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| Section | Objectives |
|---|---|
| Symmetric Cryptography | - Block Ciphers (AES, DES, 3DES) - Initialization Vectors (IV) - Stream Ciphers - Key Management |
| Cryptanalysis and Attacks | - Brute Force and Dictionary Attacks - Social Engineering Prevention - Common Attack Vectors |
| Hashing and Digital Signatures | - Hash Functions (MD5, SHA-1, SHA-256) - Message Authentication Codes (MAC) - Digital Signature Standards |
| Cryptography Fundamentals | - History and Evolution of Cryptography - Cryptographic Terminology - Symmetric vs Asymmetric Encryption |
| Asymmetric Cryptography | - Public Key Infrastructure (PKI) - RSA Algorithm - Elliptic Curve Cryptography (ECC) - Diffie-Hellman Key Exchange |
| Applied Cryptography | - Cryptographic Best Practices - VPN Security - PGP and Email Encryption - SSL/TLS Protocols |
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NEW QUESTION # 79
(Which mechanism can be applied to protect the integrity of plaintext when using AES?)
Answer: D
Explanation:
AES by itself is a symmetric block cipher that provides confidentiality, but not guaranteed integrity unless used in an authenticated mode. To protect integrity of the plaintext (ensuring it has not been altered), a Message Authentication Code (MAC) can be applied. In the classic Encrypt-then-MAC pattern, the sender encrypts the plaintext with AES and then computes a MAC (often HMAC-SHA-256 or CMAC-AES) over the ciphertext (and relevant headers). The receiver verifies the MAC before attempting decryption, preventing tampering and many padding-oracle style vulnerabilities. Alternatively, AES can be used in an AEAD mode like AES-GCM, which produces an authentication tag serving a similar purpose, but among the listed options the general integrity mechanism is "MAC." RC4 is an unrelated stream cipher and does not provide integrity.
RSA is asymmetric and not the standard integrity add-on for AES-encrypted bulk data. Kerberos is an authentication protocol and key distribution system, not a message integrity primitive. Therefore, to protect plaintext integrity when using AES, the correct mechanism is a Message Authentication Code.
NEW QUESTION # 80
(Which cipher uses shifting letters of the alphabet for encryption?)
Answer: C
Explanation:
The Caesar cipher is the classic substitution cipher that encrypts by shifting letters of the alphabet by a fixed number of positions (e.g., shift by 3: A#D, B#E, etc.). It is a monoalphabetic cipher because a single shift value is applied uniformly across the entire message, making it simple and vulnerable to frequency analysis and brute force (only 25 meaningful shifts in the Latin alphabet). Vigenere also involves shifting, but it uses a repeating keyword to vary the shift per character (polyalphabetic), whereas the question's phrasing typically points to the fundamental "shift cipher," which is Caesar.
SHA-1 is a cryptographic hash function, not a cipher. Bifid is a fractionation cipher combining Polybius square coordinates and transposition, not a direct shifting method. Therefore, the cipher that uses shifting letters of the alphabet for encryption is the Caesar cipher.
NEW QUESTION # 81
(How does Electronic Codebook (ECB) mode encryption function?)
Answer: B
Explanation:
ECB is the simplest block cipher mode: each plaintext block is encrypted independently using the same key and the block cipher primitive. There is no IV and no chaining, so identical plaintext blocks produce identical ciphertext blocks. This property leaks patterns and structure in the plaintext, which is why ECB is generally considered insecure for most real-world data beyond tiny, random-looking inputs. For example, images encrypted with ECB often reveal outlines because repeated pixel blocks map to repeated ciphertext blocks.
Option A describes CTR mode, option C describes CBC mode, and option B resembles feedback-based modes. ECB's independence also means it can be parallelized, but the pattern leakage is a severe weakness.
Modern practice prefers authenticated encryption modes (like GCM) or, at minimum, modes with IVs and chaining (like CBC with proper padding and MAC). Therefore, the correct statement is that ECB encrypts each block with the same key and each block is independent of the others.
NEW QUESTION # 82
(What is the maximum key size (in bits) supported by AES?)
Answer: A
Explanation:
AES supports three standardized key sizes: 128, 192, and 256 bits, with a fixed block size of 128 bits. The maximum of these supported key sizes is 256 bits (AES-256). Key size affects resistance to brute-force key search: larger keys exponentially increase the search space. In practice, AES-128 is already considered strong against brute force with contemporary computing capabilities, while AES-256 is often chosen for compliance requirements, conservative security margins, or to hedge against future advances. AES-512 is not part of the AES standard; if 512-bit keys are desired, systems typically use different constructions (like using AES-256 in certain key-derivation or wrapping schemes) rather than changing AES itself. Therefore, the correct maximum supported AES key size is 256 bits.
NEW QUESTION # 83
(What is an attribute of RC4 when used with WEP?)
Answer: C
Explanation:
In classic WEP deployments, RC4 was used with what is commonly called "40-bit WEP" (also labeled
"64-bit WEP" because it combines a 40-bit secret key with a 24-bit IV to form a 64-bit RC4 seed). The key attribute emphasized in many foundational descriptions of WEP is this 40-bit shared secret length, which was originally chosen due to export restrictions and legacy constraints. Although "104-bit WEP" (sometimes called "128-bit WEP," again counting the 24-bit IV) also existed, the option set here points to the historically standard and widely referenced attribute: a 40-bit key when RC4 is used in WEP.
Importantly, WEP's security failure is not only about key size; the 24-bit IV is too small and repeats frequently, and WEP's key scheduling vulnerabilities combined with IV reuse allow attackers to recover the secret key with enough captured frames. Still, among the given options, the correct attribute is the 40-bit key.
NEW QUESTION # 84
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