Kripto 2

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| By Catherine Halcomb
Catherine Halcomb
Community Contributor
Quizzes Created: 1443 | Total Attempts: 6,714,231
| Attempts: 3,063 | Pitanja: 75
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1. Sifarski sistemi mogu biti:

Explanation

The correct answer includes two options: "Simetricni" (Symmetric) and "Asimetricni" (Asymmetric). This suggests that "Sifarski sistemi" (Number systems) can be either symmetric or asymmetric.

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Kripto 2 - Quiz

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2. RSA algoritam se pojavio: (*svuda pise da se pojavio 1977.god, ali ovo su bili ponudjeni odgovori)

Explanation

not-available-via-ai

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3. Osnovna podela sifarskih sistema, prema vrsti kljuca deli se na:

Explanation

The correct answer is "Simetricne sisteme,Asimetricne sisteme". The question is asking for the basic division of cipher systems according to the type of key. The answer states that the division includes symmetric systems and asymmetric systems. Symmetric systems use the same key for both encryption and decryption, while asymmetric systems use different keys for encryption and decryption.

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4. DH algoritam za uspostavljanje deljene tajne je osetljiv na:

Explanation

The correct answer is "Napad tipa covek u sredini" (Man-in-the-middle attack). The Diffie-Hellman algorithm for establishing a shared secret is vulnerable to this type of attack. In a man-in-the-middle attack, an attacker intercepts the communication between two parties and impersonates each party to the other, allowing them to intercept and modify the messages exchanged. This can compromise the security of the shared secret established through the Diffie-Hellman algorithm.

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5. Obeliziti tacan rezultat za: 52 (mod 17) = ?

Explanation

The given expression "52 (mod 17)" represents the remainder when 52 is divided by 17. To find this remainder, we divide 52 by 17 and see what is left over. In this case, 52 divided by 17 is 3 with a remainder of 1. Therefore, the correct answer is 1.

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6. Lavinski efekat definisemo kao:

Explanation

The correct answer is "Mala promena na ulazu izazvace veliku promenu na izlazu." This statement describes the Lavinski effect, which is defined as a small change at the input causing a large change at the output. This phenomenon is commonly observed in amplifiers and electronic circuits, where a small input signal can result in a much larger output signal.

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7. Obeleziti koliko puta je RSA sporiji od AES-a:

Explanation

The correct answer is 1000 times. This means that RSA is 1000 times slower than AES. RSA is a public-key encryption algorithm that is primarily used for secure data transmission, while AES is a symmetric encryption algorithm that is commonly used for encrypting data at rest. The slower speed of RSA is due to its more complex mathematical operations, such as modular exponentiation, which require more computational resources and time compared to the simpler operations used in AES.

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8. Sifrat kod blokovskih sifara se dobije visestrukom primenom funkcije koja se naziva:

Explanation

The correct answer is "Runda". In block ciphers, the encryption process is divided into multiple rounds, where each round applies a specific set of operations to the input data. These operations typically include substitution, permutation, and mixing of the data. Therefore, the term "runda" refers to the specific step or iteration within the encryption process of a block cipher.

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9. Obeleziti: duzine kljuceva kod PKI  sistema koje su u upotrebi:

Explanation

The correct answer is the set of key lengths used in PKI systems: 1024, 2048, 4096. These key lengths are commonly used in PKI systems to ensure secure encryption and authentication. The larger the key length, the stronger the encryption and the more secure the system. Therefore, using longer key lengths such as 1024, 2048, and 4096 provides a higher level of security in PKI systems.

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10. Ako su poznati sledeci parametri (p, g, a) i Bi informacija od Boba, Alisa izracunava kljuc na sledeci nacin:

Explanation

The given answer states that the key (K) is calculated as Bi raised to the power of a, modulo p. This means that Alisa takes the value received from Bob (Bi) and raises it to the power of her own secret value (a), and then takes the remainder when divided by p. This calculation allows Alisa to generate a shared key with Bob using the Diffie-Hellman key exchange algorithm.

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11. Obeleziti tacan rezultat za: (25*13)(mod 6) = ?

Explanation

The given expression is (25*13)(mod 6). First, we multiply 25 and 13, which gives us 325. Then, we take the modulus of 325 with 6. The modulus operation gives us the remainder when 325 is divided by 6. In this case, 325 divided by 6 gives a remainder of 1. Therefore, the correct answer is 1.

