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\n\nUsing the provided binary and the encrypted file, find a way to retrieve the flag contained in “flag.enc”. Note that the binary would have been run in May 2024. Note: The flag is in the format PWNME{…}
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The challenge provides an ELF file with a file-encryption feature, along with an encrypted flag file.
\nAnalyzing this ELF shows that it seeds srand from the program’s execution time and then performs the encryption using random numbers generated by rand.
Below is the solver I wrote by reproducing the above code as custom functions.
\nIn the end, using the random numbers generated from the timestamp Thu May 09 2024 05:01:17 GMT+0900 let me obtain the key that decrypts the correct flag.
#include <stdio.h>\n#include <stdint.h>\n#include <time.h>\n\nuint64_t seed;\nuint64_t result;\nint32_t r;\nint32_t key;\nint32_t tmp;\n\nuint64_t custom_srand(s) {\n seed = s - 1;\n return 0;\n}\n\nuint64_t custom_rand(){\n result = 0x5851f42d4c957f2d * seed + 1;\n seed = result;\n return result >> 0x21;\n}\n\nint32_t gen_key(){\n r = custom_rand();\n tmp = r / 0x7f;\n key = (r - ((int8_t)(tmp << 7) - tmp)) & 0xFF;\n return key;\n}\n\nint main(void){\n struct tm time_info = {0}; // Initialize the time structure to zero\n time_info.tm_year = 2024 - 1900; // Year (years since 1900)\n time_info.tm_mon = 5 - 1; // Month (0 = January, 1 = February, ..., 4 = May)\n time_info.tm_mday = 9; // Day\n time_info.tm_hour = 22; // Hour (24-hour clock)\n time_info.tm_min = 1; // Minute\n time_info.tm_sec = 16; // Second\n setenv(\"TZ\", \"Asia/Tokyo\", 1);\n tzset();\n time_t epoch_time = mktime(&time_info);\n\n // target = 0x70 0x63 0x42 0x50 0x6a\n // for (int i = 0 ; i < 60*60*24; i++) {\n // custom_srand(epoch_time-i);\n // if (gen_key() == 0x70) {\n // if (gen_key() == 0x63) {\n // if (gen_key() == 0x42) {\n // printf(\"%d\\n\", epoch_time-i);\n // break;\n // }\n // }\n // }\n // }\n\n epoch_time = 1715198477;\n printf(\"keys = [\");\n for (int i = 0; i < 2; i++) {\n custom_srand(epoch_time-i);\n printf(\"[\");\n for (int j = 0; j < 40; j++) {\n int32_t r = custom_rand();\n int32_t tmp = r / 0x7f;\n int32_t key = (r - ((int8_t)(tmp << 7) - tmp)) & 0xFF;\n \n if (j != 39) { printf(\"%d,\", key); }\n else { printf(\"%d\", key); }\n }\n if (i != 1) { printf(\"],\"); }\n else { printf(\"]\"); }\n }\n printf(\"]\\n\");\n \n return;\n}The solver that uses the random numbers generated by the code above as the key is shown below.
\nkeys = [[112,99,66,80,106,86,57,41,90,9,43,34,111,95,73,32,112,54,87,46,112,74,53,81,36,29,8,83,76,20,114,97,55,89,121,109,60,22,22,56],[101,30,103,28,88,32,47,8,114,95,21,97,98,2,42,21,36,12,88,119,99,108,97,45,5,41,65,101,15,61,86,32,70,7,106,64,78,45,43,86]]\n\nwith open(\"flag.enc\", \"rb\") as f:\n data = f.read()\n\nfor key in keys:\n tmp = \"\"\n for i,d in enumerate(data):\n tmp += chr(d^key[i])\n if \"PWNME\" in tmp:\n print(tmp)With this, I was able to recover the correct flag.
\nI actually wanted to do something smarter, like reusing the original binary’s code via something like dlopen, but that did not work because the target was not a library file and the functions I wanted were not exported.
Still looking for a better method…
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