downloadbrowseXzzX's XzzX#CrackMe1

Download, 2 kb (password:
Browse contents of

************** XzzX#CrackMe1 **************

author : XzzX
created : 26/05/2007
language : Assembler / MASM

difficulty: 3/10
goal : write a keygen
rules : no patching - far too easy ;-)

info : no packer
no anti-debugging tricks

This is my first CrackMe but I think it's quite hard for a beginner to solve. ;-)


seeing forward for your solutions ^^

Difficulty: 3 - Getting harder
Platform: Windows
Language: Assembler

Published: 26. May, 2007
Downloads: 728


Waiting for at least 3 votes
(we have only 2).

Rate this crackme:

Send a message to XzzX »

View profile of XzzX »


Solution by frugurt, published 03. aug, 2007; download (17 kb), password: or browse.

frugurt has rated this crackme as quite nice.

Solution by TiGa, published 01. jun, 2007; download (38 kb), password: or browse.

TiGa has not rated this crackme yet.

Solution by red477, published 01. jun, 2007; download (6 kb), password: or browse.

red477 has not rated this crackme yet.

Submit your solution »

Discussion and comments

27. May 2007
Interesting and easy:D
28. May 2007
write a solution or at least send me a pm with your name and serial ;-)
28. May 2007
hmmm i dont want to give anything away but from what i can see this crackme cannot be solved. with making a keygen because the end compare is eax to a value that i will not state here. but the reason this value cannot ever be met is because it is obtained through an AND of a value that could never become that value no matter how hard you try.

EG. lets say the key is AB in binary that is 10101011 and your keys "value" is ANDed with 11110101 well lets try to figure out what value we need to put in to get the key.

OUR 1010?0?1
AND 11110101
KEY 10101011

the question marks are their because we need a 1 in the end but we donot have the needed 1 in the and. so please let me know if i am just stupid and explain to me how that could be obtained or lemme know that i am correct. because i am pretty damn confused

28. May 2007
you are fully right. but it isn't a bug. try searching deeper. ;-)

Every name has a UNIQUE serial.
28. May 2007
Sorry for my poor promt translation :)
I doubt that the checking algorithm number two has the decision for each name. But there are decisions for all names beginning a symbol 0xe8. Whether tell, please, there is a decision for name Disasm90
28. May 2007
I'm not sure, if I fully understood you.
My KeyGen gives a valid solution for "Disasm90", if you mean this.

I don't know what you want to say with:
>But there are decisions for all names beginning a symbol 0xe8
28. May 2007
Sorry Dis@sm I had to remove your post because this is almost the solution and it would take all the fun away.

I'll repost it when it is solved.

I think you know how ss,sx are generated. So just think about some maths. There is a much easier solution.

Send me you name/serial when you know it.

5 to 12 ;-)
02. Jun 2007
Original post of Dis@sm:

>Thank you.

>All names beginning with char 0xe8 have solutions like
><empty line>, 0, 0-0, 00-0, 0-00, .....
>In this case the login string will be empty

>I can get solution using code like that: (pascal)
>for i:=0 to 2^32-1 do
>for j:=0 to 2^32-1 do
>if(ss<>(i*$40+j*$c3)) then continue;
>if(sx<>(i*$55+j*$103)) then continue;

>but this would take a very long time

>I will try to get another way to solve it.

>Thank you for your work!
26. Feb 2008
Man, Congratulations. I love this Crackme1. It´s Very nice and clever.
I couldn´t solve the crackme1 alone but it gave me a lot of new info and knowledge.
Special thanks to the autor and Tiga, who makes a spectacular Solution

crackme2,3,4 here I go!!!
26. Feb 2008
Cool crackme. Quite easy, but good fun.

You may leave your comment, thoughts and discuss this crackme with other reversers here.
Acting childish will not be tolerated.
HTML and such will be left as-is, so don't try.