Upload: 8:18 AM 02/22/2019
Just want to try myself in these things.
Phemis on 1:08 PM 02/23/2019: Pass : LiL2281337
nextco on 7:59 PM 02/24/2019: Including .pdb for this entry level crackme is overkill.
erafio on 6:08 AM 02/27/2019: LiL2281337
P45H3 on 12:49 PM 02/28/2019: Perfect crackme
Neuron1k on 2:51 AM 03/06/2019: LiL2281337
Nioh on 4:00 AM 03/18/2019: Cute! Solution: gets, strcmp. Passwort is in plaintext.
Rusty on 3:44 PM 03/18/2019: Too easy, but i like it
KennyX on 6:16 PM 04/03/2019: s_LiL2281337_0041d048 XREF: FUN_00401010:00401040(*)
0041d048 4c 69 4c ds "LiL2281337"
32 32 38
31 33 33
Lucaskyy on 9:08 AM 04/12/2019: Hmm, that was easy.
Nice job though! :)
Kr0ff on 9:23 PM 04/25/2019: There was no need to put the password in the comments, just write a solution, you ruin the whole challenge !
Flyour on 2:42 AM 04/29/2019: It's very stupid CrackMe :-)
greendword on 10:29 PM 05/15/2019: LiL2281337 - constant
lurumdare on 10:11 AM 05/17/2019: 00351040 | 68 48D03600 | push stupidcrackme.36D048 | 36D048:"LiL2281337" easy pass
bazan on 12:17 AM 06/20/2019: The password is LiL2281337
Artykalamata on 7:57 AM 06/29/2019: This one is very easy, even for beginners.
0010200303 on 6:37 PM 07/11/2019: Nice and easy CrackMe just as you said
zeusX03 on 12:27 PM 07/15/2019: Nice Crackme
juansacco on 2:12 PM 10/31/2019: Hello man! It's very stupid CrackMe :-).
Nice job :-). Password found.
P4RS on 6:25 PM 11/20/2019: Pass: LiL2281337 :)
Havlicek420 on 12:47 AM 01/02/2020: pass - LiL2281337
Gaiaphage on 2:18 PM 02/11/2020: LiL2281337
robon on 1:08 AM 04/20/2020: very Easy challenge but i figure it out how its work and how it will be check password by dividing password in different pieces :) pass : LiL2281337
robon on 1:09 AM 04/20/2020: very Easy challenge but i figure it out how its work and how it will be check password by dividing password in different pieces :) pass : LiL2281337
Vbbab on 5:57 PM 05/16/2020: Really simple...
Vbbab on 5:59 PM 05/16/2020: ds "LiL2881337"
Well, that was simple...
For a second, since you were using a gets call, I thought it was vulnerable to buffer overflows and the max size for input was 12... Apparently not.
bulnku on 6:44 PM 07/30/2020: Password is: LiL2281337
You must me logged to submit a solution
Solution by lesquivemeau:Step-by-step solution using a Hex editor or a disassembler
Solution by dehent:Analysis using IDA.
Solution by lurumdare:Patch for all password. Password easy find.
Solution by 0010200303:Detailed information how to find the password in Ghidra
Solution by VOX:Detailed explanation using x64dbg, patches included.
Write-Up is in MarkDown.
Solution by Vbbab:Simple step-by-step solution by finding strcmp call.
Solution by DaveBarrett312:Solving the CrackMe by doing simple string analysis on the executable file. No debugger/disassembler required.
Share how awesome the crack me was or where you struggle to finish it ! (Stay polite)