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12. Double DES (2DES) koristi duzinu kljuca:

Explanation

Double DES (2DES) uses a key length of 112 bits. Double DES is a variant of the Data Encryption Standard (DES) algorithm that applies DES encryption twice to increase security. In Double DES, two 56-bit keys are used consecutively. Since each key is 56 bits long, the total key length becomes 112 bits. This increases the complexity of the encryption process and makes it more difficult for attackers to crack the encryption.

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13. Da bi se sprecio napad na DH algoritam neophodna je:

Explanation

Autentifikacija je neophodna kako bi se sprečio napad na DH algoritam. Autentifikacija je proces provere identiteta korisnika ili uređaja, čime se osigurava da samo legitimni korisnici imaju pristup sistemu. U kontekstu DH algoritma, autentifikacija se koristi kako bi se osiguralo da komunicirajuće strane zaista jesu one koje tvrde da jesu, čime se sprečava prisustvo neovlašćenih korisnika koji bi mogli pokušati da izvrše napad na algoritam.

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14. RSA algoritam koristi sledece kljuceve:

Explanation

RSA algoritam koristi javni ključ za enkripciju podataka i privatni ključ za dekripciju podataka. Javni ključ je dostupan svima i koristi se za enkripciju podataka koje samo vlasnik privatnog ključa može dekriptovati. Privatni ključ se čuva tajno i koristi se za dekripciju podataka koje su enkriptovane javnim ključem. Ova kombinacija javnog i privatnog ključa omogućava sigurnu razmenu podataka između korisnika RSA algoritma.

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15. Duzina bloka kod AES algoritma moze da bude:

Explanation

The length of a block in the AES algorithm can be 128 bits, 192 bits, or 256 bits. This means that the AES algorithm can process data in blocks of these sizes. The choice of block size depends on the specific application and security requirements. A larger block size generally provides stronger security but may also result in slower processing speed.

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16. Princip konfuzije:

Explanation

In a brute-force attack, all keys are equally likely to be the correct key. This means that the attacker has to try all possible keys until they find the correct one. This is because the brute-force attack method involves systematically trying every possible combination until the correct key is found. Therefore, there is no bias towards any specific key in a brute-force attack.

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17. Message Autentification Cod (MAC) se moze racunati preko sledeceg blokovskog rezima:

Explanation

CBC (Cipher Block Chaining) is a block cipher mode of operation that provides confidentiality and message authentication. In CBC mode, each plaintext block is XORed with the previous ciphertext block before encryption. This ensures that even if two plaintext blocks are identical, the corresponding ciphertext blocks will be different. Additionally, CBC mode uses an Initialization Vector (IV) to provide randomness and prevent patterns in the ciphertext. The IV is XORed with the first plaintext block before encryption. Therefore, CBC mode not only encrypts the message but also provides authentication by ensuring the integrity of the ciphertext.

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18. Bob odredjuje parametre za DH ptokol prema sledecem: Bi= g^b mod p. Javni parametri su:

Explanation

The given answer consists of the variables p and g. These variables are mentioned in the question as the public parameters for the DH protocol. The variable p represents a prime number, while g represents a generator. These parameters are essential for the DH protocol as they help in generating the shared secret key between two parties.

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19. 3DES sa EDE2 radi sa sledecom notacijom:

Explanation

The correct answer is C = E(D(E(P,K1),K2),K1). This is the correct notation for 3DES sa EDE2. It shows that the plaintext (P) is first encrypted with key K1, then decrypted with key K2, and finally encrypted again with key K1. This triple encryption process provides stronger security compared to single DES encryption.

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20. Koliko iznosi aditivna inverzija broja 3 (mod 7)

Explanation

The question is asking for the additive inverse of the number 3 (mod 7). In modular arithmetic, the additive inverse of a number is the number that, when added to the original number, gives a sum that is congruent to 0 modulo the given modulus. In this case, the modulus is 7. The additive inverse of 3 (mod 7) is 4, because 3 + 4 = 7, which is congruent to 0 modulo 7.

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21. Duzina kljuca kod AES algoritma moze da bude

Explanation

The AES algorithm allows for different key lengths to be used, including 192 bits, 128 bits, and 256 bits. These key lengths determine the level of security provided by the algorithm. A longer key length generally provides stronger security, as there are more possible combinations for the key. Therefore, 192 bits, 128 bits, and 256 bits are all valid key lengths for the AES algorithm.

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22. Ako se uporede simetricni i asimetricni sifarski sistemi obeleziti sta je tacno:

Explanation

The correct answer states that symmetric cipher systems are faster and the key of the asymmetric system is longer compared to the key of the symmetric system. This means that symmetric cipher systems are faster in terms of encryption and decryption processes. Additionally, asymmetric systems require longer keys for encryption and decryption compared to symmetric systems, which adds to the complexity and security of the asymmetric system.

