Yesterday I read in one of the polish portal (with news) an information about interesting challenge organized by the Government Communications Headquarters (GCHQ). This is a British intelligence agency responsible for providing signals intelligence (SIGINT) and information assurance to the UK government and armed forces. Based in Cheltenham, it operates under the guidance of the Joint Intelligence Committee. CESG (originally Communications-Electronics Security Group) is the branch of GCHQ which works to secure the communications and information systems of the government and critical parts of UK national infrastructure.
GCHQ, is aiming to attract the next generation of web-savvy spies by running an ad campaign that challenges computer hackers to crack a code to get an interview.
Ok so let’s look at it closer 🙂
First level
I will not analyze the security of webpage and server. I will try to discuss about pure challenge. OK so first question is what does this hexcode means? Let’s look it closer:
0xeb 0x04 0xaf 0xc2 0xbf 0xa3 0x81 0xec 0x00 0x01 0x00 0x00 0x31 0xc9 0x88 0x0c 0x0c 0xfe 0xc1 0x75 0xf9 0x31 0xc0 0xba 0xef 0xbe 0xad 0xde 0x02 0x04 0x0c 0x00 0xd0 0xc1 0xca 0x08 0x8a 0x1c 0x0c 0x8a 0x3c 0x04 0x88 0x1c 0x04 0x88 0x3c 0x0c 0xfe 0xc1 0x75 0xe8 0xe9 0x5c 0x00 0x00 0x00 0x89 0xe3 0x81 0xc3 0x04 0x00 0x00 0x00 0x5c 0x58 0x3d 0x41 0x41 0x41 0x41 0x75 0x43 0x58 0x3d 0x42 0x42 0x42 0x42 0x75 0x3b 0x5a 0x89 0xd1 0x89 0xe6 0x89 0xdf 0x29 0xcf 0xf3 0xa4 0x89 0xde 0x89 0xd1 0x89 0xdf 0x29 0xcf 0x31 0xc0 0x31 0xdb 0x31 0xd2 0xfe 0xc0 0x02 0x1c 0x06 0x8a 0x14 0x06 0x8a 0x34 0x1e 0x88 0x34 0x06 0x88 0x14 0x1e 0x00 0xf2 0x30 0xf6 0x8a 0x1c 0x16 0x8a 0x17 0x30 0xda 0x88 0x17 0x47 0x49 0x75 0xde 0x31 0xdb 0x89 0xd8 0xfe 0xc0 0xcd 0x80 0x90 0x90 0xe8 0x9d 0xff 0xff 0xff 0x41 0x41 0x41 0x41
For the first quick view we can say there is some interesting bytes. With red color I sign this bytes which can be a dump of x86 assembler instructions (of course there is more but this can be typical for shellcodes) and with the blue color I sign interesting bytes for me 🙂 – 0x41414141 and 0x424242 for me always will be connected with exploiting 😉
OK so let’s try to analyze this bytes as x86 instructions:
0x0804a040 <+0>: jmp 0x804a046 <shellcode+6> 0x0804a042 <+2>: scas %es:(%edi),%eax 0x0804a043 <+3>: ret $0xa3bf 0x0804a046 <+6>: sub $0x100,%esp 0x0804a04c <+12>: xor %ecx,%ecx 0x0804a04e <+14>: mov %cl,(%esp,%ecx,1) 0x0804a051 <+17>: inc %cl 0x0804a053 <+19>: jne 0x804a04e <shellcode+14> 0x0804a055 <+21>: xor %eax,%eax 0x0804a057 <+23>: mov $0xdeadbeef,%edx 0x0804a05c <+28>: add (%esp,%ecx,1),%al 0x0804a05f <+31>: add %dl,%al 0x0804a061 <+33>: ror $0x8,%edx 0x0804a064 <+36>: mov (%esp,%ecx,1),%bl 0x0804a067 <+39>: mov (%esp,%eax,1),%bh 0x0804a06a <+42>: mov %bl,(%esp,%eax,1) 0x0804a06d <+45>: mov %bh,(%esp,%ecx,1) 0x0804a070 <+48>: inc %cl 0x0804a072 <+50>: jne 0x804a05c <shellcode+28> 0x0804a074 <+52>: jmp 0x804a0d5 <shellcode+149> 0x0804a079 <+57>: mov %esp,%ebx 0x0804a07b <+59>: add $0x4,%ebx 0x0804a081 <+65>: pop %esp 0x0804a082 <+66>: pop %eax 0x0804a083 <+67>: cmp $0x41414141,%eax 0x0804a088 <+72>: jne 0x804a0cd <shellcode+141> 0x0804a08a <+74>: dec %eax 0x0804a08b <+75>: cmp $0x42424242,%eax 0x0804a090 <+80>: jne 0x804a0cd <shellcode+141> 0x0804a092 <+82>: pop %edx 0x0804a093 <+83>: mov %edx,%ecx 0x0804a095 <+85>: mov %esp,%esi 0x0804a097 <+87>: mov %ebx,%edi 0x0804a099 <+89>: sub %ecx,%edi 0x0804a09b <+91>: rep movsb %ds:(%esi),%es:(%edi) 0x0804a09d <+93>: mov %ebx,%esi 0x0804a09f <+95>: mov %edx,%ecx 0x0804a0a1 <+97>: mov %ebx,%edi 0x0804a0a3 <+99>: sub %ecx,%edi 0x0804a0a5 <+101>: xor %eax,%eax 0x0804a0a7 <+103>: xor %ebx,%ebx 0x0804a0a9 <+105>: xor %edx,%edx 0x0804a0ab <+107>: inc %al 0x0804a0ad <+109>: add (%esi,%eax,1),%bl 0x0804a0b0 <+112>: mov (%esi,%eax,1),%dl 0x0804a0b3 <+115>: mov (%esi,%ebx,1),%dh 0x0804a0b6 <+118>: mov %dh,(%esi,%eax,1) 0x0804a0b9 <+121>: mov %dl,(%esi,%ebx,1) 0x0804a0bc <+124>: add %dh,%dl 0x0804a0be <+126>: xor %dh,%dh 