Category: Reverse Engineering Points: 300 Solves: 12 Description:
I want a valid license for a piece of software, here is the license validation software. Can you give me a valid license for the email mzisthebest@notarealemail.com?
Note: flag is not in easyctf{} format.
Load the binary in IDA, and we see that it's obfuscated.
Obfuscation is not complex, mostly
opaque predicatethat is a) always executed, b) never executed;- jump to the middle of the command;
- mov eax, jmp_loc; jmp eax;
- antidebug
Like
.text:00401370 B8 95 66 E4+ mov eax, 4E46695h
.text:00401375 3D 36 55 ED+ cmp eax, 0F4ED5536h ; loc_401383 is bad loc
.text:0040137A 74 07 jz short loc_401383 ; and `opaque predicate` that is never executed
.text:00401327 74 03 jz short near ptr loc_40132B+1 ; jump to the middle of the command
.text:00401329 75 01 jnz short near ptr loc_40132B+1 ; and `opaque predicate` that is always executed
.text:0040132B loc_40132B: ; CODE XREF: .text:00401327j
.text:0040132B D9 B9 04 00+ fnstcw word ptr [ecx+4]
.text:0040137C B8 89 13 40+ mov eax, offset loc_401389
.text:00401381 FF E0 jmp eax ; jmp to loc_401389, but IDA doesn't see it
.text:0040106B C6 45 E7 00 mov byte ptr [ebp-19h], 0
.text:0040106F C7 45 FC 00+ mov dword ptr [ebp-4], 0
.text:00401076 9C pushf
.text:00401077 81 0C 24 00+ or dword ptr [esp], 100h
.text:0040107E 9D popf ; if program under debugger - we jump to seh at 0x0040108F
.text:0040107F 90 nop ; else - normal execution
.text:00401080 C7 45 FC FE+ mov dword ptr [ebp-4], 0FFFFFFFEh
.text:00401087 EB 14 jmp short loc_40109D
.text:00401087
.text:004010A5 FF 15 00 30+ call ds:IsDebuggerPresent
.text:004010AB 85 C0 test eax, eax
.text:004010AD 74 51 jz short deb_detected_loc
I don't know how to deobfuscate automatically, so I'll do it manually. For deobfuscation we need:
- keypatch - keystone-engine.org/keypatch/
- brain
- a bit of luck
After deobfuscation(nop all unnecessary code) and decompilation we have very beautiful code, you can even compile it, for me it working fine in vc17 :D
UPD: original source of this task
#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// int f_strlen(const char *str);
unsigned int f_check_key(int *key);
int main(int argc, const char **argv, const char **envp)
{
if (argc >= 3) // we need `email` and `key`
{
const char *email = (const char *)argv[1];
const char *key = (const char *)argv[2];
if (strlen(key) == 16) // `key` len must be 16(4x4 parts)
{
size_t email_len = strlen(email);
if (email_len >= 10) // `email` len must be >= 10
// len("mzisthebest@notarealemail.com") = 29, all is ok :D
{
unsigned int email_hash = 0xAED0DEA; // initial email_hash value
bool is_domain = false;
unsigned char byte;
for (size_t i = 0; i < email_len; ++i)
{
byte = email[i];
if (byte == '@') // start of domain in email
is_domain = true;
if (is_domain)
email_hash ^= byte; // for domain of email we xor email_hash with ord(byte)
else
email_hash += byte; // for nickname of email we add to email_hash ord(byte)
}
if (email_hash == 0xAED12F1 && !(f_check_key((int *)key) ^ 0x1AE33))// `email_hash == 0xAED12F1`
// is always rigth for `mzisthebest@notarealemail.com`
puts("correct!");
}
}
}
else
{
printf("Usage: %s <email> <license key>\n", *argv);
}
}
/*
int f_strlen(const char *str) // just strlen, we no need it and will use standart strlen
{
for (int len = 0; ; ++len)
if (!*str++)
break;
return len;
}
*/
unsigned int f_check_key(int *key)
{
unsigned int key_part;
char *endPtr;
unsigned int key_hash = 0;
for (int i = 0; i < 4; ++i)
{
key_part = key[i];
key_hash ^= strtol((char *)&key_part, &endPtr, 30);
}
return key_hash;
}We do not need to check the email (because it always passes successfully), we need key check: when f_check_key return 0x1AE33 we will win. In function f_check_key our key with len 16 is cut into 4x4 parts, convert parts from the base 30 (0_0) to base 10 and xor them among themselves.
key_hash_0 ^ key_part_0 == key_hash_1 ; key_hash_0 = 0
key_hash_1 ^ key_part_1 == key_hash_2
key_hash_2 ^ key_part_2 == key_hash_3
key_hash_3 ^ key_part_3 == 0x1AE33
After simplify this expressions we have
key_part_0 ^ key_part_1 ^ key_part_2 ^ key_part_3 == 0x1AE33
Let's write z3 script-keygen! :D
from z3_staff import * # https://github.com/KosBeg/z3_staff
import string
digs = string.digits + string.letters
def int2base(x, base = 30):
if x < 0:
sign = -1
elif x == 0:
return digs[0]
else:
sign = 1
x *= sign
digits = []
while x:
digits.append(digs[x % base])
x /= base
if sign < 0:
digits.append('-')
digits.reverse()
return ''.join(digits)
var_num = 4
create_vars(var_num, size=32, prefix = 'k')
solver()
init_vars(globals())
set_ranges(var_num, rstart=27000, rend=809999, prefix = 'k') # 27000 is 1000 in 30 base, the least digit with 4 symbols
# 809999 is TTTT in 30 base, the largest digit with 4 symbols
add_eq( k0 ^ k1 ^ k2 ^ k3 == 0x1AE33 )
# for a bit of fun :D
add_eq( k0 == 583574 ) # LICE in 30 base
add_eq( k1 == 646635 ) # NSEF in 30 base
add_eq( k2 == 672974 ) # ORME in 30 base
i = 0
start_time = time.time()
while s.check() == sat:
ans = prepare_founded_values(var_num, prefix = 'k')
_str = ''
for j in ans:
_str += int2base(j)
print _str
iterate_all(var_num, prefix = 'k')
i += 1
print('--- %.2f seconds && %d answer(s) ---' % ((time.time() - start_time), i) )PS C:\!CTF> python LicenseCheck_keygen.py
licenseformeq7eg
--- 0.01 seconds && 1 answer(s) ---
That's all :)