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23. Kod blokovskih sifri u ECB rezimu je moguc sledeci napad:

Explanation

Cut and paste is a possible attack on block ciphers in ECB mode. In this attack, the attacker intercepts the ciphertext and rearranges the blocks to create a new ciphertext. By doing so, the attacker can manipulate the plaintext without knowing the encryption key. This attack is possible because ECB mode encrypts each block independently, allowing the attacker to rearrange the blocks without affecting the decryption process. This can lead to serious security vulnerabilities, which is why ECB mode is not recommended for secure encryption.

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24. Funkcije kod AES algortima:

Explanation

The given correct answer states that the functions in the AES algorithm are invertible. This means that for every output of the function, there exists a unique input that produces that output. In other words, it is possible to reverse the function and obtain the original input from the output. This property is crucial in encryption algorithms like AES, as it allows for the decryption process to retrieve the original message from the encrypted data.

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25. DH algoritam je osetljiv na napad:

Explanation

The correct answer is "Man in the middle" because the DH algorithm is vulnerable to this type of attack. In a man-in-the-middle attack, an attacker intercepts communication between two parties and impersonates each party to the other, allowing them to intercept and manipulate the data being transmitted. This can compromise the security of the DH algorithm, as the attacker can potentially gain access to the shared secret key and decrypt the communication.

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26. PKI predstavlja sisteme sa:

Explanation

PKI (Public Key Infrastructure) predstavlja sisteme koji koriste javne i privatne ključeve. Javni ključ se koristi za šifrovanje podataka, dok se privatni ključ koristi za dešifrovanje podataka. Ova kombinacija ključeva omogućava sigurnu razmenu podataka između korisnika, jer samo osoba sa odgovarajućim privatnim ključem može dešifrovati podatke koji su šifrovani javnim ključem. Ovo omogućava autentifikaciju korisnika i bezbednu komunikaciju na internetu.

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27. Postupak sifrovanja otvorene poruke M kod RSA algoritma je:

Explanation

The given correct answer states that the encryption process for RSA algorithm is C = M^e mod N. In RSA encryption, the ciphertext (C) is calculated by raising the plaintext (M) to the power of the public exponent (e) and then taking the modulo N. This process ensures that the ciphertext is only decryptable with the corresponding private key.

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28. Kod RSA algortima u procesu digitalnog potpisivanje privatni kljuc: 

Explanation

The correct answer is "Samo strani koja digitalno potpisuje" (Only the party that digitally signs). In the process of digital signing using the RSA algorithm, the private key is only known to the party that is digitally signing the message. This ensures that the digital signature is unique to that party and verifies the authenticity and integrity of the message. The private key should not be known to any other party, including the party receiving the message or the certification authority.

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29. Fejstel sifra predstavlja:

Explanation

The correct answer is "Tip blokovske sifre" because Fejstel sifra (Feistel cipher) is a type of block cipher. In a Feistel cipher, the plaintext is divided into blocks and each block undergoes a series of transformations using a round function and a key. This type of cipher is commonly used in symmetric encryption algorithms to ensure confidentiality and security of data.

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30. Trostruki DES sa tri nezavisna kljuca koristi kljuc:

Explanation

Triple DES with three independent keys uses a key size of 168 bits. This means that each key used in the encryption process is 56 bits long. Triple DES is a symmetric encryption algorithm that applies the Data Encryption Standard (DES) algorithm three times in a row to provide enhanced security. By using three keys, it increases the key size and makes it more difficult for attackers to decrypt the encrypted data.

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31. Notacija za trostruki DES je: 

Explanation

The correct answer is C = E(D(E(P,K1),K2),K3). This answer correctly represents the notation for triple DES, where P is the plaintext, K1, K2, and K3 are the three keys used for encryption and decryption, E represents the encryption function, and D represents the decryption function. The notation shows that the plaintext P is first encrypted with K1, then decrypted with K2, and finally encrypted again with K3 to produce the ciphertext C.

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32. Broj rundi kod DES algoritma je:

Explanation

The number of rounds in the DES algorithm is 16.

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33. Tajnost Diffie-Hellman (DH) algoritma se zasniva na slozenosti izracunavanja:

Explanation

The correct answer is "Diskretnog logaritma." The secrecy of the Diffie-Hellman (DH) algorithm is based on the computational complexity of calculating discrete logarithms. In DH, two parties can exchange public keys over an insecure channel and use those keys to establish a shared secret key without directly exchanging the key itself. This is possible because calculating the discrete logarithm in a finite field is computationally difficult, making it hard for an attacker to determine the shared secret key even if they have access to the public keys.

