ksydfius's What Is This???
| Download ksydfius8.zip, 9 kb (password: crackmes.de) Browse contents of ksydfius8.zip Read the readme :)
Difficulty: 2 - Needs a little brain (or luck) | Send a message to ksydfius » View profile of ksydfius » |
Solutions
Solution by phueghy, published 24. dec, 2012; download (33 kb), password: crackmes.de or browse.
phueghy has rated this crackme as nothing special.
Discussion and comments
| congchuahoatuyet1997 28. Nov 2012 | how to check???? |
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| idid231 04. Dec 2012 | So i have to find its password and press a button or just find the password? |
| L0ngin0s 05. Dec 2012 | My solution is 1.52061 ¿Is it possible?. Sorry for my bad English ;) |
| L0ngin0s 06. Dec 2012 | Exactly 1.5206117639947453996940561267823061250557102935443782479396672938700617746736758676608778545506705188. I hope don't be wrong ;) |
| ksydfius Author 06. Dec 2012 | Sorry but thats not right. Think a bit outside the box... |
| phueghy 07. Dec 2012 | Is it even analytically possible to find a correct input value? As the algorithm is so sensitive on smallest changes of the input value I am a bit lost here. |
| fcktrl 18. Dec 2012 | My solution "3231" Right? :) |
| ksydfius Author 19. Dec 2012 | sorry, not right |
| phueghy 21. Dec 2012 | My solution is "18975". The explaining document is still in the solutions pipeline. I was only able to bruteforce it because I ws hinted to an integer solution. |
| [Wizzer] 21. Dec 2012 | ***SPOILER ALERT*** Solved too, but brute forced :( That was interesting in a sense that I discovered TI family of devices, never used before, but seen in shops :) was pretty fun playing around with emulator. Also found a website that can show source from application. Anyway, I couldn't write proper c# version of bruteforcer, turned out to be an issue of rounding difference of device and .net framework, so wrote it on device itself (and removed clock speed limit), took sometime and solution was found. program code for device to bruteforce, mention that there will be a syntax error when solution is found, cause I didn't check documentation details on how to break out of the loop, just choose "quit" and answer will be there: :1->V :While V < 1000000 :0→C :V->A :While C<100 :A+10*tan(A)→A :C+1→C :End :If A=19911.236 :Then :Disp V :Return 0; :End :V+1->V :End Thank you! |
| [Wizzer] 21. Dec 2012 | Small addition: to fix syntax error please make sure to use "Return" without any arguments and semicolon (habit) :) ..... :Disp V :Return :End ..... |
| eynstyne 21. Dec 2012 | A decent emulator: TilEm What would the inverse function be then? I have tried: A-tan^-1(A)/10->A A-10/tan^-1(A)->A |
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