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aname on 11:02 AM 12/12/2024: password is decrypted in first part, decrypt print message and get input.
thed0ublec on 3:02 PM 12/12/2024: ```python
print("omg")
```
IWDFObject_AcquireLock on 5:02 PM 12/12/2024: This was a really interesting CrackMe, and was pretty fun too. Setting BreakPoints with your debugger and looking at the values of registers (and the addresses they point to) along with a little bit of function analysis does the trick.
ByteHook on 6:57 PM 12/12/2024: pass: crackmeYG
Coti on 8:07 PM 12/12/2024: Not sure if that was the right way but I was just stepping and noticed the password RDX
alxxx on 2:26 AM 12/13/2024: good for beginners. sha256{3cc3a014b9bebc060d8df8b329750a7b054492ff835bf5fed80bea4cda94a071}
AboutSeflash on 5:08 PM 12/17/2024: OMG
SirWardrake on 12:36 AM 12/23/2024: 123 test test ;-)
re.login on 7:24 AM 12/24/2024: Thanks for challenges.
DeFexGG on 10:43 PM 12/24/2024: think difficulty harder but crackmeYG i found it in memory with cheat engine inspecting
thenotoriusflux on 1:21 AM 12/26/2024: Nice debugger challenge
yursing on 12:14 PM 12/26/2024: guys im new on this site . i would like to know, how could i get the password of zip files. pls help
Strypts on 10:30 PM 12/26/2024: the password is the site name: crackmes.one
iaakanshff on 1:50 PM 01/06/2025: was this supposed to be hard ? crackmeYG was in plain text after decrypted by the first call xD. Anyways can you give the source code i like how the compiler generated assembly using sse2 registers ! Nice
bernas198YT on 9:27 PM 01/07/2025: Password its in this adress on x64 dbg
00007FF7B99112E1| 49:0F47D6| cmova rdx,r14 | rdx:"crackmeYG"
Coti on 8:11 PM 01/11/2025: I was just stepping and found the answer.
Clyde on 6:03 PM 01/12/2025: pass: crackmeYG
triepixx_ on 7:58 PM 02/13/2025: написал hui и мне выдало фимоз лол
DozorRage on 1:40 PM 04/06/2025: 0000000BD20FFAD0 63 72 61 63 6B 6D 65 59 47 00 00 00 00 00 00 00 crackmeYG