| RE02 |
A bit harder to find:
Ch1ll1ng_w1th_Ant1-D1s4ss3mbly_t3chn1qu3_83f52144 |
2026-06-01 21:53 |
| C++ Obfuscator |
First i found the password and then when i got the password "thisisverysecret" I tried to see what happened in the program logic when the key was correct to try to patch it and I succeeded :) |
2025-04-23 18:18 |
| Password-cmd |
The password is "10" you can see in the "cmp [rsp+58h+var_38], 0Ah" is comparing our input with 0Ah (10 in hex) |
2025-02-26 21:54 |
| zun1 crackme |
Dont worth to make a keygen
because its really ez
https://pastebin.com/bN0PtZEL |
2025-01-16 22:05 |
| Find the Flag |
Really easy to crack using x64 dbg:
"meilovecats" |
2025-01-16 17:08 |
| VeryEasyCrackme |
Its not hard if you use ida pro and decompile it to c++:
the password is: "YourPass" |
2025-01-16 16:58 |
| easycrackme |
In the main function:
sub_401639 is called with v4 as target, v5 as input ("331") and a3 = 22.
This means it generates a key derived from "331" by applying an offset of 22.
For "331":
'3' → (3 + 22) % 10 = 5
'3' → (3 + 22) % 10 = 5
'1' → (1 + 22) % 10 = 3
Result/Correct Key: "553" |
2025-01-09 15:27 |
| Interesting Crackme |
My keygen: https://pastebin.com/G2w3psbT |
2025-01-09 15:09 |
| SecretKeyCrackme |
Here my solotion (it took me more work than expected)
https://pastebin.com/LmgNGD3S |
2025-01-08 22:40 |
| LSDtrip crackme! |
here my keygen for this crackme: https://pastebin.com/YRPLfEKq
|
2025-01-07 22:34 |
| crypted string crackme |
Password its in this adress on x64 dbg
00007FF7B99112E1| 49:0F47D6| cmova rdx,r14 | rdx:"crackmeYG" |
2025-01-07 21:27 |
| Crack me for beginners |
Password: secret123 to patch
the program change "JNE" to "JE" |
2025-01-05 18:36 |
| Bubo |
The password is: 1304ckletlqgjnbo
put a breakpoint here on x64dbg:
0000000000409C01 | 48:8B08 | mov rcx,qword ptr ds:[rax] | rcx:"1304ckletlqgjnbo", [rax]:"1304ckletlqgjnbo" |
2025-01-05 18:25 |
| MEDIUM |
The program basically does this:
Username: 4 characters of any type.
Password: Sum of the ASCII values of the 4 characters of the username, multiplied by 4.
Here my c++ keygen
https://pastebin.com/Mh7fiJAr |
2025-01-05 17:46 |
| ezcrackme |
The comparison at 00007FF7EAA012768138 32313430cmp dword ptr ds:[rax],30343132 checks if the first four characters match the hexadecimal value 30343132, which translates to the characters "2140" in ASCII.
The comparison at 00007FF7EAA0127E8078 04 31cmp byte ptr ds:[rax+4],31 checks if the fifth character is '1'.
Therefore, the correct password is "21401 |
2025-01-05 15:00 |
| crackme |
00401347- cmp eax,dword ptr ss:[ebp-4]
Nice crackme the password is: 19082004
|
2025-01-04 23:33 |
| first crackme by zxsrxt |
The password is "yourpass"
I Used x64 dbg |
2025-01-04 16:19 |
| UPKGC - UnsafeProductKeyGeneratorChallenge |
First Key- 5ZY4HSUIYKHTPFPN7Q30 |
2025-01-04 15:43 |
| find the encryptor |
Nice crackme password: thatflagissus3 |
2025-01-03 14:45 |