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34. Kod RSA algoritma javni kljucevi su:

Explanation

The correct answer is N,e. In the RSA algorithm, the public key consists of two parts: N and e. N represents the modulus, which is the product of two large prime numbers p and q. e represents the public exponent, which is a smaller prime number that is coprime with (p-1) * (q-1). The public key is used for encryption, while the private key (not mentioned in the question) is used for decryption.

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35. Princip difuzije:

Explanation

The principle of diffusion refers to the characteristic of encryption with a random sequence. In this principle, each bit of the ciphertext is a function of all the bits of the plaintext and all the bits of the key. This principle is historically significant and represents a characteristic of classical cipher systems. Therefore, the correct answer is that each bit of the ciphertext is a function of all the bits of the plaintext and all the bits of the key.

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36. Obeleziti blokovske sifre:

Explanation

The given answer includes a list of block ciphers, which are encryption algorithms that operate on fixed-size blocks of data. The ciphers mentioned in the answer, IDEA, AES, Blowfish, 3DES, and DES, are all well-known and widely used block ciphers in the field of cryptography. These ciphers provide varying levels of security and have been extensively studied and analyzed for their strength against various attacks. The inclusion of these ciphers in the answer suggests that they are considered reliable and effective for securing data.

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37. Kod blokovskih sifri, sifrovanje se primenjuje nad:

Explanation

In block ciphers, encryption is applied to a message divided into fixed-length blocks, and the resulting ciphertext blocks have the same length as the input message blocks. This means that each block of the plaintext message is encrypted into a corresponding block of the ciphertext, maintaining the same length.

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38. Kod digitalnog potpisivanja koristimo: (*dvosmisleno pitanje: moze biti i samo privatni kljuc ako si misli iskljucivo na potpisivanje, a ako se misli na ceo proces digitalnog potpisivanja onda je kombincija oba kljuca)

Explanation

We use the private key in digital signature. The private key is used to generate a unique digital signature for a document or message. This signature can be verified using the corresponding public key, ensuring the authenticity and integrity of the data. The private key is kept secret and known only to the owner, while the public key is shared with others for verification purposes. Therefore, the private key is the correct answer for this question.

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39. Kada Alisa i Bob primene DH algoritam oni su:

Explanation

When Alice and Bob apply the DH algorithm, they establish a secret value that can be used as a symmetric key. The DH algorithm allows two parties to agree upon a shared secret value without directly transmitting it over an insecure channel. This secret value can then be used as a symmetric key for encryption and decryption purposes, ensuring secure communication between Alice and Bob.

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40. Digitalni sertifikat moze da sadrzi:   - proveriti

Explanation

A digital certificate can contain the name and surname of the owner as well as their public key. The name and surname of the owner are important pieces of information that can be used to identify the owner of the certificate. The public key is necessary for encrypting and verifying digital signatures. Therefore, both the name and surname of the owner and the public key are commonly included in a digital certificate.

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41. Primalac poruke koja je digitalno potpisana  proverava integritet poruke tako sto:

Explanation

The correct answer is "Koristi javni kljuc posiljaoca." When a message is digitally signed, the sender uses their private key to create a unique signature for the message. The recipient can then use the sender's public key to verify the integrity of the message by comparing the signature with the message itself. By using the sender's public key, the recipient can ensure that the message has not been tampered with during transmission.

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42. Obeleziti sta je tacno za blokovske sifre:

Explanation

The correct answer is that an error in transmission does not accumulate beyond 2 blocks, and an error of 1 bit in transmission affects two blocks in the OT decoding process. This means that if there is an error in the transmission of data, it will only affect a maximum of 2 blocks and will not propagate further. Additionally, if there is a single bit error in transmission, it will impact the decoding of two blocks in the OT process.

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43. Za racunanje MAC koda za proveru integriteta poruke preko blokovske sifre:

Explanation

The correct answer is that for calculating the MAC code for message integrity verification using block cipher, the CBC mode of operation is used and the MAC represents the last block of the cipher. This means that the message is divided into blocks and each block is encrypted using the CBC mode, where the ciphertext of each block is dependent on the previous block. The MAC code, which is used to verify the integrity of the message, is then derived from the last block of the ciphertext.

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44. DES algoritam je podlozan (slab) na:

Explanation

The correct answer is "Potpunu pretragu kljuceva" (Brute-force attack). The DES algorithm is vulnerable to a brute-force attack, where an attacker systematically tries all possible keys until the correct one is found. This is because DES has a relatively small key size of 56 bits, making it computationally feasible to try all possible keys within a reasonable amount of time.

