haiklr's Scandal0us C0de
| Download Scandal0us_C0de-haiklr.zip, 115 kb (password: crackmes.de) Browse contents of Scandal0us_C0de-haiklr.zip Scandal0us C0de
Difficulty: 4 - Needs special knowledge | Send a message to haiklr » View profile of haiklr » |
Solutions
There are no solutions to this crackme yet. Have you solved it? Please write a tutorial and submit it here!
Discussion and comments
| Crosys 01. Aug 2007 | Nice one, solved it :D gj again haiklr |
|---|---|
| halsten 01. Aug 2007 | A nice one indeed. |
| evilcry1 01. Aug 2007 | Nice, truly nice, but i did not understand why it crashes on my Win 2k3. On XP works fine ;) |
| Ox87k 01. Aug 2007 | nice one haiklr, good job! btw, reverse the fpu part isn't that easy for me.. |
| Zaphod 01. Aug 2007 | I can find a number ( 12135053 ) for the first editbox that makes it 7 times through the loop from 4025AD to 4025CC, but not the 8th time. Am I on a completely wrong track? Is this a fake algo or something? |
| Ox87k 01. Aug 2007 | Zaphod, it's the real algo! The first number exists, try to do some test with a pen and a paper and... ;) |
| haiklr Author 01. Aug 2007 | Thx :] @Ox87k : i don't why it crashs on Win 2k3, sorry :s @Zaphod : no it's a true algo... 12135053 is unavailable, your number must make it 8 times through the loop ;) |
| TiGa 02. Aug 2007 | Does it crash on startup or when the button is clicked? |
| evilcry1 02. Aug 2007 | It crashes when is launched |
| TiGa 02. Aug 2007 | I've debugged the crash under Vista x64 and it happens at 004028FE call GetLocalTime The same code under XP SP2 is: 004028FE call GlobalAlloc The same thing must be happening under 2K3. |
| Zaphod 02. Aug 2007 | Yeah!! : I solved the crackme! Oh no!! : I bruteforced! Awww... |
| Ox87k 03. Aug 2007 | Zaphod, did you bruteforce the second part (the fpu's part) or the first? I don't understand how to solve the fpu.. some hint? |
| Zaphod 04. Aug 2007 | 0x87k: I am a little ashamed to say that I bruteforced both parts :( I don't see how you can logically deduct the number for the first textbox. As for the fpu-part: You take the sine for the numbers in box 2 and 3, multiply them and round the result off to an integer. This integer must be 0. That means that both sines must be very small. So I wrote a short routine that went through the numbers between 1 and 100000000 to find two numbers with very small sines. Perhaps this part can be solved by using some formula involving sin and cos, I don't know... |
| evilcry1 04. Aug 2007 | @Zaphod: It's just a mathematical inversion, if you make a little change of abstraction level you will see that: Sum = aA^2 + ... + nN^2 n = Costant N = Your Value Next you have an IDIV, but you can consider it as a MOD operation.. Make the correct inversions and you will have the solution.. Can be used also a Math Modelling program as Mathematica to do that @0x87k: Still working on it, i'll send you some hint when finished ;) Have a nice Day |
| Zaphod 04. Aug 2007 | evilcry 1: I'm afraid I don't quite understand what you mean. We are dealing with an 8-digit number, num = n1n2n3n4n5n6n7n8 which is divided with 2,3,4...9, so that: num MOD 2 = n1 num MOD 3 = n2 num MOD 4 = n3 ... num MOD 9 = n8 How do you inverse that and find num? |
| evilcry1 04. Aug 2007 | There is a way to reverse this, you need some Number Theory concept ;) Regards |
| haiklr Author 05. Aug 2007 | evilcry1 is alright, for the first part, think about arithmetic basics :) it's only logical and deduction, you don't need to calculate anything :] For the fpu part, think about this equation : sin(x) = 0 ... what can be x ? Good luck ! |
| Ox87k 06. Aug 2007 | @haiklr: I found only one couple of values after some test that it allow me to solve it. But i don't think there is only one.. |
| haiklr Author 06. Aug 2007 | @Ox87k : no there's several couples of values which are okay. However, one couple is sufficient :] If you found one, you solve the crackme ;) |
| Zaphod 08. Aug 2007 | haiklr: Yes, if sin(x) = 0, x should be PI, 2*PI, 3*PI etc., so we shall find a whole number that is very close to a multiple of PI. But it seems to me that this involves some bruteforcing unless we have some formula that can help. Do we have such a formula? |
| haiklr Author 08. Aug 2007 | @Zaphod : yes, you're okay... the aim of this crkme is to find an approximation of pi ;) Bruteforcing is forbidden, so you must to find an other way to solve this problem. However, many ways are used to approximate irrationnal numbers (exp, square root of 2...), and they don't need to bruteforce. Maybe search about continued fractions... it's a big hint ;) Good luck ! |
| evilcry1 09. Aug 2007 | continued fractions... it's a big hint ;) :D:D yeah truly big big big! |
| Zaphod 09. Aug 2007 | He! I'm sure it's a big hint - if you know what continued fractions are. Hmm - I'll try Google... |
| Zaphod 10. Aug 2007 | haiklr + evilcry 1: Thanks for directing my attention to continued fractions. There are some really good sites on this subject. Very interesting! Now I can find the numbers for box 2 and 3 using pen and paper. What part of Number Theory should I know something about to find the number for the first box? |
| Ox87k 10. Aug 2007 | Zaphod, try to read here: http://en.wikipedia.org/wiki/Continued_fraction After that, just write in a paper the fpu calcs like a phormula and finally try to tie this with the reading... Sorry for my bad english.. :( |
| Zaphod 10. Aug 2007 | 0x87k: No, I think you misunderstand - I have finished the fpu-part. Now it's the modulo-part I'm thinking about. |
| haiklr Author 10. Aug 2007 | @Zaphod : think about arithmetic basics, in particular about divisibilty. I think you must reduce step by step the different possibilities, according to the divisors. Sorry for my english :/ |
| EFLAGS 12. Aug 2007 | One of the answers: 11121217?589 80143857 16707065 589与ComputerNam的校验和有关。 sin(80143857) * 10000000 * 100000000 * sin(16707065)近似取整(frndint指令)结果为0。 |
| EFLAGS 12. Aug 2007 | Scan 3..200000000 to find x that sin(x)<=2.265e-8。 |
| Noteworthy 09. Aug 2013 | Cracking done ! Math rocks. Thanks haiklr. |
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.