Can I decrypt this hash?

Domain Cached Credentials: MD4(MD4(($pass)).(strtolower($username)))

closed as unclear what you're asking by schroeder, Mark, Xander, Eric G, Rоry McCune Jul 10 '15 at 19:04

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  • uh - it looks like an MD4 hash ... What is your question? What type is it, or how to "decrypt" the hash? – schroeder Jul 9 '15 at 2:52
  • 1
    Please note that no one can "decrypt" a hash. They are meant to be one-way only. – schroeder Jul 9 '15 at 2:53
  • Let me guess, you found that in hash-identifier? – schroeder Jul 9 '15 at 2:54
  • first i want to thank you to answer :) ,yes .. i found it in hash-identifier , and i don' t know if i can use the Hashcat to guess the hash please if you now a method which can guess this hash or if you know any information about this hash tell me please :) – Radwan Cherfy Jul 9 '15 at 13:46
  • Everything you need to know about that hash is in the code. I'm not sure what else you need to know. – schroeder Jul 9 '15 at 14:50

Hashes are not compression schemes. The password you are looking for is not stored in the hash, though it is theoretically possible that somebody has a dictionary that maps it to its original content.

Given knowledge of the scheme,


You can guess it. MD4 is decently fast to calculate, so if you have $username, all that's left is $pass.

You can either build such a thing yourself, or use something like John the Ripper, which I assume can be configured to attack this scheme.


As to what type of hash it is, you've mostly answered the question by naming how it is composed. MD4 is a rather old yet fast digest algorithm. This scheme uses that algorithm twice; first, it hashes the password and concatenates the hash with the lowercase form of the username, then it hashes that resulting string.


Sample implementation in perl:


use Digest::MD4 "md4_hex";

my $username = shift;
my $pass = shift;

print md4_hex( md4_hex($pass) . lc($username) ) . "\n";

Some outputs:

$ perl sample.pl user password
$ perl sample.pl AzureDiamond hunter2

Sample (dumb) brute force password cracker:


use Digest::MD4 "md4_hex";

my $username = shift;
my $pw_hash = shift;
my $i = 0;

sub hash_check {
  if (md4_hex( md4_hex($_) . lc($username) ) eq $pw_hash) {
    print "\r$i\nPassword found: '$_'\n";

LINE: while() {

  my @variants = ($_);
  my $v = $_;
  $v =~ s/\W//g; # strip non-word chars
  push(@variants, $v) if ($_ ne $v);
  $v = lc($_); # all lowercase
  push(@variants, $v) if ($_ ne $v);
  $v =~ s/\W//g; # all lowercase AND strip non-word chars
  push(@variants, $v) if ($_ ne $v);
  $v = uc(substr($_,0,1)).substr($_,1); # uppercase 1st char
  push(@variants, $v) if ($_ ne $v);
  $v =~ s/\W//g; # uppercase 1st char AND strip non-word chars
  push(@variants, $v) if ($_ ne $v);

  for my $pw (@variants) {
    # try followed by nothing or 1-2 digits
    for my $j ("", 0..99, qw/00 01 02 03 04 05 06 07 08 09/) {
      print "\r$i" if $i % 111 == 0; # only print every 111 (this is FAST)
      $_ = $pw . $j;
      last LINE if hash_check($_);
print "\r$i\n";

Some outputs:

$ echo password |perl sample.pl user 200677efff76ec27e716b319b611e217
Password found: 'password'
$ perl /tmp/sample.pl user a3c705c458e12e2f9982ae2682b437f8 american-english
Password found: 'password07'
$ perl sample.pl admin 4ef6c6a21c0172c8cd322dd8f78ba6c1 american-english-huge

My laptop (which ran this at 650k pw/sec) didn't find your password given these simple iterations (dictionary word with some guesses about case, 0-2 digits following it). /usr/share/dict/american-english-huge has 341,472 words and comes from the Debian package wamerican-huge version 7.1-1.

  • this is the hash 4ef6c6a21c0172c8cd322dd8f78ba6c1 and this is the $username ; that is mean the follow : MD4(MD4(($pass)).(strtolower(admin))) the problem in password can i use John The Riper to find it – Radwan Cherfy Jul 10 '15 at 2:47
  • I keep meaning to try John the Ripper, but I haven't yet. You can manually try this with a simple script (added above), but you'll have to invent all of the possible password schemes manually (it wasn't an obvious one). – Adam Katz Jul 10 '15 at 22:30

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