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SEC Consult SA-20241107-0 :: Multiple Vulnerabilities in HASOMED Elefant and Elefant Software Updater Unsafe eval() in TestRail CLI IT Security and Insecurity Portal www.waraxe.us Forum Index -> Hash related information -> What kind of hash? :O View previous topic :: View next topic What kind of hash? :O Posted: Mon Dec 29, 2008 1:25 am 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 j9%3C%A5%0B%60w%B6%14%BF g%11%0F%5E%0C%2Av b%0A%0C%AF%F3%12%A5 mI%C2k%0E%D9 s%A5%84%B5%17a+1%7F I have no idea what so ever...please help? Re: What kind of hash? :O Posted: Mon Dec 29, 2008 3:44 am tehhunter Valuable expert Joined: Nov 19, 2008 Posts: 261 10_Sec_Hero wrote: j9%3C%A5%0B%60w%B6%14%BF g%11%0F%5E%0C%2Av b%0A%0C%AF%F3%12%A5 mI%C2k%0E%D9 s%A5%84%B5%17a+1%7F I have no idea what so ever...please help? Looks like it is hex encoded, so I ran it through waraxe's url decoder on this site. I got the following: Code: j9<Â¥`w¶¿ g^*v b ¯óÂ¥ mIÂkÙ s¥„µa 1 See if that works at all. Posted: Mon Dec 29, 2008 4:18 am 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 i really doubt thats gonna work, they have to be plaintext. thx for trying Posted: Mon Dec 29, 2008 4:48 am k40t1x Regular user Joined: Dec 27, 2008 Posts: 22 10_SEC how did you get them? I dont think that you can use them at all :S Posted: Mon Dec 29, 2008 5:06 am 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 ok..i got them through sql injection from a website. i got the email address which is not encrypted and then i got these which are the passwords. Posted: Mon Dec 29, 2008 1:18 pm Henderson Valuable expert Joined: Jul 11, 2008 Posts: 58 The strings are urlencoded, but it doesn't change the fact that they're encrypted (XOR maybe?). Send me PM with URL if you want Posted: Tue Dec 30, 2008 11:18 am Henderson Valuable expert Joined: Jul 11, 2008 Posts: 58 The first letter of encrypted string is the first letter of user's e-mail address. Rest is a password XOR-ed with a key which is calculated basing on mentioned first letter. I don't know how the key is generated (probably it's some big number) but you could register an account with the same beginning letter of e-mail address as of the account you want to attack, and being able to execute SQL queries, read the encrypted password of yourself from database. Then just XOR the ciphertext with your real password and you'll get the key. Having the key, XOR victim's encrypted password with it and you'll get password in plaintext. Out of curiosity it would be cool to read the script's source using load_file() to discover the key generation algorithm. Posted: Tue Dec 30, 2008 4:47 pm 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 Cool, thank you a lot! Posted: Fri Jan 02, 2009 6:24 pm 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 I did like you said, created an account with them, this is the account I created: Code: p%8F%28%1C%08%0FR pandila_005@yahoo.com plaintext of p%8F%28%1C%08%0FR is parola I still have no idea how to crack it what do you reckon? Posted: Fri Jan 02, 2009 7:27 pm Henderson Valuable expert Joined: Jul 11, 2008 Posts: 58 Since we don't know the length of key, it would be better if you created a password of maximum length. Posted: Sat Jan 03, 2009 4:46 am 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 here we go, managed to pull out the 16 character password (alpha-numeric) for latest account created: email address used: email001@yahoo.com er6%04H%1B%95%7F%96%02%B7v%F2a%02%9E%FC and plaintext is 1234567890abcdef created a differect account couple of days ago, heres the email:password email address used: johnny23462@yahoo.com j2h%FF%16%60g and plaintext is adidas Posted: Sat Jan 03, 2009 10:07 am Henderson Valuable expert Joined: Jul 11, 2008 Posts: 58 The phrase er6%04H%1B%95%7F%96%02%B7v%F2a%02%9E%FC consists of two parts: e = same as first letter of e-mail 723604481B957F9602B776F261029EFC = ciphertext (hex) We know the plaintext, so we XOR the cipthertext with it to get the key. 31323334353637383930616263646566 = "1234567890abcdef" (hex) 723604481B957F9602B776F261029EFC XOR 31323334353637383930616263646566 = key Having this particular key you can decipher ciphertext of phrases beginning with letter "e" (because the key is calculated from it) by XOR-ing the ciphertext with it: ciphertext XOR key = plaintext Hope it's clear now. Last edited by Henderson on Tue Jan 06, 2009 12:30 pm; edited 1 time in total Posted: Sat Jan 03, 2009 5:31 pm 10_Sec_Hero Advanced user Joined: Oct 22, 2008 Posts: 52 Yeah I got it now, thanks man! 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