0x0804a0c0 <+128>: mov (%esi,%edx,1),%bl 0x0804a0c3 <+131>: mov (%edi),%dl 0x0804a0c5 <+133>: xor %bl,%dl 0x0804a0c7 <+135>: mov %dl,(%edi) 0x0804a0c9 <+137>: inc %edi 0x0804a0ca <+138>: dec %ecx 0x0804a0cb <+139>: jne 0x804a0ab <shellcode+107> 0x0804a0cd <+141>: xor %ebx,%ebx 0x0804a0cf <+143>: mov %ebx,%eax 0x0804a0d1 <+145>: inc %al 0x0804a0d3 <+147>: int $0x80 0x0804a0d5 <+149>: nop 0x0804a0d6 <+150>: nop 0x0804a0d7 <+151>: call 0x804a079 <shellcode+57> 0x0804a0dc <+156>: inc %ecx 0x0804a0dd <+157>: inc %ecx 0x0804a0de <+158>: inc %ecx 0x0804a0df <+159>: inc %ecx 0x0804a0e0 <+160>: add %al,(%eax)
So this is it. This code make sense and this was good way of analyzing. I sign by read color this instruction which always cause exit() syscall and half of the code won’t be executed (as we will see further even more). But first, at the beginning this code jump over 2 next instruction (so they are never executed) and than allocate memory which is filled by natural numbers. Next they are converted to some more interested values and finally there is static jump to the code which cause syscall exit() – red colour. That’s all, so what next? As we can see directly after jump instruction, the program tries to get the new value for the stack pointer exactly from the stack. It gaves us an information that smth should be changed there 😉 Also as we can see further (blue colour) from the stack is popped also value for %%eax register and compared with the 0x41414141 value. Next this value is decremented by one and again compared but now with the 0x42424242 value. Logically it makes no sense. If first compare will be true than next will be bad – 0x41414141 – 1 = 0x41414140 so it will never be 0x42424242. If we want to pass all this checks and executed further code we must change a lot 😉
First change
Ok let’s come back to the syscall exit(). We don’t want to stop the execution flow but continue, and we know that further code expect the new stack pointer in the stack. Also we know that after this operation program tries to get new value for %%eax register from the new stack and compare with the value 0x414141. As we can see in the end of the shellcode we have instructions:
0x0804a0dc <+156>: inc %ecx 0x0804a0dd <+157>: inc %ecx 0x0804a0de <+158>: inc %ecx 0x0804a0df <+159>: inc %ecx
this is exactly the value 0x414141:
(gdb) x/x 0x0804a0dc 0x804a0dc <shellcode+156>: 0x41414141 (gdb)
so here we go with answer 😉 Like in the oldschool technique of getting current stack pointer used in viruses – let’s change syscall exit() to the call which gave us back the flow to the shellcode. Old code:
0x0804a0d5 <+149>: nop 0x0804a0d6 <+150>: nop 0x0804a0d7 <+151>: call 0x804a079 <shellcode+57> # value: 0x80cd
New code:
0x804c095: nop 0x804c096: nop 0x804c097: call 0x804c039 # value: 0x45eb
OK – works. First compare is passed but of course second is not and again game over. But If we think again about it, decrementing instruction can be also overwrite to ours and as we know the new stack pointer is now _after_ the shellcode. We are able to add new bytes after this shellcode and change the assembler instruction which decrements value in register %%eax to pop new value from the stack 🙂 This is in fact the answer 😉
Second change
Old instruction:
0x0804a08a <+74>: dec %eax # byte: 0x48
New instruction:
0x804c04a: pop %eax # byte: 0x58
Perfect. Now we have another problem – which data should be added in the end of shellcode? We can manually add 0x42424242 value to pass the compare check but what about further code? Maybe this 0x42424242 is a tip? In further code the value for %%edx register is also popped from the stack. And next bytes are used to copy in temporary place and manipulate them. So of course all of this is a tip. Lets come back again to the main page of the crack site. Bytes which we used to create shellcode are not in the site as text but as image. We were frustrating to rewrite them manually not just simply copying. But wait a minute why this is an image?
