Number of crackmes:
Number of writeups:
Comments:
| Name | Author | Language | Arch | Difficulty | Quality | Platform | Date | Downloads | Writeups | Comments |
|---|
| Crackme | Date | Infos |
|---|---|---|
| InDuLgEo V3-B — The DOS Intro Challenge | 2025-07-04 08:43 |
| Crackme | Comment | Date |
|---|---|---|
| Existential Vending Machine | Too easy, so i don't give the right answer. One breakpoint and viewing memory is enough to solve. | 2026-01-12 01:19 |
| Multithread P-Code Crackit | Password : needles | 2026-01-12 00:46 |
| Find correct key | Password : 23623e23623e23623e23623e23623e23623e23623e23623e Wait more than 100 seconds for response on i7-14700K. | 2026-01-10 23:02 |
| Insane Crackme - The Chaos VM | insanecrackme.exe F9,53,7B,F9,55,7B,F9,43,7B,F9,43,7B,F9,45,7B,F9,53,7B,F9,53,7B,F9,5f,7B,F9,56,7B,F9,4d,7B,F9,5f,7B,F9,31,7B,F9,33,7B,F9,33,7B,F9,37,7B Good work, good crackme. | 2025-12-31 20:09 |
| binsafe | Username : nightxyz Key : 5542327fcf2aca5a | 2025-12-31 07:56 |
| Ez Crackme | password is = YYYYMMDD + 4777 For example : 20251229 + 4777 = 20256006 | 2025-12-29 15:22 |
| decode me (hard) | Password: youhavedecodedit! I'm a little late :) | 2025-12-19 13:38 |
| [INSANELY HARD] Uncrackable.exe | Not very hard. Spoiler : I_CRACKED_IT_************** | 2025-12-19 11:38 |
| Easy crackme | Too easy. | 2025-12-15 17:27 |
| puzzle | Access Granted! | 2025-11-21 00:29 |
| VeryHard 99% you can't do it | Password : sPåqSDâçàLMEwåLHåPtL_éÇéåü! | 2025-11-09 16:17 |
| Easiest | We used to avoid sharing crackme solutions with difficulty levels 1 or 2, even if we knew the solutions, to avoid discouraging newcomers. It's been a while since a crackme was released, but if someone still hasn't found the solution, then it's acceptable. | 2025-11-04 00:33 |
| medium crackme | Key : KeDUORLIE | 2025-10-10 20:42 |
| unpackme | Password : SecretPass123 But Flag is strange. | 2025-10-02 21:56 |
| try to debug | Without using any debugger, i executed program at cmd prompt and it says "Enter key: [Guard] DETECT: INT3 found in RX[Guard] DETECT: INT3 in RX". I am using Windows 10 x64 22H2. | 2025-09-30 17:09 |
| logic trap | Password : 00 | 2025-09-29 11:52 |
| Patch protect lite | I set breakpoint at createthread api function and changed rsp+20 from 0 to 4 and thread started in suspended mode. Then it never bothered me. After Cpuid function, there is some mathematics, so i wrote a program in turbo pascal to achive my own unique password which related CPU information. Lots of password has found and i got 'great Job' message from all of them.Thanks karabatik for 0xDEADC0DE exit code information. | 2025-09-28 19:09 |
| by litvin | Password : 789326321 | 2025-09-24 18:53 |
| crack me test | After started main function, second call is anti-debug (inside this call, there is a comment about anti-debug) and after return from this call, you check AL register. I set it so i bypassed it. Then, after getting password from user (randomly entered from me) there is a call which constructs password. I stepped every instruction and password started to become visible in comment section of X64DBG program. RAX register points that password address. End of the function, constructed password is placed [RBP-40] address. You should disable debug information. In anti-debug section, you could set a global variable and could use inside password construction function. So bypassing anti-debug function could change the results. | 2025-09-18 09:13 |
| crack me test | Too easy. It took 10 minutes. | 2025-09-17 13:20 |
| MISSION IMPOSSIBLE | License key : 3f39999999999999 | 2025-09-01 22:31 |
| The Deceiver | Password : 1152 | 2025-08-29 19:18 |
| KeygenMe again! (fix) | You can use any length username but i used 4 characters long username. For example username=AAAA pseudo code: mov eax,username ; eax becomes=41414141 xor eax,AC4C6B37 ; eax becomes=ED0D2A76 password is KGM-xxxxxxxx (totaly 12 chars length) xxxxxxxx is 4 bytes of xor'ed username reverse order.That is 762A0DED password final result is KGM-762A0DED username=AAAA Password=KGM-762A0DED | 2025-08-28 17:29 |
| Non-cryptographically secure hash (my first crackme!) | You didn't mention in the title that you don't accept brute force. There are many crackme authors who hashed the password with sha256 and asked us to find the password. I didn't even think about buffer overflow in your crackme program. I saw that it was performing some mathematical operations, and I automated it. The fact that the password can only be found using numbers is a weakness of the program. The program also provided other passwords besides the one I provided, but I only wrote one. | 2025-08-18 18:17 |
