Public Key Cryptography Questions and Answers

Answer: A

The values are n = 437, φ = 396.

\[ \begin{aligned} n &= p \times q = 19 \times 23 = 437 \\ \phi(n) &= (p - 1)(q - 1) = 18 \times 22 = 396 \end{aligned} \]

n is the public modulus, and \(\phi(n)\) is used to choose the public exponent e (coprime to 396) and compute the private exponent d. The other options do not come from these formulas: 38 and 52 are neither product, and 19 and 23 are just p and q.

Answer: C

The reason is that public-key methods avoid sharing a secret beforehand.

With symmetric encryption both parties need the same secret key, and getting it to the other side safely is the problem. Public-key cryptography solves it: a sender can encrypt a fresh session key with the recipient's public key, or the two sides can run Diffie-Hellman, establishing a shared secret over an open channel without any prior secret. The bulk data is then encrypted with the fast symmetric key, the hybrid scheme used by TLS.

Public-key methods are much slower than symmetric ones, which is why they are kept to key exchange. Symmetric keys can be distributed in other ways, by courier or pre-shared keys, just less conveniently. Public keys do expire, usually with their certificates.

Answer: C

The one that is not public-key cryptography is AES.

The Advanced Encryption Standard (the Rijndael cipher, adopted by NIST in 2001) is a symmetric block cipher: the same secret key encrypts and decrypts, with 128-bit blocks and 128, 192 or 256-bit keys.

The rest are all public-key (asymmetric) schemes. RSA rests on the difficulty of factoring large numbers, and ECC (elliptic curve cryptography) on the elliptic-curve discrete logarithm problem. Kyber is a lattice-based key-encapsulation mechanism designed to resist quantum computers, standardised by NIST in 2024 as ML-KEM.

Answer: B

Euler's phi of n is 60.

In RSA, n = pq, and for two distinct primes Euler's totient is the count of numbers below n that share no factor with it:

\[ \begin{aligned} n &= 7 \times 11 = 77 \\ \phi(n) &= (p - 1)(q - 1) \\ &= 6 \times 10 = 60 \end{aligned} \]

The public exponent e is then chosen coprime to 60, and the private exponent d is its inverse modulo 60. 77 is the modulus n itself, not φ(n), and 7 and 11 are just the primes.

Answer: D

In RSA with p = 3 and q = 11 the modulus is n = pq = 33 and \(\phi(n) = (p-1)(q-1) = 20\). The private exponent d = 7 needs a public exponent e with \(ed \equiv 1 \pmod{20}\), and since \(3 \times 7 = 21 \equiv 1\), e = 3. The cipher text is then

The options 33 (the modulus itself), 5 (the plain text) and 25 do not satisfy this calculation. The keyed answer is 26.