Rating: 4.0
| Category | Points | Solves |
|---|---|---|
| Misc | 50 | 76 |
Don't forget. This is a MISC challenge.
This is a misc challenge with some reversing. We noticed the DOS code section of the binary was not an standard one, so we load the binary as an MS-DOS executable on the disassembler and then we see the xor instruction being applied over two strings, that would be a good indicator of an incoming flag:
seg000:0000 public start
seg000:0000 start proc near
seg000:0000 mov ax, seg dseg
seg000:0003 mov ds, ax
seg000:0005 assume ds:dseg
seg000:0005 lea dx, unk_10030 ; "This program cannot be..."
seg000:0009 mov ah, 9
seg000:000B int 21h ; DOS - PRINT STRING
seg000:000B ; DS:DX -> string terminated by "$"
seg000:000D mov cx, 0
seg000:0010
seg000:0010 loc_10010:
seg000:0010 cmp cx, 39h ; '9'
seg000:0013 jz short loc_1002C
seg000:0015 lea bx, unk_1005B ; "WWE3>...\x0c\x00..." first string
seg000:0019 add bx, cx
seg000:001B mov dl, [bx]
seg000:001D lea bx, unk_10094 ; "\x15"1\x13VirQy..." second string
seg000:0021 add bx, cx
seg000:0023 xor dl, [bx]
seg000:0025 mov ah, 2
seg000:0027 int 21h ; DOS - DISPLAY OUTPUT
seg000:0027 ; DL = character to send to standard output
seg000:0029 inc cx
seg000:002A jmp short loc_10010
seg000:002C ; ---------------------------------------------------------------------------
seg000:002C
seg000:002C loc_1002C:
seg000:002C mov ah, 4Ch
seg000:002E int 21h ; DOS - 2+ - QUIT WITH EXIT CODE (EXIT)
seg000:002E start endp ; AL = exit code
seg000:002E
seg000:002E seg000 ends
seg000:002E
The answer is the XOR of the two strings:
57 57 45 33 3E 0C 00 34 5E 61 62 55 37 65 6E 64 76 02 57
20 65 52 62 79 30 27 0A 76 32 16 72 67 66 49 05 10 57 20
0B 51 19 25 3B 6D 03 64 60 07 30 36 72 6D 48 55 73 3A 0F
^
15 22 31 13 56 69 72 51 79 12 42 34 17 03 02 05 11 22 07
63 31 14 19 18 44 78 66 45 53 23 06 38 12 21 36 62 64 53
54 35 29 15 56 32 73 54 12 73 43 69 42 03 17 31 43 49 72
=
But here's a flag PCTF{at_l3a5t_th3r3s_d00m_p0rts_0n_d0s}
PCTF{at_l3a5t_th3r3s_d00m_p0rts_0n_d0s}
