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Final rank: 112/775 with 7 challenges solved

Table of Contents

[Web] Warmup


Warmup (Web, Baby, 10 pts)

Author: George Zaytsev (groke)


Just get the flag


When browsing to the link in the sentence, we were instantly redirected to /final.html, which displays a very long text but no flag.

Using Burpsuite, we could intercept the redirection and grab the flag lcoated in /index.html.

Warmup flag

Once base64-decoded, we could read the following flag: cybrics{4b646c7985fec6189dadf8822955b034}.

[Web] Bitkoff Bank


Bitkoff Bank (Web, Easy, 50 pts)

Author: Alexander Menshchikov (n0str)

Need more money! Need the flag!



I’m pretty sure I didn’t solve this challenge the intended way because my method was pretty dumb. Judge by yourself! :wink:

After registering an account, you were greeted with the following php page:

Bitkoff home

Everytime you clicked on “MINE BTC”, 0.0000000001 was added to your BTC counter. It was, then, possible to convert BTC to USD in order to buy the flag when you reached 1$.

An auto-miner costed 0.01$ and would click for you every second… not so worth it.

What I did is pretty simple: I’ve automated the mining request thanks to a python script and ran 6 instances of it in parallel until I got enough BTC to buy the flag. Absolutely nothing was optimized but I had a few other things to do so I’ve left the script run in background for approximately 2 hours and I could buy the flag!

import requests,re

payload = {'mine': '1'}
cook = {'name' : 'boiteaklou', 'password' : 'boiteaklou'}

while 1:
    r ='',data=payload,cookies=cook)
    btc = re.findall('Your BTC: <b>([^<]*)</b>',r.text)
    print("BTC: %s"%btc[0])

Here is the cheapest bitcoin mining farm ever:

Mining farm

A minor difficulty consisted in the fact that we could not enter a value lower than 0.0001 in the change field because of some HTML client-side check. However, we could forge the request using Burpsuite and it worked like a charm.

Bitkoff 1 dollar

And the flag:

Bitkoff flag

flag: cybrics{50_57R4n93_pR3c1510n}

[Web] Caesaref


Caesaref (Web, Hard, 50 pts)

Author: Alexander Menshchikov (n0str)

There is an additional one: Fixaref

This web resource is highly optimized:


After register a new account, we were greeted with the following web page where we could ask questions to the support:

Caesaref home

At first, I lost a lot of time trying to redirect the support guy to my website via XSS payloads like <img src="http://mywebsite" />.

Actually, it was not necessary. We only had to paste an HTTP link in the text box and a bot would visit it instantly.

Using this information, we could paste the link to a webhook instance and surprisingly find the PHPSESSID cookie of the bot sent with the request.

Caesaref request

Then, we could replace our own PHPSESSID cookie by the retrieved one and refresh the page in order to access the bot account. Once connected, a new button was here to give us the flag.

Flag: cybrics{k4Ch3_C4N_83_vuln3R48l3}

[Network] Sender


Sender (Network, Baby, 10 pts)

Author: Vlad Roskov (vos)

We’ve intercepted this text off the wire of some conspirator, but we have no idea what to do with that.


Get us their secret documents


The given text file shows a SMTP trace from which we could extract some credentials as well as the password of an archive.

220 ESMTP Postfix (Ubuntu)
EHLO localhost
250-SIZE 10240000
250 DSN
334 VXNlcm5hbWU6      
334 UGFzc3dvcmQ6                        # Username
Q29tYmluNHQxb25YWFk=                    # Password
235 2.7.0 Authentication successful
250 2.1.0 Ok
250 2.1.5 Ok
354 End data with <CR><LF>.<CR><LF>
From: fawkes <>
To: Area51 <>
Subject: add - archive pw
Content-Type: text/plain; charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0

=62=74=77=2E=0A=0A=70=61=73=73=77=6F=72=64 =66=6F=72 =74=68=65 =61=72=63=
=68=69=76=65 =77=69=74=68 =66=6C=61=67=3A =63=72=61=63=6B=30=57=65=73=74=
250 2.0.0 Ok: queued as C4D593E8B6
221 2.0.0 Bye

Base64-decoded credentials: fawkes / Combin4t1onXXY

Mail content (quoted-printable decoded):


password for the archive with flag: crack0Weston88vertebra


A NMAP scan showed us that the pop3 port (tcp 110) was open so we could connect to it and authentify ourselves using the retrieved credentials.

$ telnet 110
Connected to
Escape character is '^]'.
+OK Dovecot ready.
USER fawkes
PASS Combin4t1onXXY
+OK Logged in.
+OK 1 messages:
1 138808
+OK 138808 octets
Return-Path: <>
Received: by sender (Postfix, from userid 1000)
        id B83843EBFF; Thu, 18 Jul 2019 16:41:23 +0000 (UTC)
Date: Thu, 18 Jul 2019 16:41:23 +0000
From: fawkes <>
To: Area51 <>, fawkes <>
Subject: interesting archive
Message-ID: <>
MIME-Version: 1.0
Content-Type: multipart/mixed; boundary="J2SCkAp4GZ/dPZZf"
Content-Disposition: inline
User-Agent: Mutt/1.5.24 (2015-08-30)

Content-Type: text/plain; charset=us-ascii
Content-Disposition: inline

take a look. dont share. secret.

