![]() ![]() We find that n is divisible by every odd integer in that range, meaning that the modulus is composed of many prime factors and that we need to compute phi(n) in a manner other than the traditional (p-1)(q-1) for dual-prime moduli. Just to be sure, we compute n mod every integer between 1 and 100 by the following python snippet: Despite this, we don't know whether or not the modulus is composed of exactly two primes, and it seems unlikely given its size. Thus, we can conclude that 5 is a factor of the modulus. From first glance, this seems impossible - for how could we factor a coprime modulus of such magnitude in a reasonable amount of time? However, upon closer inspection, we find something interesting about n: it ends in. Without a private key d, the only option is to factor n, take the totient phi(n), and find the modular inverse of e with phi(n). Opening best_rsa.txt, we find an overwhelmingly large modulus n and ciphertext c, and a public exponent e (65537). ![]() Please refer to the file titled best_rsa.txt for the numerical values. *Note: the numbers used for this challenge, due to their size (~157000 digits), have been omitted from this writeup. >We don't believe his cyber is as secure as he says it is.ĭetermine the RSA plaintext given ciphertext, public exponent, and modulus >The more bits I have, the more secure my cyber, and my modulus is YUUUUUUUUUUUUUGE At his last rally, Trump made an interesting statement: ![]()
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