DJ1hAD0's DJ1hAD0's newbie decryptme1
| Download DJ1hAD0_s_hewbie_decryptme_1.zip, 9 kb (password: crackmes.de) Browse contents of DJ1hAD0_s_hewbie_decryptme_1.zip It's a very easy small decryptme for absolute newbies, like me:)
Difficulty: 1 - Very easy, for newbies | RatingWaiting for at least 3 votes View profile of DJ1hAD0 » |
Solutions
Solution by Astrocyt, published 20. oct, 2016; download (50 kb), password: crackmes.de or browse.
Astrocyt has rated this crackme as nothing special.
Solution by pixelpirat, published 20. oct, 2016; download (1 kb), password: crackmes.de or browse.
pixelpirat has not rated this crackme yet.
Solution by beaver, published 20. oct, 2016; download (1 kb), password: crackmes.de or browse.
beaver has not rated this crackme yet.
The submission of solutions is closed.
Discussion and comments
| sn4pf1sh 29. Aug, 15:36 | Can anyone give me a hint? |
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| beaver 30. Aug, 12:31 | sn4pf1sh, 4013A9 fgets call |
| luxdav 30. Aug, 14:40 | Was fun! Thank you! "Congratulations! You won, Dude! Please, send your solution." |
| sn4pf1sh 30. Aug, 20:22 | I don't know what to do with the hashed string ""W{zsfu`axu`}{zg54M{a4c{z84Papq54Dxqugq84gqzp4m{af4g{xa`}{z:4OP^%|UP$I"" @beaver: Thx, but this didn't help, maybe i'm missing something? |
| luxdav 31. Aug, 00:01 | spoiler alert: The first thing you can do is decrypting the string without the magic word. The first part of the code checks if your magic word is correct, the second part decrypts the string (second part starts at 401401). If you solved the decryption, think about how the password has to look like, you want to enter the second part. |
| beaver 31. Aug, 06:14 | sn4pf1sh, just XOR(wierd_string, 0x14). 14 is the result of the subtraction. |
| sn4pf1sh 31. Aug, 22:11 | Can anyone please post a solution? This CrackMe is quite interesting and would help me for understanding the (ASM) decrypting logic. Thank you anyway @beaver ;) |
| luxdav 01. Sep, 00:22 | I actually found my own password "luxdav__|" |
| luxdav 01. Sep, 01:10 | This is my uncommented solution: https://github.com/LuxXx/reverse-engineering/tree/master/DJ1hAD0 But really, try it yourself again, looking up my solution will ruin your fun. |
| sn4pf1sh 01. Sep, 13:17 | Ok, thank you guys, especially @luxdav and @beaver. This helped me a lot. Now I'm able to understand control flow of the logic. I wrote the decrypting logic for myself in ANSI C and debugged this in x64dbg. Now I understand a lot more. Sorry for my bad english! :) |
| luxdav 06. Sep, 18:18 | I had some time and wrote a tool that generates some passwords: https://github.com/LuxXx/reverse-engineering/tree/master/DJ1hAD0/solution/password/bruteforce It wasn't necessary but I wanted to do it. |
| herbie 28. Sep, 19:29 | My own password: herbie{c} |
| rentf 06. Oct, 19:23 | My password is: rentf{Zc} #include <stdio.h> int main() { char in[10]; int a = 0; fgets(in, 10, stdin); for (int i=1; i<=9; i++) a += in[i]; if (a != 866) printf("%i>866", a); else printf("Bingo!"); } |
| rickolous 28. Nov, 02:44 | I wrote a very thorough walkthrough for this problem. Here is a link to my github.io page [https://rickolous.github.io/2016/11/27/very-easy-1-windows/] |
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