INFINITE_FREE_TRIAL
We've decided to make an app specially for flag hoarding, can you make sure no one can crack it?
NOTE: The flag is a valid registration key
- File : ift.zip
This is a small application in trial mode asking us the registration key
REGISTRATION
CRC8 check
A crc check by block of 6 chars is done on the key for 6 blocks so the key must be 36 chars.
┌ 148: sym.do_crc_check (int64_t arg1);
│ ; var int64_t var_18h @ rbp-0x18
│ ; var int64_t var_5h @ rbp-0x5
│ ; var signed int64_t var_4h @ rbp-0x4
│ ; arg int64_t arg1 @ rdi
│ 0x00002c9a 55 push rbp
│ 0x00002c9b 4889e5 mov rbp, rsp
│ 0x00002c9e 4883ec20 sub rsp, 0x20
│ ; DATA XREF from sym.step_timer @ 0x24df
│ 0x00002ca2 48897de8 mov qword [var_18h], rdi ; arg1
│ 0x00002ca6 c745fc000000. mov dword [var_4h], 0
│ ┌─< 0x00002cad eb49 jmp 0x2cf8
│ │ ; CODE XREF from sym.do_crc_check @ 0x2cfc
│ ┌──> 0x00002caf 8b55fc mov edx, dword [var_4h]
│ ╎│ 0x00002cb2 89d0 mov eax, edx
│ ╎│ 0x00002cb4 01c0 add eax, eax
│ ╎│ 0x00002cb6 01d0 add eax, edx
│ ╎│ 0x00002cb8 01c0 add eax, eax
│ ╎│ 0x00002cba 4863d0 movsxd rdx, eax
│ ╎│ ; DATA XREF from sym.register_tm_clones @ 0x2334
│ ╎│ 0x00002cbd 488b45e8 mov rax, qword [var_18h]
│ ╎│ 0x00002cc1 4801d0 add rax, rdx
│ ╎│ 0x00002cc4 be06000000 mov esi, 6
│ ╎│ 0x00002cc9 4889c7 mov rdi, rax
│ ╎│ 0x00002ccc e8f1feffff call sym.crc8
│ ╎│ 0x00002cd1 8845fb mov byte [var_5h], al
│ ╎│ 0x00002cd4 0fb645fb movzx eax, byte [var_5h]
│ ╎│ 0x00002cd8 4898 cdqe
│ ╎│ 0x00002cda 488d153f0400. lea rdx, obj.crccheck ; 0x3120
│ ╎│ 0x00002ce1 0fb61410 movzx edx, byte [rax + rdx]
│ ╎│ 0x00002ce5 8b45fc mov eax, dword [var_4h]
│ ╎│ 0x00002ce8 4898 cdqe
│ ╎│ 0x00002cea 488d0daf2400. lea rcx, obj.crcout ; 0x51a0
│ ╎│ 0x00002cf1 881408 mov byte [rax + rcx], dl
│ ╎│ 0x00002cf4 8345fc01 add dword [var_4h], 1
│ ╎│ ; CODE XREF from sym.do_crc_check @ 0x2cad
│ ╎└─> 0x00002cf8 837dfc06 cmp dword [var_4h], 6
│ └──< 0x00002cfc 7eb1 jle 0x2caf
│ 0x00002cfe ba07000000 mov edx, 7
│ 0x00002d03 488d053a0500. lea rax, [0x00003244] ; "w1nR4rs"
│ ; DATA XREF from entry0 @ 0x22d8
│ 0x00002d0a 4889c6 mov rsi, rax
│ 0x00002d0d 488d058c2400. lea rax, obj.crcout ; 0x51a0
│ 0x00002d14 4889c7 mov rdi, rax
│ 0x00002d17 e8d4f3ffff call sym.imp.memcmp
│ 0x00002d1c 85c0 test eax, eax
│ ┌─< 0x00002d1e 7507 jne 0x2d27
│ │ 0x00002d20 b801000000 mov eax, 1
│ ┌──< 0x00002d25 eb05 jmp 0x2d2c
│ ││ ; CODE XREF from sym.do_crc_check @ 0x2d1e
│ │└─> 0x00002d27 b800000000 mov eax, 0
│ │ ; CODE XREF from sym.do_crc_check @ 0x2d25
│ └──> 0x00002d2c c9 leave
└ 0x00002d2d c3 ret
We can see that the crc of a block is used as an offset in a table obj.crccheck and the output is in obj.crcout So :
