Cryptoverse CTF 2022 Writeups

This past weekend, I joined Shellphish for a last minute CTF called Cryptoverse

While I’m still a crypto n00b, I spent a few hours gathering some flags for the team.

Crypto

Warmup 1

293 solves / 107 points

Decode the following ciphertext: cGlwZ3N7cG5yZm5lXzY0X3Nnan0=.
Remember: CyberChef is your friend.

The ciphertext for this challenge had the distinct appearance of base64. The output format was correct, perhaps a simple rotation cipher such as ROT13 will help. This gets us our flag.

flag: cvctf{caesar_64_ftw}

Warmup 2

274 solves / 113 points

First described by Giovan Battista Bellaso in 1553, this cipher is easy to understand and implement, but it resisted all attempts to break it until 1863, three centuries later.
Remember: The Key to success is determination.
fzvxw{hqtegmfr_lw_msf_scrslg_kvwlhyk_fpr_kxg?}

If the given key and flag didn’t give it away, a quick search could tell you this is likely a Vigenère cipher.

Throw that in CyberChef and get the flag.

flag: cvctf{vigenere_is_too_guessy_without_the_key?}

Warmup 3

313 solves / 102 points

You should recognize this instantly.
-.-. …- -.-. - ..-. – —– .-. ….. …– .. … -. —– - ….. —– ..-. ..- -.
Note: Add { and } around the flag. Flag is all lower case.

Dots and dashes scream Morse code. dcode.fr is a great tool for all things decoding, including morse code.

flag: cvctf{m0r53isn0tfun}

Warmup 4

73 solves / 387 points

Last warmup. You should get it fast if you use any social media.
Ｉn sｃｉencｅ fіctіοｎｍetａνerse ｉs ｉtｅrａtiｏｎоf the Iｎternｅｔ as a sⅰｎgｌe, universal and immersive virtual world that is facilitated by the use of virtual reality and augmented reality headsets.
Note: Wrap the message you decoded in cvctf{}.
Note 2: This challenge involves some steganography tool related to a social media.

For this challenge, I tried several things before ultimately being successful. I thought maybe it would have to do with zero-width characters and tried tools like Stegsnow with no luck.

Eventually I found a Twitter Secret Message tool that gave the flag.

flag: cvctf{secretsaretobeh1dd3n}

Substitution

240 solves / 129 points

Substitution is a cryptographic technique where a plaintext is replaced by a ciphertext. The ciphertext is a substitution of the plaintext.
Here is a very simple CTF-related substitution cipher. Find out the flag.
Hxpkdiz kcz Osxe ja x apzhjxs ljvr go jvogimxkjgv azhdijkf hgmpzkjkjgva. Kcziz xiz kcizz hgmmgv kfpza, Uzgpxirf, Xkkxhl Rzozvhz xvr mjyzr. Jo fgd cxwz ojedizr gdk kcz xqgwz mzaaxez, cziz ja fgdi osxe, pszxaz xrr hdisf qixhlzka qzogiz adqmjaajgv: hwhkoxwzifajmpszadqakjkdkjgv

For this challenge, I realized hwhkoxwzifajmpszadqakjkdkjgv likely was the flag, hence the first hwhko should map to cvctf. After a little trial and error targeting 1 or 2 letter words, I decrypted with the following:

Plaintext: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
Ciphertext: sxhugyocritkmxfpbdlxxnvaxeXXXXXXXCXITXXXFXXDXXJXXAXX

Later, I found out quipquip can solve this for us.

flag: cvctf{averysimplesubstitution}

CyberMania

42 solves / 443 points

I got this piece of ciphertext from my friend who is a Cyber Mania. It must have hidden something…
file: secret.txt

secret.txt contains:

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

After a bit of CyberChef magic…

These emoji’s seemed suspicious, so I found cryptoji and it popped out the flag.

flag: cvctf{3m0j1_c4n_L34K_7h1ng5}

RSA 1

183 solves / 182 points

The n is so large that it’s not possible to factor it. Or is it?
n = 0x7c05a45d02649367ebf6f472663119777ce5f9b3f2283c7b03471e9feb1714a3ce9fa31460eebd9cd5aca7620ecdb52693a736e2fcc83d7909130c6038813fd16ef50c5ca6f491b4a8571289e6ef710536c4615604f8e7aeea606d4b5f59d7adbec935df23dc2bbc2adebbee07c05beb7fa68065805d8c8f0e86b5c3f654e651
e = 0x10001
ct = 0x35b63f7513dbb828800a6bcd708d87a6c9f33af634b8006d7a94b7e3ba62e6b9a1732a58dc35a8df9f7554e1168bfe3de1cb64792332fc8e5c9d5db1e49e86deb650ee0313aae53b227c75e40779a150ddb521f3c80f139e26b2a8880f0869f755965346cd28b7ddb132cf8d8dcc31c6b1befc83e21d8c452bcce8b9207ab76e

factordb allows us to “factorize” n to find p and q. The rest is plug and chug.

