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Challenge Description :

John, an ordinary office worker, receives a mysterious email filled with seemingly random numbers, letters and codes. Little does he know, it holds the key to uncovering a hidden treasure left behind by his eccentric uncle who loved ciphers and encryptions. To claim his unexpected inheritance, John must decipher the cryptic message using his newfound skills in cryptography.

source.py

from Crypto.Util.number import*
from gmpy2 import *
from secret import e,b,hint,msg,d
p = getPrime(512)
q = getPrime(512)
n = p*q
m = bytes_to_long(msg)
h = bytes([i^b for i in hint])
print(f"h = {hex(bytes_to_long(h))}")
ct = pow(m,e,n)
de = pow(ct,d,n)
assert(m == de)
print("ct = ",ct)
print("p = ",p)
print("q = ",q)

output.txt

h = 0x6f535e1b5e1b061b0c020f0b0b10134f535e1b4852555c575e1b59424f5e1b4f535a4f1b4c5a481b4354495e5f121b0112
ct =  90411409551177819360717236462351545237822367597930505531741437834918499125195272674859389978951589180632146502190429979348445123366914000167832349866368754227474060832624537550600921894849466284315037863094795265822884392628050584343158613338754532642964368052098136565157343201877382609610774291396944124354
p =  10425866553433272288676977376976736493869099145622614885498170561565122111495807572631609087909399078701783905493563029715011322065331636751277834978526061
q =  9215753518399683669080201592666232851634627861957009698720674021492716071355990364002777325458055207969176695525292834842774295594232711456066623178861093

Author : im._.a.p

Solution:

• Values of 'p','q','ciphertext','h'(encrypted hint which has the value of e) is given.
• Th value of the modulus can be found using 'p' and 'q' as p*q = n.
• From the given code, we understand that 'hint' is xored with a random byte 'b' to get the ciphertext h.
• To find 'hint', 'h' is xored with each single byte (bruteforced) to get 'hint', thus getting the value of e.

from string import printable
h= '6f535e1b5e1b061b0c020f0b0b10134f535e1b4852555c575e1b59424f5e1b4f535a4f1b4c5a481b4354495e5f121b0112'
m= bytes.fromhex(h)
for i in range(256):
  x=bytes(i^j for j in m)
  try:

    x = x.decode()
    for j in x:
      assert j in printable
    print(i,x)
  except:
    pass

59 The e = 79400+(the single byte that was xored) :)
e = 79459

• With the values 'e','n','p','q', the ciphertext ct can be decrypted to get the message.

e = 79459
n = p*q
phi = (p-1)*(q-1)
d = invert(e,phi)
print(d)
hint = pow(ct,d,n)
print(long_to_bytes(hint))

b"Here is your reward 'vvrkxuqgi{r0i43m0r_f0_hu3_u3gtu3!!!}' You can ask 'Doraemon' to help you with this. Bye!!"
vinegere cipher key : doraemon

The message has the encrypted flag and key which is to be decrypted.(vinegere cipher).

Flag : shaktictf{d0r43m0n_t0_th3_r3scu3!!!}