Archived Challenges - October 2025

skull urself

ictf{it_starts_again._welcome_to_round_58!}

just use hashcat or john the ripper to extract and crack the hash from the encrypted zip

ictf{5_minute_chal_that_takes_5_seconds_to_solve}

decoded the values as unicode, the characters are chinese (mostly I couldn't find good ones for v or y so they're japanese) homoglyphs for english characters

ictf{horseramblingstego}

idk man

ictf{a_ramp_was_built_in_the_breeding_shed_to_allow_Northern_Dancer_to_service_mares_who_were_taller_than_him}

The horse's name is "shishamo" it's written on his feed bucket in katakana, use jbis to find his racing record

ictf{shishamo_20-10-2022_10-09-2025_salt_mill}

reverse image search horse.jpg and you'll see its skibidi rizz, for the second image you can see the words groupe Lucien Barriere, this was the main sponsor of the prix de l'arc triomphe between 1999 and 2007, there are 14 horses in the starting area, the only year that this occurred is 1999. In the 1999 race the only jockey wearing bright yellow with a blue horizontal stripe was riding El Condor Pasa. The most recent common ancestor between El Condor Pasa and Skibidi Rizz was Mr. Prospector.

ictf{el_condor_pasa_mr._prospector}

stalk the owner's socials you'll see that shishamo is still living at his trainer's stable at the kasamatsu racecourse, right next to the course is kotohira shrine

ictf{kotohira_shrine}

yeah don't trust linear operators https://cybersharing.net/s/b6ee099be3efeee1688c9aeead42f125

ictf{4dD1ng+m0re_op3raTIon5~1sNT_alwAyS_stR0ngEr}

The modulus is not secure, it has a bunch of tiny factors and 1 big factor, this makes it computationally easy to factor. The biggest factor is also a smooth number, so you can easily compute discrete logs with pholig-hellman. Use the 2 plaintext pairs to recover the state of the prng (solve the rsa problem and the discrete log).

https://cybersharing.net/s/2e0ff23d9ae0a67fef4c7b6e72ddbb0c

ictf{a_doubly_smooth_number_in_this_economy?!?}

haru_cipher(a,b,c) is equivalent to a*b, its just multiplication lol

ictf{heavy_duty_multiplication_for_heavy_duty_cryptography}

`` from Crypto.Util.number import long_to_bytes from ast import literal_eval

M = literal_eval(open("output.txt").read()) flag_bits = "".join(str((c - 1) ^ 1) for c in M[0][1:]) flag = int(flag_bits, 2) print(long_to_bytes(flag)) ``

ictf{l3@Rn_s0m3_Tr0pic4l_s3M1R1n9}

just brute force the timestamp

ictf{unix_timestamps_are_just_delta_chrono_genesis}

from base64 import *


def dumb_encode(data):
    out = b''
    for i in range(len(data)):
        temp = b64encode(data[i:i+1])
        if i % 2:
            temp = temp[::-1]
        out += temp
    return out

ictf{verxinas_owner_is_a_famous_baseball_player}

easy rail fence cipher use dcode or cyberchef

ictf{thishorseabsolutelyhatesfencesforsomereason}

very bad implementation of merkle-hellman, don't even need lattices to break it, just divide every public key component by the previous to obtain the private key and decrypt

ictf{yukio_okabe_wrote_smut_about_riding_him}

Just use the determinant bro,

ictf{I_gave_up_but_I_am_ashamed_that_he_didnt}

This looks like a knapsack but it's linear algebra. Multiplication by a constant over GF(2^N) is just a linear map over the coefficient vector, Find the map corresponding to each of the rs in the public key and they'll form an overdetermined system, solve and get the flag.

ictf{I_stroked_Rudolphs_Body_his_coat_as_gentle_as_touching_the_finest_velvet}

https://cybersharing.net/s/b993ebe2402e17751dc90f0234fe0292

ictf{Meishi_Doto_famously_hates_ryuji_wada_but_liked_TM_Opera_O}

This binary embeds a modified RC4 stream cipher inside a tiny stack-based VM. Reverse the VM to recover the keystream and the flag. Alternatively, since RC4 encryption and decryption are identical, you can simply feed the ciphertext as input (with a debugger to modify the memory) and set a breakpoint at the comparison step — the memory there will already contain the decrypted flag.

ictf{4_vm_4nd_Rc4_for_fuN}

LCG discrete log, the LCG in question only has a period of 2**48, easy baby step giant step or kangaroo once you put the generator in matrix form, source: https://cybersharing.net/s/2089b3a3837fcb24c3489f47672afa35

ictf{ACosmicBitFlipSimulatorLol}

basic hamming code chal, source https://cybersharing.net/s/168c1ea77cb0df5a6047f2b72f1b817b

ictf{hmmmm_hamming_sandwich_code}

don't use the password hash to encrypt the data lmao, source https://cybersharing.net/s/f84f37345d1d4380174762350d2e6171

ictf{unf41r_70_c0mp4r3_p3rf0rm4nc35_wh3n_7h3yr3_c0mp371n6_4641n57_7h3_1999_60ld3n_63n3r4710n}

just a huffman code challenge easy to decode, src https://cybersharing.net/s/1b3973b8ac812ff2c2a0e7e8d7b36248

ictf{sakura_star_O_might_be_the_single_greatest_tragedy_in_jra_history..._anyways_where_is_he_cygames}