| Non-cryptographically secure hash (my first crackme!) | Pasword : 4530395016 I wrote c++ program to make brute-force. | 2025-08-18 11:59 |
| DLL MEGA HARD v3 | Password : 2--P4F4564865464564564 after P character, it is randomly enter by me. A little try needed to solve. | 2025-08-11 21:56 |
| Welcome to the Rice Fields! (fixed) | Without debugger, i try to solve questions while trying. After some levels later, CPU usage goes high extremely. What makes this console program so the cpu throttles ? I suspected and decided not to continue. | 2025-08-11 15:38 |
| pixy | I think any jpeg file which name is "~~~.jpeg" in the same folder is enough to get the success message. | 2025-08-07 08:30 |
| GoodLuck | crackmes.one site changed some characters in my previous post. If you want to get key, enter "Password is correct!!" as key value. | 2025-08-04 12:52 |
| GoodLuck | Key : gf|`%ygdgdf&n-5ix)c&p Crackme already gives proper key whatever you enter. For example this one give "Password is correct!!". You use SHA256 hash encryption which is impossible to crack.You also made sha256 hash of input key but never use and delete immediately without storing sha256 hash value. | 2025-08-04 12:51 |
| DLL MEGA HARD v3 | I am interesting. Couldn't find the trigger point of messagebox. csrss.exe handles it. Ordinary messagebox, createwindow, dialogbox functions don't work. I see createwindow inside thread but breakpoint inside thread doesn't work because of packer. Anti-debug also bothers me. There are lots of obfuscation codes and decoding mechanism to decode function names. I didn't understand the crackmes name relation with DLL. I didn't encountered any DLL structure inside exe file. Messagebox comes to screen after some delay, so this makes it difficult to detect. I am sure it is inside thread but didn't find yet. Because there are lots of threads. | 2025-07-24 23:44 |
| Custom Packer | The program doesn't work properly anyway. Sometimes it works after a second double-click. Sometimes it says Access Denied, sometimes it displays mixed ASCII characters instead of "access denied" message, and sometimes it displays a message with blank characters. So, i didn't struggle with your custom PE packer. In x64dbg, after some unpacking routines, i came successfully accessed the unpacked file entry point which comes after initterm, argc and argv routines. | 2025-07-22 10:44 |
| Custom Packer | No need to unpack it. Password is HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Cryptography\MachineGuid value. Each user has unique MachineGuid. | 2025-07-21 20:54 |
| dll is so hard v2 | @Ultrazvukoff You made shr cx,1A , which makes almost cx=0. So this section doesn't affect result. If it could affect, it would be harder to calculate and solve. Best regards... | 2025-07-18 18:03 |
| dll is so hard v2 | Password : aaaaabmpc There are lots of passwords which meets requirement. I made brute force because of mathematics. | 2025-07-18 14:24 |
| Dll is so hard! | @ultrazvukoff No, it was not too easy. In fact, this is the first time I've encountered a program that exits after a certain amount of time after calling fgets. Although I didn't fully understand how fgets triggered the function addressed by beginthread, I was able to debug it easily by bypassing it. It's a nice crackme. There are even methods that bypass ScyllaHide. | 2025-07-17 02:44 |
| Dll is so hard! | Excellent crackme. File has embedded DLL file. Die64 detects it. -----SPOILER----- Password : ******96 -----SPOILER----- | 2025-07-16 16:19 |
| Anti-debug fixed | Then my crackme works normally. I bypassed some of anti-debug techniques already like NtSetInformationThread and etc.. | 2025-07-12 13:44 |
| Anti-debug fixed | What is the aim of the crackme ? It always shows "OK" string. If you enter one char, it shows two "OK"s. iy you enter two char, it shows three "OK"s. Under debugger or without debugger, it works like this. | 2025-07-12 13:16 |
| crappy code, crackMe\patchMe\keyMe | JD-GUI failed but CFR succeeded in its mission. But it asks path, i enter c:\users\profilename\desktop and it gives "error occured". However, key works. Key is 7 characters long. | 2025-07-11 21:48 |
| txrbo's basic crackme 1 | @balazsofficial You are right. I forgot to mention it. | 2025-07-10 08:28 |
