Tags: rsa

Rating:

Peoblem Statement:

Dr. Xernon made the mistake of rolling his own crypto.. Can you find the bug and decrypt the message? Connect with nc 2018shell2.picoctf.com 3609.

Solution :
i got the following data :

c: 19474669958697898483768380499982060217038494985407624948963596074034125281014421
n: 25132405989001317956087767778734083918050854186063214627718037281542891163684097
e: 65537

Problem Statement(425 points):

Wow, he made the exponent really large so the encryption MUST be safe, right?! Connect with nc 2018shell2.picoctf.com 29661.

Solutions:

I tried all attacks but can't get any result.So in the i used https://www.alpertron.com.ar/ECM.HTM and cracked the factor of n.

After 12 min 55 sec i got the facors :

158 754754 453572 149934 540516 805858 365529 (39 digits) × 158309 627169 945915 479670 296299 307350 953193 (42 digits)

So we got our p and q

p=158754754453572149934540516805858365529
q=158309627169945915479670296299307350953193
t=(p-1)*(q-1)
import gmpy2
d = gmpy2.invert(e,t)

# Decryption
decimalmessage = pow(c,d,n)
print(decimalmessage)
hexmessage= hex(decimalmessage)
print(hexmessage)
print(hexmessage[2:].decode("hex"))
print(bytes.fromhex(hexmessage).decode())

Python script here
The program returned the flag:

picoCTF{[email protected]_pr1m3\$_1335}

Original writeup (https://github.com/d4rkvaibhav/PICOCTF-2018/tree/master/Cryptography/Super%20safe%20RSA).