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BlueHensCTF 2023 - RSA School 5th Grade

Authors

RSA School 5th Grade

Table of Contents

  1. Source
  2. Solution

Source

Fifth_Grade.py file content :

from Crypto.Util.number import *
msg=b"UDCTF{REDACTED}"
pt=bytes_to_long(msg)
p=getPrime(1024)
q=getPrime(1024)
N=p*q
e=3
ct=pow(pt,e,N)

print(N)
print(e)
print(ct)

output.txt file content :

19071553514906413228005623880868413172589438760530345745552708038769515697875361787053550188848159274987925247955174211167277615747329764460652862539122337714189780686582390326881171096308885109154336023212767779863472386169665627283720649094479648444588259600544834704143105214853522264311830387911281263299214052701109619722665736303738110883886917231219876629681611411323913511707032906816948757362133848480976586951323342448069343747851239877539085111823678094070778241732994351072251605007909682674187665596109353312252881532685577047967768366217935948525094732268620589271065304471832191222326947334404799847563
3
270903177796878498388304376598565799121492331770875203351555502784804760985678087802688162298096409297508110557051747972509915173895153270896299567072600809265143377905255294763705268648639628042173298874918538565864469546919085252896111245679898930789

Solution 

import gmpy2

# Given values
N = 19071553514906413228005623880868413172589438760530345745552708038769515697875361787053550188848159274987925247955174211167277615747329764460652862539122337714189780686582390326881171096308885109154336023212767779863472386169665627283720649094479648444588259600544834704143105214853522264311830387911281263299214052701109619722665736303738110883886917231219876629681611411323913511707032906816948757362133848480976586951323342448069343747851239877539085111823678094070778241732994351072251605007909682674187665596109353312252881532685577047967768366217935948525094732268620589271065304471832191222326947334404799847563
e = 3
ct = 270903177796878498388304376598565799121492331770875203351555502784804760985678087802688162298096409297508110557051747972509915173895153270896299567072600809265143377905255294763705268648639628042173298874918538565864469546919085252896111245679898930789

# Check if ct is a perfect cube
cube_root, exact = gmpy2.iroot(ct, e)
if exact:
    print(bytes.fromhex(hex(cube_root)[2:]).decode('utf-8'))
else:
    print("Ciphertext is not a perfect cube.")

Flag : UDCTF{0k_m4yb3_d0nt_u5e_e_3qu4l5_3}