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| Section | Objectives |
|---|---|
| Topic 1: Hash Functions and Message Authentication | - Cryptographic hash functions (e.g., SHA family concepts) - MAC and HMAC mechanisms |
| Topic 2: Cryptographic Protocols and Applications | - Secure communication design principles - TLS/SSL conceptual overview |
| Topic 3: Key Management and PKI | - Key exchange and lifecycle management - Certificates, certificate authorities, and PKI structure |
| Topic 4: Asymmetric Encryption | - Public key cryptography principles - RSA and ECC fundamentals |
| Topic 5: Foundations of Cryptography | - Core concepts of confidentiality, integrity, authentication, non-repudiation - Historical and modern cryptography principles |
| Topic 6: Symmetric Encryption | - AES and legacy algorithms (e.g., DES conceptually) - Block and stream ciphers |
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NEW QUESTION # 88
(Why did the National Institute of Standards and Technology (NIST) choose Ascon for lightweight cryptography?)
Answer: D
Explanation:
NIST's lightweight cryptography effort targets environments like IoT and embedded systems where CPU, memory, energy, and bandwidth are constrained, yet strong security is still required. Ascon is an authenticated encryption with associated data (AEAD) family designed to be efficient in both hardware and software with small footprint, making it well-suited for constrained devices. NIST selected Ascon because it offers a strong security design with good performance and implementability under tight resource budgets, while providing modern protections (confidentiality + integrity) through AEAD. That aligns with option C: secure and efficient encryption for resource-constrained devices. The selection was not primarily about authenticating users (that is typically handled by protocols and identity systems, not an AEAD primitive). It was also not mainly about legacy compatibility; lightweight cryptography aims at new and constrained deployments rather than preserving outdated stacks. And while Ascon can certainly be used to protect data at rest, that is only one application; the core reason for the choice is its suitability for constrained environments and robust, efficient authenticated encryption.
NEW QUESTION # 89
(How can auditing enhance an organization ' s cryptographic practices?)
Answer: B
Explanation:
Auditing improves cryptographic practice by systematically evaluating whether cryptographic controls are correctly selected, implemented, configured, and maintained. Through audits, an organization can discover weak algorithms (e.g., deprecated hashes), improper key lengths, unsafe modes (e.g., unauthenticated CBC), missing integrity controls, poor certificate validation, and operational problems such as key reuse, weak randomness sources, inadequate rotation, or overly permissive access to key material. Audits also assess compliance with internal policy and external standards, ensuring crypto is used consistently across systems and that exceptions are documented and risk-managed. Importantly, auditing does not guarantee that incidents will never happen; it reduces risk by finding gaps before attackers do. It also does not eliminate the need for updates-audits often reveal that policies must evolve as threats and best practices change. Employee training can be recommended as an outcome of auditing, but audits do not automatically ensure training. Thus, the most accurate benefit is that auditing identifies weaknesses and drives corrective action, strengthening cryptographic posture over time.
NEW QUESTION # 90
(How does adding salt to a password improve security?)
Answer: A
Explanation:
A salt is a unique, random value stored alongside a password hash and combined with the password during hashing. Its main security benefit is that it ensures identical passwords do not produce identical hashes across different accounts or systems. If two users choose the same password, their stored hashes will differ because their salts differ, which directly prevents attackers from spotting shared passwords by comparing hashes. Salts also defeat precomputation attacks such as rainbow tables, because an attacker would need to regenerate tables for each possible salt value-a task that becomes infeasible when salts are large and unique per password. Salt does not enforce password complexity rules (that's a policy/validation function), does not guarantee users choose different passwords, and does not prevent password reuse across sites. The correct statement is that salt makes the resulting hash different even for the same password, improving resistance to offline cracking at scale and eliminating the "same hash = same password" shortcut attackers rely on.
NEW QUESTION # 91
(How is Public Key Infrastructure (PKI) commonly utilized in web browsers?)
Answer: D
Explanation:
Web browsers rely on PKI to establish trust in secure connections, primarily through X.509 certificates and a built-in set of trusted root Certificate Authorities (CAs). When a browser connects to an HTTPS site, the server presents a certificate chain. The browser validates that chain up to a trusted root, checks that the certificate is valid for the domain (SAN/CN matching), confirms validity dates, and may check revocation status. This PKI process allows browsers to authenticate the website's identity and negotiate encrypted session keys for TLS, enabling confidentiality and integrity for the connection. In practical terms, the browser' s PKI components include certificate stores, validation logic, and mechanisms for handling intermediates, trust policies, and revocation. While PKI supports authentication as an outcome, the best description of how browsers utilize PKI is that they manage and validate digital certificates and associated keys to establish trust.
PKI is not about compressing messages or encrypting data at rest; it is about identity binding and trust chains that make secure web communication possible.
NEW QUESTION # 92
(What is the value of 23 mod 6?)
Answer: A
Explanation:
The expression 23 mod 6 asks for the remainder when 23 is divided by 6. Modular arithmetic is foundational in cryptography, especially in public-key systems (RSA, Diffie-Hellman, ECC) where operations occur in finite rings or fields. To compute 23 mod 6, identify the largest multiple of 6 that does not exceed 23.
Multiples of 6 are 6, 12, 18, 24. Since 24 is greater than 23, the largest valid multiple is 18. Subtract: 23 # 18
= 5, so the remainder is 5. Therefore, 23 mod 6 = 5, which corresponds to option "05." Modular reduction keeps numbers within a fixed range (0 to modulus#1), enabling stable arithmetic under wraparound behavior.
In cryptographic protocols, this wraparound property is essential for defining groups and ensuring operations remain bounded and consistent.
NEW QUESTION # 93
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