crcout[i] = crccheck[CRC8(key[i:i+6])]
and crcout must be equal to w1nR4rs to pass the memcmp @ 0x00002d17 and return 1
Here is a small script solving a key that passes the CRC check, this is not the flag but it will allow us to debug the second verification stage
- File : solve_crc.py
import itertools
import struct
import string
import crc8
crccheck = [
0x464dd4d6,0xc73ecd53,0x8a506d41,0x8e2cbf22,
0x55019c09,0xc5f43510,0x4fd8686b,0xa81315d5,
0x3242d308,0xa1940654,0xffadfbe0,0x82319e5f,
0xf21eca02,0x47e2d74a,0x14806648,0x2d27da67,
0x1140e862,0x81842123,0xcebe1774,0x0eb5929b,
0xf799f0c6,0x763adfa6,0xf6d17cdd,0x07b7e9a9,
0x7ec27a97,0x304cb390,0x8545fd5d,0xf3e375a3,
0x380dbd49,0xfab98bb4,0x2bb259aa,0xe60bcf6a,
0xbc3c6305,0x887987e5,0x433403a5,0x897d1def,
0xb13358f1,0x7f958378,0xf5b67bdb,0x37ba2f1b,
0xd012188d,0x703fe773,0x640a0ca7,0xae6c719f,
0xb896eb28,0x868f19a2,0xc9dc0fd9,0xab5e39f9,
0x25c1cb51,0xee446520,0x1fa43b5c,0xc829afcc,
0x61ac602a,0x4b5bf85a,0x9d8cec93,0x98dec3a0,
0xeae436bb,0xb03d0072,0x6f774e24,0x1ac0fe52,
0x2e566991,0x04fc169a,0x571c26e1,0xc46ed2ed]
crccheck = b''.join([ struct.pack('<I',elt) for elt in crccheck ])
crc_block = [[ hex(i)[2:] for i,elt in enumerate(crccheck) if elt == c ] for c in b'w1nR4rs' ]
crc_block = [ elt[0] for elt in crc_block ]
# bruteforce crc8
out = [None]*7
charset = string.ascii_letters
# 1st block
out[0] = b'rarctf'
# 2nd block
for test in itertools.combinations(charset.encode(),5):
test = b'{'+bytes(test)
crc_fx = crc8.crc8(test)
if crc_fx.hexdigest() == crc_block[1]:
out[1] = test
# last block
for test in itertools.combinations(charset.encode(),5):
test = bytes(test)+b'}'
crc_fx = crc8.crc8(test)
if crc_fx.hexdigest() == crc_block[-1]:
out[-1] = test
# other blocks
for test in itertools.combinations(charset.encode(),6):
test = bytes(test)
crc_fx = crc8.crc8(test)
if crc_fx.hexdigest() in crc_block[2:-1]:
for i,elt in enumerate(crc_block):
if elt == crc_fx.hexdigest():
out[i] = test
if all([ elt != None for elt in out]):
break
out = b''.join(out)
print(out.decode())
Giving us : rarctf{RUXYZabceRSabceRUabceEKabcezWMUVXZ}
XOR check
Here a XOR is done between 6 chars block and is compared with the obj.xorcheck table :
┌ 161: sym.do_xor_check (int64_t arg1);
│ ; var int64_t var_18h @ rbp-0x18
│ ; var signed int64_t var_4h @ rbp-0x4
│ ; arg int64_t arg1 @ rdi
│ 0x00002d2e 55 push rbp
│ 0x00002d2f 4889e5 mov rbp, rsp
│ 0x00002d32 4883ec20 sub rsp, 0x20
│ 0x00002d36 48897de8 mov qword [var_18h], rdi ; arg1
│ 0x00002d3a c745fc000000. mov dword [var_4h], 0
│ ┌─< 0x00002d41 eb56 jmp 0x2d99
│ │ ; CODE XREF from sym.do_xor_check @ 0x2d9d
│ ┌──> 0x00002d43 8b55fc mov edx, dword [var_4h]
