Archived Challenges - May 2025

take the current month, multiply by the utc hour we release challenges at, then subtract the number of active board members you get 57, this is the least convoluted challenge we're gonna release this month

ictf{welcome_to_round_57_of_imaginaryCTF}

least convoluted statistics problem, if n isn't a power of 2 this doesn't shuffle uniformly so you can differentiate it from a real random shuffle, see solve https://cybersharing.net/s/777599ff768ad61533ed087cb8513539

ictf{is_n=8_possible_though?}

This is a Feistel cipher where each key is the same in the key scheduling, this allows you to turn the encryption oracle into a decryption oracle by just reversing the chunks in the output

'' from Crypto.Util.number import long_to_bytes, bytes_to_long

A = long_to_bytes(int(input("print flag result?"))) A = A[16:] + A[:16] print("Enter this into encrypt message",bytes_to_long(A))

B = long_to_bytes(int(input("encrypt message result?"))) print(B[16:] + B[:16]) ''

ictf{using_1_k3y_1n_f31573l_lol}

'' import base64 encrypted = bytearray(base64.b64decode('FAoXEh0PHAEaLDYaLD4+LwIXEQANJiwHERcXChsNCRUBOycXHS08CwkNADM9FwErPg8XDhYYCiw4CAALOzUFDT03CRcTOj4+OQoNBho=')) encrypted[0] = ord('i') for i in range(1,len(encrypted)): encrypted[i] ^= encrypted[i - 1] print(encrypted) ''

ictf{this_is_a_pretty_stereotyped_xor_chall_but_anyways_good_job_have_a_flag}

https://cybersharing.net/s/05f6b2290e2e6d54a34f131c3e65fb37

ictf{devious_mersenne_prime_makes_root_finding_easy_actually}

it's called recurring because intended solution draws on recurrence ideas https://cybersharing.net/s/c08b2bfc679007e034223e3cad0a7127

ictf{wh4t_ev3n_i5_@_r34l_w0r1d_4ppl1c4ti0n_9OoYVHHxYhQG6teVZXHC}

Vigenere cipher with known key length, easy solve using https://www.dcode.fr/vigenere-cipher, correct key is BALSAMICVINEGAR

ictf{is_this_what_they_mean_by_an_unbalanced_oil_and_vinegar_scheme}

solve script: https://cybersharing.net/s/8ead4e12a3214b930beba35e12b8b5bb brief writeup: a%n = a-k*n, hence a%n1-a%n2 = a-k*n1-a+k*n2. By the following theorem: A Linear Diophantine equation of the form ax+by=c has a solution (where x and y are integers) if and only if c is a multiple of the greatest common divisor of a and b., a%n1-a%n2 | gcd(n1,n2). Find 2 polynomials with polynomial GCD | x-1, then factor n using Pollard's p-1 factorisation

ictf{l1ne4r_di0phan+1n3_equat1on!}

username as \' OR 1=1;-- and password as whatever backslashes are not escaped so you can inject a backslash that escapes the "sanitizer" backslash

ictf{baby's_first_time_writing_web!_NDy3MOvyplcw}

bypass the referer thing using <meta name="referrer" content="never">, then use the fact that firefox supports stylesheet in the link header to load an arbitrary css file and perform css injection to extract the flag from the div

ictf{firefox_yupeee!}

The program encodes a string as a tree by taking successive bit pairs and either encodes the remaining bits recursively across one or both sides of the tree. Easy to recover the flag once you understand the encoding by simple parsing. https://cybersharing.net/s/88de20c9167680ebf073da28c0328878

ictf{S7r1nG5_a5_Tr3e5_4s_5tr1Ng5}

So basically we can notice this is a version of subleq but instead of checking for ⇐ it checks for equality. If we decompile it we notice there is a nested vm with a special instruction set. The . instruction decrements the current input char and - checks if the current value minus 0x23 is equal to zero. From this information we can get the flag by extracting the nested vm code, splitting it using (parts = code.split('-')), and then the flag is for part in parts: flag += chr(part.count('.')+0x23)

from pwn import * 
with open('code.bin', 'rb') as f:
    code = f.read()
raw = code[0x8000//2:]
data = b''
for i in range(0, len(raw), 8):
    data += bytes([u64(raw[i:i+8])])
flag = '' 
for part in data.split(b'-'):
    if len(part)==0: continue 
    flag += chr(len(part)+0x23)
print(flag)

ictf{fun_w17h_f4k3_350l4n65_a55b1dfe}