Number of crackmes:
Number of writeups:
Comments:
| Name | Author | Language | Arch | Difficulty | Quality | Platform | Date | Downloads | Writeups | Comments |
|---|---|---|---|---|---|---|---|---|---|---|
| ExploitPack-Challenge | juansacco | C/C++ | x86 | 2.0 | 4.0 | Windows | 2019-11-01 12:00 | 19 | 0 | 14 |
| DragonCrack | juansacco | C/C++ | x86 | 2.0 | 4.0 | Windows | 2019-07-18 11:07 | 14 | 0 | 6 |
| CrackMe J1 | juansacco | C/C++ | x86 | 2.0 | 5.0 | Windows | 2019-01-09 10:32 | 17 | 2 | 22 |
| Crackme | Date | Infos |
|---|
| Crackme | Comment | Date |
|---|---|---|
| Badgercracker | Enter your license key: X4A9Z-82JQK-47L6P-1N2TB License key is valid. Welcome! | 2025-02-17 13:39 |
| Bubo | uridk42ks1d43c | 2023-12-07 11:10 |
| CrackMe1 | public static void bzoLCpGWzMFbU() { jOloNtfoGORHw.add(Integer.valueOf(5256)); } | 2023-09-12 09:19 |
| Crack1 easy find | Here is the key 8J3NYK-XY6021-LJ1hJB-591TT0 | 2023-09-12 09:10 |
| robot[1]: find key | Enter the key: 0b2d key is correct, robot is happy ^_^ thank you for prototype 1 test participation! | 2022-12-15 14:23 |
| Illogical Logic | Yes, its bugged FLAG{hehe_you_solved_an_illogical_mathematical_logic_xD} | 2022-12-15 09:18 |
| TryCrackMe | Put the key: 34407373373234353336 [+] Correct key! | 2022-12-14 08:46 |
| my_f1rs7_cr4ckm3 | Enter digit:1337 u can do this, congrut well done go next | 2022-12-01 10:32 |
| FindThePassword1 | + ------------------- + | Find the Password 1 | | by Xeeven | + ------------------- + Password: 8675309 + *** *** *** *** *** + | Congratulations! | + *** *** *** *** *** + | 2022-11-30 07:57 |
| First crackme | FLAG{ThisShouldNotBeHard} | 2021-11-11 14:58 |
| CrackMe_01_01_x86 | (@vailedre 2021) PASSWORD: Th16_is_tH3_PassW0rd FLAG: veiledre{--------{---(@} | 2021-07-10 08:16 |
| Basis static | CTF{Y0U_d1D_1t_n1C3} | 2021-03-24 07:04 |
| FindMySecret | Enter the secret number5837 Success! You have completely reverse engineered and found the secret number! | 2021-02-14 09:27 |
| FindMySecret | __int16 *result; // eax double input_key; // [esp+10h] [ebp-8h] if ( !byte_406034 ) { real_key = real_key * 10000.0; byte_406034 = 1; } input_key = (double)*a1; if ( real_key | 2021-02-14 09:26 |
| crackme | nice! 19082004 | 2021-01-22 14:56 |
| VERY EASY VERY SIMPLE C CODE | simple strcmp, thanks for sharing! welcome to my crack me ---------------------------------------------------------- enter the password:password123 congrats you cracked the password | 2021-01-18 18:29 |
| crackme dady v2 | Nice thanks! what is your username(do not add spaces to make it easier) ::juan -1- what is the password ::fq]j congrats you got it!!... the password was really 'fq]j' | 2021-01-18 18:25 |
| EscapeTheDunge0n - Expl0it | I agree is a good exercise for who's starting with reversing | 2020-11-25 10:20 |
| EscapeTheDunge0n - Expl0it | 1 2 1 2 788960 | 2020-11-25 10:20 |
| Crypt0 - Beginner CrackMe | Username: Nick password: 4ACE00F | 2020-11-25 09:59 |
| bouzu.exe | strcmp.. bouzu. this should be level 1 or -1 | 2020-10-13 08:10 |
| no strings attached | Enter password: encrypted-c-string CORRECT | 2020-08-29 10:15 |
| whoami | Who am I? : Dad RIGHT! Press any key to continue . . . | 2020-08-26 06:50 |
| Simple Crackme | VAL: BXXGYYYBGIBXX RCX: 0551777016055 | 2020-07-06 09:40 |
| Simple CrackMe | Key please: Jn0192mMkqpskO91jsjJajiUPSJn Congratz!! u in. Thanks for sharing! | 2020-05-11 07:22 |
| Find password | .text:004B58E5 cmp eax, [ebp+var_34] ; Compare Two Operands .text:004B58E8 jnz short loc_4B5890 ; Jump if Not Zero (ZF=0) inserisci la password per accedere al programma djejie benvenuto | 2020-04-24 07:26 |
| KeyCheck_86 | ----------------------------------------------------- [+] Find Out The Password [+] ----------------------------------------------------- Password: hellocracker ::: Congratulations! ::: KEY: SEVSRSwgVEhJUyBJUyBZT1VSIEtFWSA6IFEyOXVaM0poZEhWc1lYUnBiMjV6SVE | 2020-03-12 09:10 |
| Cracking / thinking challenge | Just remove it. This crackme doesn't work. Added the DLL and still crashes | 2020-03-09 09:24 |
| CrackMe #2 - register me in your name | Registered to: Juan :D | 2020-03-09 09:21 |
| ExploitPack-Challenge | It's not a buffer overflow as someone said below! But a crack-me and indeed you can use conditional debuggers to avoid some protections! Glad you like this one! | 2019-12-18 20:33 |
| ExploitPack-Challenge | To bypass for example isdebuggerpresent without patching you can use conditional breakpoints for example, great job madlogik! | 2019-11-13 18:50 |
| StupidCrackMe | Hello man! It's very stupid CrackMe :-). Find password. Password: LiL2281337 Nice job :-). Password found. | 2019-10-31 14:12 |
| Crackme 1: Get The Password | Enter password: HVVf3z8M22 Password is correct! ;) | 2019-10-22 08:26 |
| easy keyg3nme | It's possible to write a keygen :D | 2019-10-17 12:36 |
| easy keyg3nme | Enter your key: 1223 Good job mate, now go keygen me. { return a1 % 1223 == 0; } There is nothing to keygen here. key is hardcoded | 2019-10-17 12:27 |
| VIP_access_me | Are you a vip member? (y/n): y What is your name?: joe Please enter your vip code: f5gz51xyxy6ggj96j1mlgz21j Welcome vip member joe! ####################################### # _ # # -=\`\ # # |\ ____\_\__ # # -=\c``) # # `~~~~~/ /~~` # # -==/ / # # # ####################################### Press any key to continue . . . | 2019-10-10 13:49 |
| Input Magic | ENTER LICENSE: ? Congrats! This license is valid! | 2019-10-04 20:53 |
| Input Magic | The crackme only checks for the serial to start with ? Does not check for size, smaller or larger than, so some of the explainations here are wrong. xor edx, edx ; Logical Exclusive OR mov ecx, 3Ch ; ' | 2019-10-04 20:53 |
| crackme_3 by br0ken | Name : juan Serial : br0-276568-192288476-ken Serial is valid. Now make a keygen :) | 2019-09-17 08:52 |
| OldSoft's KeyGenMe #2 -- Upgraded from DOS by wolverine2k | Enter your name: juan Enter a serial number: 10-452 Good Job. You have cracked OldSoft's KeyGenMe #2 Press any key to continue . . . | 2019-09-04 08:57 |
| asm_is_fun by qhf | Name: juan Key: 6364z07qd Thanks for registering! | 2019-09-04 08:44 |
| Brainfuscated | What's the key? Hyper Congratz! You got it! Added a breakpoint on the call to get.. from there found the key on memory during the compare. ( just a bit harded because of the code from brainfuck) | 2019-08-29 11:44 |
| mexican | flag{M3x1c4nMl4lw4r3_pl3rro} Thanks for sharing! | 2019-08-28 13:35 |
| ice9 by tripletordo | Was found on the comparisson here: .text:004011F5 push offset String2 ; lpString2 .text:004011FA push offset String1 ; lpString1 juan password: 109549323n | 2019-08-27 12:28 |
| DragonCrack | @unc4nny Here is the VM detection for VMWare used on this crackme: All the protections on this crackme were implemented in a way that could be bypassed by setting as in example in IDA something like: SetRegValue("ZF,0x1/0x0) __try { __asm { push edx push ecx push ebx mov eax, 'VMXh' mov ebx, 0 // any value but not the MAGIC VALUE mov ecx, 10 // get VMWare version mov edx, 'VX' // port number in eax, dx // read port // on return EAX returns the VERSION cmp ebx, 'VMXh' // is it a reply from VMWare? setz[rc] // set return value pop ebx pop ecx pop edx } } | 2019-08-12 08:51 |
| Save Scooby | Hi Scooby !! Where are you?? here You won a medal Scooby !! | 2019-08-05 09:07 |
| Save Scooby | Alternative solution to getcwd: Break on-runtime here: loc_559B769418B2: cmp [rbp+var_C], 0 | 2019-08-05 09:07 |
| Parkour | 1. Kernel32 - IsDebuggerPresent() 2. Bool 3. Flag was found using strings THIRD STAGE Please, input the password to the system:flag{th3_cr3d3n14ls_4r3_s4f3} Congratulations, you have logged into the system. | 2019-07-30 08:26 |
| DragonCrack | zip password: crackmes.one | 2019-07-30 08:04 |
| Easy Peasy | nice one! I enjoyed it thanks for sharing Please, login with your credentials. Username:iwonderhowitfeelstobeatimetraveler Now, please insert the password. Password:heyamyspaceboardisbrokencanyouhelpmefindit? You have successfully logged into the system. | 2019-07-15 11:34 |
| Crackme not main | kawaii-flesh: Yes on my case I did got the file: CNP.7z. I have enjoyed your crackme, keep them coming! | 2019-07-11 09:23 |
| Crackme not main | Cracked it using IDA Pro and got the key "1785". After the initial call to scanf there are 4 more after that, and indeed prints "Good key!" if you got it right. | 2019-07-10 08:01 |
| CrackMe J1 | HINT: This crack me uses antidebugging techniques such as: VM Detection , Traps and IsDebuggerPresent | 2019-01-17 10:26 |
| CrackMe J1 | @K2000 Besides other protections this crack me checks if you are running it inside a VM and if it does, then it quits, you need to bypass this in order to make it run | 2019-01-17 10:26 |
| CrackMe J1 | @K2000 you need to bypass the protection so the program doesn't quits or restart your PC hehe | 2019-01-16 15:40 |
| CrackMe J1 | Password for the crackme: crackmes.one | 2019-01-15 16:53 |
| CrackMe J1 | There are some other protections than isDebuggerPresent() to bypass :-) Keep trying! | 2019-01-15 09:15 |
| visual c++ crackme | Found it! Username: aaaa Password: 1132463 .text:00401215 lea esi, [esp+110h+password] ; Load Effective Address .text:00401219 mov eax, offset password | 2019-01-09 11:36 |