downloadbrowsemopy's Simple math keygenme ;P

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I think this little keygenme is to hard for you :P
It uses simple math instructions,but in a hard order :D

-no patching

on success:
-write a keygen + solution

Difficulty: 2 - Needs a little brain (or luck)
Platform: Windows
Language: Assembler

Published: 05. Jun, 2012
Downloads: 389


Votes: 7
Crackme is quite nice.

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Solution by MR.HAANDI, published 20. jun, 2012; download (147 kb), password: or browse.

MR.HAANDI has rated this crackme as quite nice.

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Discussion and comments

06. Jun 2012
Take the 0x40-character serial number. Of the serial number are selected 8 characters and are divided into 5 groups.
With 2, 4, 7 and 8 group perform arithmetic operations.
At the end of the calculations in EAX should be 0x48A86FC4...

How? Bruteforce?
06. Jun 2012
bruteforce is for the last is the problem(2math involved is as checksum with div/imul cmp) maybe as not solved by math simples(and, xor, add), only bruteforce...
<as was see>
i think there the serial as idea
not will work you must use
x= 0x40 char as say with serial number
post check
there that eax, is influenced by a add(or and)
you will see there add example 2+7 and save in stack xor 1(35059FC5), thinking there 35xx is a value of add , now is xor 2 =eax
004011E8 |. 334424 28 XOR EAX,DWORD PTR SS:[ESP+28] (35059FC5 xor 7DADF001(number+other/anded before 11258023)=value eax)

the real problem is if and is "AND with 11258023", not must exist a value more of that and max xor is 24201FE6, and will be influenced by the add ...really math problem an not a simple algo...
maybe the bruteforce will be in only digits in leng with x40, and for start only with 2 places to cmp,the edi'reg place

and post 2+2 and next there integrated all values with the parcial solution to check what's limit of values is valid...
or taking a year for check the all numbers posible in 0x40 and saving to analize (maybe only with cuda can done some amazing work)

(but are a and for the next area not loss)
post will see xor ebx, stack value, and save
that value with xor, for me crash the idea of BF
...(not have time) br.Apuromafo
07. Jun 2012
The serial number as stated above, is divided into 8 groups.
GR1, GR2, GR3, GR4, GR5, GR6, GR7, GR8. Each group should be in the range from 00000000 to FFFFFFFF.

There are five tests with the use of mathematical operations:
1. GR_2 \ GR_4 \ GR_7 \ GR_8... EAX should be 0x48A86FC4...
2. GR_1 \ GR_3 \ GR_5 \ GR_6... EDI should be 0x489AC581...
3. GR_4 \ GR_7... EDX should be 0x6582138E...
4. Not reached yet... EDI should be equal ECX...
5. Not reached yet... EBP should be equal EAX...

I passed the first two tests ...
Serial is: 00000000000000001C32C3996C8870224748C1E6000000007DADF00100000000
07. Jun 2012
Not enough time! I'm not going to continue to research! I'm sorry!
20. Jun 2012
Just some bit shuffling and basic math:
The math instructions are not really in a "hard order". In my keygen they are basically in the same order.
Nevertheless a fun keygenme.
20. Jun 2012
MR.HAANDI you're great :)

Now upload please your keygen+(source)+good tutorial :]
25. Jun 2012
Very professional solution MR.HAANDI :)

Thank you :)

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