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[unpackme][keygenme] Keygenme 17


ChOoKi

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Looks like it's time for another challenge. This one comes with a mix of ideas so take your time reversing them and do not expect hints any time soon ^_^

 

Basic info/rules:

Compiler: Borland Delphi
Level: Designed to provide maximum fun and to test your skills, so kindly allow other members to enjoy solving it on their own
Packer: Packed with free (non-commercial) and
relatively easy Compressor/Protector (credit to author) to help you unpack it yourself or simply pm GIV :DGood luck to all and happy reversing

kgm#17.7z

  • Like 5
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@


Romulous


I guess this is a crack.


You don't have a valid name/serial combination.


If so is worthless.


@


EvOlUtIoN


I think all options are checked before protecting.


:)


As far i see OEP is stolen, imports are stolen etc....


 


Edit.


Better go direct to the Serial verification routine @ 004700DC

Edited by GIV
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I would say that this is a OEP, looks like one:



00406648 /. 53 PUSH EBX
00406649 |. 8BD8 MOV EBX,EAX
0040664B |. 33C0 XOR EAX,EAX
0040664D |. A3 88A74700 MOV DWORD PTR DS:[47A788],EAX
00406652 |. 6A 00 PUSH 0
00406654 |. E8 2BFFFFFF CALL 00406584
00406659 |. A3 F4374800 MOV DWORD PTR DS:[4837F4],EAX
0040665E |. A1 F4374800 MOV EAX,DWORD PTR DS:[4837F4]
00406663 |. A3 94A74700 MOV DWORD PTR DS:[47A794],EAX
00406668 |. 33C0 XOR EAX,EAX
0040666A |. A3 98A74700 MOV DWORD PTR DS:[47A798],EAX
0040666F |. 33C0 XOR EAX,EAX
00406671 |. A3 9CA74700 MOV DWORD PTR DS:[47A79C],EAX
00406676 |. E8 C1FFFFFF CALL 0040663C
0040667B |. BA 90A74700 MOV EDX,0047A790
00406680 |. 8BC3 MOV EAX,EBX
00406682 |. E8 99DDFFFF CALL 00404420
00406687 |. 5B POP EBX

But it seems some part of it is missing because it is being emulated by the packer.


Edited by xSRTsect
  • Like 1
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I would say that this is a OEP, looks like one:

00406648 /. 53 PUSH EBX

00406649 |. 8BD8 MOV EBX,EAX

0040664B |. 33C0 XOR EAX,EAX

0040664D |. A3 88A74700 MOV DWORD PTR DS:[47A788],EAX

00406652 |. 6A 00 PUSH 0

00406654 |. E8 2BFFFFFF CALL 00406584

00406659 |. A3 F4374800 MOV DWORD PTR DS:[4837F4],EAX

0040665E |. A1 F4374800 MOV EAX,DWORD PTR DS:[4837F4]

00406663 |. A3 94A74700 MOV DWORD PTR DS:[47A794],EAX

00406668 |. 33C0 XOR EAX,EAX

0040666A |. A3 98A74700 MOV DWORD PTR DS:[47A798],EAX

0040666F |. 33C0 XOR EAX,EAX

00406671 |. A3 9CA74700 MOV DWORD PTR DS:[47A79C],EAX

00406676 |. E8 C1FFFFFF CALL 0040663C

0040667B |. BA 90A74700 MOV EDX,0047A790

00406680 |. 8BC3 MOV EAX,EBX

00406682 |. E8 99DDFFFF CALL 00404420

00406687 |. 5B POP EBX

But it seems some part of it is missing because it is being emulated by the packer.

You miss the stolen OEP bytes.

 

004B0D4E    55              PUSH EBP                                 ; <------------------ Prima instructiune

004B0D52    8BEC            MOV EBP,ESP                              ; <----------------------- A doua instructiune

004B0D57    83C4 F0         ADD ESP,-0x10                            ; <------------------ A treia intructiune

004B0D5D    B8 1C8C4700     MOV EAX,kgm#17.00478C1C                  ; <------------------ A patra intructiune

004B0D65    68 6F0D4B00     PUSH kgm#17.004B0D6F                     ; <------------------ A cinceaintructiune

 

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Yeah I have just noticed it, a while ago. Seems this is going to give some work, since I think none of my Imprec plugins seem to do a fine job recovering the API's etc... I'd rather have to code a plugin myself.


 


eedc82c26a.png


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1. You must reconstruct OEP first instructions.

