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| Section | Objectives |
|---|---|
| Hashing and Digital Signatures | - Digital Signature Standards - Message Authentication Codes (MAC) - Hash Functions (MD5, SHA-1, SHA-256) |
| Cryptography Fundamentals | - History and Evolution of Cryptography - Cryptographic Terminology - Symmetric vs Asymmetric Encryption |
| Applied Cryptography | - PGP and Email Encryption - VPN Security - Cryptographic Best Practices - SSL/TLS Protocols |
| Symmetric Cryptography | - Stream Ciphers - Key Management - Initialization Vectors (IV) - Block Ciphers (AES, DES, 3DES) |
| Cryptanalysis and Attacks | - Common Attack Vectors - Brute Force and Dictionary Attacks - Social Engineering Prevention |
| Asymmetric Cryptography | - Diffie-Hellman Key Exchange - RSA Algorithm - Public Key Infrastructure (PKI) - Elliptic Curve Cryptography (ECC) |
>> Introduction-to-Cryptography Fragenkatalog <<
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41. Frage
(What is the value of 23 mod 6?)
Antwort: B
Begründung:
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.
42. Frage
(What is the correlation between the number of rounds and the key length used in the AES algorithm?)
Antwort: C
Begründung:
In AES, the number of rounds is explicitly tied to the key length. AES-128 uses 10 rounds, AES-192 uses 12 rounds, and AES-256 uses 14 rounds. The purpose of additional rounds is to increase diffusion and confusion, strengthening resistance against cryptanalysis as the key schedule and state transformations iterate more times. Although key length primarily affects brute-force resistance, AES's designers and standardization parameters link longer keys with more rounds to maintain security margins across variants, especially considering differences in the key schedule structure. Thus, as key length increases from 128 to 192 to 256 bits, the number of rounds increases correspondingly from 10 to
12 to 14. This relationship is fixed by the AES specification and does not vary dynamically at runtime.
Therefore, the correct correlation is that the number of rounds increases as the key length increases.
43. Frage
(Which symmetric encryption technique uses a 112-bit key size and a 64-bit block size?)
Antwort: D
Begründung:
3DES (Triple DES) is a symmetric block cipher that retains DES's 64-bit block size while increasing effective security by applying DES multiple times. The common "two-key 3DES" variant uses two independent 56-bit DES keys (K1 and K2) in an Encrypt-Decrypt-Encrypt (EDE) sequence: Encrypt with K1, Decrypt with K2, then Encrypt again with K1. Because each DES key is 56 bits (ignoring parity bits), the total keying material is 112 bits. This matches the question's "112-bit key size and 64- bit block size." Plain DES uses only a 56-bit effective key and a 64-bit block size, so it does not match the 112-bit key size. AES has a 128-bit block size and key sizes of 128/192/256. IDEA uses a 64-bit block size but has a 128-bit key. Therefore, the correct algorithm is 3DES. Although 3DES improved on DES, it is now considered legacy due to its small 64-bit block size (birthday-bound issues for large data volumes) and performance overhead compared to AES.
44. Frage
(How does adding salt to a password improve security?)
Antwort: C
Begründung:
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.
45. Frage
(Which type of exploit involves looking for different inputs that generate the same hash?)
Antwort: B
Begründung:
A birthday attack targets hash functions by exploiting the birthday paradox: collisions (two different inputs producing the same hash output) can be found much faster than brute-forcing a specific preimage. For an n-bit hash, the expected work to findanycollision is on the order of 2
2026 Die neuesten Zertpruefung Introduction-to-Cryptography PDF-Versionen Prüfungsfragen und Introduction-to-Cryptography Fragen und Antworten sind kostenlos verfügbar: https://drive.google.com/open?id=14lSRb1FcsOsdNgB7wv6eqs2wL9ZBkOmh