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1.1 KiB
1.1 KiB
Quick Math
The main idea finding the flag is decrypting the RSA exponentiation.
Step-1:
Given statement
Ben has encrypted a message with the same value of 'e' for 3 public moduli -
n1 = 86812553978993 n2 = 81744303091421 n3 = 83695120256591 and got the cipher texts -
c1 = 8875674977048 c2 = 70744354709710 c3 = 29146719498409. Find the original message.
Step-2:
This article is quite renowned: https://www.johndcook.com/blog/2019/03/06/rsa-exponent-3/
Step-3:
So, I wrote exploit.py script to get the flag.
from sympy.ntheory.modular import crt
N = [86812553978993, 81744303091421, 83695120256591]
c = [8875674977048, 70744354709710, 29146719498409]
x = crt(N, c)[0]
print("Hex String:")
print(round(x ** (1. /3)))
After running the script by python3 exploit.py, I got a hex string.
Step-4:
I converted it online to ASCII here.
Step-5:
Finally the flag becomes:
csictf{h45t4d}