ezman's easy keyg3nme

Author:
ezman

Language:
C/C++

Upload:
2019-10-13 12:55

Download

Platform:
Unix/linux etc.

Difficulty:
1.3

Quality:
4.9

Arch:
x86-64

Downloads:
29

Size:
16.28 KB

Writeups:
5

Comments:
32

Description

easy, you just need to figure out the logic behind key validation. this should be fairly easy even with an ugly debugger. i'm new here, so the difficulty ranking could be a little off.

Comments (32)
Writeups (5)

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mrT4ntr4 on 2019-10-14 09:41: [Click to reveal]

rootabeta on 2019-10-17 02:06: To the author - This was a pretty fun crackme, and I apologize in advance for the comment I'm about to leave for the previous commenter. None of my criticism is for you. Now on the other hand, @mrT4ntr4 - What is wrong with you? What convinced you it was a good idea to post the answer to a crackme right in full view of anyone planning to download it? You wanna post a writeup somewhere else, that's fine. You even wanna link to that writeup? Go ahead. But there is no justification for waving the flag in the face of anyone else. That completely defeats the point of a crackme - to learn. You rob the player of that opportunity, and that is something I abhor.

juansacco on 2019-10-17 12:27: [Click to reveal]

juansacco on 2019-10-17 12:36: It's possible to write a keygen :D

mrT4ntr4 on 2019-10-18 13:15: Hey @rootabeta, What I wanted to say was that it doesn't really need a keygen, also you'll find many crackmes which would lead you to the solution right away, so you just have to look for spoilers (don't read my comment). PS @juansacoo thanks for posting another one :XD

qu3st1on on 2019-11-06 13:08: good for beginners :D everythind with mod 1223 = 0 i put 2446

evilyach on 2019-11-08 09:11: I'd say it's pretty good exercise to get to know your disassembler, so obviously it is pretty easy, but if you are at the very beginning it's fine. If you want harder challenges, there are plenty on the Internet, guys!

BitFriends on 2019-11-11 16:11: This is not very easy. My decompiler displays the key_validate funtion like this: int validate_key(int arg0) { if (arg0 == ((SAR(HIDWORD(arg0 * 0x1acb0aad), 0x7)) - (SAR(arg0, 0x1f))) * "_registerTMCloneTable") { rax = 0x1; } else { rax = 0x0; } return rax; } Please help me.

cipherhater on 2019-11-14 11:02: Hi, $ ./keyg3nme Enter your key: 42424234 Good job mate, now go keygen me. $ ./keyg3nme Enter your key: tertrret Good job mate, now go keygen me. --- 1213: 29 c1 sub %eax,%ecx 1215: 89 c8 mov %ecx,%eax 1217: 85 c0 test %eax,%eax - 1219: 75 07 jne 1222 (0x1222) + 1219: 90 nop + 121a: 90 nop 121b: b8 01 00 00 00 mov $0x1,%eax 1220: eb 05 jmp 1227 (0x1227) 1222: b8 00 00 00 00 mov $0x0,%eax --- ))) Best regards

Asm0d3us on 2019-11-17 12:30: check - (param_1 % 0x4c7 == 0); so simply 0 as key will work --------------------------------- Enter your key: 0 Good job mate, now go keygen me. ---------------------------------

l@m@l on 2019-11-26 05:18: How would you go about solving this solely using IDA/a debugging tool? I figured out the function that was interesting and the result I needed, but manually reversing the assembly "equation" seems like an inefficient way... Any help/suggestions would be greatly appreciated (I solved by using an alternate program)!

m0xz on 2019-12-04 07:57: having a lot of fun with this one. Also, to everyone else, please don't post solutions in the comment section :/ Thanks ezman, welcome to the community!

