Post

2023 ImaginaryCTF - web

  • Points: 100

Description

We recovered this file from the disk of a potential threat actor. Can you find out what they were up to?

Attached

web.zip

Analyzation

Check the login.json file

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
{"nextId":2,
"logins":[
    {"id":1,
    "hostname":"https://yoteachapp.com",
    "httpRealm":null,
    "formSubmitURL":"https://yoteachapp.com",
    "usernameField":"",
    "passwordField":"",
    "encryptedUsername":"MDIEEPgAAAAAAAAAAAAAAAAAAAEwFAYIKoZIhvcNAwcECJs6PTFwzrMiBAiRmXcD4tn3bw==",
    "encryptedPassword":"MGIEEPgAAAAAAAAAAAAAAAAAAAEwFAYIKoZIhvcNAwcECBZPCW+NjkpUBDieso9w5lPvD85RNcErLbGTXdamyji7ZKcL9FHxjnvt1WqwcVCsOETgCWCgwCg1jJmAW/MYugOoqQ==",
    "guid":"{8ee7f027-974b-48cb-b9aa-29fc5a728c39}",
    "encType":1,
    "timeCreated":1688943236140,
    "timeLastUsed":1688943236140,
    "timePasswordChanged":1688943236140,
    "timesUsed":1,
    "encryptedUnknownFields":null}],
"potentiallyVulnerablePasswords":[],
"dismissedBreachAlertsByLoginGUID":{},
"version":3}

We need to login to yoteachapp.com by the given encoded account.

Solution

Use firefox_decrypt.py to decrypt the account

1
py firefox_decrypt/firefox_decrypt.py ./.mozilla/firefox/
1
2
3
Website:  https://yoteachapp.com
Username: ''
Password: 'UeMBYIbgPqNiSWzOVguTbccMOnLirDoEGTjgiqNrbOvwzynbyN'

Login and find the flag.

The flag is

1
ictf{behold_th3_forensics_g4untlet_827b3f13}
This post is licensed under CC BY 4.0 by the author.