| CrackMe with password |
well i think you forgot to encrypt ur password isnt it?
strcpy(Str2, "rootaccess1337"); Its stored right in the strings,
also i think the idea was to compare my xored input in VM with hardcoded xor values that represents the correct password?
Also the idea with VM its kinda cool ngl.
something like this -
hash = 0xCBF29CE484222325uLL;
index = 0;
do
{
index_copy = index;
v7 = 13 * (index / 13);
v8 = index++;
hash = __ROL8__(0x100000001B3LL * (password[v8] ^ hash), (index_copy - v7 + 3) & 0x3F);
}
return VM(crypted_bytecode, bytecode_len, data, input_byte_hash_dest, key) && strcmp(user_input, password) == 0;
|
2026-01-28 17:46 |
| Time Rift 2026 |
uint64_t foo(const char *buf) {
uint32_t hash = 5381;
const char *ptr = buf;
while (*ptr != '\0') {
hash = 33 * hash + (uint8_t)(*ptr);
ptr++;
}
return hash;
}
int main(void) {
char buf[32];
printf("Enter a string: ");
if (fgets(buf, 10, stdin) == NULL) {
printf("Error reading input.\n");
return 1;
}
size_t len = strlen(buf);
if (len > 0 && buf[len - 1] == '\n') {
buf[len - 1] = '\0';
}
uint64_t hash = foo(buf);
hash = (hash ^ 0x2026) * 132;
printf("Hash of '%s' is: 0x%llX; int: %u\n", buf, hash, hash);
return 0;
} |
2026-01-12 04:58 |
| keygen |
Enter username: asd
Enter license-key: FOYWVUTSRQPO
Success! 2026 CrackMe Solved.
big asf password generation algorithm |
2026-01-07 04:44 |
| Medium-Hard crackme |
\xDD\xBF\xDE\xAD\x9C\xEC\x9E cp866 crackme.exe |
2025-12-30 01:14 |
| makeKEYgen |
https://github.com/generin0/Keygens/blob/master/keygen3.c easy one |
2025-12-09 00:50 |
| simple crackme |
input =^ input_size;
input != l?xo`r0e ? fail : pass |
2025-11-29 23:09 |
| decode me |
the password is - (Pa__-3\\-L@_) || 0 - '_', 1 - '', 2 - '\' |
2025-11-27 14:31 |
| KeyGen APP |
https://github.com/generin0/Keygens/blob/master/keygen1.c
:) |
2025-11-26 14:49 |
| KeyGenMeV3 |
https://github.com/generin0/Keygens, cool task |
2025-11-26 13:56 |
| crack the points |
i cant send it properly for some reason but you can change the value in brackets and output is gonna change |
2025-11-20 14:54 |
| crack the points |
(0000001FB6BFF738 (11) 00 00 00 00 00 00 00 70 41 1E 9E 51 01 00 00
000000A23AFFF738 (11) 00 00 00 00 00 00 00 70 41 A1 C5 2C 02 00 00 |
2025-11-20 14:53 |
| crack the points |
(0000001FB6BFF738 (11) 00 00 00 00 00 00 00 70 41 1E 9E 51 01 00 00
000000A23AFFF738 (11) 00 00 00 00 00 00 00 70 41 A1 C5 2C 02 00 00 |
2025-11-20 14:52 |
| by litvin |
password - 789326321
00007FF68A74F120 | cmp dword ptr ss:[rsp+4C],eax | |
2025-11-07 05:49 |
| Patch protect |
@PL45M4: Yes i did implement most of ur recommendations in v2, but i forgot to crypt some of strings unfortunately, btw nice writeup! |
2025-10-04 06:12 |
| Patch protect v2 |
@zm0d: ur welcome :) |
2025-09-22 16:11 |
| Patch protect v2 |
@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 |
2025-09-19 15:38 |
| Patch protect v2 |
@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 |
2025-09-19 15:34 |
| Patch protect |
@bytehandler: Thx for your explanation, it would be helpful for beginners |
2025-09-19 01:22 |
| Patch protect |
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 |
2025-09-18 00:05 |
| XOR crackme |
pass - XXE\DZFYPTaX\RTC(,/"7 |
2025-09-10 12:17 |
| MISSION IMPOSSIBLE |
key is 3f39999999999999, my explanaition gets bugged, very sad :( |
2025-09-08 12:28 |
| MISSION IMPOSSIBLE |
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 |
2025-09-08 12:27 |
| beginner friendly |
crackmes.one |
2025-08-30 09:08 |