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45. 3DES sa EDE2 koristi kljuc duzine 

Explanation

3DES sa EDE2 koristi ključ dužine 112 bita. 3DES (Triple Data Encryption Standard) je kriptografski algoritam koji koristi blokovsku šifru DES tri puta za povećanu sigurnost. EDE2 (Encrypt-Decrypt-Encrypt 2) je način implementacije 3DES algoritma koji koristi dva koraka šifriranja i jedan korak dešifriranja. Ključ dužine 112 bita se sastoji od tri ključa od po 56 bita, što ukupno čini 168 bita. Međutim, samo prvih 112 bita se koristi za šifriranje i dešifriranje podataka.

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46. U hibridnom sifarskom sistemu, obeleziti sta su prihvatljivi parametri (parametri koji odgovaraju).

Explanation

In a hybrid cipher system, the acceptable parameters are a secret key of length 256 bits, a hash value of length 256 bits, and a private and public key of length 2048 bits. This combination ensures a strong level of security for the encryption and decryption process.

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47. Obeleziti regularne blokove DES algoritma:

Explanation

The correct answer is the order in which the blocks are listed: P box, S box, K box, E box. This is the correct order of the regular blocks in the DES algorithm. The P box is responsible for permuting the input bits, the S box is responsible for substituting groups of bits, the K box is responsible for generating the round keys, and the E box is responsible for expanding the input bits.

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48. Duzina kljuca kod DES algoritma je:

Explanation

The length of the key in the DES algorithm is 56 bits. DES (Data Encryption Standard) is a symmetric encryption algorithm that uses a 56-bit key to encrypt and decrypt data. This key length was chosen to strike a balance between security and efficiency. However, due to advances in computing power, a 56-bit key is now considered relatively weak and vulnerable to brute-force attacks. As a result, DES is no longer widely used and has been replaced by more secure encryption algorithms with longer key lengths.

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49. Sifrovanjme otvorenog teksta obezbedjujemo:

Explanation

The correct answer is "Poverljivost poruke" (Confidentiality of the message). This means that encrypting the plaintext ensures that the message is kept confidential and cannot be accessed or understood by unauthorized individuals.

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50. Simetricni sifarski sistemi koriste:

Explanation

Simetricni sifarski sistemi koriste isti tajni kljuc za sifrovanje i desifrovanje. U ovakvim sistemima, isti kljuc se koristi kako bi se podaci sifrovali i desifrovali. Ovo omogucava efikasno sifrovanje i desifrovanje podataka, ali zahteva bezbedan nacin deljenja tajnog kljuca izmedju komunikacionih strana.

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51. Duzina podkljuca kod DES algoritma u svakoj rundi je:

Explanation

The correct answer is 48 bits. In the DES algorithm, the key size used in each round is 48 bits. This key is derived from the original 64-bit key through a process called key schedule. The key schedule generates 16 different 48-bit keys, one for each round of the algorithm. These keys are used to perform various transformations and substitutions during the encryption process.

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52. Obeleziti poznate matematicki teske probleme na kojima pocivaju asimetricni sifarski sistemi:

Explanation

The correct answer is to mark the known difficult mathematical problems on which asymmetric cipher systems are based: the discrete logarithm, the Euclidean algorithm, number factorization, and Euler's function.

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53. Postupak desifrovanja sifrata C kod RSA algoritma je:

Explanation

The correct answer is M = C^d mod N. In the RSA algorithm, the process of decrypting the ciphertext involves raising the ciphertext (C) to the power of the private key exponent (d), and then taking the modulo N. This calculation results in the original plaintext message (M).

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54. Sertifikaciono telo potpisuje sertifikat:

Explanation

The correct answer is "Privatnim kljucem CA" (Private key of the CA). A certification authority (CA) signs a certificate using its private key. The private key is used for encryption and decryption, and in this case, it is used to sign the certificate to ensure its authenticity and integrity. The private key is kept secret by the CA to maintain the security of the certification process.

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55. Ukoliko koristimo CBC rezim rada kod AES algoritma:

Explanation

When using the CBC mode of operation in the AES algorithm, the Initialization Vector (IV) needs to be random and secret. This is because the IV is used to initialize the first block of plaintext before encryption, and if it is predictable or known to an attacker, it can lead to security vulnerabilities. Additionally, using an IV is mandatory in CBC mode to ensure the uniqueness of the ciphertext and prevent patterns from emerging. Therefore, the correct answer is that the IV needs to be random, secret, and its usage is mandatory.