Steganography
Analyzing image can be hard 😉 But I’m lazy buster and usually before I move to real hard job like analyze or full RE I try to get as much information as I can in as simple way as it can be. So let’s run strings command 😉 There is 983 lines (not small image) but one line from the top is very interesting:
$ strings cyber.png |head IHDR sRGB pHYs tIME ]iTXtComment QkJCQjIAAACR2PFtcCA6q2eaC8SR+8dmD/zNzLQC+td3tFQ4qx8O447TDeuZw5P+0SsbEcYR 78jKLw==2 IDATx .^cwuW $
Here you go 😉 First impression is – this is base64. Let’s check in one of the online sites if there is logic in this string. After decoding to the ASCII we see that hexdump should be done _but_ look for the first 4 bytes in this string:
BBBB2∅∅∅Øñmp :«gÄûÇfüÍÌ´ú×w´T8
ëÃþÑ+ÆïÈÊ/?ÿØ∅∅
Yes, this is exactly 0x42424242 value. So probably this is what we need 😉 After adding this hex in the end of the shellcode and all of our changes and adding the code for dumping the memory after whole process of executing shellcode (this dump function write by yourself) we will see this beautiful message from the memory:
GET /15b436de1f9107f3778aad525e5d0b20.js HTTP/1.1
Interesting, isn’t it? 🙂 Yep the first level is done.
Second level
What does this link have?
//--------------------------------------------------------------------------------------------------
//
// stage 2 of 3
//
// challenge:
// reveal the solution within VM.mem
//
// disclaimer:
// tested in ie 9, firefox 6, chrome 14 and v8 shell (http://code.google.com/apis/v8/build.html),
// other javascript implementations may or may not work.
//
//--------------------------------------------------------------------------------------------------
var VM = {
cpu: {
ip: 0x00,
r0: 0x00,
r1: 0x00,
r2: 0x00,
r3: 0x00,
cs: 0x00,
ds: 0x10,
fl: 0x00,
firmware: [0xd2ab1f05, 0xda13f110]
},
mem: [
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],
exec: function()
{
// virtual machine architecture
// ++++++++++++++++++++++++++++
//
// segmented memory model with 16-byte segment size (notation seg:offset)
//
// 4 general-purpose registers (r0-r3)
// 2 segment registers (cs, ds equiv. to r4, r5)
// 1 flags register (fl)
//
// instruction encoding
// ++++++++++++++++++++
//
// byte 1 byte 2 (optional)
// bits [ 7 6 5 4 3 2 1 0 ] [ 7 6 5 4 3 2 1 0 ]
// opcode - - -
// mod -
// operand1 - - - -
// operand2 - - - - - - - -
//
// operand1 is always a register index
// operand2 is optional, depending upon the instruction set specified below
// the value of mod alters the meaning of any operand2
// 0: operand2 = reg ix
// 1: operand2 = fixed immediate value or target segment (depending on instruction)
//
// instruction set
// +++++++++++++++
//
// Notes:
// * r1, r2 => operand 1 is register 1, operand 2 is register 2
// * movr r1, r2 => move contents of register r2 into register r1
//
// opcode | instruction | operands (mod 0) | operands (mod 1)
// -------+-------------+------------------+-----------------
// 0x00 | jmp | r1 | r2:r1
// 0x01 | movr | r1, r2 | rx, imm
// 0x02 | movm | r1, [ds:r2] | [ds:r1], r2
// 0x03 | add | r1, r2 | r1, imm
// 0x04 | xor | r1, r2 | r1, imm
// 0x05 | cmp | r1, r2 | r1, imm
// 0x06 | jmpe | r1 | r2:r1
// 0x07 | hlt | N/A | N/A
//
// flags
// +++++
//
// cmp r1, r2 instruction results in:
// r1 == r2 => fl = 0
// r1 < r2 => fl = 0xff
// r1 > r2 => fl = 1
//
// jmpe r1
// => if (fl == 0) jmp r1
// else nop
throw "VM.exec not yet implemented";
}
};
//--------------------------------------------------------------------------------------------------
try
{
VM.exec();
}
catch(e)
{
alert('\nError: ' + e + '\n');
}
//--------------------------------------------------------------------------------------------------
As we can read this level is completely different from the previous 🙂 Short overview:
- We must implement own Virtual Machine(!)
