Using the new md5($salt.$pass.$salt) mode - Printable Version +- hashcat Forum (https://hashcat.net/forum) +-- Forum: Deprecated; Previous versions (https://hashcat.net/forum/forum-29.html) +--- Forum: Old oclHashcat Support (https://hashcat.net/forum/forum-38.html) +--- Thread: Using the new md5($salt.$pass.$salt) mode (/thread-3742.html)

Using the new md5($salt.$pass.$salt) mode - protocol - 10-15-2014 Hello, May I have some pointers on how to use the recently added new hash mode -m 3810 = md5($salt.$pass.$salt)? For example if I have a test hash created from the following; salt1cf2a23 pass:test salt2:ab7e343 md5($salt1.$pass.$salt2) -> d55edafe94ce15a50739a765d4d3eff1 If using mode 20 = md5($salt.$pass), could solve this way; oclHashcat-1.31\oclHashcat64.exe -a 3 -m 20 -n 256 -u 1024 d55edafe94ce15a50739a765d4d3eff1cf2a23 "?a?a?a?aab7e343" It seems mode -m 3810 = md5($salt.$pass.$salt) is perfect for the situation. How does the $salt2 value get specified in a command line using -m 3810? Thanks RE: Using the new md5($salt.$pass.$salt) mode - undeath - 10-15-2014 it only takes one salt as far as i can see. RE: Using the new md5($salt.$pass.$salt) mode - forumhero - 10-15-2014 the example for m 3810 shows $pass:$salt i guess you only need the 2nd salt? http://hashcat.net/wiki/doku.php?id=example_hashes RE: Using the new md5($salt.$pass.$salt) mode - epixoip - 10-15-2014 3810 is one salt that is used twice, not two unique salts. RE: Using the new md5($salt.$pass.$salt) mode - protocol - 10-20-2014 thanks guys