Content-Type: application/zip
Content-Disposition: attachment; filename=""
Content-Transfer-Encoding: base64



Then, we could extract the content of the retrieved archive using the previously found password: 7z e -pcrack0Weston88vertebra

And we could finally read the flag inside the extracted PDF: cybrics{Y0uV3_G0T_m41L}.

[Network] Paranoid


Paranoid (Network, Easy, 113 pts)

Author: Vlad Roskov (vos)

Added at 14:40 UTC: to save some guessing, flag is the current password. Flag format is still cybrics{…}, so you’ll know when you find it.

My neighbors are always very careful about their security. For example they’ve just bought a new home Wi-Fi router, and instead of just leaving it open, they instantly are setting passwords!

Don’t they trust me? I feel offended.

Can you give me their current router admin pw?


The zip archive contained a pcap that I opened in Wireshark. Since the capture was quite big, I used Statistics > Protocol Hierarchy in order to get the big picture.

The packet capture was composed of 802.11 traffic and thanks to the protocol hierarchy, I could spot some HTTP requests.

As the statement was mentioning a password change, I decided to examine HTTP POST requests only thanks to the wireshark filter: http.request.method == "POST".

Inside the payload of HTTP POST request n°19173, we could find WLAN_AP_WEP_KEY1=Xi1nvy5KGSgI2&. Then, I added this wep key to wireshark decryption keys and it allowed us to find more HTTP requests.

Still filtering HTTP POST requests, I found a new password change request with this payload: WLAN_AP_WPA_PSK=2_RGR_xO-uiJFiAxdA33-PsdanuK& that I immediately set as WPA decryption key.

Once again, we had access to more decrypted HTTP traffic inside which the flag was located.

Paranoid flag

Flag: cybrics{n0_w4Y_7o_h1d3_fR0m_Y0_n316hb0R}

[Misc] ProCTF


ProCTF (CTB, Baby, 10 pts)

Author: Vlad Roskov (vos)

We Provide you a Login for your scientific researches. Don’t try to find the flag.

ssh pro@ Password: iamthepr0


After connecting to the machine via SSH, we were trapped inside a SWI-Prolog interactive interpreter. We could verify this assumption by pressing TAB twice, which would display the list of available functions.

$ ssh pro@
pro@\'s password:
Welcome to Ubuntu 19.04 (GNU/Linux 5.0.0-15-generic x86_64)

 * Documentation:
 * Management:
 * Support:

  System information as of Sat Jul 20 12:28:39 UTC 2019

  System load:                    4.15
  Usage of /:                     2.4% of 220.08GB
  Memory usage:                   10%
  Swap usage:                     0%
  Processes:                      508
  Users logged in:                3
  IP address for enp1s0:
  IP address for docker0:
  IP address for br-62bc0c6d2f97:

84 updates can be installed immediately.
48 of these updates are security updates.

WARNING: Your kernel does not support swap limit capabilities or the cgroup is not mounted. Memory limited without swap.
abort                           built_in_procedure              current_output                  erf
abs                             busy                            cut                             erfc
access                          byte                            cut_call                        error
access_level                    c_stack                         cut_exit                        eval
acos                            call                            cut_parent                      evaluable
acosh                           call_continuation               cycle                           evaluation_error
active                          callable                        cycles                          event_hook
acyclic_term                    canceled                        cyclic_term                     exception
add_import                      case_insensitive                date                            exclusive
address                         case_preserving                 db_reference                    execute
... skipped 54 rows

After some googling, I found an easy way to get a shell and to display the flag:

?- shell('sh').
$ cd /home
$ ls
$ cd user
$ ls
$ cat flag.txt

Flag: cybrics{feeling_like_a_PRO?_that_sounds_LOGical_to_me!____g3t_it?_G37_1T?!?!_ok_N3v3Rm1nd...}

[Stegano] Honey, Help!


Honey, Help! (rebyC, Baby, 10 pts)

Author: Vlad Roskov (vos)

Added at 10:50 UTC: there was a typo in the flag. Please re-submit.


I was working in my Kali MATE, pressed something, AND EVERYTHING DISAPPEARED!



This challenge was very easy but a bit painful for my eyes because I solved it late in the night.

The idea is to compare the clear and the encoded output in order to establish a match for each character.

Using this technique and a tiny bit of guessing (because it’s stega), I could build the following matching table and reconstruct the flag.

240C : c
< : y
2409 : b
23BC : r
240B : i
23BD : s
Pi : {
2424 : h
half-T : l
23BB : p
Cross : n
low T : w
|- : t
Pound : }
grey square : a


The CTF was pretty fun, thanks CyBRICS for the event!

BoiteAKlou :hammer: