Archived Challenges - April 2024

Guess that the password is also ctf and solve the babypwn challenge.

ictf{Round_45_Sanity_Check}

Write a good essay idk. or guess the flag.

Excerpt from a good essay: Throughout this era in human history, Liao derives his claims of community voices from three main factors. His likablity as a polticial figure, his confidence as a public speaker, and his god-like genetics. Likeability, confidence, and genetics (LCG) allow the public to see through his validity.

• written by Cayden Liao

jctf{rеd_flags_and_fakе_flags_form_an_еquivalеncе_class}

The challenge code just adds 0x1337 to each character. Convert the hex to binary, use .decode() on it and then subtract 0x1337 from each character to get the flag

ictf{c3r71f13d_1337_h4x0r!}

Look into Lua source code to figure out that it uses xoshiro256** to generate random numbers. As a transition from state n to state n+1 is a linear transformation, it is possible to calculate its standard matrix. Now it's trivial to quickly calculate a transition through many states quickly by representing the state of the PRNG as a vector and left-multiplying it with the standard matrix raised to the appropriate power. Now all that's left is bruteforcing the timestamp (there are 60 possibilities). A Sage script solving the challenge is available here: https://imaginaryctf.org/f/OoCLY#solve.sage

ictf{xoshiro_jumps_in_lua_like_a_pro}

The challenge is seeding the current time and then getting random bytes. We can brute-force this! Solve script at: https://fdow.nl/8yJ6n

ictf{1_c4n_bru73f0rc3_t1m3_1ts3lf!}

It uses os.urandom to generate the first byte of the key. It seeds this byte and calculates the next byte and so on. If we know the first byte we can get the entire key. Just brute-force the first byte! Solve at: https://imaginaryctf.org/f/FAEBu

ictf{1_n33d_t0_1mpr0v3_s3cur17y}

Get 624 random numbers, feed it into randcrack, profit! Solve: https://fdow.nl/iCsUY

ictf{1_4m_cr4ck1ng_7h3_g4m3_g3t_1t?}

jwt sign/encrypt confusion attack, use https://github.com/SecuraBV/jws2pubkey to recover the public key from two jwts. if the solve-script takes too long, just kill it and run it again. Solve.py: https://fdow.nl/23LEn

ictf{i_had_too_much_fun_styling_the_html}

Eyes (i) represent 0, Els (L) represent 1. The attachment is converted into binary representation, and then use "from binary" and "from Base64" recipe on Cyberchef. Partially inspired by John Hammond's video on plaintext QR code that bypasses email filters.

Recipe: https://gchq.github.io/CyberChef/#recipe=From_Binary('Space',8)From_Base64('A-Za-z0-9-_',true,false)

ictf{DotTheEyes_and_WhatDoWeDoForTheEls}

As the challenge description suggests, this is bacon cipher. Eyes (i) represent 'A' or zero and Els (L) represent 'B' or one. Decoding this from bacon cipher (Using Standard mode in cyberchef. The other one does not work), we get our flag! Cyberchef solve: https://gchq.github.io/CyberChef/#recipe=Find_/_Replace(%7B'option':'Regex','string':'I'%7D,'0',true,false,true,false)Find_/_Replace(%7B'option':'Regex','string':'l'%7D,'1',true,false,true,false)Bacon_Cipher_Decode('Standard%20(I%3DJ%20and%20U%3DV)','0/1',false)&input=bElJbEkgSUlJSUkgbElJSWwgbElJbEkgbElsbEkgSUlJSWwgSUlJSUkgSUlJbEkgSWxsSWwgSWxsSUk

ictf{tastybacon}

You can use this website: https://www.ilovepdf.com and then go to "Unlock PDF" and it will unlock it for you. Alternatively, you can use PDFcrack with rockyou to crack the PDF.

ictf{n0t_s0_s3cur3_p4ssw0rd}

Just run strings on it!

ictf{7h1s_1s_7h3_4c7u4l_fl4g!}

Searching the image on Google Images, you can find that it is "Charity Glacier". The flag would be ictf

ictf{Charity_Glacier}

Decompile the class file (using e.g. Bytecode Viewer) and figure out that it xors the flag with a 2-byte key "\x42\x24", base64-encodes the result & compares it to a static string. base64-decode the static string and xor it with "\x42\x24" to get the flag. https://gchq.github.io/CyberChef/#recipe=From_Base64('A-Za-z0-9%2B/%3D',true,false)XOR(%7B'option':'Hex','string':'4224'%7D,'Standard',false)&input=SzBjMlFqa1ZjUmQxZXlGV2NVQXJVREY3TlJRd1VDcDdja0lkUjNKUk1GZHhWejg9&oeol=VT

ictf{1337_cr3dits_w0rth_0f_c0urs3s}

The binary is a self-corrupting program which does so by calling itself recursively twice. The flag is divided between the two recursive calls. The starting point is to identify that characters ELF are in their usual place (as they should be in a typical ELF file) and is XOR'd by a three-character key.

ictf{maybe_SelfcoRRuptiON_isTheAnswer}

Extract libinterop.so.enc from the jar and decrypt it using the algorithm from the apply method in the Main class. Next analyse this shared library and realize that it encrypts the input using textbook RSA and compares it to a constant number. Finally decrypt the modulus of the RSA key, use Sage or http://factordb.com/ to find out the primes (one of them is very small, so it's easy to factor this number) and decrypt the flag.

ictf{w0w_m4ny_l4ngu4g3s_4nd_br0k3n_r54}

ictf{r3t2DLR3s0lve_D0cumEnT4t10n_FTW!!}

ictf{U5ing_l1BC_T0_dr0P_4_sh33ll}

writeup.md: https://fdow.nl/ulFsV

ictf{UAF_ftw_7fde1a2b6}

There are basically no useful rop gadgets for pop rdi, but since gets puts a pointer to _IO_stdfile_0_lock (which is in a writable region in libc) in rdi, we can just use gets to read into that area and then call system (with our command in rdi). For whatever reason gets or some other IO operation will mess up the string, use sh # so that the invalid part that gets added is commented out.

ictf{_IO_stdfile_0_lock_xd_c35a49e9}

No bounds checks on registers (it has 16 gen. purpose regs but you can use the -3 or 999 reg); leak ptr to the vm's memory, calc libc base from it; no bounds checks on memory accesses, arb r/w. Use it to overwrite main's ret ptr with a ROP chain, then HALT.

ictf{m0v_r999_0x1337_f0r_th3_w1n}