Depends on both attribute and checksum.
Attribute: 01 00 00 00 (little endian, $00-03)
Checksum: 03 1f (little endian, $06-07)
$08-87:
d3 c8 19 e2 c4 0d 27 c5 e5 d8 08 0d 73 f6 7a b5 f4 73 f4 2b b3 34 35 56 e1 41 bd fd 08 bd ca 05 bb 1f 9a e0 9f 71 f4 79 43 e6 4f 90 6b f6 0c 0b ac 14 9e 93 91 a3 9b 60 a0 63 fe 0c 05 3a 2f 53 6c 84 68 e2 96 3e 93 17 f5 91 d9 71 87 1e 95 06 08 ee bb 3f 65 7c 57 97 be ab 2f 62 bb 79 fc 5d 70 95 1c 8a d4 4a d9 23 da 93 d1 ef 82 32 d7 95 17 01 db 7f e0 57 a6 f6 2c 16 56 34 4c 9f 3d 29
Attribute: 0a 96 21 8b (little endian, $00-03)
Checksum: 03 1f (little endian, $06-07)
$08-87:
30 c8 fb e2 26 0d c5 c5 07 d8 ea 0d 91 f6 98 b5 16 73 16 2b 4c cb 35 56 e1 41 bd fd 08 bd ca 05 ba 1f 9a e0 9f 71 f4 79 43 e6 4f 90 6b f6 0c 0b ac 14 9e 93 91 a3 9b 60 a2 63 fe 0c 05 3a 2f 53 8e 84 8a e2 74 3e 71 17 17 91 3b 71 65 1e 77 06 ea ee 59 3f 9a 83 57 97 bc ab 2f 62 bb 79 fc 5d 70 95 1c 8a d4 4a d9 23 da 93 d1 ef 82 32 d7 95 17 01 db 7f e0 57 a6 f6 2c 16 56 34 4c 9f 3d 29
Notice the similarity.
So...anyone think he can crack this?
The only similarity that I notice is that attribute:
0A 96 21 8B XOR
01 00 00 00 MOD
03 1F / 2 = E2 is.
If you XOR following values with E2
0x0A, 0x0C, 0x0E, 0x10, 0x12, 0x14, 0x16, 0x18, 0x1A (9-times, every 2nd byte)
0x48, 0x4A, 0x4C, 0x4E, 0x50, 0x52, 0x54, 0x56, 0x58, 0x5A (10-times, every 2nd byte)
you get the opposite files values (differences)
But there are exceptions, see complete list of changes:
0x08 = E3 (difference 01)
0x0A = E2
0x0C = E2
0x0E = E2
0x10 = E2
0x12 = E2
0x14 = E2
0x16 = E2
0x18 = E2
0x1A = E2
0x1C = FF (difference 1D)
0x1D = FF (difference 1D)
0x28 = 01 (difference 1F)
0x40 = 02 (difference 20)
0x48 = E2
0x4A = E2
0x4C = E2
0x4E = E2
0x50 = E2
0x52 = E2
0x54 = E2
0x56 = E2
0x58 = E2
0x5A = E2
0x5C = FF (difference 1D)
0x5D = FF (difference 1D)
0x60 = 02 (difference 20)
Maybe someone else can figure out something more... good night folks!