plikan on 2026-06-21 06:27:
[Click to reveal]Pass: CR4CKM3_1234
qmemcpy(v7, "CR4CKM3_1234", 12);
while ( *((_BYTE *)v7 + v5) == ((unsigned __int8)((v4[v5] ^ 0x55) + 7) ^ 0x33) )
{
if ( ++v5 == 12 )
{
sub_140001480("correct! You cracked it.\n");
return 0;
}
}
plikan on 2026-06-21 06:33:
[Click to reveal]oops, im sorry. pass is 3C 0F 55 3C 24 22 AC 30 AE AF AC 55 in hex or b'<\x0fU<$"\xac0\xae\xaf\xacU' in python
SirWardrake on 2026-06-21 23:50:
Hm, I don't like this crackme. Solution is easy, BUT:
A cmd parameter is required, but the solution contains non-printable characters (e.g., 0x0f)
vandoam on 2026-06-22 04:14:
[Click to reveal]CR4CKM3_1234
while (segredo[contador] == (senha_usuario[contador] ^ 0x55) + 7 ^ 0x33)
????????
misha_2007 on 2026-06-24 19:41:
[Click to reveal]PS C:\vscode\Crack_ME\1> ./crackme.exe
Enter password: SbsTQqy38w34Pr4XrpI1m8qIg2x4NqZz
congrats from tenzo! can you tell me how you cracked it?
PS C:\vscode\Crack_ME\1>
misha_2007 on 2026-06-24 19:48:
Ops, I miss. I chose the wrong crackmes