│ ╎│ 0x00002d46 89d0 mov eax, edx
│ ╎│ 0x00002d48 01c0 add eax, eax
│ ╎│ 0x00002d4a 01d0 add eax, edx
│ ╎│ 0x00002d4c 01c0 add eax, eax
│ ╎│ 0x00002d4e 4898 cdqe
│ ╎│ 0x00002d50 488d15692400. lea rdx, obj.xorout ; 0x51c0
│ ╎│ 0x00002d57 4801c2 add rdx, rax
│ ╎│ 0x00002d5a 8b45fc mov eax, dword [var_4h]
│ ╎│ 0x00002d5d 8d4801 lea ecx, [rax + 1]
│ ╎│ 0x00002d60 89c8 mov eax, ecx
│ ╎│ 0x00002d62 01c0 add eax, eax
│ ╎│ 0x00002d64 01c8 add eax, ecx
│ ╎│ 0x00002d66 01c0 add eax, eax
│ ╎│ 0x00002d68 4863c8 movsxd rcx, eax
│ ╎│ 0x00002d6b 488b45e8 mov rax, qword [var_18h]
│ ╎│ 0x00002d6f 488d3401 lea rsi, [rcx + rax]
│ ╎│ 0x00002d73 8b4dfc mov ecx, dword [var_4h]
│ ╎│ 0x00002d76 89c8 mov eax, ecx
│ ╎│ 0x00002d78 01c0 add eax, eax
│ ╎│ 0x00002d7a 01c8 add eax, ecx
│ ╎│ 0x00002d7c 01c0 add eax, eax
│ ╎│ 0x00002d7e 4863c8 movsxd rcx, eax
│ ╎│ 0x00002d81 488b45e8 mov rax, qword [var_18h]
│ ╎│ 0x00002d85 4801c8 add rax, rcx
│ ╎│ 0x00002d88 b906000000 mov ecx, 6
│ ╎│ 0x00002d8d 4889c7 mov rdi, rax
│ ╎│ 0x00002d90 e8a8feffff call sym.xor_block
│ ╎│ 0x00002d95 8345fc01 add dword [var_4h], 1
│ ╎│ ; CODE XREF from sym.do_xor_check @ 0x2d41
│ ╎└─> 0x00002d99 837dfc05 cmp dword [var_4h], 5
│ └──< 0x00002d9d 7ea4 jle 0x2d43
│ 0x00002d9f ba24000000 mov edx, 0x24 ; '$'
│ 0x00002da4 488d05750400. lea rax, obj.xorcheck ; 0x3220 ; "\t\x16\x17\x0f\x17V\x16D:\x18So\x14\x03*\x06o1\x1cG*\x06-_Q\x1b"
│ 0x00002dab 4889c6 mov rsi, rax
│ 0x00002dae 488d050b2400. lea rax, obj.xorout ; 0x51c0
│ 0x00002db5 4889c7 mov rdi, rax
│ 0x00002db8 e833f3ffff call sym.imp.memcmp
│ 0x00002dbd 85c0 test eax, eax
│ ┌─< 0x00002dbf 7507 jne 0x2dc8
│ │ 0x00002dc1 b801000000 mov eax, 1
│ ┌──< 0x00002dc6 eb05 jmp 0x2dcd
│ ││ ; CODE XREF from sym.do_xor_check @ 0x2dbf
│ │└─> 0x00002dc8 b800000000 mov eax, 0
│ │ ; CODE XREF from sym.do_xor_check @ 0x2dc6
│ └──> 0x00002dcd c9 leave
└ 0x00002dce c3 ret
The main loop is done 6 times and the xor block is done in the xor_block function.
Now we can see that the first step was not necessary because the flag can be recover with the xorcheck table if the first key block is known..... -_-
Yet the first key block is known : rarctf
Here is a script that solve the xor check from xorcheck table
- File : solve_xor.py
import struct
xorcheck = [
0x0f171609,0x44165617,0x6f53183a,0x062a0314,
0x471c316f,0x5f2d062a,0x46001b51,0x5504004a,
0x4c015066,0x526e3177,0x00737234,0x3b031b01]
xorcheck = b''.join([ struct.pack('<I',elt) for elt in xorcheck ])
key = 'rarctf'
for i,elt in enumerate(xorcheck[:0x24]):
key += chr(ord(key[i])^elt)
print(key)
Giving us : rarctf{welc0m3_t0_y0ur_new_tr14l_281099b9}
flag : rarctf{welc0m3_t0_y0ur_new_tr14l_281099b9}