My cryptographer friend Raczhzzze has created his own encryption algorithm. Your task is to crack the scheme without the private key.

The flag will be 'flag{sha256 of the concatenated digits of the plain text}'.

Cipher: 1445, 2897, 1438, 2890, 1444, 2901, 1439, 2891, 1444, 1435, 2902

Public Key: 2909

Cryyyyyyyyypto_a57edc42994dfc69c25ae5b058122926.py

# Python code for CTF question.
import random

d = 10
max_input = 255

# min input is 2

def valid_primes_sieve():
    bound = ((d+2)*max_input)
    isprime = [True for x in range(bound)]
    for i in range(2, bound):
        for j in range(2, bound/i):
            isprime[i*j] = False;
    prime_options = {}
    count = 0
    primes = isprime
    for i in range(len(isprime)):
        if isprime[i] is True:
            prime_options[count] = i
            count = count + 1
    return prime_options

def sieve():
    bound = ((d+2)*max_input)
    isprime = [True for x in range(bound)]
    for i in range(2, bound):
        for j in range(2, bound/i):
            isprime[i*j] = False;
    return isprime

def factorize(input):
    factors = []
    for i in range(1, input):
        if input%i is 0:
            factors.append(i)
    return factors


input = # Input List

print "Encrypting..."

# Step 1: Inflate input difficulty(d)+1 times
inflated_input = [(d+1)*x for x in input]
prime_options = valid_primes_sieve()
priv_key = 2
while priv_key <= (d+1)*max_input:
    priv_key = prime_options[random.randint(1,len(prime_options))]

pub_key = 2
while pub_key <= (d+1)*max_input or pub_key == priv_key:
    pub_key = prime_options[random.randint(1,len(prime_options))]

print "Private Key: ", priv_key
print "Public Key: ", pub_key

check = [0 for x in range(pub_key+1)]
for x in range(1, 256):
    check[(priv_key*(d+1)*x)%(pub_key)] = check[(priv_key*(d+1)*x)%(pub_key)] + 1
    if check[(priv_key*(d+1)*x)%(pub_key)] > 1:
        print "error"

cipher = []
for x in input:
    cipher.append((priv_key*(d+1)*x)%(pub_key)+random.randint(1,3))

print "Cipher Text: ", cipher