2. Recover API's

3. See what could go wrong next..

:)

 

 

Are you sure that is the way to unpack? I didn't know that, In fact I thought that the proper way to unpack anything was to throw the exe into the air, and expect it to fall upon my desk fully unpacked and functional...

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@ChOoKi: really nice one! :)


 



kao : 17A126194EF743CB6237887C578FFC88



I'm not going to make a keygen - I have all the components, but I'm too lazy to put them together in human readable form or to write a tutorial. ;)


  • Like 3
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@kao: Way to go man! :thumbsup:


@GIV: I couldn't agree more with your posts here :devil:


@njkermk: Omg! are you really [going to keygen this shit!] :bleh:


@xSRTsect: To pass the last check you'll have to get my nickname right this time around :^


 


So glad to see many of you enjoying this challenge, hopefully we get to see a kg or a tut at the end to share with the rest. Keep up the good work all.


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Here comes the keygen (still no tutorial, though).

Releasing complete keygen source is quite pointless, as 60% of it is ripped code, 30% - reversed bit operations and the rest are calls to crypto/bignum libs and UI.

The main part is this:

KeygenProc PROC uses edi hWnd: HWND 	invoke GetDlgItemText, hWnd, IDC_NAME, offset userName,20h	; zero all buffers	mov	edi, offset buffer1	mov	ecx, 50h	xor	eax, eax	rep	stosb	; ripped code (004700DC - uses HAVAL hash with different count of rounds and hash size)	invoke	lstrlen, offset userName	mov	ecx, eax	mov	eax, offset userName	call	make_buffer1_buffer2	; ripped code (00470408 - bit operations)	invoke	make_buffer4	; we have calculated haval hash and buffer4, reverse the rest	; bruteforce the reverse of 004704F0. In less than 128 iterations it is really fast	invoke	bruteforce_chooki	; undo method at 00470460. This is ugly -----	; bswap loop (004704B8..004704EA)	xor	edx, edxbswaploop:	mov	eax, dword ptr [buffer3+edx]	bswap	eax	mov	dword ptr [buffer5+edx], eax        add	edx,4        cmp	edx,10h        jb 	bswaploop	;reverse xor with ror (004703EE..004703F2)	lea	eax, buffer4	lea	edx, buffer5	invoke	undo_xor_with_ror	; extra xor (004703E8..004703EC)	xor	dword ptr [buffer5], 0D4h	; recursively reverse loop (004703B7..004703E6)	invoke	untangle, 10h, 80h, 0	; bswap it again (00470463..0047049F)	xor	edx, edxbswaploop2:	mov	eax, dword ptr [buffer5+edx]	bswap	eax	mov	dword ptr [buffer3+edx], eax        add	edx,4        cmp	edx,10h        jb 	bswaploop2	; finally do simple bigint math (00470606..0470664)	call	bigintMath	invoke SetDlgItemText, hWnd, IDC_SERIAL, offset serialBuffer@exit:	retKeygenProc ENDP

keygen.zip

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@kao: Well done mate :1a: and thank you for taking the time to solve this challenge.


Now let's see who else is finished with his kg and maybe even a tut :type:


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i'm stuck at big number math. These numbers are so big.


289190426366822397640236384991278098977 * x mod 340282366920938463463374607431768211297 = 46974954106352480795018069738232571914


x = ?


 I have no other ideas besides bruteforce them for million years. There must be a special functions like modinv or onebignumwhatisitmodtwobignuminvert which I do not know.


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This is real basic bignum stuff. I'm sure there are good tutorials about it, just have to find them.

In short: you need to solve x * y = z modulo q where y,q are constants and z comes from name.

Solving for x: x = z / y modulo q

It is the same as x = z * (1 / y) modulo q

Since y is constant, so is 1 / y modulo q.

You can calculate 1 / y modulo q using modinv function in any bignum lib or just use Wolfram Alpha.

Once you have that, keygen needs to do just one simple mulmod operation.

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@njkermk - how the hell did you solve my crackme #C4N number 2? Its the exact same problem, and you coded ExtendedGCD operation to solve it...


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Hello ChOoKi,

 

it took a long time but finally I finished the Keygen.
It has cost me a lot of time and energy for recognizing, researching and coding the Cipher you have used.
But I must say it made real fun.Again, thank you for the Keygenme and see you ... maybe.....on your next one....... :sneaky2: :sneaky2: :prop: :prop: :prop:

Solution_to_ChoOki#17.rar

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