| txrbo's basic crackme 1 | Best tool is ILSpy to decompile it. | 2025-07-08 15:02 |
| txrbo's basic crackme 1 | --- SPOILER --- Password length should equal or greater than 8 chars. Password should contain al least one Capital Letter. Password should contain at least one lowercase letter. --- SPOLIER --- | 2025-07-08 15:00 |
| Professional Crackme | Password : R0b0_Su22y13378Pussy59283 This one got easy because i discovered "Rand" function from Watermelon crackme. | 2025-07-05 10:19 |
| Watermelon Crackme | Password : N!993R_K1ll3r It was hidden inside "rand" function. After several days, finally i found. | 2025-07-05 10:06 |
| Basic | The password for every crackme file on this site is "crackmes.one" (without quotes). As for this crackme file, it was specified as Multiplatform by the author of the program, but it was wrong. In fact, it is a Linux file and works under Linux derivatives. | 2025-07-04 19:48 |
| InDuLgEo V3-B — The DOS Intro Challenge | @InDuLgEo Thank you. I have used Softice, Sourcer, Turbo Debugger, Debug at old good DOS days. If i see anything related to DOS, it attracts me. At first, i only uploaded COM file but site manager rejected it because it doesn't have any description. Later, i uploaded solution but system only allows one file uploading. I wish I had compressed the 2 files into one and uploaded them. I asked the moderator but he didn't. After I uploaded the file, the system didn't allow me to upload the file again. I will try to upload again. | 2025-07-04 17:25 |
| InDuLgEo V3-B — The DOS Intro Challenge | @Elvis, author want patched file also but i could only uploaded pdf file. Can you combine my first uploaded COM file with this pdf file ? | 2025-07-04 08:47 |
| InDuLgEo V3-B — The DOS Intro Challenge | There are 3 sections in the program. In the first section, there is a blue image with a waving shape. After that, there is a fire effect in the second section. In the last section, there is a fractal but it is not an animated image. In the top rows, the messages are fixed and the phone ringing sound is heard intermittently. In the second section with the fire effect, is it to print text on the screen or to change the messages on the last screen that the programmer printed? | 2025-07-03 08:49 |
| InDuLgEo_CrackME_V2 | Name : nightxyz Serial : 691604A0FA6FA1C99BCED0013AE788A2B39B56AFC0A96867BD2A1C31BE5DC9CE-OMGWTFBBQ | 2025-07-01 12:09 |
| Impossible(ish) CrackMe challenge V2 | I guess I'll make you angry again but this one didn't even take 10 minutes to crack. After seeing the part that checks the length of the password to 10 characters, I entered a random password. After the size check, after a few more steps, x64dbg showed a string as "awesome". I didn't even look at the memory area. I tried that and it worked. The previous crackme was harder in my opinion. If the password in the previous crackme file had been encrypted, my job would have been harder. The plus of this was that while debugging this, if I wanted to run another program directly, the debugger would detect it and the program would exit. And it wasn't even running under the debugger. Your obfuscated codes are very similar to the programs written by "nado" called "watermelon crackme" and "professional crackme", but I haven't been able to crack them yet. There is too much math and obfuscated code. | 2025-06-29 07:02 |
| unpack_me_plzzz | Ida didn't work properly so i used X64dbg. ---SPOILER--- password is 47 chars long. ---SPOILER--- | 2025-06-27 17:04 |
| Impossible(ish) CrackMe challenge | @hmx78912 It's a nice crackme. It was hard at first, I looked at the same places over and over. I examined the anti-debug sections over and over. Then, while tracing line by line, I was looking at the registers and I found the place where the first character of the random password I entered was compared to the "c" character. Then, when I dumped it to the rsi+4 address, I saw the text crackme diagonally. There is a 36 byte difference between each letter. The following part of the code helped me find the result. 00007FF611D52473 | 8BC3 | mov eax,ebx 00007FF611D52475 | 44:3A7E 04 | cmp r15b,byte ptr ds:[rsi+4] 00007FF611D52479 | 0F94C0 | sete al 00007FF611D5247C | FFC0 | inc eax r15b is the address of the password I entered rsi+4 is the address of the real password. I am waiting v2, best regards... | 2025-06-24 18:26 |