- VM must emulate segmented memory model with 16-byte segment size (notation seg:offset)
- There is defined own assembler with own simple architecture
- CPU have 8 registers: 4 general-purpose registers (r0-r3), 2 segment registers (cs, ds equiv. to r4, r5), 1 flags register (fl), and of course IP register (Instruction Pointer)
- We know how instruction encoding looks like.
- We know how to manipulate flags register.
- 8 instructions are defined
- Memory dump is available which must be used to execute our emulator
To solve this level we must know that segment-offset architecture has always shifted memory etc. If we look closer we can find that emulated %%ds register has value 0x10 (because its 16 bits architecture).
I won’t post here my implementation of this VM, but after all again we must dump the memory how it is changed. Here it is:
1^D3<AA>@^B<80>^CR^@r^As^A<B2>P0^T<C0>^A<80>^@^P^P^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@2^@u^L32@^B<80>^CR^@r^As^C<B2>^@ð^@0C0>^A<FF>^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@u^P^A^@^@^@^@^@^@^@^@^@^@^@<CC>^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@}^_^U`MMR}^N'm^PmZ^FVG^TB^N<B6><B2><B2><E6>봃<8E><D7><E5><D4><D9><C3><F0><80><95><U+4209A><BD><95><A4><8D><9A>+0iJieU^\{i^\n^Dt5!&/`^CN7^^3T9溴<A2><AD><A4>ŕ<C8><C1><E4><8A><EC>璋<E8><81><U+2D624><D0><C0><8D><AC>"Re~'+Z^Ra
^Azk^]gGET /da75370fe15c4148bd4ceec861fbdaa5.exe HTTP/1.0^@^@^@^@^@^@^@^@^@^@^@^@^@^@7z^G^Q^_^]h%2w^^b#[GUS0^QB<F6><F1><B1><E6><C3><CC><F8><C5><E4><CC><C0>Ӆ<FD><9A><E3>恵<BB><D7>͇<A3><D3>k6oofU0^VE^ t\?)+f=^M
^M^B0(5^U ^U<DD><EC><B8><E2><FB><D8><CB><D8>ыՂٚ<F1><92><AB><E8><A6><D6>Ќ<AA>Ҕ<CF>EFg }D^TkEmT^C^W`bUZJfa^QWhu^Eb6}^B^P"B2<BA><E2><B9><E2>ֹ<FF><C3>銏<C1><8F>Ḥ<96><U+4F071><8D><89><CC><D4>xvar>7#Vsqyc^Q iz^Th^E!^^2'Y<B7>ϫ<DD><D5>̗<93><F2><E7><C0><EB><FF>飿<A1><AB><8B><BB><9E><9E><8C><A0><C1><9B>Z//NN^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@
As we can see there is hidden link for the next level 😉 Now we must probably play with binary 😉
Btw. to be sure that our emulator is correct, last instruction must be ‘hlt’ – halt.
