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Dec
04

iCTF 2011. Challenge 29 ($800)

Program is not packed x86 executable PE file.
binary

After analyze winmain function, you can see, that it is mostly anti-debug code:

IsDebugProcess = 0;
 kernel32_base = 0;
 rdtsc_low = 0;
 rdtsc_high = 0;
 rdtsc_high2 = 0;
 kernel32_base = getKernel32Base();
 printf("Enter key: ");
 for ( i = 0; i < 0x30; ++i )
 {
   stdin = _iob_func();
   new_symb = fgetc(stdin);
   if ( (signed int)(unsigned __int8)new_symb < '0' || 
        (signed int)(unsigned __int8)new_symb > '9' )
   {
     if ( (signed int)(unsigned __int8)new_symb < 'a' || 
          (signed int)(unsigned __int8)new_symb > 'f' )
       break;
   }
   pass[i] = new_symb;
 }
 AntiDebugRdtsc(&rdtsc_high, &rdtsc_low);
 IsDebugProcess = (int (*)(void))GetIsDebugger(0xA36DC676u, 
                                               kernel32_base, 
                                               LoadLibraryW, 
                                               GetProcAddress);
 if ( i == 0x30 && IsDebugProcess() <= 0 )
 {
   v3 = 245 * (IsDebugProcess() + 3);
   i = (unsigned int)dword_40F000;
   AntiDebugRdtsc(&rdtsc_high2, &rdtsc_low);
   i += 32 * (rdtsc_high2 - rdtsc_high + 3);
   do
     result = i + 337;
   while ( i == 0xAF );
 }
 else
 {
   printf("Invalid key");
   result = -1;
 }

The main check function begin at 0x004011CD

And go to 0x4042e6

After that long mutation code begins. One function un pack another and so on and so far:



It is continuing until 0x40b3e3 where you will find main check algorithm:

So from that we can reconstruct code and got the back account:

C:\ctf\ictf\challenge_29>reverse1.exe
Enter key: 529a1427074a1208b9ca1c4564738291dc3f91455763ccb1
Bank account: 2526390575284-60846167886

2 comments

  1. Franjo Stipanovic fritz says:

    Nice! Just keep it coming :-)

  2. Solutions to the side challanges of the iCTF2011 | USS says:

    […] More (Challenges 29, 30, 31, 32, […]

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