Number of crackmes:
Number of writeups:
Comments:
| Name | Author | Language | Arch | Difficulty | Quality | Platform | Date | Downloads | Writeups | Comments |
|---|
| Crackme | Date | Infos | Actions |
|---|---|---|---|
| zun1 crackme | 2024-12-31 19:14 | I hope you understand my explanation | View |
| Crackme | Comment | Date |
|---|---|---|
| RE02 | [Click to reveal]A bit harder to find: Ch1ll1ng_w1th_Ant1-D1s4ss3mbly_t3chn1qu3_83f52144 | 2026-06-01 21:53 |
| C++ Obfuscator | [Click to reveal]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 | [Click to reveal]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 | [Click to reveal]Really easy to crack using x64 dbg: "meilovecats" | 2025-01-16 17:08 |
| VeryEasyCrackme | [Click to reveal]Its not hard if you use ida pro and decompile it to c++: the password is: "YourPass" | 2025-01-16 16:58 |
| easycrackme | [Click to reveal]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 | [Click to reveal]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 | [Click to reveal]Password: secret123 to patch the program change "JNE" to "JE" | 2025-01-05 18:36 |
| Bubo | [Click to reveal]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 | [Click to reveal]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 | [Click to reveal]00401347- cmp eax,dword ptr ss:[ebp-4] Nice crackme the password is: 19082004 | 2025-01-04 23:33 |
| first crackme by zxsrxt | [Click to reveal]The password is "yourpass" I Used x64 dbg | 2025-01-04 16:19 |
| UPKGC - UnsafeProductKeyGeneratorChallenge | [Click to reveal]First Key- 5ZY4HSUIYKHTPFPN7Q30 | 2025-01-04 15:43 |
| find the encryptor | Nice crackme password: thatflagissus3 | 2025-01-03 14:45 |