Academy of Investigation - Administration - FCSC 2020
This section is the write-up for the Academy of Investigation forensics-challenge (FCSC 2020).
We have a live memory dump named mem.dmp and an encrypted file flag.txt.enc.
Ce poste administre un serveur distant avec le protocole SSH à l’aide d’une authentification par clé (clé protégée par mot de passe). La clé publique a été utilisée pour chiffrer le message ci-joint (flag.txt.enc).
Analysis of the encrypted file
The file flag.txt.enc has a 2048-bits length. So, the RSA protocol had been used.
We have to find this key to decrypt flag.txt.enc
Extraction of the key
With rsakeyfind, I can extract it :
./rsakeyfind mem.dmp
FOUND PRIVATE KEY AT c64ac50
version =
00
modulus =
00 d7 1e 77 82 8c 92 31 e7 69 02 a2 d5 5c 78 de
a2 0c 8f fe 28 59 31 df 40 9c 60 61 06 b9 2f 62
40 80 76 cb 67 4a b5 59 56 69 17 07 fa f9 4c bd
6c 37 7a 46 7d 70 a7 67 22 b3 4d 7a 94 c3 ba 4b
7c 4b a9 32 7c b7 38 95 45 64 a4 05 a8 9f 12 7c
4e c6 c8 2d 40 06 30 f4 60 a6 91 bb 9b ca 04 79
11 13 75 f0 ae d3 51 89 c5 74 b9 aa 3f b6 83 e4
78 6b cd f9 5c 4c 85 ea 52 3b 51 93 fc 14 6b 33
5d 30 70 fa 50 1b 1b 38 81 13 8d f7 a5 0c c0 8e
f9 63 52 18 4e a9 f9 f8 5c 5d cd 7a 0d d4 8e 7b
ee 91 7b ad 7d b4 92 d5 ab 16 3b 0a 8a ce 8e de
47 1a 17 01 86 7b ab 99 f1 4b 0c 3a 0d 82 47 c1
91 8c bb 2e 22 9e 49 63 6e 02 c1 c9 3a 9b a5 22
1b 07 95 d6 10 02 50 fd fd d1 9b be ab c2 c0 74
d7 ec 00 fb 11 71 cb 7a dc 81 79 9f 86 68 46 63
82 4d b7 f1 e6 16 6f 42 63 f4 94 a0 ca 33 cc 75
13
publicExponent =
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 01 00 01
privateExponent =
62 b5 60 31 4f 3f 66 16 c1 60 ac 47 2a ff 6b 69
00 4a b2 5c e1 50 b9 18 74 a8 e4 dc a8 ec cd 30
bb c1 c6 e3 c6 ac 20 2a 3e 5e 8b 12 e6 82 08 09
38 0b ab 7c b3 cc 9c ce 97 67 dd ef 95 40 4e 92
e2 44 e9 1d c1 14 fd a9 b1 dc 71 9c 46 21 bd 58
88 6e 22 15 56 c1 ef e0 c9 8d e5 80 3e da 7e 93
0f 52 f6 f5 c1 91 90 9e 42 49 4f 8d 9c ba 38 83
e9 33 c2 50 4f ec c2 f0 a8 b7 6e 28 25 56 6b 62
67 fe 08 f1 56 e5 6f 0e 99 f1 e5 95 7b ef eb 0a
2c 92 97 57 23 33 36 07 dd fb ae f1 b1 d8 33 b7
96 71 42 36 c5 a4 a9 19 4b 1b 52 4c 50 69 91 f0
0e fa 80 37 4b b5 d0 2f b7 44 0d d4 f8 39 8d ab
71 67 59 05 88 3d eb 48 48 33 88 4e fe f8 27 1b
d6 55 60 5e 48 b7 6d 9a a8 37 f9 7a de 1b cd 5d
1a 30 d4 e9 9e 5b 3c 15 f8 9c 1f da d1 86 48 55
ce 83 ee 8e 51 c7 de 32 12 47 7d 46 b8 35 df 41
prime1 =
00
prime2 =
00
exponent1 =
00
exponent2 =
00
coefficient =
00
Interpretation of the key
The private key (d) is privateExponent. The cipher modulus (n) is modulus. The cipher file (c) is his content.
The decrypt RSA-algorithm can be computed by m = c^d mod n :
$ cat flag.txt.enc | xxd -p | tr -d '\n'
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
With a Python script, we decipher the encrypted file :
c = int(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)
d = int(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)
n = int(0xd71e77828c9231e76902a2d55c78dea20c8ffe285931df409c606106b92f62408076cb674ab55956691707faf94cbd6c377a467d70a76722b34d7a94c3ba4b7c4ba9327cb738954564a405a89f127c4ec6c82d400630f460a691bb9bca0479111375f0aed35189c574b9aa3fb683e4786bcdf95c4c85ea523b5193fc146b335d3070fa501b1b3881138df7a50cc08ef96352184ea9f9f85c5dcd7a0dd48e7bee917bad7db492d5ab163b0a8ace8ede471a1701867bab99f14b0c3a0d8247c1918cbb2e229e49636e02c1c93a9ba5221b0795d6100250fdfdd19bbeabc2c074d7ec00fb1171cb7adc81799f86684663824db7f1e6166f4263f494a0ca33cc7513)
t = pow(c,d,n)
print(hex(t))
Finally, we have to decode the hexa to ascii :
$ python3 script.py
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
$ echo -n '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' | xxd -r -p | strings -n 10
FCSC{ac5cad66114d4866a4b55e43cb8896cc4947855241b5af8d2f8a123c36083d98}
Written on June 11, 2020