| Impossible(ish) CrackMe challenge | @hmx78912 While debugging with X64dbg, I saw that the first character of the real password was the character "c". Then, when I looked at the memory area where this "c" character came from, the letters appeared as "c r a c k m e" in a crisscross pattern, and when I tried that password, I saw that it was correct. So, as you can see, I didn't do brute force. Also, I'm not the first person to directly explain the password in the comments section of this site. If you look at the comments sections of other crackme files, you can see hundreds of examples. I sent a solution once, but I couldn't make the management like it either. Then I gave up sending a solution. | 2025-06-24 16:54 |
| Basic | Password : Qk*{i9}6 | 2025-06-23 22:44 |
| Impossible(ish) CrackMe challenge | Password : crackme | 2025-06-23 21:59 |
| The Obfuscator's Riddle | Password : ElementaryMyDr!! | 2025-06-20 20:31 |
| XOR crackme | Password is 21 characters long. It gets windows usernames like DESKTOP-XXXXXXX/Username and makes some calculations. I am busy to solve mathematics but i found my own password. | 2025-06-17 13:03 |
| Hard | Password : e4444 | 2025-06-08 10:19 |
| virtual machine | Password : ppppp | 2025-06-07 11:37 |
| bym1das | Password : M1d3s19 | 2025-06-06 17:59 |
| MyFirst | I found first pair manually using calculator to get 7326. Numbers 66 x 111 is first pair. Then started with oBo.. but didn't pass criteria. Then started with Boo... and worked. I asked gemini other pair, it gaved me J and c char but i din't calculate. So, i manually calculated every step. Only AI gavr me second pair but i didn't use. | 2025-05-23 08:06 |
| MyFirst | Boozy :) | 2025-05-21 20:20 |
| simple crackme | Password : 4921 | 2025-04-21 19:28 |
| matrix_breach | Correct pattern : 121212121 | 2025-04-21 06:10 |
| C++ Obfuscator | Password : thisisverysecret In x64dbg, a few lines later, letters are shown vertically. So, your obfuscators doesn't work very well. | 2025-04-18 18:49 |
| locked box | password : correct | 2025-04-08 23:36 |
| Robot Crackme | Lots of passwords. One of them is : BP!9$d!! | 2025-04-06 10:17 |
| LabyrinthDEAD | I tested with 66 characters long password and didn't get DEADC0DE code with anyway. Did you test this crackme before publish ? | 2025-02-11 17:15 |
| xorkey crackme | Password : Alon Alush | 2025-01-26 15:48 |
| first crakme xor | Password : iiiiiii} | 2025-01-26 13:51 |
| ktoto | Username : zzzzzzzzzz Password : 1220 Sum of the username letters decimal equivalent. For example if you enter only A char for username, then password is decimal equivalent 65. username : A password : 65 | 2024-12-13 17:34 |
| LSDtrip crackme! | @Mattpackman Patch is not allowed otherwise specified. | 2024-12-04 15:07 |
| LSDtrip crackme! | @Enhancer Password is 4 chars. Total is decimal 895. It also adds Line Feed character as a fifth char. So, 5 x 0ah = decimal 50. Now result is 945. | 2024-11-29 08:30 |
| LSDtrip crackme! | Password = 3p(smaller than sign)n Password is 4 characters long. | 2024-11-21 01:51 |
| LSDtrip crackme! | Password : 3p | 2024-11-21 01:49 |
| First Crack Me | Key = ThisPasswordIsRandomAsFuck | 2024-11-15 17:56 |
| Interesting Crackme | @gimi001 Read Faq at menu. | 2024-11-12 13:54 |
| TryCrackMe (v1) | pseudo code: var1 = length of name looptimes = var1 repeat looptimes { var1 = (var1 * 19660Dh) + 3C6EF35Fh char result array[] = (var1 mod 5Eh) + 21h } split result with xxxx-xxxx-xxxx... format or xxxx-xx if odd length of input name. if result hex value contains letter like 7A, 4B etc.. then add 0Ch to that value. For example 7A becomes 7M, 3B becomes 3N. 3B645B becomes 3N64-5N | 2024-11-08 20:57 |
| Fatmike's Crackme #6 | I only used getdlgitemtexta breakpoint and used F7, F8 keys. | 2024-11-05 07:32 |
| Fatmike's Crackme #6 | Interestingly, i debuged program with x32dbg in my notebook and same password worked. But without debugger, password doesn't work in my notebook. In my desktop, password works without debugger. | 2024-11-05 07:29 |
| Fatmike's Crackme #6 | @fatmike I don't have discord account and discord unavailable in my country temporarily. On my desktop computer, I deleted zip file and exe file. Also turn off my computer and start again. Download zip file again and extracted exe file. Started exe file and entered my password. It says "Well Done". Something is wrong with your crackme because my password doesn't work on my notebook which is same operating system windows 10 x64. | 2024-11-05 00:11 |