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#!/usr/bin/env python3
from Crypto.Util.number import inverse, long_to_bytes

# Given:
n = 0x7c05a45d02649367ebf6f472663119777ce5f9b3f2283c7b03471e9feb1714a3ce9fa31460eebd9cd5aca7620ecdb52693a736e2fcc83d7909130c6038813fd16ef50c5ca6f491b4a8571289e6ef710536c4615604f8e7aeea606d4b5f59d7adbec935df23dc2bbc2adebbee07c05beb7fa68065805d8c8f0e86b5c3f654e651
e = 0x10001
ct = 0x35b63f7513dbb828800a6bcd708d87a6c9f33af634b8006d7a94b7e3ba62e6b9a1732a58dc35a8df9f7554e1168bfe3de1cb64792332fc8e5c9d5db1e49e86deb650ee0313aae53b227c75e40779a150ddb521f3c80f139e26b2a8880f0869f755965346cd28b7ddb132cf8d8dcc31c6b1befc83e21d8c452bcce8b9207ab76e

# FactorDB the given 'n'
p = 8156072525389912369788197863285751656042515380911795404436333529629416084362735262281722179416240983448945672233749861517470671156357917601583268804973543
q = 10678085245814899631026086851678635624044610674331494223434578587048178556659016077336866548714220013612176819608742965144962254942844889066032236627832071
assert n == p*q

phi = (p-1)*(q-1)
d = inverse(e, phi)
m = pow(ct, d, n)
print(long_to_bytes(m))

flag: cvctf{f4c70rDB_15_p0w3rfu1}

Reverse

Baby Reverse

239 solves / 130 points

Baby warmup.
files: chall

After running the program, it gave the output:

$ ./chall
You do not really need to run the program...
The flag is: cvctf{7h15_15_4_f4k3_fl4g}

So, I ran:

strings chall | grep cvctf

This gave the flag.

flag: cvctf{r3v3r53_15_4w350m3}

Basic Transforms

97 solves / 337 points

Basic JS for you to tackle.
file: app.js

app.js contains:

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var readline = require('readline');
var Crypto = require('vigenere');

var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
    terminal: true
});

rl.on('line', function(line){
    if (line.length == 20 && line.startsWith("cvctf{") && line.endsWith("}")) {
        var cat = Crypto.encode(line.substring(6, line.length - 1), "nodejsisfun").split('').map(function(c) {
            return String.fromCharCode(c.charCodeAt(0) + 1);
        }).join('');
        if (Buffer.from(cat.split("").reverse().join("")).toString('base64') == "QUlgNGoxT2A2empxMQ==") {
            console.log("Correct!");
        }
    }
});

For this challenge, I simply used CyberChef again and reversed the cipher.

However, I wasn’t getting the full flag upon decoding the Vigenère cipher. So, I reverted back to js using the runkit environment.

flag: cvctf{0bfu5_N0d3_H@}

Misc

iKUN1

https://www.youtube.com/watch?v=w8bidbK0PI
Find the flag for iKUN 1, already wrapped in flag format.
Hint: Please find me (CryptoverseCTF) on THE platform for code repositories hosting.
Note: The YouTube link is unrelated to solving the challenge.

THE platform seemed likely to be Github. I was quickly able to find the account, then found a repo with a cvctf{fake_flag_but_correct_repo} bio. I checked the commit history and found the flag:

flag: cvctf{git_reveals_everything}

Not Georainbolt

37 solves / 451 points

Georainbolt is able to guess the location on Google Map in 0.1 seconds. Can you do it in a second?
Given an IP address or a coordinate, tell me which city it is in. All answers are case-insensitive. You will need a space between words, such as “new york city”.
For simplicity, I will make it easy. There are 50 questions and you only need to achieve an accuracy of more than 50% to get the flag!
nc 137.184.215.151 22606

After connecting, the challenge will prompt the user to guess the city based on IP or Coordinate (as shown below). If you answer in more than 1 second, it’s automatically “Wrong.”

$ nc 137.184.215.151 22606
...
IP: 2.145.91.238
City:

Coordinate (lat, lon): 45.3747, -75.5647
City:

We can script this to solve!

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#!/usr/bin/env python3
import time
import ipinfo
from pwn import *
from geopy.geocoders import Nominatim

# init ipinfo
at = '<REDACTED>'
h = ipinfo.getHandler(at)

# init geo locator
geolocator = Nominatim(user_agent="geoapiExercises")

r = remote('137.184.215.151', 22606)
r.recvuntil(b'wrong.\n')
log.info('Solving...')
for i in range(50):
    time.sleep(0.2)
    q = r.recvuntil(b'City:')
    if b'IP' in q:
        ip = q.split(b'IP: ')[-1].split(b'\n')[0].decode()
        d = h.getDetails(ip)
        r.sendline(d.city.encode())
    else:
        coor = q.split(b'lon): ')[-1].split(b'\n')[0].decode()
        location = geolocator.reverse(coor)
        try:
            city = location.raw['address']['city']
        except:
            city = 'failed'
        r.sendline(city.encode())
r.interactive()

flag: cvctf{4r3_y0u_4_R34L_Ge@r41nB0L7?}

Crypto#

Warmup 1#

293 solves / 107 points#

Warmup 2#

274 solves / 113 points#

Warmup 3#

313 solves / 102 points#

Warmup 4#

73 solves / 387 points#

Substitution#

240 solves / 129 points#

CyberMania#

42 solves / 443 points#

RSA 1#

183 solves / 182 points#

Reverse#

Baby Reverse#

239 solves / 130 points#

Basic Transforms#

97 solves / 337 points#

Misc#

iKUN1#

Not Georainbolt#

37 solves / 451 points#