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56. Kod RSA privatni kljucevi su:

Explanation

The correct answer is "d" because in the RSA encryption algorithm, the private key consists of the values "d", "p", and "q". "d" is the private exponent, which is used for decrypting the encrypted messages. "p" and "q" are prime numbers used in the generation of the private key. The values "N" and "e" are part of the public key, used for encrypting the messages.

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57. Asimetricnim sifarskim sistemom mogu se obezbediti sledeci servisi:

Explanation

The correct answer is 1. Poverljivost, 2. Integritet, 3. Neporecivost. These three services can be provided by an asymmetric cryptographic system. Poverljivost refers to ensuring that the information is kept confidential and only accessible to authorized parties. Integritet ensures that the data remains unchanged and unaltered during transmission or storage. Neporecivost guarantees that the sender cannot deny sending a message, providing evidence of the message's origin and authenticity.

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58. Obeleziti sta je tacno za CA

Explanation

The correct answer is "Treca strana od poverenja, Izdaje i potpisuje sertifikat, Format sertifikata je X.509". This answer is correct because it mentions the third-party trust, which is an important aspect of certificate authorities (CA). CAs are responsible for issuing and signing certificates, which is also mentioned in the answer. Additionally, the answer correctly identifies the format of the certificate as X.509, which is a widely used standard for digital certificates.

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59. Ukoliko koristimo CBC rezim rada kod AES algoritma:

Explanation

If we use CBC mode of operation in the AES algorithm, the ciphertexts are dependent on each other. This means that the encryption of each block of plaintext depends on the previous ciphertext block. This dependency ensures that even if the same plaintext block is encrypted multiple times, the resulting ciphertext will be different, providing a higher level of security. Therefore, the correct answer is that the ciphertexts are mutually dependent on each other.

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60. Za funkciju poverljivosti (tajnosti) standardno se koriste:

Explanation

Symmetric encryption systems are commonly used for confidentiality (secrecy) purposes. In symmetric encryption, the same key is used for both encryption and decryption of the message. This means that the sender and receiver must have the same key to communicate securely. Symmetric encryption is efficient and fast, making it suitable for encrypting large amounts of data. However, the challenge lies in securely distributing the key to the intended recipient.

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61. Obeleziti sta je karakteristicno za kriptografske jednosmerne funkcije

Explanation

Cryptographic one-way functions are characterized by the fact that it is computationally complex to calculate the inverse function, while it is easy to calculate in the direct direction. This means that it is difficult and time-consuming to find the original input from the output of the function, providing a level of security for cryptographic systems. Additionally, the fact that these functions are computationally complex makes it difficult for attackers to reverse-engineer the original input, further enhancing the security of cryptographic systems.

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62. Rezimi rada blokovskih sifri su

Explanation

The correct answer is CTR, CBC, ECB. These are different modes of operation for block ciphers. CTR (Counter) mode turns a block cipher into a stream cipher, where each plaintext block is XORed with the output of the block cipher. CBC (Cipher Block Chaining) mode uses the output of the previous block as the input for the next block, adding an initialization vector (IV) to the first block. ECB (Electronic Codebook) mode encrypts each plaintext block separately with the same key, making it less secure as patterns in the plaintext can be easily identified.

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63. Osnovna tri nacina upotrebe kriptografije sa javnim kljucevima:

Explanation

The correct answer is "Encryption, Exchange of symmetric key, Digital signature." This is because encryption is a fundamental use of public key cryptography where data is encoded in such a way that only authorized parties can access it. The exchange of symmetric key refers to the process of securely sharing a symmetric key between two parties using public key cryptography. Digital signature is another important use of public key cryptography where a digital signature is used to verify the authenticity and integrity of a message or document.

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64. Kod asimetricnih sifarskih sistema:

Explanation

The correct answer is "Kljuc za desifrovanje i sifrovanje je razlicit, Sifrovanje se vrsi javnim kljucem primaoca." In asymmetric encryption systems, the key used for encryption is different from the key used for decryption. This ensures that only the intended recipient, who possesses the private key, can decrypt the message. Additionally, in asymmetric encryption, encryption is performed using the recipient's public key, which can be freely shared, while decryption is done using the recipient's private key, which is kept secret.

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65. RSA algoritam mozemo koristiti za:

Explanation

RSA algoritam se može koristiti za generisanje sertifikata, digitalno potpisivanje i šifrovanje. Generisanje sertifikata je proces stvaranja digitalnog identiteta koji se koristi za potvrdu autentičnosti entiteta na internetu. Digitalno potpisivanje se koristi za verifikaciju integriteta i autentičnosti digitalnih dokumenata. Šifrovanje se koristi za zaštitu podataka tako da se mogu preneti sigurno preko nesigurnih mreža. RSA algoritam omogućava sve ove funkcionalnosti kroz upotrebu javnih i privatnih ključeva.