Third level (last)
First of all, this is Windows binary so I will use VM to play with it (I use VirtualBox with Win XP SP3 + IDA of course ;p). We need to install cygwin to run the binary because of the libcrypt which was used to compile the binary. After all when we run the binary we will be asked to give an argument
usage: keygen.exe hostname
Let’s look inside of the binary…
mov [esp+78h+var_74], offset aR ; "r" mov [esp+78h+var_78], offset aLicense_txt ; "license.txt" call _fopen64 mov [ebp+var_4C], eax cmp [ebp+var_4C], 0 jnz short loc_401120 mov [esp+78h+var_78], offset aErrorLicense_t ; "error: license.txt not found\n" call printf mov [ebp+var_50], 0FFFFFFFFh jmp loc_401204
So program is looking for the license.txt file
lea eax, [ebp+var_38] mov [esp+78h+var_70], eax mov [esp+78h+var_74], offset aS ; "%s" mov eax, [ebp+var_4C] mov [esp+78h+var_78], eax call fscanf
Get the string from this file (here is by the way stack overflow bug :D). Next:
cmp [ebp+var_38], 71686367h jnz short loc_4011CF
program checks the first 4 bytes if they are equal to 0x71686367 which in fact is equal to the string “gchq” – so the pattern of The Government Communications Headquarters 🙂 Next:
mov eax, dword_402000 mov [esp+78h+var_74], eax lea eax, [ebp+var_38] add eax, 4 mov [esp+78h+var_78], eax call crypt mov edx, eax mov eax, dword_402000 mov [esp+78h+var_74], eax mov [esp+78h+var_78], edx call strcmp test eax, eax jnz short loc_4011A5
Here is small trick 🙂 For me very clever, so we can see call for the crypt() function and next compare and jump. If we analyze it closer we can see that argument for the crypt() is the same like one argument for the strcmp() function. So crypt() must return exactly the same value like one which was passed as argument. Pseudocode can look like:
char *tmp="hqDTK7b8K2rvw";
if (strcmp(crypt(bufor+4, tmp), tmp)) {
...
This is very interesting situation because we must find string which after pass it to the crypt() function with the salt “hqDTK7b8K2rvw” will be encrypted to exactly the same value as salt (“hqDTK7b8K2rvw”) 🙂
Two ways can be used to find this string – bruteforce and rainbow tables. I’ve done both 😀 In fact bruteforcing won’t take as much time. Anyway the answer is….
cyberwin
🙂 Yes, we have it 🙂 Now we know that license key should looks like:
gchqcyberwin
Whatever we add after this string it won’t change the value of encrypted string (because of the salt). So let’s try to run the code with exactly this license.txt file:
_> da75370fe15c4148bd4ceec861fbdaa5.exe www.canyoucrackit.co.uk keygen.exe loading stage1 license key(s)... loading stage2 license key(s)... request: GET /hqDTK7b8K2rvw/0/0/0/key.txt HTTP/1.0 response: HTTP/1.1 404 Not Found Content-Type: text/html; charset=us-ascii Server: Microsoft-HTTPAPI/2.0 Date: Thu, 01 Dec 2011 23:55:05 GMT Connection: close Content-Length: 315 <HTML body>
So what we know now… File tries to connect to the server given as the argument in command line and tries to GET a key.txt file from the URL:
/hqDTK7b8K2rvw/0/0/0/key.txt
So again we have some question. Why there is three values of 0 (zero) number? Is the server given in command line is correct? When the program is running it prints:
loading stage1 license key(s)... loading stage2 license key(s)...
Why he do that? If we look again to the java script code from the previous level we can read:
// stage 2 of 3
Is it somehow connected to this string? OK so… if we add anything after the magic string in license.txt file program tries to dump it as hex value and put in the URL, so:
_> da75370fe15c4148bd4ceec861fbdaa5.exe www.canyoucrackit.co.uk keygen.exe loading stage1 license key(s)... loading stage2 license key(s)... request: GET /hqDTK7b8K2rvw/41414141/42424242/43434343/key.txt HTTP/1.0 response: HTTP/1.1 404 Not Found Content-Type: text/html; charset=us-ascii Server: Microsoft-HTTPAPI/2.0 Date: Thu, 01 Dec 2011 23:58:05 GMT Connection: close Content-Length: 315 <HTML body>
Again we cannot put too long string because we can make stack overflow bug 😉 Of course we do not need to do that 😉 What is interesting we must find correct three 4bytes values and this is the main goal – find a correct path for the key.txt file – as I said in first level there is unused 4 bytes – use it + in 2 stage there was also 2 unused bytes but written directly in the .js file 😉
UPDATE: Some ppl didn’t believe I know what bytes should be used. But as I said in previous sentence – all bytes are in previous stages. Here is the correct link:
http://www.canyoucrackit.co.uk/hqDTK7b8K2rvw/a3bfc2af/d2ab1f05/da13f110/key.txt
So first value as I pointed in all previous sentence (when I described first level also I point it specially) is from the unused code from the shellcode. Two next bytes are the firmware from the second stage.
