| Impossible CrackMe |
g3chantr |
It's a good challenge :) The solution contains a detailed analyse and a C keygen ! |
2022-02-20 15:57 |
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| crackme2-be-D4RK_FL0W |
KeiTachikawa |
Tried again and fixed the solution. |
2022-02-16 23:46 |
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| CTF #4 by VladMetz |
ThePhilosopher |
A comprehensive solution |
2022-02-15 21:02 |
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| flag_eater_crackme KeyGen |
pjenik@seznam.cz |
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2022-02-14 22:28 |
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| Twiddling |
pjenik@seznam.cz |
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2022-02-14 21:52 |
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| Anti Crackme |
g3chantr |
Good chall to understand basic AntiDebug techniques ! the solution contains a analyse and a keygen |
2022-02-10 19:42 |
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| revv |
ne0n_c4lic0 |
Binary exploration and reversing. |
2022-02-08 17:54 |
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| New years crackme v02 |
g3chantr |
This solution contains a detailed analyse and a functional keygen for a given password :) |
2022-02-05 19:52 |
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| Legion |
husky151 |
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2022-02-04 00:40 |
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| Easy CrackMe - Creeper215 |
Dante777 |
Solution |
2022-02-03 19:55 |
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| Basic |
g3chantr |
Nice chall !! Thanks |
2022-02-01 18:07 |
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| PowerFul_Crackme |
D4RKFL0W |
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2022-01-30 19:04 |
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| SecureSoftware v1.5 |
g3chantr |
A really nice chall !!
This solution contains a detailled analyse and a python code to generate the key ! |
2022-01-30 17:37 |
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| Eat Sleep Trace Repeat |
Elvis |
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2022-01-30 04:06 |
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| CrackMe1 |
DarkKnights |
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2022-01-28 15:02 |
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| gugus the first |
ne0n_c4lic0 |
This walkthrough contains reversing and a little debugging for this challenge. |
2022-01-27 15:01 |
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| Confusion |
dev0 |
Thank you bl4ack! |
2022-01-26 13:36 |
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| cm001 |
husky151 |
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2022-01-26 06:05 |
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| Another KeygenMe |
g3chantr |
Thank's for this chall !!!
This solution contains a detailed analyse and a little kegen for a given username |
2022-01-19 23:10 |
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| kgenme1.0 |
g3chantr |
Really nice crackme !!! Thanks a lot !
This solution contains a details solution using IDA and a keygen script (in python language) :) |
2022-01-18 19:18 |
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| UltimateKeyGenMe |
g3chantr |
Thank s a lot for this chall ! This solution contains a detailed analyse and a python script to generate a password (in the markdown file) |
2022-01-16 19:43 |
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| Grandfather Clock |
MM |
Write-up for the challenge with disassembly by radare2 |
2022-01-16 17:59 |
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| Keypad |
logerse |
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2022-01-16 14:38 |
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| SerialActivation |
g3chantr |
This solution contains a code analyse and a python script to generate a activation key depending of the serial key.
It's a nice chall !Thanks |
2022-01-15 20:05 |
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| 4N0NY31TY's First Crackme |
ne0n_c4lic0 |
This is a walkthrough pdf for this crackme. |
2022-01-13 13:17 |
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| Find the pass |
g3chantr |
Nice chall ! Thank's |
2022-01-12 19:22 |
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| CrackMe_V4_Marquire |
g3chantr |
Thank's for this chall |
2022-01-11 17:37 |
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| Easy CrackMe - Creeper215 |
Johga |
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2022-01-10 14:38 |
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| Simple crackme |
KeiTachikawa |
Very fun, loved it. |
2022-01-10 14:06 |
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| x64_crackme_keygen |
amir9339 |
The solution and the keygen are in the same python file.
Thanks for the challenge! |
2022-01-07 15:33 |
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| crackme_15lug2015 by shindanaide |
yoursleep |
Basically, this isn't just checking like getUsersKey() == key, but its not that hard at all. |
2022-01-06 01:56 |
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| Indexianado |
7cherubin |
- The program calls GetUserNameA to get the logged on user's name.
- It converts the username into all capitalized letters.
