1

I have a custom application that monitors and logs a user's SMTP session, and came across this spammer trying to use my MTA as a relay.

Is it possible to determine what username they are trying to log in as?

Waiting for a connection... Connected!


SmtpReceiveTestAgent_OnEhloCommand
DisableStartTLS: False
Domain: OWNEROR-KTATDUI
Spambypass False
AuthenticationSource Anonymous
HelloDomain
IsConnected True
IsExternalConnection True
IsTls False
LastExternalIPAddress 5.9.32.178
LocalEndPoint 10.10.10.242:25
RemoteEndPoint 5.9.32.178:2648
SessionId 634767757514516172
Waiting for a connection... Connected!


SmtpReceiveTestAgent_OnAuthCommand
AuthenticationMechanism:
Spambypass False
AuthenticationSource Anonymous
HelloDomain OWNEROR-KTATDUI
IsConnected True
IsExternalConnection True
IsTls False
LastExternalIPAddress 5.9.32.178
LocalEndPoint 10.10.10.242:25
RemoteEndPoint 5.9.32.178:2648
SessionId 634767757514516172
Waiting for a connection... Connected!


SmtpReceiveTestAgent_OnReject
Command: TlRMTVNTUAADAAAAGAAYAH4AAABSAVIBlgAAAAAAAABYAAAACAAIAFgAAAAeAB4AYAAAAAA
AAADoAQAABYKIogYBsR0AAAAP1fXonCW+WU07L/KUILITX3QAZQBzAHQATwBXAE4ARQBSAE8AUgAtAEs
AVABBAFQARABVAEkAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAdkWHUUPIHk2TIK1nq2Rj8QEBAAAAAAA
Acvo4+TJYzQFloTpuXluwygAAAAACABAAUgBFAEwAQQBZADMANgAwAAEAGABDAE8ATgBZAEMARQBYADM
ANgAwADAAMgAEABwAcgBlAGwAYQB5ADMANgAwAC4AbABvAGMAYQBsAAMANgBDAE8ATgBZAEMARQBYADM
ANgAwADAAMgAuAHIAZQBsAGEAeQAzADYAMAAuAGwAbwBjAGEAbAAFABwAcgBlAGwAYQB5ADMANgAwAC4
AbABvAGMAYQBsAAcACABy+jj5MljNAQYABAACAAAACAAwADAAAAAAAAAAAAAAAAAwAAC0ykOxCYthQLJ
DgBWZ1QybmTgAin969Z+a+/3oBg6+MwoAEAAAAAAAAAAAAAAAAAAAAAAACQAQAFMATQBUAFAAUwBWAEM
ALwAAAAAAAAAAAAAAAAA=
Original Arguments:
Parsing Status: Error
SMTP Response: 535 5.7.3 Authentication unsuccessful
Spambypass False
AuthenticationSource Anonymous
HelloDomain OWNEROR-KTATDUI
IsConnected True
IsExternalConnection True
IsTls False
LastExternalIPAddress 5.9.32.178
LocalEndPoint 10.10.10.242:25
RemoteEndPoint 5.9.32.178:2648
SessionId 634767757514516172
Waiting for a connection... Connected!


SmtpReceiveTestAgent_OnHeloCommand
Helo Domain: 8.8.8.65
Spambypass False
AuthenticationSource Anonymous
HelloDomain
IsConnected True
IsExternalConnection True
IsTls False
LastExternalIPAddress 114.43.5.69
LocalEndPoint 10.10.10.242:25
RemoteEndPoint 114.43.5.69:11968
SessionId 634767757514516612
Waiting for a connection... Connected!


SmtpReceiveTestAgent_onMailCommand
Auth:
BodyType: NotSpecified
DSN requested: NotSpecified
EnvelopeID:
FromAddress: [email protected]
Oorg:
Size: 0
Spambypass False
AuthenticationSource Anonymous
HelloDomain 8.8.8.65
IsConnected True
IsExternalConnection True
IsTls False
LastExternalIPAddress 114.43.5.69
LocalEndPoint 10.10.10.242:25
RemoteEndPoint 114.43.5.69:11968
SessionId 634767757514516612
Waiting for a connection... Connected!


SmtpReceiveTestAgent_OnReject
Command: RCPT TO: <[email protected]>
Original Arguments:
Parsing Status: Error
SMTP Response: 550 5.7.1 Unable to relay
Spambypass False
AuthenticationSource Anonymous
HelloDomain 8.8.8.65
IsConnected True
IsExternalConnection True
IsTls False
LastExternalIPAddress 114.43.5.69
LocalEndPoint 10.10.10.242:25
RemoteEndPoint 114.43.5.69:11968
SessionId 634767757514516612
Waiting for a connection...

1 Answer 1

7

It seems they tried to login as test. This is the dissection of the field Command:

NTLMSSP identifier: NTLMSSP
NTLM Message Type: NTLMSSP_AUTH (0x00000003)
Lan Manager Response: 000000000000000000000000000000000000000000000000
NTLM Client Challenge: 0000000000000000
NTLM Response: 7645875143c81e4d9320ad67ab6463f10101000000000000...
NTLM Client Challenge: 65a13a6e5e5bb0ca
Domain name: NULL
User name: test
Host name: OWNEROR-KTATDUI
Session Key: Empty
Flags: 0xa2888205
Version 6.1 (Build 7601); NTLM Current Revision 15
MIC: d5f5e89c25be594d3b2ff29420b2135
3
  • How did you parse this? Commented Jul 2, 2012 at 16:28
  • 7
    With wireshark. I decoded the base64 text, converted it to ascii hex dump and imported it as dummy tcp packet from wireshark. Then I applied the ntlmssp dissector to the dummy packet.
    – Zzz
    Commented Jul 2, 2012 at 18:38
  • +1 for the initiative and explanation in the preceding comment!
    – dotancohen
    Commented Jan 27, 2014 at 11:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .