| Keygen Crackme |
[Click to reveal]stage 1. takes machineguid from registry
stage 2. deletes all letters from it
stage 3. (stage2 % 5)
stage 4. multiplying by two from right to left with stage3, with the transfer of one to the next digit when necessary
stage 5. '0' -> x, digits from 1 - 9 converting to letters (password[i] = 'a' + (digit - 1);)
Enter Your Verification Code: bhhdbfbdebifihchdfchfeacxafadehx
Good Job Bro!
|
2026-03-31 23:57 |
| WhoKnocks ? |
[Click to reveal]#include <stdio.h>
#include <string.h>
#include <stdint.h>
static const uint8_t magic[16] = {
0x20, 0x1B, 0x08, 0x4C, 0x5F, 0x7F, 0x32, 0x06,
0x58, 0x1A, 0x58, 0x45, 0x59, 0x52, 0x52
};
static void xor(uint8_t *output, const char *username, size_t length) {
size_t ulen = strlen(username);
size_t mod = ulen + 1;
for (size_t i = 0; i < length; i++) {
output[i] = (uint8_t)username[i % mod] ^ magic[i];
}
output[length] = 0;
}
static void encode(char *out, const uint8_t *in, size_t len) {
for (size_t i = 0; i < len; i++) {
uint8_t b = in[i];
if (b & 1) {
out[i] = 'K';
} else if (b % 3 == 0) {
if (b & 3)
out[i] = 'f';
else
out[i] = 'V';
} else {
out[i] = 'b';
}
}
out[len] = '\0';
}
int main(void) {
char username[17] = {0};
uint8_t xored[17] = {0};
char password[128] = {0};
printf("Enter username: ");
if (!fgets(username, sizeof(username), stdin)) return 1;
size_t ulen = strlen(username);
if (ulen > 0 && username[ulen - 1] == '\n') username[--ulen] = '\0';
if (ulen == 0) {
return 1;
}
if (ulen > 16) {
return 1;
}
xor(xored, username, 16);
size_t xlen = strlen((char *)xored);
encode(password, xored, xlen);
printf("Username : %s\n", username);
printf("Password : %s\n", password);
return 0;
}
|
2026-03-05 18:26 |
| Continental |
[Click to reveal]@karabatik thx for you comment, 100% accepted solution, gj, btw i know about this problem in the hash and i was trying to fix it, but how you can see it didnt happend, i was confused how is that happens and left it as it is because i dont have enough knowledge i think, in the future im gonna do better one, i hope so :P
|
2026-02-09 21:14 |
| CrackMe with password |
[Click to reveal]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 |
[Click to reveal]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 |
[Click to reveal]Enter username: asd
Enter license-key: FOYWVUTSRQPO
Success! 2026 CrackMe Solved.
big asf password generation algorithm
|
2026-01-07 04:44 |
| Medium-Hard crackme |
[Click to reveal]\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 |
[Click to reveal]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 |
[Click to reveal]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 |
[Click to reveal]pass - XXE\DZFYPTaX\RTC(,/"7
|
2025-09-10 12:17 |
| MISSION IMPOSSIBLE |
[Click to reveal]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 |
[Click to reveal]crackmes.one
|
2025-08-30 09:08 |