csictf 2020

CTF Writeup - https://ctftime.org/event/1081

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21 July 2020

Quick Math

by anishbadhri

Ben has encrypted a message with the same value of ‘e’ for 3 public moduli

and got the cipher texts

Find the original message. (Wrap it with csictf{})


The given problem is a typical example of hastad attack. Given 3 pairs of n and c with the same value of e=3, the original message can be decoded.

Solution Script:

from pwn import remote
from sympy.ntheory.modular import crt
from gmpy2 import iroot

e = 3
N = [86812553978993, 81744303091421, 83695120256591]
C = [8875674977048, 70744354709710, 29146719498409]

resultant, mod = crt(N,C)
value, is_perfect = iroot(resultant,e)


tags: Crypto