| Fatmike's Crackme #6 | I completely closed my x32dbg and executed exe file. My password works only on my computer. Did you try it on another computers ? | 2024-11-04 13:19 |
| Fatmike's Crackme #6 | @fatmike In my computer, it works without debugger. But it doesn't work on another computer. You used rand function inside program, maybe it doesn't work properly. | 2024-11-04 12:07 |
| Fatmike's Crackme #6 | Name : 1234 Serial : 11111111-bbc39b72-94229d8a | 2024-11-04 09:40 |
| Fatmike's Crackme #3 | I simply removed with cffexplorer but in x64dbg, i couldn't bypass. So, i decided not to struggle with more. | 2024-10-13 17:08 |
| Fatmike's Crackme #3 | Name = FATMIKE Key = x8hpx8hpx8hpx8h | 2024-10-13 09:42 |
| Juggler v2 | I used X64dbg , ScyllaHide Basic profile. First, i found the point where user key gets. After that, i traced step by step. I bypassed Int 4 trap. I "traced into" every "call juggler_v2.7FF72EDE9860" because "Trace over" failed. After loading all libraries, I continued to stepping. Somewhere in loop, I saw "curly bracket" next some register. The other register shows my entered key first character. I remembered the password structure for first revision of this crackme and I guessed that this crackme has same type password. I executed loop 32 times and write everey character to a paper. Finally it worked. I can't write solution but i solved with this way and a little bit luck :) | 2024-10-12 23:35 |
| Juggler v2 | Key = {7h3_h4nd_0f_90d_h0v321n9_480v3} | 2024-10-12 10:13 |
| StillAching's CrackMe | @StillAching I used x64 debugger and made 3 breakpoints until getting password. Then continued running step by step. Somewhere, there was comparison using like test eax,eax. I changed zero flag, it said "coreect password". Then i return a few commands back and it compares rcx and rdx register contents. One of them holds password which user entered and the other one holds real password. I tried that password and it worked 😀 | 2024-10-06 00:31 |
| StillAching's CrackMe | StillAching@CrackMes.one | 2024-10-04 16:16 |
| Fort Knox | I solved it, but it is requested to write keygen. | 2024-04-23 13:19 |
| HardBaby (1.5 kb crackme) | Password : UZZZ Password can be multiple value, so there is no one password. | 2024-04-18 13:55 |
| AsTinyAsHard (2.5 kb crackme) | @nopx64 You've probably either made a runtime debugging or made a patch to get that message. Patching is not allowed unless otherwise specified, You need to find the real password. | 2024-04-15 07:04 |
| First crackMe in C++ | ----- spoiler ------- yallGayAsf ----- spoiler ------- | 2024-04-09 08:57 |
| AsTinyAsHard (2.5 kb crackme) | Solved. -----Spoiler------ Starts with "x" character. -----Spoiler------ | 2024-04-09 01:54 |
| self-modifying crackme | @cnathansmith As you said, there is no input even with commandline or argv ,argc etc.. | 2024-03-20 17:40 |
| self-modifying crackme | @cnathansmith Yes, it is burried inside binary and i think it doesn't execute it. I examined the code and several function calls adds numbers like sub_691170. i manually changed EIP to .text.00691614 marking with Ctrl-N key on IDA Pro after sub_6915D0 executed. Then pressing F8 key displayed the Flag. .text:00691614 call sub_691170 .text:00691619 lea eax, [esp+28h+flOldProtect] .text:0069161D push eax ; lpflOldProtect .text:0069161E push 40h ; '@' ; flNewProtect .text:00691620 push 80h ; dwSize .text:00691625 push offset sub_691170 ; lpAddress .text:0069162A call ds:VirtualProtect .text:00691630 push offset asc_6931DC ; "..." .text:00691635 lea ecx, [esp+2Ch+Block] ; void * .text:00691639 call sub_691720 .text:0069163E call sub_6914B0 .text:00691643 call sub_6914E0 .text:00691648 call sub_691510 .text:0069164D call sub_691540 .text:00691652 call sub_691570 .text:00691657 call sub_6915A0 .text:0069165C call sub_6915D0 | 2024-03-20 17:39 |
| self-modifying crackme | ZAYOTEM{FACTS_CAN_BE_SO_MISLEADING} | 2024-03-19 20:33 |
| CFFlat | ANSWER:MORDOR | 2024-03-19 19:29 |
| Bobby | @cnathansmith Good algorithm to solve it. I manually calculated every digits 😀 | 2024-03-18 08:46 |
| KeygenMe(Second)_SWD | @cnathansmith Thank you. I have already found the password equivalent of usernames such as "AAAAA" or "aa". Later, I thought about making a table for all the letters, but I gave up because it was a long job and I was struggling with cracking the password of the crackme file named Bobby. | 2024-03-16 00:09 |
