08-21-2017, 07:58 PM
This reply is a bit late, i never posted it. but:
1. They are not identical. -m 1710 is 128:10 and -m 1720 is 128:13. You are, however, using the incorrect hashcat mode. You should be using mode 9600. SHA512 was first utilized in Office 2013 according to my source. https://en.wikipedia.org/wiki/Microsoft_...protection
"Office 2013 uses 128-bit AES, however hash algorithm has been updated to SHA-2 class, and it is SHA-512 by default."
You can see earlier version are using sha-1 class algorithms, as well as md5.
2. It does not give you a salt either, what you are obtaining is the KDF for the 128-bit AES that you have to step through to start initialization, which is why it is so slow. Also this is not a static variable. If you made an exact copy of your spreadsheet and password protect it with the same password you will get a different KDF value.
All of this is irrelevant to you though. You do not need to get anything other than the extracted string, from the proper tool which i have give you information about below.
3. Yes, According to https://msdn.microsoft.com/en-us/library...e.14).aspx - "Iterations to Run Hashing Algorithm.Represents the attribute in schema: spinCount"
What you should be doing is using office2john.py (which is obtained from JtR's github).
Example (using hashcat as my password)
1. They are not identical. -m 1710 is 128:10 and -m 1720 is 128:13. You are, however, using the incorrect hashcat mode. You should be using mode 9600. SHA512 was first utilized in Office 2013 according to my source. https://en.wikipedia.org/wiki/Microsoft_...protection
"Office 2013 uses 128-bit AES, however hash algorithm has been updated to SHA-2 class, and it is SHA-512 by default."
You can see earlier version are using sha-1 class algorithms, as well as md5.
2. It does not give you a salt either, what you are obtaining is the KDF for the 128-bit AES that you have to step through to start initialization, which is why it is so slow. Also this is not a static variable. If you made an exact copy of your spreadsheet and password protect it with the same password you will get a different KDF value.
All of this is irrelevant to you though. You do not need to get anything other than the extracted string, from the proper tool which i have give you information about below.
3. Yes, According to https://msdn.microsoft.com/en-us/library...e.14).aspx - "Iterations to Run Hashing Algorithm.Represents the attribute in schema: spinCount"
What you should be doing is using office2john.py (which is obtained from JtR's github).
Example (using hashcat as my password)
Code:
$ ./o2j.py hashcat_password.xlsx
hashcat_password.xlsx:$office$*2013*100000*256*16*331c08dcd32601d2xxxxxxxxxxxx8fe7*2e1xxxxxxxxxxxx7e866bd6262c3ba4c*bc6bccc7d215a5824xxxxxxxxxxxx2947fc59a4e19c46a107df91b57be1349e2