor of 3). By removing these numbers I effectively divide 20! by 10000 and I'm left with a number with no zeroes.

So I was left with factors 1,3,4,6,7,8,9,11,12,13,14,16,17,18,19 and another 3. I started multiplying the units digits:
1*3=3
3*4=12, I take the units digit 2
2*6=12 (2)
2*7=14 (4)
4*8=32 (2)
2*9=18 (8)
8*11=88 (8)
8*12=96 (6)
6*13=78 (8)
8*14=112 (2)
2*16=32 (2)
2*17=34 (4)
4*18=72 (2)
2*19=38 (8)
8*3=24 ...so the units digit should be 4.
(I realized later that for 11-19 I could just multiply by units digit, i.e., 1-9).

The units digits of 20!/10000 should be the rightmost non-zero digit of 20!.

Problem is - there is no such answer choice!
So I calculated 20! and got 2432902008176640000.
Did I go wrong? Or is it the question?

I do the same way like you (but I eliminate 1,11,10,20). Is somebody could solve this problem with usage the root? I couldn't