CR3 CTF 2024

Chall	Category	Total Solved
getting-closer-dang	crypto	46 Solved
donut	forensic	36 Solved
jscripting	web	13 Solves

This only me + n2l team writeup ( because i also play in another team called Huntik ) . im not played in N2L so N2L got 119th place.

getting-closer-dang

855B

crypto_getting-closer-damn.7z

archive

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chall.py

import os, json, random, math
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from Crypto.Util.number import getPrime, GCD
from hashlib import sha256
#from secret import FLAG

def get_prod():
    return math.prod([random.choice(pool) for _ in range(3)])

FLAG = b'cr3{???????????????????????????????}'

N = getPrime(512)

pool = [getPrime(9) for _ in range(10)]
a, b = [get_prod() for _ in range(2)]

g = GCD(N**a - 1, N**b - 1) # pros of having a quantum computer ^-^

key = sha256(str(g).encode()).digest()[:16]
iv = os.urandom(16)
cipher = AES.new(key, AES.MODE_CBC, iv)
ct = cipher.encrypt(pad(FLAG, 16))
out = {'iv': iv.hex(), 'ct': ct.hex()}

with open('output.txt', 'w') as f:
    f.write(f'{N = }\n')
    f.write(f'{out = }')

program enkripsi aes cbc sederhana di mana, pada variabel g terdapat penghitungan GCD antara (N^a)-1 dengan (N^b)-1 output.txt

N = 13185232652915309492470700885494158416479364343127310426787872363041960044885500175769971795951432028221543122753022991035176378747918784016983155565886369
out = {'iv': '6f69ac380715dbf9b00ef32ca8c204bb', 'ct': 'e654237a76d61d3bf97b315af1c2a517797b34b8eca270dbc8132dda1f425065b7b84690d4c21cdaf2ab17c2876738ed'}

mekanisme enkripsi:

1. Generate N prime random sebesar 512 bit

2. Generate nilai array pool

3. Menghitung faktor secara random yang terdapat pada array pool dan di simpan pada variabel a dan b

4. Menghitung GCD(FPB) dari (N^a) - 1 dengan (N^b) - 1)

5. Menyimpan hasil perhitungan GCD kemudian dijadikan sebagai key untuk mengencrypt

mekanisme decrypt menurut gwhj: Merecover nilai a,b -> tidak mungkin karena tidak disediakan nilai g, karena hal tersebut saya menemukan opsi lain yakni penghitungan nilai g dari GCD akan sama dengan persamaan berikut (N^GCD(a,b))-1, karena hal tersebut sama maka tinggal bruteforce saja nilai GCD(a,b)

solver.py

import os, json, random, math
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from Crypto.Util.number import getPrime, GCD
from hashlib import sha256
import sys
sys.set_int_max_str_digits(0)
#from secret import FLAG
N  = 13185232652915309492470700885494158416479364343127310426787872363041960044885500175769971795951432028221543122753022991035176378747918784016983155565886369
iv = b'oi\xac8\x07\x15\xdb\xf9\xb0\x0e\xf3,\xa8\xc2\x04\xbb'
ct = b'\xe6T#zv\xd6\x1d;\xf9{1Z\xf1\xc2\xa5\x17y{4\xb8\xec\xa2p\xdb\xc8\x13-\xda\x1fBPe\xb7\xb8F\x90\xd4\xc2\x1c\xda\xf2\xab\x17\xc2\x87g8\xed'
g = ((N**(439))-1)
#g = [((N**i)-1) for i in range(1000)]
key = sha256(str(g).encode()).digest()[:16]
#key = [sha256(str(g).encode()).digest()[:16] for g in g]
#iv = os.urandom(16)
cipher = AES.new(key, AES.MODE_CBC, iv)
#cipher = [AES.new(key, AES.MODE_CBC, iv) for key in key]
fl = cipher.decrypt(ct)
#fl = [i.decrypt(ct) for i in cipher]
#ct = cipher.encrypt(pad(FLAG, 16))
#out = {'iv': iv.hex(), 'ct': ct.hex()}

print(f"""
g = {g}
key = {key}
N  = {N}
flag = {fl.decode('latin-1')}""")

by : gr3yr4t

donut

186KB

for_donut.7z

archive

Open

chall source

after we extract it we can see there is a .git files and my terminal shows it git

we can see we in branch master

bet the git seems broken after i use command log or status

so i directly use grep or strings like this:

cd .git
grep -roa 'flag'

and we can see in the head we can strings to see the log of commit

grep -roa 'flag'
output : 
index:flag
HEAD:flag
hooks/fsmonitor-watchman.sample:flag
logs/HEAD:flag
logs/HEAD:flag
logs/HEAD:flag
COMMIT_EDITMSG:flag

strings logs/HEAD
e1cf8e13bfe7f2b9522642a878368677621349b1 e1cf8e13bfe7f2b9522642a878368677621349b1 dxqwww <[e-mail]> 1713541575 +0300	checkout: moving from master to flag
e1cf8e13bfe7f2b9522642a878368677621349b1 2a461812df0b80851da28cf6e69f3a2d84129b01 dxqwww <[e-mail]> 1713541687 +0300	commit: added flag
2a461812df0b80851da28cf6e69f3a2d84129b01 e1cf8e13bfe7f2b9522642a878368677621349b1 dxqwww <[e-mail]> 1713541710 +0300	checkout: moving from flag to master

we can see the hash of the added flag we directly use git show

git show 2a461812df0b80851da28cf6e69f3a2d84129b01

cr3{w3_4r3_411_10v3_d0n7T5!1

jscripting

3KB

web_jscripting.7z

archive

Open

referr to this Bypassing force return & proccess null

cr3{W0vv53Rs_1n5q3c70R_G4dg3T!!1}