ctf-writeup/2023/BlueHens CTF 2023/RSA School 2nd Grade

RSA School 2nd Grade

Ok a little tougher

About the Challenge

We were given 2 files, Second_Grade.py and output.txt. Here is the content of First_Grade.py

from Crypto.Util.number import *
n=166045890368446099470756111654736772731460671003059151938763854196360081247044441029824134260263654537
e=65537
msg=bytes_to_long(b'UDCTF{REDACTED}')
ct=pow(msg,e,n)
print(n)
print(e)
print(ct)

And here is the content of output.txt

166045890368446099470756111654736772731460671003059151938763854196360081247044441029824134260263654537
65537
141927379986409920845194703499941262988061316706433242289353776802375074525295688904215113445883589653

How to Solve?

In this case im using X-RSA to recover the plaintext, and then choose the first option

UDCTF{pr1m3_f4ct0r_the1f!}