⚠️⚠️⚠️ This is a solution to the challenge. This post will be full of spoilers.
Download the binaries here: https://github.com/mlesterdampios/huntress-2024-binary-challenges
In this challenge, I learned that this binary is somewhat similar to Ghostpulse where it hides payload on the PNG. I was able to uncover this by checking some functions and saw that the binary loads a PNG resource file and it do some decryption routine.
My solution to this is somewhat weird.
As stated in the article above, it do some crc and hashing checking to the PNG parts to identify location of the encrypted locations.
I first put a breakpoint on the cmp dword ptr [rsp+238h+var_1E8+0Ch], 0AAAAAAAAh. It seems like 0AAAAAAAAh is like an index for an encrypted message that the programs want to print out. You will notice this also in other parts, such as 0AAAAh, 0AAAAAh, 0AAAAAAh.
In the first breakpoint hit, you will see this:
Since, 0AAAAh is not equals to 0AAAAAAAAh, then it will just skip and proceed to the next iteration of loop. But what we can do here is to control the rip to proceed with the decryption block instead of reiterating the loop.
It did leaked the first encrypted message from the PNG file.
We just repeat these step to leak others as well.
We could go on to look other message, I think there are 12 encrypted messages there. But to cut short, this message is interesting.
Here are the IPs that we extracted.
10 25 3 103
10 5 13 54
10 185 7 102
172 21 29 54
172 20 20 51
172 30 27 54
192 168 34 57
192 168 71 6
10 76 2 97
10 199 9 97
192 168 245 16
172 25 31 54
192 168 226 0
10 215 6 57
192 168 41 1
10 212 10 49
10 119 16 50
10 0 0 102
172 30 21 57
192 168 43 2
192 168 113 16
172 26 24 100
192 168 89 12
172 21 33 101
192 168 37 125
172 17 19 49
10 169 8 52
10 179 4 123
172 29 22 50
192 168 180 8
172 28 26 97
172 24 23 50
192 168 40 18
172 16 30 98
10 13 1 108
192 168 42 0
172 16 17 102
10 105 11 55
192 168 36 49
172 30 18 56
172 24 25 99
192 168 100 12
192 168 35 97
172 30 28 99
172 27 32 53
192 168 58 18
10 184 15 101
192 168 50 15
10 129 5 53
10 126 12 98
10 32 14 57Then lets sort it based on the 3rd octet
10 0 0 102
10 13 1 108
10 76 2 97
10 25 3 103
10 179 4 123
10 129 5 53
10 215 6 57
10 185 7 102
10 169 8 52
10 199 9 97
10 212 10 49
10 105 11 55
10 126 12 98
10 5 13 54
10 32 14 57
10 184 15 101
10 119 16 50
172 16 17 102
172 30 18 56
172 17 19 49
172 20 20 51
172 30 21 57
172 29 22 50
172 24 23 50
172 26 24 100
172 24 25 99
172 28 26 97
172 30 27 54
172 30 28 99
172 21 29 54
172 16 30 98
172 25 31 54
172 27 32 53
172 21 33 101
192 168 34 57
192 168 35 97
192 168 36 49
192 168 37 125
192 168 40 18
192 168 41 1
192 168 42 0
192 168 43 2
192 168 50 15
192 168 58 18
192 168 71 6
192 168 89 12
192 168 100 12
192 168 113 16
192 168 180 8
192 168 226 0
192 168 245 16Then extract the 4th octet.
102
108
97
103
123
53
57
102
52
97
49
55
98
54
57
101
50
102
56
49
51
57
50
50
100
99
97
54
99
54
98
54
53
101
57
97
49
125
18
1
0
2
15
18
6
12
12
16
8
0
16Convert to ASCII then remove the non-printable characters.
GGz!