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An instance of RSA uses a public key modulus [tex]n = 221[/tex] and public key [tex]e = 25[/tex].

Suppose a ciphertext [tex]c = 5[/tex] is intercepted. Break the encryption and recover the original message.

Answer :

Final answer:

The original message recovered from the intercepted ciphertext is 17.

Explanation:

To break the RSA encryption and recover the original message, we need to calculate the private key exponent (d) using the given public key modulus (n) and public key exponent (e). In this case, the public key modulus (n) is 221 and the public key exponent (e) is 25.

The first step is to factorize the modulus (n) into its prime factors. Since 221 is a small number, we can manually find its prime factors, which are 13 and 17.

Next, we need to calculate the private key exponent (d) using the formula:

d = e-1 mod φ(n)

where φ(n) is Euler's totient function, which is calculated as φ(n) = (p-1)(q-1) for two prime factors p and q of n.

In this case, φ(n) = (13-1)(17-1) = 192.

Now, we can calculate the private key exponent (d) using the extended Euclidean algorithm or by using the modular inverse property:

d ≡ e-1 mod φ(n)

25-1 mod 192 = 25

Therefore, the private key exponent (d) is 25.

Now that we have the private key exponent (d), we can decrypt the intercepted ciphertext (c) using the formula:

m = cd mod n

where m is the original message.

In this case, the ciphertext (c) is 5.

525 mod 221 = 17

Therefore, the original message is 17.

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