Category: Crypto

Problem: description:

One of our developers has created a new way to encrypt any text via Namta (Multiplication). But we think the encryption is kinda weak.

Can you break it and decode the encrypted text.

Input: CTF_BD{#####################################}

Input: 14

Output:

798, 1596, 3612, 5656, 8190, 4452, 8330, 11200, 13734, 7980, 7546, 16632, 21840, 14504, 10080, 19712, 16898, 20664, 13034, 22680, 19404, 25872, 31234, 16128, 28700, 44408, 33642, 43904, 48720, 29820, 42966, 50176, 46200, 32368, 33810, 50904, 24864, 26068, 26754, 54880, 67158, 51744, 73444, 48048, 73710, 52164, 32242, 37632,

The following code is also given as attachment.

# namta.py
import base64

encryption_string = input("Enter Text to Encrypt: ")
multi = int(input("Enter Number to Multiply: "))

encryption_string_bytes = encryption_string.encode("ascii")
base64_bytes = base64.b64encode(encryption_string_bytes)

crypt_data = base64_bytes.decode("ascii")

lol = ""
lol2 = ""

for i in range(multi):
    lol += crypt_data[multi-1]
    l = list(crypt_data)
    del(l[multi-1])
    crypt_data = "".join(l)

new_crypt = lol + crypt_data

for i in range(multi):
    lol2 += new_crypt[-multi]
    l = list(new_crypt)
    del(l[-multi])
    new_crypt = "".join(l)

new_crypt = lol2 + new_crypt

x = new_crypt
chunks, chunk_size = len(x), len(x)//2
y = [x[i:i+chunk_size] for i in range(0, chunks, chunk_size)]

new_crypt = y[1][::-1] + y[0][::-1]

full_crypt = ""

for i in range(len(new_crypt)):
    print(f"{multi} * {i+1} = {multi*(i+1)}")
    full_crypt += f"{(multi*(i+1))*ord(new_crypt[i])}, "

with open("output.txt", "w") as enc_out:
    enc_out.write(full_crypt)

# print(full_crypt)

Solution:

After inspection of the code, we can observe the following:

1. Flag is first base64 encoded
2. 14 characters from index 14 are brought to the front
3. 14 characters from index 14 from the last is brought to the front in reverse order
4. The string is reversed
5. Each character of the string becomes the multiplication of 14, its index+1 and it’s ascii code

So, the following code reverses the logic and prints out the flag:

import base64

multi = 14

input_file = open("input.txt", "r")
str = input_file.readline()
str = str.split(", ")
str[len(str)-1] = str[len(str)-1].replace(",", "")
str = [int(i) for i in str]

crypt_data = ""

for i in range(0, len(str)):
    char = str[i]/(multi * (i+1))
    char = int(char)
    crypt_data = crypt_data + chr(char)

crypt_data = crypt_data[::-1]

data_lenght = len(crypt_data)
lol2 = crypt_data[0:14]
lol2 = lol2[::-1]

crypt_data = crypt_data[14:35] + \
    lol2 + crypt_data[35:data_lenght]

crypt_data = crypt_data[14:27] + crypt_data[0:14] + crypt_data[27:data_lenght]

print(crypt_data)

result = base64.b64decode(crypt_data)
print(result.decode("ascii"))

Flag : CTFBD{Mul7iplic4ti0n_3nC_1s_fuNny}