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aname on 2024-12-12 11:02: password is decrypted in first part, decrypt print message and get input.
thed0ublec on 2024-12-12 15:02: ```python
print("omg")
```
IWDFObject_AcquireLock on 2024-12-12 17:02: 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 2024-12-12 18:57: pass: crackmeYG
Coti on 2024-12-12 20:07: Not sure if that was the right way but I was just stepping and noticed the password RDX
alxxx on 2024-12-13 02:26: good for beginners. sha256{3cc3a014b9bebc060d8df8b329750a7b054492ff835bf5fed80bea4cda94a071}
AboutSeflash on 2024-12-17 17:08: OMG
SirWardrake on 2024-12-23 00:36: 123 test test ;-)
re.login on 2024-12-24 07:24: Thanks for challenges.
DeFexGG on 2024-12-24 22:43: think difficulty harder but crackmeYG i found it in memory with cheat engine inspecting
thenotoriusflux on 2024-12-26 01:21: Nice debugger challenge
yursing on 2024-12-26 12:14: guys im new on this site . i would like to know, how could i get the password of zip files. pls help
Strypts on 2024-12-26 22:30: the password is the site name: crackmes.one
iaakanshff on 2025-01-06 13:50: 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 2025-01-07 21:27: Password its in this adress on x64 dbg
00007FF7B99112E1| 49:0F47D6| cmova rdx,r14 | rdx:"crackmeYG"
Coti on 2025-01-11 20:11: I was just stepping and found the answer.
Clyde on 2025-01-12 18:03: pass: crackmeYG
triepixx_ on 2025-02-13 19:58: написал hui и мне выдало фимоз лол
DozorRage on 2025-04-06 13:40: 0000000BD20FFAD0 63 72 61 63 6B 6D 65 59 47 00 00 00 00 00 00 00 crackmeYG