标题: Novell Netware 4.1/4.11 - SP5B Remote.NLM Weak Encryption EDB-ID: 19364 CVE: 1999-0470 类型: REMOTE 平台: NETWARE 发布日期: 1999-04-09 漏洞利用脚本: source: https://www.securityfocus.com/bid/482/info

The encrypted passwords for Remote.NLM are remotely accessible to anyone with the ability to view SYS:System\LDRemote.NCF. The password encryption algorithm for Remote.NLM has been broken and can be decrypted with pencil and paper.

The password will look something like this:
AF8CBBF48CA9955F5ADAFDADAA23

The structure of the password is as follows:
AF8CBBF48CA99 55F5ADAFDADAA - 23

The first section contains the low-order bits, and the second, the high-order bits. 23 is the time byte, which is decremented by the server once every two seconds, from FF to 02, then back up to FF, etc.

Step 1) Realign the low-order bits and high-order bits.
This is extremely simple to do. The high-order bits are in order from the first character to the last, and so are the low-order bits.
Example:
Password: AF8CBBF48CA99 - 55F5ADAFDADAA,
Output: 5A 5F F8 5C AB DB AF F4 D8 AC DA A9 A9
At this point, ignore 5A 5F F8 5C, or the first four bytes. They are appended somewhere during encryption, and decrypt to "%*@$". It was a TERRIBLE idea for Novell to implement those four characters into every single password, as those are what helped me rebuild their hash table from scratch. Also, if the length of the password is 10, the password is automatically decryptable to nul.

Step 2) Match each of the password characters (group of two hex characters) to the hash table below. Use their position from the beginning of the table to determine the value of the pre-hash encrypted password. Example: F4, the 8th character of the password, matches the hash table at 95. This means that 95 is the pre-hash value of F4. Thus far, (ignoring the first four characters) the password was:
AB DB AF F4 D8 AC DA A9 A9
and now the password is:
98 A0 9B 95 A1 9D A6 9C 9C

Remote.NLM Hash Table

00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F
00 5B 58 5E 5F 59 5C 5A 5D-73 70 76 77 71 74 72 75
10 13 10 16 17 11 14 12 15-7B 78 7E 7F 79 7C 7A 7D
20 53 50 56 57 51 54 52 55-03 00 06 07 01 04 02 05
30 1B 18 1E 1F 19 1C 1A 1D-0B 08 0E 0F 09 0C 0A 0D
40 2B 28 2E 2F 29 2C 2A 2D-63 60 66 67 61 64 62 65
50 83 80 86 87 81 84 82 85-3B 38 3E 3F 39 3C 3A 3D
60 8B 88 8E 8F 89 8C 8A 8D-33 30 36 37 31 34 32 35
70 93 90 96 97 91 94 92 95-6B 68 6E 6F 69 6C 6A 6D
80 9B 98 9E 9F 99 9C 9A 9D-A3 A0 A6 A7 A1 A4 A2 A5
90 F3 F0 F6 F7 F1 F4 F2 F5-AB A8 AE AF A9 AC AA AD
A0 DB D8 DE DF D9 DC DA DD-FB F8 FE FF F9 FC FA FD
B0 23 20 26 27 21 24 22 25-B3 B0 B6 B7 B1 B4 B2 B5
C0 CB C8 CE CF C9 CC CA CD-BB B8 BE BF B9 BC BA BD
D0 C3 C0 C6 C7 C1 C4 C2 C5-D3 D0 D6 D7 D1 D4 D2 D5
E0 43 40 46 47 41 44 42 45-E3 E0 E6 E7 E1 E4 E2 E5
F0 4B 48 4E 4F 49 4C 4A 4D-EB E8 EE EF E9 EC EA ED

Step 3) Subtract the length (the number of groups of hex characters, excluding the time character) of the full password from each encrypted password character. Now you have the ACTUAL pre-hash encrypted password. If the subtracted value is negative then simply continue from FF down to the negative value. Example: if the password character is at 04, and the length is 6, the value of the password character will be FF.

The length is 13 (D in hex), so the password was:
98 A0 9B 95 A1 9D A6 9C 9C
and is now:
8B 93 8E 88 94 90 99 8F 8F

Step 4) Get the time var, in this situation 23 (hex), and subtract it from FF. This new character is for use in Step 5. Example: FF-23=DC.

Step 5) Finally, XOR each character (group of 2 hex characters) of the encrypted password with the new time character, and you now have the decrypted password!

The password was:
8B 93 8E 88 94 90 99 8F 8F (before the XOR)

Now, the decrypted password is:
57 4F 52 54 48 4C 45 53 53
"WORTHLESS"

The exploit program does all this automatically. 

https://gitlab.com/exploit-database/exploitdb-bin-sploits/-/raw/main/bin-sploits/19364.zip URL: https://www.exploit-db.com/exploits/19364