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AFAIK, the compound file format identifier is D0 CF 11 E0 A1 B1 1A E1, which is located in the offset 0 of the header, which has a length of exactly 512 bytes.

However that is not enough to get the boundaries of the md5 hash of the password. After that I plan to find a preimage in order to recover the document.

Can someone help me please? Thank you in advance.

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migrated from crypto.stackexchange.com Jan 18 '13 at 16:48

This question came from our site for software developers, mathematicians and others interested in cryptography.

    
I've downloaded this specification, but I am not sure on how to extract the hash: wvware.sourceforge.net/word97.zip – user4772 Jan 17 '13 at 17:45

You link the word97 specification for your 2000 document. I'm not sure if the file format changed, but luckily the encryption is the same.

I don't think you need the md5 hash (if there is one), you just want to break the encryption. It looks like there are word 97/2000 decryptors on the web (paid or freeware). My basic google search indicated that it's easier to break the proprietary hash + encryption (partially rc4) than to brute-force the password.

I wasn't going to link the page, but it wasn't hard to find.

Initially a unicode password is taken and some bytes appended, this password is passed through a varient of the standard md5 hash algorithm.

The non standard hash is tricky, and Im far from sure what benefit over standard md5 the modified md5 hash has.

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Not sure on how to break it, but must not be impossible since there are tools that do it.

I see in this post that a zip crack tool can be used: http://www.forensicswiki.org/wiki/Word_Document_%28DOC%29

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