Yes, there were 2 missing. What about that:
28h60dfb6c611_23435d8fc2e2277c2a145037c95e1ce0
6 9 8a 6 e 7144 788 e 40c
e 0 30 8
b e
What is your guess? Do you have a tool to try the proposed subset of characters?
28b60dfb6a611223435d6fc2e2277c2a145087c95e4ce0
"h" does not exist in the hexadecimal system. It most resembles a b.
"b6c6" is actually "b6a6" (yes they are both 6's: 6 and 0 both have a narrower base and top than 3 and 8 ) because of the circular curve present at the top right which is not present in c and e letters. There's also a faint black curve at the top stretching from top right and making an arc at the top before terminating at the bottom left (c and e are connected at the top left and don't stop there).
The _ is either a 2 or 3 - there is a slight curved blackness at the top right and a slight patch of black at the bottom right which looks like the endpoint of a straight line. I went with 2 anyhow.
"23435d8" - is actually "d6" because there is a sharp vertical black edge at the bottom of the top-right corner similar to how 6's terminate. which
0 characters are narrower than 3's and 8's - which are the same width btw. 3's have a flat tail end at the bottom left but it doesn't curl up which is what an 8 would do, hence the "145037" sequence at the end is actually "145837".
Finally at the very end, the "1" is actually a 4 because there is a touch of black on the left, right next to the shaded pink.
It's hard to solve this case, because there are still so many characters to solve. and this is really hard for me. but later I will try to solve it. but the question for me and others reading this thread is are you not afraid if someone can solved it and they will steal your assets? are you not worried about that?
It's not hard at all. I live and breathe Helvetica when I'm not typing away in Monospace programming.
P.S. OP (original poster / topic starter) did not reveal the first 18 characters.