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| Section | Objectives |
|---|---|
| Symmetric Encryption | - Block and stream ciphers - AES and legacy algorithms (e.g., DES conceptually) |
| Cryptographic Protocols and Applications | - Secure communication design principles - TLS/SSL conceptual overview |
| Asymmetric Encryption | - RSA and ECC fundamentals - Public key cryptography principles |
| Foundations of Cryptography | - Historical and modern cryptography principles - Core concepts of confidentiality, integrity, authentication, non-repudiation |
| Key Management and PKI | - Certificates, certificate authorities, and PKI structure - Key exchange and lifecycle management |
| Hash Functions and Message Authentication | - MAC and HMAC mechanisms - Cryptographic hash functions (e.g., SHA family concepts) |
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問題 #38
(An administrator has configured a Virtual Private Network (VPN) connection utilizing IPsec transport mode with Encapsulating Security Payload (ESP) between a server in the corporate office and a client computer in the remote office. In which situation can the packet content be inspected?)
答案：B
解題說明：
With IPsec ESP in transport mode, the payload of the original IP packet (typically the transport-layer segment and higher) is encrypted and integrity-protected between the two endpoints-here, the corporate server and the remote client. Because encryption is applied by the sending endpoint and removed only by the receiving endpoint, intermediate routers, switches, and monitoring devices in either network cannot view the protected payload while it is in transit. They may see outer IP headers and certain metadata needed for routing, but not the encrypted content protected by ESP. As a result, the packet's contents are inspectable only at the endpoints: before encryption on the sender (plaintext exists in memory/stack before IPsec processing) and after decryption on the receiver (plaintext is restored for the application). This is true whether the traffic traverses internal networks or the Internet; the cryptographic boundary is between the endpoints participating in the IPsec SA. Therefore, inspection of the actual content is possible only on the devices at headquarters and offsite, before sending and after receiving, not by in-transit networks.
問題 #39
(What is a component of a one-time password (OTP) that is needed to guess future iterations of passwords?)
答案：A
解題說明：
OTP systems (such as HOTP and TOTP) generate a sequence of passwords using a shared secret and a moving factor (counter or time). The critical secret that underpins the ability to compute past or future OTP values is the seed (also called the shared secret key). In HOTP, the seed is used with an HMAC function and an incrementing counter; in TOTP, the seed is used with HMAC and a time-step value. If an attacker obtains the seed and knows the algorithm and moving factor, they can compute future OTPs. The "function" and
"encryption algorithm" are typically standardized and public; security relies on keeping the seed secret. An initialization vector is not a standard OTP component in HOTP/TOTP generation. Therefore, the component needed to predict future OTP values is the seed. Protecting the seed is essential: it should be stored securely (e.
g., hardware token secure storage) and transmitted only through controlled provisioning processes. If compromised, OTP becomes predictable and no longer serves as a strong second factor.
問題 #40
(What are the roles of keys when using digital signatures?)
答案：A
解題說明：
Digital signatures provide integrity, authenticity, and typically non-repudiation by using an asymmetric key pair. The signer uses the private key to create a signature over a message (usually over a hash
/digest of the message). Because the private key is kept secret, only the legitimate signer should be able to produce a valid signature. Anyone who has the corresponding public key can then validate the signature: they verify that the signature matches the message digest under the public key and that the signed data has not been altered. This is why the public key can be widely distributed (often inside an X.
509 certificate) while the private key must be protected by the signer. If a public key were used to sign, anyone could forge signatures; if a private key were required for validation, only the signer could validate, defeating the purpose of public verifiability. Therefore, the correct key roles are private key for signing and public key for signature validation.
問題 #41
(What is a component of a one-time password (OTP) that is needed to guess future iterations of passwords?)
答案：A
解題說明：
OTP systems (such as HOTP and TOTP) generate a sequence of passwords using a shared secret and a moving factor (counter or time). The critical secret that underpins the ability to compute past or future OTP values is the seed (also called the shared secret key). In HOTP, the seed is used with an HMAC function and an incrementing counter; in TOTP, the seed is used with HMAC and a time-step value. If an attacker obtains the seed and knows the algorithm and moving factor, they can compute future OTPs. The "function" and "encryption algorithm" are typically standardized and public; security relies on keeping the seed secret. An initialization vector is not a standard OTP component in HOTP
/TOTP generation. Therefore, the component needed to predict future OTP values is the seed.
Protecting the seed is essential: it should be stored securely (e.g., hardware token secure storage) and transmitted only through controlled provisioning processes. If compromised, OTP becomes predictable and no longer serves as a strong second factor.
問題 #42
(What describes how Counter (CTR) mode encryption functions?)
答案：C
解題說明：
CTR mode turns a block cipher (like AES) into a stream-like construction by generating a keystream from successive encryptions of a changing input block. Specifically, CTR forms input blocks using a nonce (unique per message) combined with an increasing counter. Each nonce||counter block is encrypted with the block cipher under the shared key, producing a pseudorandom output block. That output is then XORed with plaintext to yield ciphertext (and XORed with ciphertext to recover plaintext). This design enables parallelization (blocks can be generated independently), efficient random access decryption, and avoids chaining dependencies seen in modes like CBC. Option B describes CFB-like behavior; option C describes ECB; option D describes CBC. CTR's security critically depends on never reusing the same nonce/counter sequence with the same key, because reuse would repeat keystream blocks and expose plaintext relationships.
Therefore, the correct description is that CTR converts the block cipher into a stream cipher using a counter value and a nonce.
問題 #43
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