brangelito's KeymakeMe #1
| Download KeymakeMe.zip, 21 kb (password: crackmes.de) Browse contents of KeymakeMe.zip KeymakeMe #1
Difficulty: 2 - Needs a little brain (or luck) | Send a message to brangelito » View profile of brangelito » |
Solutions
Solution by onepatop, published 23. aug, 2010; download (721 kb), password: crackmes.de or browse.
onepatop has rated this crackme as quite nice.
Discussion and comments
| example 18. Jun 2010 | example 16253-21229 (or "example" and "16253addingSome-21229chars") will work. don't know whether i will write a keygen. don't know that much about finding divisors in a modular ring... the assembler code is _very_ easy to read but style-wise it's very ugly.... storing the loop-counter on the stack and the likes... |
|---|---|
| brangelito Author 19. Jun 2010 | Good job! :) If you don't want to learn the math behind it, you are allowed to bruteforce that part. |
| fenoloji 19. Jun 2010 | name: fenoloji serial: 19375-10773 Very nice keygenme. Modular Arithmetic ... May be soon tutorial... |
| brangelito Author 19. Jun 2010 | Well done! I'm glad you enjoyed it. ;-) |
| Mathe 25. Jun 2010 | How do you find the serial ? |
| vptrlx 28. Jun 2010 | very easy one. and the easiest solution is empty username and 0-0 pass :) And the keygen is just calculate name hash (call it n), multiply it by 1336^-1 modulo that number to get the second part of serial, calculate modinverse of it (if it's not zero, then you don't need to do anything :) ), multiply by n^2 and you get the first part. |
| brangelito Author 28. Jun 2010 | So.. Where's the solution? ;) |
| brangelito Author 28. Jun 2010 | http://pastie.org/1021644 Here's a dirty algo for mod-inverses I wrote. |
| vptrlx 28. Jun 2010 | i looked through and didn't get the thing :) The Fermat's little theorem says that a^(p-1) = 1 mod p, hence the easiest way to find mod-inverse is to calculate a^(p-2) mod p ;) |
| brangelito Author 28. Jun 2010 | Ah, that's much simpler. :D |
| GeroZZZ 03. Jul 2010 | seems like, that a mathematical solution without a BigNumbers-Lib is not really possible, am i right? I just posted some "optimized" bruteforce solution. |
| brangelito Author 03. Jul 2010 | I'm not sure what you're getting at, but all arithmetic is made under a quite small modulo. |
You may leave your comment, thoughts and discuss this crackme with other reversers here.
Acting childish will not be tolerated.
HTML and such will be left as-is, so don't try.