Ending
When we solve all problems and enter correct string in the main page which is:
-
Pr0t3ct!on#cyber_security@12*12.2011+
you will be redirect to this page:
http://www.canyoucrackit.co.uk/soyoudidit.asp
Here is screenshot:
After you click the button you are going to be redirected here:
http://www.gchq-careers.co.uk/cyber-jobs/
Here is screenshot:
and final redirection is here:
https://apply.gchq-careers.co.uk/fe/tpl_gchq01ssl.asp?newms=jj&id=35874
Here is some screenshot with salary:
Conclusion
This is very nice challenge and requires in fact huge skills. Very nice training for our brains 😉 Anyway few points which I must write now:
- To apply for this job you must have English citizen – I don’t have so automagically I can’t apply even when I finished this challenge 😉
- This challenge requires a lot of skills and the salary(!) is very low for the ppl with this skills so I understand why they cannot find good ppl 😉
- I’m living and working now in London and I can say its quite nice place 😉
UPDATE 2: More information about second level + my implementation of VM can be found here:
http://blog.pi3.com.pl/?p=268
Best regards,
Adam Zabrocki
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Justice League » Blog Archive » UK Spooks’ Recruiting Tactic: Very Low Pound to Genius Ratio on 12.02.2011
[…] The details of how I solved it are on my personal blog. […]
UK Spooks’ Recruiting Tactic: Very Low Pound to Genius Ratio | Cigital on 12.02.2011
[…] details of how I solved it are on my personal blog. This entry was posted in Cyber Security Software Security. Bookmark the permalink. […]
Gerwazy on 12.03.2011
Nice job mate,
£25K getting new employee in our NOC and they don’t have to have big knowledge.
“To apply for this job you must have English citizen” and if you read more, your father need to be UK citizen as well for at least 10 years.
In US you only need to my citizen to get job for government that is why many good engineers which came from abroad after getting citizenship getting job. And this make sence.
lazydaemon on 12.03.2011
great job!
the GCHQ salary is indeed a joke 😉
» Złam kod, zostań szpiegiem (jak rekrutuje brytyjski wywiad) -- Niebezpiecznik.pl -- on 12.04.2011
[…] Aktualizacja pi3 przygotował rozwiązanie krok po kroku. […]
Redford on 12.06.2011
Nice article, but there’s a few things that intrigue me. I read this post after solving challenge. First thing is that there isn’t “dec eax” in this code, you made a mistake typing 0x48 (dec eax) instead of 0x58 (pop eax). Look at the picture with binary code 😉
This code works without any modifications, you only have to insert decoded (from base64) data at end of this code. This bytes comes from PNG comment (can be viewed in eg Gimp), you don’t have to read all lines printed by ‘string’ 😉
There isn’t any trick in crypt() use. It’s DES hash. Only first 2 letters are salt, next letters are hash 😛
It’s strange, but on my system the exe from the last stage always gives me 404, even with correct address, but on firefox this url works. I checked with wireshark differences in request, maybe it’s caused by older HTTP version, i dunno. But it’s their mistake 🙂
Please correct me if I’m wrong somewhere.
Thanks for the article 🙂
Redford
admin on 12.06.2011
Thanks! 🙂 Yes you’ve right I made mistake on rewriting hex to the program 😉 That’s why I bit more difficulty to pass 1st challenge 😉
About 404 in fact I was testing addresses directly in the browser because it was faster – after discovering an idea how URL is prepared 🙂
Thanks for comment 🙂
lukasz on 12.11.2011
Redford:
I also checked (using wireshark) the request performed by their exe against a working script and i seems that if you want to get the password, you have to:
1) provide the host name either by using http/1.1
GET /hqDTK7b8K2rvw/a3bfc2af/d2ab1f05/da13f110/key.txt HTTP/1.1
Host: canyoucrackit.co.uk
2) OR embed the host name in the http/1.0 get request
GET http://canyoucrackit.co.uk/hqDTK7b8K2rvw/a3bfc2af/d2ab1f05/da13f110/key.txt HTTP/1.0
Either way, their own exe cannot do the required request correctly which is a real shame 😛
Good Will Hunting on 12.14.2011
Why Shouldn’t I Work for the NSA? (Good Will Hunting):
http://www.youtube.com/watch?v=UrOZllbNarw
w0lf on 04.12.2012
How did you convert the shellcode to ASM dump? Did you use a script or did you write it by yourself ?
admin on 04.13.2012
gdb