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66. U kom rezimu rada blokovskih sifri se koristi inicijalizacioni vektor (IV):

Explanation

In the given options, the correct answers for the mode of operation of block ciphers that use an initialization vector (IV) are CTR and CBC.

CTR (Counter) mode operates by encrypting a counter value and then XORing it with the plaintext to produce the ciphertext. The IV is used as the initial counter value.

CBC (Cipher Block Chaining) mode operates by XORing each plaintext block with the previous ciphertext block before encrypting it. The IV is used as the initial value for the XOR operation.

Both CTR and CBC modes require an IV to ensure the uniqueness and randomness of the encryption process.

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67. Kod hibridnih sifarskih sistema obeleziti sta je ispravno:

Explanation

The correct answer is that confidentiality is achieved through symmetric cipher systems, authentication is performed using asymmetric cipher systems, and data integrity is achieved through asymmetric cipher systems.

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68. Kod RSA algoritma kada sifrujemo koristimo:

Explanation

When using the RSA algorithm for encryption, we use the public key. The RSA algorithm is an asymmetric encryption algorithm, which means it uses a pair of keys - a public key and a private key. The public key is used for encryption, while the private key is used for decryption. In this case, since we are encrypting, we use the public key.

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69. Sifrovanje simetricnim sifarskim sistemom obezbedjuje sledece servise:

Explanation

Symmetric encryption ensures both confidentiality (tajnost) and integrity (integritet) of data. Confidentiality means that the information is protected and only authorized parties can access it. Integrity ensures that the data remains unchanged and uncorrupted during transmission or storage. Therefore, the correct answer includes both tajnost and integritet.

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70. Ukoliko koristimo ECB rezim rada kod AES algoritma:

Explanation

When using the ECB mode of operation in the AES algorithm, the same plaintext block will always produce the same ciphertext block. This lack of randomness can be exploited by attackers, as they can identify patterns in the ciphertext and potentially deduce information about the plaintext. This vulnerability makes the encryption susceptible to "Cut and Paste" attacks, where an attacker can manipulate the ciphertext by cutting and pasting blocks to create new valid ciphertexts. Therefore, the correct answer states that the same plaintext block produces the same ciphertext block, and the encryption is sensitive to "Cut and Paste" attacks.

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71. Obeleziti blokovske sifre:

Explanation

The given list includes various block ciphers used for encryption. The correct answer includes 3DES, IDEA, Blowfish, DES, RC6, and AES. These are all well-known and widely used block ciphers in the field of cryptography. They provide secure encryption algorithms for protecting sensitive data.

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72. Kod AES algoritma:

Explanation

The correct answer is that the length of the input message blocks is independent of the length of the key. This means that regardless of the length of the key used in the AES algorithm, the length of the input message blocks remains the same. The AES algorithm operates on fixed-size blocks of data, typically 128 bits, and the key length does not affect the size of these blocks.

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73. DH protocol namenjen je za:

Explanation

DH (Diffie-Hellman) protocol je namenjen za uspostavljanje simetričnog ključa. Ovaj protokol omogućava dvema stranama da bezbedno razmene informacije i dogovore se o zajedničkom tajnom ključu koji će se koristiti za simetrično šifrovanje podataka. DH protokol je asimetrični protokol koji koristi matematičke operacije kako bi se generisao zajednički tajni ključ, koji će biti poznat samo stranama koje učestvuju u razmeni. Ovaj ključ se zatim koristi za šifrovanje i dešifrovanje podataka.

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74. Obeleziti karakteristike simetricnih sifarskih sistema

Explanation

The correct answer is "Racunarski su sigurni, Koriste tajni kljuc, Tajnost im se zasniva na tajnosti kljuca." This is because symmetric encryption systems rely on the secrecy of the key used for encryption and decryption. These systems use the same key for both encryption and decryption, which means that anyone who knows the key can easily decrypt the message. Therefore, the security of the system depends on keeping the key secret. Additionally, symmetric encryption systems are considered computationally secure, meaning that it is computationally infeasible to decrypt the message without knowing the key.

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75. Obeleziti karakteristike asimetricnih sifarskih sistema:

Explanation

The correct answer choices explain the characteristics of asymmetric cryptographic systems. These systems rely on the complexity of mathematical functions for secrecy, have different keys for encryption and decryption, are used for confidentiality of transmitted messages, and are also used for digital signing.

