| Impossible PatchingMe V2 |
Also for the vm entry part you have initialized 19 part to iterate. Final state must be 24 and session byte must be 23 |
2026-02-14 14:37 |
| Impossible PatchingMe V2 |
This looks like a production crypto system not a ctf ve berk bunu sen mi yaptın bilmiyorum ama umarım bir sistemin içi değildir bize crackletmeye çalıştığın :D. This program has base64 encoded commands for VM as you can see from the CVM1, It uses Argon2id,Blake2b with g1 key(guard-1 hash) targeting = 4195c09a24468d6016be58df7254fc34d85f913465a88c497b0a751a8f49b2ac
and g2 key(key[0:8]) targeting = 1ec02a1d2fc8be8598a3398cb6557d27a725476760f70d0ee4909b81b5ac28b1
Inside the vm it uses SESSKEY1, Final_Key and ChaCha20_Key
ChaCha20_Decrypts and sends it to the Base32(SESSKEY_hash) if its correct it will show the flag.
And yes it is impossible to bruteforce because it is using Argon2id
I analyzed it like that if something wrong someone can fix it.
Also it is too long to find key i just left. I don't have time for that. |
2026-02-14 14:34 |
| Rose's love |
Encrypted hex value: a050d611ea376bad520b967a43087843, Program itself decrypting it, after executing returns 0 (success). I dont get the idea. You used White box Aes encryption and decryption. The string is hardcoded in binary no input or anything, self encryption and decryption. Did i earn her hearth :D ? |
2026-02-11 09:08 |
| EasiestEver |
hahaha it was so fun to crack it, the password was in the hexadecimal(decimal depends) range, and it took a while to notice that.
After founding range you should look for the ASCII table to get some values from the range. Between these ranges everything can be your password. |
2026-02-03 11:48 |
| CrackMe with password |
rootaccess1337 found in dump |
2026-01-31 15:25 |
| ZEXOR-v0.1 |
It was so good and was fun to find the real password :D "AEXORRBSHA36325S33" besides that it was my first experience of cracking program and i enjoyed so much. |
2026-01-28 18:02 |