Kostya's Mini-Crypto Math SerialMe
| Download MiniKeyMe.zip, 3 kb (password: crackmes.de) Browse contents of MiniKeyMe.zip ..plz, view readme file!!
Difficulty: 3 - Getting harder | RatingWaiting for at least 3 votes View profile of Kostya » |
Solutions
Solution by aluigi, published 24. apr, 2007; download (5 kb), password: crackmes.de or browse.
aluigi has not rated this crackme yet.
Discussion and comments
| Kostya Author 07. Oct 2006 | very easy, for newbies :) |
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| DaBookshah 07. Oct 2006 | Interesting. It's easier to keygen than it looks i think, have a go at it later. A harder problem would be if the hard-coded text for the second half of the serial was the encoded version of the first half. might not have any solutions though. |
| Kostya Author 08. Oct 2006 | It is easy to find the correct key, I made it specially for this, so if I'll change some conditions in algo, nobody will find the correct key. |
| DaBookshah 08. Oct 2006 | Ok, admittedly its midnight here, but if my head is still working, the problem is to: <img src="http://users.tpg.com.au/adslm66g/latex.png"> We know the ai's, and the xi's are the input bytes....but brutforcing might take a while, I wonder if theres a smarter way. |
| DaBookshah 08. Oct 2006 | Oh right, when it said html will be left as-is, it meant.....not shown. That's sorta confusing. |
| Kostya Author 08. Oct 2006 | Yeah! Ur image is real to this problem!!! So i'm interested in your ways of solving it. Brute!? :) |
| crp- 09. Oct 2006 | hmm, you are sure there are non bruteforce methods to solve this? |
| DaBookshah 09. Oct 2006 | Yeah, I am thinking the same thing. Probably not..... |
| Kostya Author 09. Oct 2006 | I think it's clear, that there are no non-brute methods to solve it. Try to brute it, if u can do it! |
| Kostya Author 11. Oct 2006 | Thank u for trying this crypto math algo! Nobody will! It's hard as i think! |
| macabre 12. Oct 2006 | Do you have a valid serial for this? It would seem to me that if the check is an OR EAX,EAX...JNZ then EAX would have to be 0 for it to work...and the only way to do that is to have the value of EAX rollover.... no? |
| upb 13. Oct 2006 | or eax, eax sets the zero flag when eax == 0 jnz jump if zero flag not set so the jump is taken when eax != 0 |
| macabre 17. Oct 2006 | Yes but if I'm not mistaken we do not want it to take the jump. If we take the jump it sets the retun eax to 0 and we want it to set it to a 1. So in our case for success we need it to be a zero...correct? |
| Anasazi 19. Oct 2006 | I know this was'nt what you're looking for; Offset: 00401075 Old Bytes : 77 18 New bytes: EB 27 ^_^ |
| Kostya Author 20. Oct 2006 | macabre, if u want to understand the algo, enter somthing like "12345" & just trace it with Olly, and u'll see everything u'll need to see! ;) 12345 <-- it's wrong code as u can see :) |
| macabre 20. Oct 2006 | Kostya, I think understand the algo. It didn't seem very difficult it was the end comparison that I wasn't understanding how that would work. After you add the computed bytes together you then do an OR operation which I believe needs to be zero. My confusion comes from several ADDs that become 0... rollover? |
| Kostya Author 23. Oct 2006 | Yep! U r right! :) I Think that everything seems to be clear from this image http://users.tpg.com.au/adslm66g/latex.png |
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