Crackmes.de – josamont’s j666

This crackme was published December 2nd, 2014. It is rated “3 – Getting harder. The description reads:

Find the pass

This is the main routine:

code.png

The crackme starts by calling sum_up_code, shown next:

sum_up_code.png

This code sums up opcode dwords starting from the beginning of the subroutine, until an opcode dword is less than 804819Ah. The result is stored in code_check_sum. By doing this, the crackme can check for the presence of software breakpoints, which change the code by adding INT 3. I don’t plan on using a debugger, so I ignore this subroutine. Next the password is read by a call to sys_read. The value is then converted from hex to binary using the subroutine hexToInt:

hexToInt-700x1024-1.jpg

This subroutine starts with an anti patching check: the offset 80480E8 is compared to F3 A6 74 11 – this is the location and opcode for:

F3 A6     repe cmpsb
74 11     jz   short loc_80480FD

If you want to patch this check, you also need to patch the check in hexToInt. Since I don’t plan on patching the crackme, I can ignore this check. The rest of the routine is interpreting the password as hex and converting the value to binary, storing the result in password_value. There are two interesting code sequences in hexToInt:

4F    dec edi
4B    dec ebx
0A 00 or al, [eax] 

and

4E    dec esi
6F    outsd
0A 00 or al, [eax] 

The offset of these code snippets is used as the goodboy and badboy message, 4F 4B 00 and 4E 6F 00 decode to the null-terminated strings “OK” and “No” respectively.
After hexToInt there is a call to xor_code:

xorcode.png

The snippet changes 9*4 bytes starting at offset 80491B8. The routine affects neither our entered password, nor the hidden password. I don’t know what the purpose of this routine is. Finally, we get to the password check:

LOAD:080480D9                 mov     ecx, 4
LOAD:080480DE                 mov     esi, offset loc_8048096
LOAD:080480E3                 mov     edi, offset password_value
LOAD:080480E8
LOAD:080480E8 loc_80480E8:                            ; DATA XREF: hexToInt+6o
LOAD:080480E8                 repe cmpsb
LOAD:080480EA                 jz      short loc_80480FD
LOAD:080480EC                 mov     ecx, offset loc_8048157
LOAD:080480F1                 mov     edx, 4
LOAD:080480F6                 call    sys_write
LOAD:080480FB                 jmp     short loc_804810C
LOAD:080480FD ; ---------------------------------------------------------------------------
LOAD:080480FD
LOAD:080480FD loc_80480FD:                            ; CODE XREF: start+59j
LOAD:080480FD                 mov     ecx, offset loc_8048142
LOAD:08048102                 mov     edx, 4
LOAD:08048107                 call    sys_write
LOAD:0804810C
LOAD:0804810C loc_804810C:                            ; CODE XREF: start+6Aj
LOAD:0804810C                                         ; hexToInt+14j
LOAD:0804810C                 mov     eax, 1
LOAD:08048111                 xor     ebx, ebx        ; status
LOAD:08048113                 int     80h             ; LINUX - sys_exit
LOAD:08048113 start           endp

The check compares four bytes at offset 8048096 to the password that we entered, if they match, we get the “OK” string , otherwise the “No” message (both are hidden in hexToInt. At offset 8048096 we find:

LOAD:08048096                   loc_8048096:                            ; DATA XREF: start+4Do
LOAD:08048096 B9 9A 91 04 08                    mov     ecx, offset aCrackme666Jose ; "Crackme 666 Josep\n"

So the entered password needs to match B9 9A 91 04. Because it’s little endian, we need to enter the reverse:

04919AB9

This gives you the OK message:

$ ./j666
Crackme 666 Josep
Password: 04919AB9
OK

You can also leave the leading zero.

comments powered by Disqus