| Bobby | Bobby's medicine is : 07014620352040506000012238200008007020113308080020070008 | 2024-03-15 00:17 |
| simple password crack me (not simple) | We will see, who will pull the shortest time value out of his ass ? After all, we don't know the truth. | 2024-03-11 18:45 |
| KeygenMe(Second)_SWD | Found valid usernames/passwords but couldn't solve algorithm yet. | 2024-03-08 21:22 |
| simple password crack me (not simple) | 9 characters beginning with capital letter. 2 hours 20 minutes with manual unpacking using x64dbg. | 2024-03-08 16:46 |
| KeygenMe_SWD | Use latest ILSpy. YAY - GOOD JOB | 2024-03-03 02:51 |
| Heaven's Gate & Impossible Disassembly | @rrookie You are right, there is no loop. Only four characters are loaded. As you see, there is no print function also. I dedoded rest of the bytes with same xor value by manually. I thought that author wants this secret message. | 2024-02-21 08:16 |
| Heaven's Gate & Impossible Disassembly | @Programista After your code snippet, Process enters 64-bit mode and your debugger doesn't follow it. After the "ret far" command, there is a code snippet which ends with "ret far" again. The codes in between are 64-bit instructions. To decode it easily, copy that code hex equivalent and paste it to online x64 disassembler. For example this online disassembler : https://defuse.ca/online-x86-assembler.htm Hint : EB instruction jumps next byte in 64-bit mode as you will see in online disassembler. So, delete EB and redisassemble it. Also search on this site with name "heaven" and there are two other heaven's gate related crackmes which has solution on it. | 2024-02-20 21:31 |
| Find The Password | What is the challenge ? I found password at memory and enteree it. It gives success message. It creates random password at every start the crackme file so there is no fixed password. | 2024-02-20 01:01 |
| Heaven's Gate & Impossible Disassembly | @rrookie Yes,0x153168 is address of flag data which begins just after "did you capture the flag" message. It is xor'ed with 0x68657974 value four by four. | 2024-02-20 00:57 |
| Heaven's Gate & Impossible Disassembly | ZAYOTEM{r3v3r53_3n61n33r}eyt | 2024-02-12 10:22 |
| Heaven's Gate & Impossible Disassembly | ZAYOTEM | 2024-02-12 10:13 |
| Easy Password Reverse | Username : Any name Password : Sum of the ascii number of username characters. | 2024-02-03 10:39 |
| Try and patch me | @cnathansmith Check your system because this file is trojan. Check with virustotal. | 2024-01-31 18:08 |
| Try and patch me | Too easy. patched. | 2024-01-30 09:17 |
| Really Very Easy | It was really easy. | 2024-01-22 19:45 |
| Lab2 | Password : HkVf3z8MS2 | 2024-01-12 13:04 |
| lab1 | "Knight's Tour" problem in Chess Board. Password : 00473049321562172950450261181334460148311433166351280344196035120443245508392059275207402356113642055425380958215326410657223710 | 2024-01-11 22:24 |
| im crackme | @sporta778 Protectors like Enigma, Themida, VMProtect etc.. use obfuscation, anti-debug and many methods like viruses and trojans. So, those antiviruses give false alarms for these crackmes especially. | 2024-01-05 23:30 |
| im crackme | @sporta778 if you scan with virustotal.com, lots of crackmes give false alarm because trojans use same methods like anti-debugging etc.. | 2023-12-26 23:56 |
| messy crackme | @JMac2 We gave argued with you previously. I posted a solution and administrators rejected and even you didn't like my solution. So there is no need to effort making a solution. If you are not blind other posts, no everybody post solution, just gives passwords in the comment section. If admins makes forbidden to everyone, it is okay for me. So, go your way. | 2023-12-16 14:29 |
| messy crackme | @sporta778 You are right. I am the first to reveal the password in the comment section on this site!!!. I look at other crackme files and a lot of people explained the password in the comment section and no one said anything, but whenever I do it, a crack head immediately comes and criticizes me. Take a look at yourself before criticizing me. Just the other day, you wrote the answer to the crackme file whose password was "pizza" in the comment section. Don't lose motivation, work harder. | 2023-12-14 15:52 |
| messy crackme | User ID : admin Password : 328c5620c90182b4de6d21cfca38a4ec | 2023-12-13 22:35 |
| hacktooth crackme #3 | @hacktooth No. I found serial in dump view and the others are brute force and some luck :) | 2023-12-01 22:39 |
