You can find the rest of the Crackme Challenge series here. Part 6 The code for logical segment 6 is as follows: 0040181F |. B9 10000000 mov ecx,10 00401824 |. 8DB424 C0000000 lea esi,dword ptr ss:[esp+C0]sta 0040182B |. 8D7C24 10 lea edi,dword ptr ss:[esp+10] 0040182F ...

CrackMe Part 5: Logical Code Segments Continued The code in logical code segment 4 additionally changes the stack at address [esp+70]. The code is presented here: 004017E5 |. B8 3F000000 mov eax,3F 004017EA |. 8D4C24 70 lea ecx,dword ptr ss:[esp+70] 004017EE |. 8BFF mov edi,edi 004017F0 |> 8B55 ...

First we must take a look at the following piece of code that will be presented in the code segment 5: 004017FC |. B8 40000000 mov eax,40 00401801 |. 33C9 xor ecx,ecx 00401803 |> 8B940C C000000>/mov edx,dword ptr ss:[esp+ecx+C0] 0040180A |. 3B540C 70 |cmp edx,dword ptr ...

If we take our predicate that we've seen in the end of part 2 into account and input at least 64 bytes (0x40) into the Key 1 field and leave the Name field at a value AAAAAAAA, a second message box is displayed as can be seen in the ...

The First Message Box Let's start our unpacked program with OllyDbg, run it, input eight A's into the Name and Key 1 field and press CHECK STAGE 1. What happens is that a warning message is displayed saying that the key is invalid as is presented in the ...

Presenting the Problem The CrackMe challenge was first observed on the ESET CrackMe web page, which looked like the the picture below: On the web page we can observe that the challenge is being held at Black Hat 2012. The winner will receive free entry into one of the conferences ...