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Sifarski sistemi mogu biti:
RSA algoritam se pojavio: ...
Osnovna podela sifarskih sistema, prema vrsti kljuca deli se na:
DH algoritam za uspostavljanje deljene tajne je osetljiv na:
Obeliziti tacan rezultat za: 52 (mod 17) = ?
Lavinski efekat definisemo kao:
Obeleziti koliko puta je RSA sporiji od AES-a:
Sifrat kod blokovskih sifara se dobije visestrukom primenom funkcije...
Obeleziti: duzine kljuceva kod PKI  sistema koje su u upotrebi:
Ako su poznati sledeci parametri (p, g, a) i Bi informacija od Boba,...
Obeleziti tacan rezultat za: (25*13)(mod 6) = ?
Double DES (2DES) koristi duzinu kljuca:
Da bi se sprecio napad na DH algoritam neophodna je:
RSA algoritam koristi sledece kljuceve:
Duzina bloka kod AES algoritma moze da bude:
Princip konfuzije:
Message Autentification Cod (MAC) se moze racunati preko sledeceg...
Bob odredjuje parametre za DH ptokol prema sledecem: Bi= g^b mod p....
3DES sa EDE2 radi sa sledecom notacijom:
Koliko iznosi aditivna inverzija broja 3 (mod 7)
Duzina kljuca kod AES algoritma moze da bude
Ako se uporede simetricni i asimetricni sifarski sistemi obeleziti sta...
Kod blokovskih sifri u ECB rezimu je moguc sledeci napad:
Funkcije kod AES algortima:
DH algoritam je osetljiv na napad:
PKI predstavlja sisteme sa:
Postupak sifrovanja otvorene poruke M kod RSA algoritma je:
Kod RSA algortima u procesu digitalnog potpisivanje privatni...
Fejstel sifra predstavlja:
Trostruki DES sa tri nezavisna kljuca koristi kljuc:
Notacija za trostruki DES je: 
Broj rundi kod DES algoritma je:
Tajnost Diffie-Hellman (DH) algoritma se zasniva na slozenosti...
Kod RSA algoritma javni kljucevi su:
Princip difuzije:
Obeleziti blokovske sifre:
Kod blokovskih sifri, sifrovanje se primenjuje nad:
Kod digitalnog potpisivanja koristimo: ...
Kada Alisa i Bob primene DH algoritam oni su:
Digitalni sertifikat moze da sadrzi:   - proveriti
Primalac poruke koja je digitalno potpisana  proverava integritet...
Obeleziti sta je tacno za blokovske sifre:
Za racunanje MAC koda za proveru integriteta poruke preko blokovske...
DES algoritam je podlozan (slab) na:
3DES sa EDE2 koristi kljuc duzine 
U hibridnom sifarskom sistemu, obeleziti sta su prihvatljivi parametri...
Obeleziti regularne blokove DES algoritma:
Duzina kljuca kod DES algoritma je:
Sifrovanjme otvorenog teksta obezbedjujemo:
Simetricni sifarski sistemi koriste:
Duzina podkljuca kod DES algoritma u svakoj rundi je:
Obeleziti poznate matematicki teske probleme na kojima pocivaju...
Postupak desifrovanja sifrata C kod RSA algoritma je:
Sertifikaciono telo potpisuje sertifikat:
Ukoliko koristimo CBC rezim rada kod AES algoritma:
Kod RSA privatni kljucevi su:
Asimetricnim sifarskim sistemom mogu se obezbediti sledeci servisi:
Obeleziti sta je tacno za CA
Ukoliko koristimo CBC rezim rada kod AES algoritma:
Za funkciju poverljivosti (tajnosti) standardno se koriste:
Obeleziti sta je karakteristicno za kriptografske jednosmerne funkcije
Rezimi rada blokovskih sifri su
Osnovna tri nacina upotrebe kriptografije sa javnim kljucevima:
Kod asimetricnih sifarskih sistema:
RSA algoritam mozemo koristiti za:
U kom rezimu rada blokovskih sifri se koristi inicijalizacioni vektor...
Kod hibridnih sifarskih sistema obeleziti sta je ispravno:
Kod RSA algoritma kada sifrujemo koristimo:
Sifrovanje simetricnim sifarskim sistemom obezbedjuje sledece servise:
Ukoliko koristimo ECB rezim rada kod AES algoritma:
Obeleziti blokovske sifre:
Kod AES algoritma:
DH protocol namenjen je za:
Obeleziti karakteristike simetricnih sifarskih sistema
Obeleziti karakteristike asimetricnih sifarskih sistema:
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