| @PL45M4: Yes i did implement most of ur recommendations in v2, but i forgot to crypt some of strings unfortunately, btw nice writeup! |
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| @zm0d: ur welcome :) |
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| @zm0d: Btw, theres a passwords with 7 chars that should be valid, and probably with 6 and lower (wasnt testing), so the brute-force may took even less than 1 minute |
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| @zm0d: Theres is many passwords you can find that matches the hash, but you know that you cant reverse the hash, half of the data of correct password is just disappears in the hash algorithm, you can try brute-force to find some collisions, its should take around 1 min if you will use CUDA, if you gonna keep trying to solve that on ur own, it will not lead you to the finish |
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| @bytehandler: Thx for your explanation, it would be helpful for beginners |
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| Zailox, yes it counts as a successful patch, but this crackme wasnt suppose to be hard, i just wanted to test my patch protect a little bit, i wanted to weed out easy methods of bypass or make them more difficult to use, thats the reason i made a v2 |
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| pass - XXE\DZFYPTaX\RTC(,/"7 |
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| key is 3f39999999999999, my explanaition gets bugged, very sad :( |
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| Thats a nice one,
__int64 __fastcall sub_1400014B0(__int64 a1)
{
int j; // [rsp+20h] [rbp-38h]
int i; // [rsp+24h] [rbp-34h]
char v4[32]; // [rsp+38h] [rbp-20h] BYREF
qmemcpy(v4, "3f3", 3);
for ( i = 3; i |
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| crackmes.one |
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