Really S1mp(l3) Algorithm

Files Attached:

challenge.txt

Solution:

Tools that would have been helpful:

Taking a look at the file, we see 4 variables p, q, e and ct defined, and further taking inspiration from the initals of this challenge, we can infer that this challenge has to deal with RSA (and decrypting the ciphertext).

With some quick googling, I found this page and its accompanying code to decrypt our ciphertext. Since in our case the components of the prime p and q have been given to us, we do not need to factorsie anything (which the site recommeneded factordb for.)

A quick edit to the code later, we end up with this:

#!/usr/bin/python3
from Crypto.Util.number import inverse
p = 89677525768054651799811339393073439598902155753884683756875620558829954902529

q = 84438062750417241222463359000671279258755386034358736244893269480285225711041

e = 65537

ct = 2006481391032768131106572837821981606814991526844317992631122301790966354846642917808142106585149537517169168108747108233658443914026685951141453359021205

phi = (p-1)*(q-1)
d=inverse(e, phi)
m=pow(ct,d,p*q)
print(hex(m)[2:-1])

Running the program after that gives us a hexadecimal string, which we can then throw into CyberChef to convert to our flag.

(Note: I did realise when runnning the python program that there was a little bit of a weird ending to the flag – I’m pretty sure that if I bothered to look at the code more I could have fixed it, but in my case I just removed the trailing period with a closing curly brace)

Flag:

WH2022{1t5_r34lly_s1mpl3_4m_1_r1ght}