| aname crackme | @sporta778 Greats. I couldn't find time to solve those xor's , adding 6, etc.. I was waiting for someone to figure this out so I could get rid of it :) | 2023-12-01 16:11 |
| hacktooth crackme #3 | username and password are 6 characters long and must be identical. Serial : 0000-0000-0000- No need to write a keygen. | 2023-12-01 16:08 |
| Bubo | Solved. Password is 16 characters long. | 2023-11-29 22:01 |
| XOR crackmes | @JMac2 I think he made a mistake and kept us busy for nothing. He wants to cover up his mistake by saying that he forgot to say "You can patch". A virtuous person should accept his mistake and say it clearly. | 2023-11-27 23:41 |
| third | Multiple answers. solved. | 2023-11-21 10:15 |
| asm | Part 1 : 15 Part 2 : 66 After solved, i checked solution text file inside zip file. | 2023-11-21 00:45 |
| im crackme | Password : IM_GOOD | 2023-11-20 06:01 |
| func crackme | Password : 7FEB3EB8230 | 2023-11-15 19:24 |
| Trappy Crack me | It was too easy and it doesn't deserve this level of difficulty. | 2023-11-09 09:54 |
| find the encryptor | "Yoniaga yrT .ekatsim a edam u!!" | 2023-11-08 10:29 |
| crackme0 | Password = jajajaja | 2023-11-08 09:44 |
| XOR crackmes | @SUPERNOVA Yes, it gives repeatedly same values. So, i asked to the author whether he checked it or not. | 2023-11-04 18:55 |
| XOR crackmes | Did you check out ? does it work ? | 2023-11-03 20:59 |
| ROT13 crackme | Key : 5418 | 2023-11-01 17:41 |
| crackme | Solved. Leet Hacker :) | 2023-10-21 19:57 |
| illusion.exe | @cnathansmith Thank you for taking the time to share your findings. | 2023-09-14 07:55 |
| illusion.exe | @cnathansmith Did you solve ? There are antidebugs, obfuscations, self modifiging codes, tlscallbacks, 2 extra threads, cryptos. After getting registry key, i didn't found any location where it uses this information. | 2023-09-13 18:42 |
| timotei crackme#12 | @JMac2 There can be no justification for such a refusal. If you examine the solution file that they accepted, there is almost no byte left in the file that the man did not change. I spent so much effort and solved it with only 2 bytes of change. Maybe you don't like my way of explaining, but as you exaggerate, the use of software is not more than a crack solution. Since the Debugx program is not as popular as Ida pro, dnspy, ghidra, etc., I explained some commands for the convenience of the reader for the solution. | 2023-09-05 15:44 |
| GOLDBOX | @frank2 Too many mathematics... I have no time to solve and write a keygen but i can supply second key : NNNN-NNNN-UQYU-NKHF There is no keygen requirement in challenge summary. Maybe someone else digs deeper. Password is 19 characters long included - sign. First 8 characters are taken and derived second 8 characters using character array "OFCKANLUPEQDHXTYWBMImqXagNiZJWlEFSydocHP". I found second part from memory before go/nogo intersection. | 2023-09-04 18:05 |
| timotei crackme#12 | @mshark1729 I explained the reason above, but you did not understand. If you examine the comments section of other crackme files, you will see that many people do as well as me. Comments section does not require approval, solution section requires approval. Understand this difference. I gave you the link above, I wrote the solution file there in the comment section because the administration rejected it. I reproached some kind of management. It's not ego, if you work hard you can do it too. Even though I knew the password, I didn't send it, I just gave a hint. | 2023-09-02 22:48 |
| timotei crackme#12 | @mshark1729 This is solution which management has rejected. https://crackmes.one/crackme/5ab77f5833c5d40ad448c3a6 | 2023-09-02 21:19 |
| timotei crackme#12 | @mshark1729 We can't make the management like the solution. They refuse without giving a reason. I guess they want fancy pdf, they don't accept txt file. Nowhere is it specified how the solution file should be.That's why it's easier to write the result here instead of wasting effort to prepare a solution. | 2023-09-02 21:15 |
| heaven.exe | brb{stair_ret_way_to_heaven} | 2023-09-02 09:20 |
| GOLDBOX | OOOO-OOOO-TTPL-LHNE | 2023-09-01 23:55 |
| timotei crackme#12 | Cracked. Hint: "Sum of all proper divisors of a natural number". Don't include natural number itself. | 2023-09-01 08:53 |
| FirstCrackMe | Code : 263132 | 2023-08-31 08:49 |
| My First Crack Me | By cracking .Net programs, I usually use dnSpy. | 2023-08-18 06:06 |