BeachyHead on 2019-12-13 18:07: thanks a lot. You were my first. :-P

dav1nc1 on 2019-12-14 03:57: it judge whether input % 0x4c7 ?= 0

rockstardevil on 2019-12-15 18:17: https://youtu.be/3EH5gMRs7lo --- writeup dont forget to subs

hkr0x0a on 2019-12-21 10:50: This was a fun challenge. I am new to this community and I absolutely loved it! Thanks @ezman!

phant0master on 2019-12-25 19:33: Simple keygen: #include int main (int argc, char** argv) { int fixed = 0x4c7; for(int i = 0; i

phant0master on 2019-12-25 19:34: for(int i = 0; i

phant0master on 2019-12-25 19:35: for(int i = 0; i

catzrc00l on 2019-12-30 00:45: #include int main() { for(int param = -2147483648; param != 2147483647; param++) { if (param == ((int)(((long)param * (long)0x1acb0aad) 39) - (param 0x1f)) * 0x4c7) { std::cout

NightChips on 2020-01-03 20:42: Nice

Maksim on 2020-01-05 11:17: keygen: 0x1222 mov eax, 0 - mov eax, 1 seems worked, it was interesting, thanks

ezman on 2020-01-29 14:48: First of all, hello to everyone! I've got a couple things to say. Most of you seem to keep saying things like "too easy" and some of you seem to make fun of me for that, but the thing is if you just look at the difficulty rating you'll see that it's rated very easy - and I kid you not, easy crackmes are supposed to be easy. Anyway, thanks for all the (positive and not-so-positive) feedback, Personally I thought this would be a good introduction for absolute beginners and maybe a good exercise for returners, and some of you seem to agree. Again, I'm still pretty new to this community so it's not like I know a lot either. Anyway, hope y'all enjoyed this little thing :)

B0gg3 on 2020-01-31 13:48: Sorry the crackmes password does not work. Sorry for disturbing you, I am new here.

Carbon on 2020-02-04 18:06: Well this is way too easy. The key has just to be a multiple of 1223. A simple keygen could be: int key = 1223 * rand(1755914); But this could be extended to negative values.

arianizadi on 2020-03-24 07:28: My first time I ever made a keygen! Thanks for the fun crackme!!!

faznc on 2020-07-25 19:54: python keygen for i in range(1223, 99999): if i%1223 == 0: print(i)

Darky on 2021-06-10 13:17: Over Easy =)

hackingmaterials on 2021-07-21 10:29: ./keyg3nme Enter your key: 1223 Good job mate, now go keygen me.

ShadowRoot18 on 2021-12-29 20:33: param1 % 1223 = 0 param1 = multiples of 1223 e.g. 1223,2446 etc..

jimsktsmith on 2022-02-06 13:28: [Click to reveal]

blackdveil on 2023-10-23 09:24: ./keyg3nme Enter your key: test Good job mate, now go keygen me. ./keyg3nme Enter your key: hi Good job mate, now go keygen me. ./keyg3nme Enter your key: h Good job mate, now go keygen me. What the heck, Even i put random text it says good job mate

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Solution by m0xz on 2019-12-04 08:09:
The key validation appears to be hardcoded; the validation function does a simple check against the input to see if it equal 0x4c7, which is 1223 in dec. The application is validates both hex and decimal input. My keygen for this is simply the following one-liner, using python: python -c "print int('0x4c7', 16)" | ./keyg3nme Thanks ezman, this was fun!

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Solution by Kryptoenix on 2021-09-05 19:06:

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Solution by MANA624 on 2021-10-15 02:20:
Thanks for the puzzle! Real simple solution; just check the comparison at the end

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Solution by perXhide on 2024-06-27 08:00:
it's easy to start reverse engineering, just get started

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Solution by Snxhit on 2025-02-12 19:03:
This crackme is fairly simple, it just checks for to see if the entered key is divisible by 1223. I also made a three line python script to output a random key which will work, since I didn't know how exactly this was to be keygen'd. It is a good crackme for beginners to the world of reverse engineering such as myself. Had fun, thanks Ezman!

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ezman's easy keyg3nme