downloadbrowseveneta's Crackme v22 (mini-psyho)

Download Crackme_v22_(mini-psyho)_veneta.zip, 54 kb (password: crackmes.de)
Browse contents of Crackme_v22_(mini-psyho)_veneta.zip

litle simple crackme with some crypto ... a litle bit tricky (i think)

Difficulty: 3 - Getting harder
Platform: Windows
Language: C/C++

Published: 19. Aug, 2005
Downloads: 776

Rating

Votes: 8
Crackme is quite nice.

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Solutions

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Discussion and comments

Orphankill
20. Aug 2005
Ok, I'm confused. I set a breakpoint on GetDlgItemTextA and I'm in. However, first you run all the letters of the name through some loop which does a whole bunch of stuff to each
letter, then you run each letter of the entered serial through countless loops and then it goes on and on and on.... How much time do you think we have? Or maybe I missed something? A little nudge in the right direction would be nice :)
Knight
21. Aug 2005
If i'm correct (and if i can post this here) this crackme uses ElGamal (or somethink like that). Just problem is that numbers are too big, to solve it.

P.S. I think those loops u are talking about are SHA-1.
TQN
21. Aug 2005
Yes, Knight. It is SHA-1, use Miracl library.
Tymon
21. Aug 2005
ElGamal-128 + SHA1. You have to compute a and M. One question... This message 'how nice you've defeat..' means that serial is good or not ? coz there is the second message in crackme 'My Congrats Serial correct now you...'. What do you think ?
veneta
Author
21. Aug 2005
"My Congrats Serial correct now you..." is correct :)
1.ElGamal sign
2.surprise (my favourite board game ;)
veneta
Author
21. Aug 2005
i think it's sufficient answer :)
pdrill
30. Aug 2005
This does not look like Elgamal and you dont need to solve DLP for it.
Knight
31. Aug 2005
3 people said that it is ElGamal (one of them is this crackme author) and u still think that it is not? :D
pdrill
31. Aug 2005
Elgamal sig is: y^a * a^b == g^M where a and b is the sig and M is the message to be signed. This is not what is going on here. This crackme computs y^a * a^M == g^b. Set a=p-1 and you get y^a == 1 mod p. The order of p-1 mod p = 2 so a^M can have two solutions, 1 or p-1. This gives us a 50% chanse to set y^a * a ^M == 1 mod p for anny message M. Now, if you set b=p-1 you get g^b == 1 mod p. So,, does this looks like Elgamal? I can see you discuss the solution here so I hope I dont do anything wrong by posting this. This will only get you like 30% into the solution so there should be much more to solve for you.
pdrill
01. Sep 2005
Finaly I got the correct messagebox :) Have you received a solution from anyone?
veneta
Author
02. Sep 2005
great work pdrill :) your solution is first and at the time only solution :)
Zuma555
05. Dec 2009
I got it, and i submited a solution
jB_
06. Dec 2009
Zuma555: your solution does not include the 2nd part of the crackme. You only solved the crypto problem.
_pusher_
06. Dec 2009
jB: i checked quickly.. he seemed to have gotten the valid serials.. what did i miss ? :x
Zuma555
06. Dec 2009
The only problem of this crackme was the crypto problem... What else was there to solve?
Zuma555
06. Dec 2009
Btw Hello jB! ;)
jB_
07. Dec 2009
_pusher_: check the veneta message : "2.surprise (my favourite board game ;)"
There is a reversi game in the crackme. The ElGamal signature is the easy part.
_pusher_
07. Dec 2009
right, think i have to reject your solution Zuma.. sorry :x
Still.. i cant find any signs of any game at all :)
cyclops
Moderator
07. Dec 2009
"how nice
you've defeat E****** s******** (heh you know what ;))" is only half part. If you have a complete solution it will show "My Congrats Serial correct now you..." ;)

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