Investigative Reversing 0

Problem

We have recovered a binary and an image. See what you can make of it. There should be a flag somewhere. Its also found in /problems/investigative-reversing-0_4_950a47cfcfc9b661c36603148c77df3d on the shell server.

• Binary

• Image

Solution

1. Run file mystery which shows its is a ELF executable: mystery: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld...

2. Reverse the binary file using Ghidra (cheat sheet). Open it and in the symbol tree click on main. The decompiled main function will show on the right.

    void main(void)
   
    {
    FILE *flag_stream;
    FILE *image_steam;
    size_t sVar1;
    long in_FS_OFFSET;
    int i;
    int local_50;
    char local_38 [4];
    char local_34;
    char local_33;
    char local_29;
    long local_10;
   
    local_10 = *(long *)(in_FS_OFFSET + 0x28);
    flag_stream = fopen("flag.txt","r");
    image_steam = fopen("mystery.png","a");
    if (flag_stream == (FILE *)0x0) {
        puts("No flag found, please make sure this is run on the server");
    }
    if (image_steam == (FILE *)0x0) {
        puts("mystery.png is missing, please run this on the server");
    }
    sVar1 = fread(local_38,0x1a,1,flag_stream);
    if ((int)sVar1 < 1) {
                        /* WARNING: Subroutine does not return */
        exit(0);
    }
    puts("at insert");
    fputc((int)local_38[0],image_steam);
    fputc((int)local_38[1],image_steam);
    fputc((int)local_38[2],image_steam);
    fputc((int)local_38[3],image_steam);
    fputc((int)local_34,image_steam);
    fputc((int)local_33,image_steam);
    i = 6;
    while (i < 0xf) {
        fputc((int)(char)(local_38[i] + '\x05'),image_steam);
        i = i + 1;
    }
    fputc((int)(char)(local_29 + -3),image_steam);
    local_50 = 0x10;
    while (local_50 < 0x1a) {
        fputc((int)local_38[local_50],image_steam);
        local_50 = local_50 + 1;
    }
    fclose(image_steam);
    fclose(flag_stream);
    if (local_10 != *(long *)(in_FS_OFFSET + 0x28)) {
                        /* WARNING: Subroutine does not return */
        __stack_chk_fail();
    }
    return;
    }

3. We can see that the program opens the flag file, and places an encoded version of it at the end of the image file.

    $ xxd -g 1 mystery.png | tail
    0001e7f0: 82 20 08 82 20 08 82 20 08 82 20 64 1f 32 12 21  . .. .. .. d.2.!
    0001e800: 08 82 20 08 82 20 08 82 20 08 42 f6 21 23 11 82  .. .. .. .B.!#..
    0001e810: 20 08 82 20 08 82 20 08 82 20 64 1f 32 12 21 08   .. .. .. d.2.!.
    0001e820: 82 20 08 82 20 08 82 20 08 42 f6 21 23 11 82 20  . .. .. .B.!#.. 
    0001e830: 08 82 20 08 82 20 08 82 20 64 1f 32 12 21 08 82  .. .. .. d.2.!..
    0001e840: 20 08 82 20 08 82 20 08 42 f6 21 23 11 82 20 08   .. .. .B.!#.. .
    0001e850: 82 20 08 82 20 08 82 20 64 17 ff ef ff fd 7f 5e  . .. .. d......^
    0001e860: ed 5a 9d 38 d0 1f 56 00 00 00 00 49 45 4e 44 ae  .Z.8..V....IEND.
    0001e870: 42 60 82 70 69 63 6f 43 54 4b 80 6b 35 7a 73 69  B`.picoCTK.k5zsi
    0001e880: 64 36 71 5f 35 32 36 36 61 38 35 37 7d           d6q_5266a857}

4. The encoding works as follows:

   1. Add bytes 0 to 5 (inclusive) to the image.

   2. Loop through the bytes of the flag from 6 to 0xf (from 6 to 14, inclusive) and add '\x05' (5) to each bytes.

   3. Add the 15th byte minus 3 (probably the 15th byte, I am assuming based on the code structure).

   4. Loop through and add the remaining bytes (16 to 25, inclusive).

5. We can reverse this using the script.py and get the flag.

Flag

picoCTF{f0und_1t_5266a857}