| CrackMe Hard 2 |
@VladMetz Thanks, you can solve my newest crackme called CrackMe Hard 5 |
2026-06-07 12:39 |
| NUCLEAR PROJECT 2 |
the python script is following:
gen.py:
def gen_password(username: str) -> str:
result = []
for c in reversed(username):
result.append(c)
n = int(str(ord(c))[::-1])
if 32 <= n <= 126:
result.append(chr(n))
else:
result.append("x")
return "".join(result)
def main():
username = input("Enter username: ")
if len(username) < 4:
print("Username must be at least 4 characters long.")
return
password = gen_password(username)
print(f"Password: {password}")
if __name__ == "__main__":
main() |
2026-06-02 05:57 |
| NUCLEAR PROJECT 2 |
@j2cks1337 ez by lbx
crackmes.exe
===============================
===NUCLEAR HIPPIE MONEY GANG===
===============================
Enter username: liboxin
Enter password: nxixxxoobYixlx
Authentication successful! Welcome, liboxin!
|
2026-06-02 05:16 |
| CrackMe Hard 4 |
@MouadDS CrackMe |
2026-06-01 04:42 |
| CrackMe Hard 4 |
@zqdev @karabatik yes, you both did a good job, and thanks for your advice, and if you don't mind, could you leave any social media for me, so we can talk about reverse engineering. |
2026-05-31 13:27 |
| CrackMe Hard 3 |
@ogurets8235 VHARD{VMX-T1M3-L0CK-9F2} is the right answer, I try again bro, did you patch the program??? |
2026-05-30 11:34 |
| CrackMe Hard 3 |
@KasperskyLabsEngineer fantastic，could you make a write-up to share how you solve it for us,I am looking forward to it. |
2026-05-29 16:39 |
| CrackMe Hard 3 |
@KasperskyLabsEngineer fantastic，could you make a write-up to share how you solve it for us,I am looking forward to it. |
2026-05-29 16:39 |
| CrackMe Hard 3 |
@jhon_crack could you share how you did it?? |
2026-05-29 03:02 |
| CrackMe Hard 2 |
@jhon_crack could you share how you did it?? |
2026-05-29 03:01 |
| A real test for a professional |
@misha_2007 yes,I do know that there should be a space at the beginning and the end.If I didn't know that, the program wouldn't print the success info, find the password is very easy by xdbg in dynamic debug.
I want to ask you that by my static analysis, your password is to read the path /a/ and /b1_H/ file to a buffer, first compare the size, and by a simple xor operation to decrypt the cihiper to the memory and call the memcpy to compare the input and decrypted buffer is same or not.Could you tell my opinion is right or not. |
2026-05-28 11:58 |
| A real test for a professional |
@misha_2007
Enter the password: eUIBPa98aIOJve647adsMKEWATq8qev
Correct password
I might get the right password, but I found the file in c/data is auto modified but still encrypted, which is unreadable for human.So could you give some Hint?
|
2026-05-28 09:10 |
| first golang crackme |
Enter password: simba123
CORRECT!
mov r8,33323161626D6973 == (0x73, 0x69, 0x6D, 0x62, 0x61, 0x31, 0x32, 0x33) == simba123
cmp qword ptr ds:[rdx],r8
ez by lbx |
2026-05-28 08:34 |
| CRACKME |
Enter password: simba123
CORRECT! |
2026-05-28 08:21 |
| Simple VM Crackme |
@talhasa ez by lbx
Please enter the key: c0rRect!
Key is correct! Good job! |
2026-05-26 12:27 |
| CrackMeBaby2 |
@igr0t7 ez by liboxin
==============================================================
Input the pass:
flag:igr0t-theking
You are good , but...
Say my name...
igr0t
You Win.. (;
1."lfkm0cmx:~'~boacdm" every char xor 0xA == flag:igr0t-theking
2.author name == igr0t
then pass the procedure |
2026-05-22 03:45 |
| PhantomGate |
@AdminX by the way, what is your discord nickname/username ??? |
2026-05-19 10:44 |
| PhantomGate |
@AdminX yes, I have discord, my discord nickname is golden_unicorn_89655, you can search and add me,and we can talk with reverse engineering |
2026-05-19 04:46 |
| PhantomGate |
@AdminX your crackme is fun, the information strings are encoded by a func, and decoding while running, and the core check func has VM check and Z3 symbol check and XXXXX-XXXXX-XXXXX-XXXXX format check, but luckily, I finally solved this, and I wrote a write-up here, and you can try to read it |
2026-05-18 11:54 |
| PhantomGate |
@AdminX
PhantomGate v1.0 -- Reverse Engineering Challenge
Enter serial (XXXXX-XXXXX-XXXXX-XXXXX): 4QJCG-P4R7Y-3VMY7-XRWHM
[+] Access Granted! The gate yields. Well played, Reverser.