| My First Crack Me | Username: Any username Password: Ftwer^*!@$))*54^YUUJHssd | 2023-08-18 06:05 |
| timotei crackme#10 | @timotei_ We can't make the management like the solution. They refuse without giving a reason. I guess they want fancy pdf, they don't accept txt file. Nowhere is it specified how the solution file should be.That's why it's easier to write the result here instead of wasting effort to prepare a solution. | 2023-08-18 05:57 |
| Crack Me v2.- | Username : Any username Password : 996486sdfusernamejnfn$%^&%^hds for example: Username : TOUCH Password : 996486sdfTOUCHjnfn$%^&%^hds | 2023-08-17 22:58 |
| FindMe | USER-ID: user0x1 Name: Any strings... | 2023-08-17 22:43 |
| timotei crackme#11 1K-Edition :-) | Password is "|2]+3567439235 or |2]+3567439235 according to use. Usage at cmd command prompt : cm#11.exe "|2]+3567439235 Desktop shortcut : "|2]+3567439235 | 2023-08-17 22:25 |
| timotei crackme#11 1K-Edition :-) | I will upload password today night. Password stayed on my computer. I am not at home now. First 4 character and 10 digit numbers. Total 14 digits. | 2023-08-17 11:44 |
| Native AOT CrackMe | Password is : This code was compiled natively using .NET 7 AOT compilation, and you found the password! Well done. | 2023-08-17 10:01 |
| Auth Pin | 3924 | 2023-08-17 08:48 |
| timotei crackme#11 1K-Edition :-) | what is gg ? | 2023-08-17 08:47 |
| timotei crackme#11 1K-Edition :-) | Cracked. | 2023-08-17 00:59 |
| BadPin WPF Edition | I tried that password to this crackme and it worked. You shared source code of this crackme. | 2023-08-14 12:22 |
| BadPin WPF Edition | You have been shared the password two weeks ago. | 2023-08-14 06:57 |
| pavler3 | Username : pavler Password : qbwmfs | 2023-08-11 18:24 |
| TinyCrackMe | @bang1338 You are right. I made lots of effort to prepare a solution. But management rejects without making any description. So, i write keys, passwords, solutions to the comment section. Maybe, they can change their policy. | 2023-08-11 12:15 |
| TinyCrackMe | @bang1338 You made same things on your previous comments. | 2023-08-11 10:50 |
| Very Hard Crackme | Any clue ? Can we manipulate system date as cnathansmith said ? | 2023-08-09 15:45 |
| timotei crackme#10 | Name : 1234 Serial : 1234178575008 | 2023-08-08 18:59 |
| Spoof me! (network validation) | Virustotal warns about file as trojan. About 20 of 70 virus scanners. Is it normal ? Or is it because of some ptotection, packers etc.. ? | 2023-08-05 17:56 |
| patchit by sas0 | I used Debugx for disassembly and used hexworkshop for patching. I traced every instruction by using T key inside debugx. I choosed 32-bit registers by typing "rx" inside debugx. I arrived the location where "Nop this isn't the right msg" string displaying code via 09h function of int 21h. Address was 0088h. Then i used -d command for memory dump to the screen. Other good message "Good work you did it" address was 006E. I noted these addresses. I restarted the program and i traced with -t command and realized that xor'ing , adding , ror'ing mechanism 4 bytes 4 inside "CALL 0087". This is the decryption function. Before it leaves function (0087h) , last operation : it adds 30393632h to the eax value , then ror 4 times and xor with 48AC697Bh. Last dword was : D12CD6CA ; it stores this in EAX. 30393632 ; it adds this constant. ------------ 01660CFC it rors 4 times. C01660CF ; result is. 48AC697B ;xor with this constant value. -------------- 88BA09B4 ; result is. BA 88 means = mov dx,0088 (note that zero section is in previous 4 bytes block). B4 09 means = mov ah,09 88 is bad string address. So i changed it with 6E and reversed the mathematic. 6EBA09B4 48AC697B xor constant. ------------- 261660CF result. rol 4 times. 61660CF2 ;result. 30393632 ; subtract this constant. ------------------------------------ 312CD6C0 ; result. Finally , using HexWorkShop i changed the bytes and saved. Totally only 2 bytes changed at offset 027C CA D6 2C D1 old value. C0 D6 2C 31 new value. | 2023-08-03 09:19 |
| timotei crackme#9 | @BeginnerCracker123 There is no stored password inside file.There is an algorithm which you must solve to get valid passwords. I wrote pascal program to find a valid password using algorithm and brute force and finally found one of them is 141157CM | 2023-07-14 13:33 |
| my first | Software crashes when enter 12-characters wide password. Probable strcpy_s misusage. | 2023-07-11 20:21 |
| timotei crackme#9 | 141157CM | 2023-07-11 13:36 |