- It generates a key by using each letter of the username as an index, based on its ascii code, against an offset pointer to a secret key string ("ThisIsAStringOfLength26MW2"). *It should be noted that, instead of being a normal sequence of bytes, each letter is offset by four bytes.*
- This aforementioned string begins at the hardcoded address .data.00421084 and is offset by the ascii code of the letter 'A' which is 0x41. Thus, the first letter of the string begins at .data.421000.
- This key is generated *after* the user is asked to provide a key of their own.
- If the user inputs the same key as the one that was generated, the program enters its success branch.
- It should be noted that the program does not take into account usernames that contain characters other than letters since the characters of the secret string correspond only to the ascii characters of the capitalized letters. Thus, any other character in the username would point to randomized memory *around* the secret string. |
2022-01-03 20:19 |
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| Keyg3n_M1#1 |
KeiTachikawa |
Loved it, check the decompilation and compare it with assembly. It uses structs so its nice to have some understanding of structs in assembly |
2022-01-03 16:05 |
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| FindMySecret |
7cherubin |
The program calls a function which generates the secret number. time(0) is called to get a seed for its pseudorandomizer algorithm. Then it performs a few simple calculations against the seed and converts it into a floating point number. It proceeds to perform some further calculations and returns and stores the number. The program then requests input from the user. Another function is called where the guess is converted into a floating point value. If the secret number and the user's guess are the same, as a truncated value, then the program ends up at the success branch whithin a separate "decision" function which prints the result.
The aforementioned solution, however, does not explain the strange number that is printed before the program exits. Out of curiosity I decided to investigate and found out that the program simply prints that address. Maybe an easter egg? 420... blaze it? 69... best position? 18... yeah... |
2022-01-03 15:22 |
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| user-pass. |
ILoveMyWii |
My Solution on how to find the username & password it might be a bit complex |
2021-12-30 14:47 |
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| Crackme_OF |
7cherubin |
There is a bug with the implementation of the second scanf used to get the user's comment due to it allowing an arbitraty amount of characters to be inputted.
This bug allows for a buffer overflow attack to be executed if the input is exactly 8 characters.
The input overwrites a value that decides whether the program will follow the success or failure branch. |
2021-12-29 14:59 |
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| Ultimatum CrackMe |
7cherubin |
Simply step through the program until a bunch of comparisons are performed at which point it is revealed that the password has been hardcoded in plain text and compared in an obfuscated (?) way against the user's input.
*Password: VGhpcyBpcyB0aGUgY29ycmVjdCBwYXNza2V5LCB3ZWxsIGRvbmUh* |
2021-12-28 11:48 |
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| Visual C++ crackme :D |
easlog |
1/10 |
2021-12-21 06:33 |
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| Crypt0 - Beginner CrackMe |
Genzen |
My first Crackme, pretty hard for me to explain it well so apologies in advance. |
2021-12-21 04:29 |
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| Easiest crackme |
pareidolia |
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2021-12-21 02:30 |
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| f-crackme |
KikoRiko |
Made a write-up on how to crack the application using 0 modification to the application.
https://github.com/yesitskiko/tk334-s-f-crackme-write-up |
2021-12-19 13:00 |
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| 20$ If you crack me :) |
prev |
simple doc to explain the process of how this program got cracked |
2021-12-19 01:56 |
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| f-crackme |
RelationalAlgebra |
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2021-12-17 06:28 |
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| EZ crackme |
g3chantr |
It a simple chall but nice one ! Thanks |
2021-12-16 11:52 |
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| Keygen me Quick! |
goldenboy |
my solution |
2021-12-13 22:48 |
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| hvm.exe |
Elvis |
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2021-12-11 14:15 |
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| Basic |
ten-ten |
Python solver script to restore the original key. |
2021-11-30 17:59 |
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| MY FIRT CRACKME |
ten-ten |
Python solver script for "MY FIRT CRACKME by Dojas" |
2021-11-30 17:58 |
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| Easiest crackme |
RelationalAlgebra |
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2021-11-28 08:28 |
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| Crack me! (or you can't do that) |
spoo |
Can be read as a txt file as well. Just change the extension to `.txt`. |
2021-11-23 21:50 |
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