Press ENTER to exit... |
2026-05-17 17:29 |
| Find the decryption key | Custom packer,no virtualization #2 |
@deo I've dumped the file by using x64dbg, but it seems like the dumped file is not totally unpacked and it's very hard to find the logic of the decryption,is there any one solved this crackme, if someone solved, please share your solution and write-up,it's a very crazy crackme, and now I'm going to be crazy |
2026-05-13 13:43 |
| Smidge's crackme |
@Smidge
interesting,your crackme use MT19937 to gen a random string Code, and the password is (username + randomCode + reverse(username)), here is the result:
Input user: liboxin
Input password: liboxinXhbIUNFEZjnixobil
Congrats
Welcome in ! |
2026-05-12 17:14 |
| CrackMe#1 |
@iseey0u-re
An interesting Crackme, when the exe file under the debugger, the xor key will be changed from 93 to 109, and I finally solved this
Enter the password: crackmes
Success! |
2026-05-12 12:26 |
| InputPrinter |
@khaledddd a good pwn challenge，I've pwned your elf file by writing the shellcode and overflowing the buffer
liboxin@ubuntu:~/pwn$ ls
exp.py flag vuln
liboxin@ubuntu:~/pwn$ python3 exp.py
[+] Starting local process './vuln': pid 82795
[+] buf = 0x7ffd6fb42630
[*] Switching to interactive mode
$ ls
exp.py flag vuln
$ cat flag
flag{pwned_by_liboxin}
$ |
2026-05-09 16:16 |
| KeygenMe_3_SWD |
Username: liboxin
Secret code: 0G8A-03U06-F0M
Verification PIN: 1030886717
Secret code -> PASS
Verification PIN -> PASS
a good crackme, it stuck me sometime, but I finally made it |
2026-05-08 12:39 |
| VaultBreaker |
Op3nS3same |
2026-05-07 05:35 |
| DebugMe |
@Ben_Lolo so the solution of the crackme is to write the hook code to hook the debugger-checker API to pass all the verification stages instead of patching the file or register, right ? |
2026-05-06 15:21 |
| Cultures Puzzle |
@zqdev could you tell me how to find the key of stage 2, bro, I was stucked in here |
2026-05-04 16:22 |
| DebugMe |
@Ben_Lolo your file should be being under debugged to get the right solution on Linux,being under the debugger, the file would be triggerred into the verify procedure, is that right??? |
2026-05-04 16:16 |
| DebugMe |
$ ./DebugMe_patched 0
DebugMe stage 1: Pass
DebugMe stage 2: Pass
DebugMe stage 3: Pass
Well done :)
|
2026-05-03 13:46 |
| Tenzo Crackme |
== Tenzo Crackme ==
Enter password: wowyoufoundit
Congrats! Access granted. You found the right password.
Press Enter to exit... |
2026-04-30 15:59 |
| CrackMe Hard |
@DevVolodya that's OK, I'm waiting here to see your write-up |
2026-04-30 04:40 |
| kernix! |
C:\Users\lenovo\Desktop\69eac2a53fba64e45dcea8a1>kernix.exe --sign correct_test.bin
signed. hash = 0xCDCC2520
kernix.exe is now sealed. ship it.
[sign] ok |
2026-04-29 13:31 |
| CrackMe Hard |
@qobyce you did a good job bro, can you share your solution or make a write-up for others |
2026-04-29 11:31 |
| aola |
b.exe
Enter password: Yippie-Ki-Yay
ok |
2026-04-29 11:29 |
| JustTry crackme |
username:123
Serial:f01e16c07344b098 |
2026-04-23 06:16 |
| Simple login crackme |
User: 123
Pass: 49307604
Access granted |
2026-04-17 05:33 |
| Simple login crackme |
User: 123
Pass: 49307604
Access granted |
2026-04-17 05:33 |
| Find The Key |
Enter the magic number : 4
Enter Activation Key: ZW-ZZ
Z+{Act1va7i0n}
You must input like %%-%%(% -> the char you input), your input string length must be 5, and input[3] - input[1] == 3,and each char you input must be a capital letter, like AA-DA, ZA-DA,etc |
2026-04-16 13:52 |