Archived Challenges - March 2025

sanity

ictf{sanity_check_round_55_means_flag_is_55_characters}

As far as I know, solving this with a legit solution (probably) isn't possible without having *, but what is possible is gambling 🤑. Only 5% of the numbers between 100M and 1B are prime, so the odds of picking a random number from that range 100 times and having none of them be prime is (1-5/100)^100 = 0.0059205 or just about 1/169 odds of winning by sending something like n-n. So set up a simple script to keep connecting and sending that until it gives you the flag!

ictf{what?_it_was_gambling_all_along?}

floating point arithmetic is kinda weird, this operation actually masks out the lower 21 bits of the float, so just unpack it and read the bits. https://cybersharing.net/s/26a75ffff3982ed6c5c07c9bc333fa6d

ictf{5p1ky_d157r1bu710n5_c4n_l00k_fl47_fr0m_4_d1574nc3}

we've been doing this for 56 rounds kinda crazy to think about huh

ictf{welcome_to_round_56}

If you google image search the photo you'll see it was taken at umekoji park in Kyoto Japan. You can then go on street view and find that it was taken right next to 34.9859301,135.7484277

ictf{34.98593,135.74842}

This is just the scream cipher described in https://xkcd.com/3054/

ictf{THIS_CIPHER_IS_VERY_LOUD_AND_INSECURE}

First construct a 2^33-collision of md5 by chaining fastcoll, then use crc's linearity to solve a linear system to collide on crc. https://cybersharing.net/s/660599dbef535298c1e9d5c133adbf60

ictf{easy_peasy_hash_collision_from_linearity}

This is a polygramic cipher called the morbit cipher. I've given you the keyword 'SECRETKEY' however as the flag points out this isn't really necessary due to how small the keyspace for this cipher is.

ictf{POLYGRAMICCIPHERWITHVERYTINYKEYSPACECANBEBRUTEFORCEDALSOMORBIUSCIPHERLOL}

python -c 'print("a"*200000)' | nc [ip] [port]

ictf{why_us1ng_tr1pIe_x0r_1n_Crypt0syst3m}

https://cybersharing.net/s/18cc819b830847e7c1c183844ab8f7a9

ictf{XOR_m337_in_th3_m1ddl3}

This is a program written in John Conway's FRACTRAN esolang. If you run the program it will output 613^12203154003490548175721013914001941979229278133070782051717611318191896301489789, the exponent of which is the encoded flag

ictf{The_Very_Basics_of_FRACTRAN}

This is a very shitty RSA implementation reverse engineer the communication process then answer the server's math questions to get the flag see source for details: https://cybersharing.net/s/064fbd92c3e498e21b713d3369eec832

ictf{The_R_in_RSA_Stands_For_Rev}

This is just a basic randcrack challenge with the twist that the first 1000 values have been xored using the C rand() function. srand was called with the current time so just use the date modified time of the encrypted.txt +- a second or two. source and solve: https://cybersharing.net/s/f63860a4f325375bc85d7b1ff563f268

ictf{1dk_h4v3_4_10_m1nu73_r4nd_cr4ck_pr0bl3m}

pretty simple stream cipher based on 2x2 matrices over Z(256), see source for details https://cybersharing.net/s/633796204069d521b5419ec346629cf1

ictf{4_l177l3_h4rd3r_7h4n_v163n3r3}