Mancity - CryptoCTF - writeup

Wed, 24 Dec 2025 00:00:00 +0000

This challenge is another RSA-style challenge, in which we need to factor n. We are given the files mancity.py and output.txt. This is the main code:

nbit = 256
p, q = keygen(nbit)
m = bytes_to_long(flag)
assert m < n
e, n = 1234567891, p * q
c = pow(m, e, n)

print(f'n = {n}')
print(f'c = {c}')

And these, the functions used to generate the prime numbers:

def man(n):
    B = bin(n)[2:]
    M = ''
    for b in B:
        if b == '0':
            M += '01'
        else:
            M += '11'
    return int(M, 2)

def keygen(nbit):
    while True:
        p = getPrime(nbit)
        r = man(p)
        B = bin(p)[2:] + '1' * nbit
        q = int(B, 2)
        if isPrime(q) and isPrime(r):
                return q, r

In the keygen function, a prime p is generated, and these two primes are generated from it:

Vinad - CryptoCTF - writeup

Tue, 23 Dec 2025 00:00:00 +0000

This challenge is an RSA-type challenge in which we need to factor n. We are given a file vinad.py and its output in output.txt. This is the RSA key generation method implemented in vinad.py:

def parinad(n):
    return bin(n)[2:].count('1') % 2

def vinad(x, R):
    return int(''.join(str(parinad(x ^ r)) for r in R), 2)

def genkey(nbit):
    while True:
        R = [getRandomNBitInteger(nbit) for _ in range(nbit)]
        r = getRandomNBitInteger(nbit)
        p, q = vinad(r, R), getPrime(nbit)
        if isPrime(p):
            e = vinad(r + 0x10001, R)
            if GCD(e, (p - 1) * (q - 1)) == 1:
                return (e, R, p * q), (p, q)

The parinad function computes the parity of the hamming wright of a given number. The vinad function takes a number x and an array R and computes a number by applying the parinad function to all the pairs.

Factorization on manbolq's Blog

Mancity - CryptoCTF - writeup

Vinad - CryptoCTF - writeup