> crackme1

– First of all, execute the binary ( ‘coz I know that’s not a malware) :

$ ./crackme1
usage: ./crackme1 
$ ./crackme1 thegame
Bad password! Try again running as root maybe?

– Well, let’s try with gdb !

$ gdb ./crackme1
gdb-peda$ start thegame
gdb-peda$ ni
// snip //
0x80486c0: add eax,0x4
0x80486c3: mov eax,DWORD PTR [eax]
0x80486c5: mov DWORD PTR [esp],eax
=> 0x80486c8: call 0x80484a0 <strlen@plt>
0x80486cd: mov DWORD PTR [esp+0x2c],eax
0x80486d1: add DWORD PTR [esp+0x2c],0x1337
0x80486d9: cmp DWORD PTR [esp+0x2c],0x1342
0x80486e1: je 0x80486fc
Guessed arguments:
arg[0]: 0xffffd6e5 ("thegame")

– That’s very interesting ! This part of assembler code means :
– Call the function strlen
– Compare to 0x1337 (4919) + return(strlen) and 0x1342 (4930)
– Jump if egal

– To sum up, that means if (4919 + strlen(password) != 4930; => exit(). So,
I know that the password MUST BE a 11 bytes length string.

– Let’s have a look with this condition :

gdb-peda$ start aaaaaaaaaaa
gdb-peda$ ni
// snip
=> 0x80486e1: je 0x80486fc
| 0x80486e3: mov eax,ds:0x804a048
| 0x80486e8: mov DWORD PTR [esp],eax
| 0x80486eb: call 0x8048470 <puts@plt> | 0x80486f0: mov DWORD PTR [esp],0x1
|-> 0x80486fc: mov ebx,DWORD PTR ds:0x804a04c
// The JUMP is taken, as expected ! \o/
gdb-peda$ ni
0x80485f2: call 0x80484d0 <calloc@plt>
// A memory space is allocated
// snip
0x804862c: shl ebx,0x2
0x804862f: add edx,ebx
0x8048631: mov edx,DWORD PTR [edx]
0x8048633: xor edx,ecx <== That is very interesting !!
0x8048635: mov DWORD PTR [eax],edx

– Now, I know that my input is XORED !

gdb-peda$ ni
0x8048719: call 0x8048430 <strcmp@plt>

– After that, if my XORed input password equals to the string hardcoded, I get the
flag. If not, I have the following message :

0xffffd480 --> 0x80487ec ("Bad password! Try again running as root maybe?")

– Well, many things are possibles now :
– ltrace the binary (That’s what I did)
– Use ld_preload to load my own strcmp()
– Coffee time (haha)

– Right now, let’s have a look to the binary with ltrace :

$ ltrace ./crackme1 $(python -c 'print "a"*11')
__libc_start_main(0x804864a, 2, 0xffec8674, 0x8048740, 0x80487b0
calloc(1, 12)
strcmp("&\016\016\005A\003\016\030A[H", ">_\032;SR05\023\t\036")
puts("Bad password! Try again running "...Bad password! Try again running as \ root maybe?) +++ exited (status 0) +++ 

Ok, as expected, I have the strcmp() display. The first argument is my input
password (aaaaaaaaaaa) XORed; the second argument is the harcoded string.

– Some theorie about XOR :

A ^ B = C, so A ^ C = B

Or, I know my input password, I know my input password XORed. I just need to XOR
them together (I use ipython) to find the key :

In [14]: input = 'a'*11
In [16]: xored = "&\016\016\005A\003\016\030A[H"
In [18]: ''.join(chr(ord(a) ^ ord(b)) for a, b in zip(input, xored))
Out[18]: 'Good boy :)'

– Well, « Good boy 🙂 » is the key that’s used to XOR. So, I have to take the second
argument (the hardcoded string) of the strlen() and XOR it with the key !

In [19]: key = "Good boy :)"
In [20]: hardcoded = ">_\032;SR05\023\t\036"
In [21]: ''.join(chr(ord(a) ^ ord(b)) for a, b in zip(hardcoded, key))
Out[21]: 'y0u_s0_L337'

– Look very good ! It seems to be the expected input password.
Let’s try :

$ ./crackme1 y0u_s0_L337
Good boy 🙂

Pwned \o/
Flag : y0u_s0_L337


-> crackme1-w32 (thx to @Tishrom for his help)

– This writeup will not be detailed as the first (no time)
We have a function called WindowProc (sub_401579) which is the callback of the

– This function is waiting for WM_KEYDOWN (that means a touch was pushed)

cmp eax, 100h
jz short loc_4015A9

– With the different conditions, I reconstructed the expected keystroke :

mov eax, [ebp+arg_8]
cmp eax, 4Dh // M
jz loc_40179A

– After that, there is a comparison with the variable ‘i’. If it’s bad, it begins
Keystroke :


– That means :

Pwned \o/
Flag : BR34K_MY_K3YL0G

Enjoy !
Notfound (Boris P.)


Bonus :

Let’s have a look to the amazing write-up from tlk about the crackme3 :


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