Network-Security
Question 1 |
In an RSA cryptosystem, the value of the public modulus parameter n is 3007. If it is also known that Φ(n) = 2880, where Φ() denotes Euler's Quotient Function, then the prime factor of n which is greater than 50 is ______.
107 | |
97 | |
45 | |
92 |
Question 1 Explanation:
It can be solved by Hit and trial method in less time.
n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),
where p, q are prime factor of n.
The unit place of n is 7, it is a prime number and factor will be
1.7=7
11*17
21*37
31*47
….
31*97 =>3007
n = 3007 => 31*97
Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.
So, 97 is the correct answer.
Other methods:
When ϕ(n) is given when n=pq where p and q are prime numbers, then we have
ϕ(n) = (p−1)(q−1) = pq−(p+q)+1
But pq=n,
therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).
Now, p and q are the roots of the equation,
x2 − (p+q)x + pq = (x-p)(x-q)
Substituting for p+q and pq in the above equation
x2 - (n+1-ϕ(n))x + n
n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),
where p, q are prime factor of n.
The unit place of n is 7, it is a prime number and factor will be
1.7=7
11*17
21*37
31*47
….
31*97 =>3007
n = 3007 => 31*97
Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.
So, 97 is the correct answer.
Other methods:
When ϕ(n) is given when n=pq where p and q are prime numbers, then we have
ϕ(n) = (p−1)(q−1) = pq−(p+q)+1
But pq=n,
therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).
Now, p and q are the roots of the equation,
x2 − (p+q)x + pq = (x-p)(x-q)
Substituting for p+q and pq in the above equation
x2 - (n+1-ϕ(n))x + n
Question 2 |
The value of 351 mod 5 is ______.
3 | |
5 | |
2 | |
1 |
Question 2 Explanation:
351 mod 5
⇒ 31 = 3 ⇒ 3 mod 5 = 3
32 ⇒ 9 mod 5 = 4
33 ⇒ 27 mod 5 = 2
34 ⇒ 81 mod 5 = 1
35 ⇒ 243 mod 5 = 3
For every four numbers sequence is repeating.
So, (51 % 4) = 3
⇒ 33 = 27
⇒ 27 mod 5 = 2
⇒ 31 = 3 ⇒ 3 mod 5 = 3
32 ⇒ 9 mod 5 = 4
33 ⇒ 27 mod 5 = 2
34 ⇒ 81 mod 5 = 1
35 ⇒ 243 mod 5 = 3
For every four numbers sequence is repeating.
So, (51 % 4) = 3
⇒ 33 = 27
⇒ 27 mod 5 = 2
Question 3 |
A sender is employing public key cryptography to send a secret message to a receiver. Which one of the following statements is TRUE?
Sender encrypts using receiver’s public key | |
Sender encrypts using his own public key | |
Receiver decrypts using sender’s public key | |
Receiver decrypts using his own public key |
Question 3 Explanation:
Sender can encrypts using the receiver public key and receiver decrypts it using his own private key.
Question 4 |
Which of the following statement(s) is TRUE?
1. A hash function takes a message of arbitrary length and generates a fixed length code.
2. A hash function takes a message of fixed length and generates a code of variable length.
3. A hash function may give the same hash value for distinct messages.
1 only | |
2 and 3 only | |
1 and 3 only | |
2 only |
Question 4 Explanation:
(1) A hash function takes a message of arbitrary length and generates a fixed length code. So this is correct.
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
Question 5 |
Your are given the following four bytes :
10100011 00110111 11101001 10101011Which of the following are substrings of the base 64 encoding of the above four bytes?
zdp | |
fpq | |
qwA | |
oze |
Question 5 Explanation:
You are given the following four bytes:
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64 we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.
Now, the longest substring will be from checking option is 'fpq'.
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64 we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.
Now, the longest substring will be from checking option is 'fpq'.
Question 6 |
Consider the following two statements:
(i) A hash function (these are often used for computing digital signatures) is an injective function.
(ii) An encryption technique such as DES performs a permutation on the elements of its input alphabet.
Which one of the following options is valid for the above two statements?
Both are false | |
Statement (i) is true and the other is false | |
Statement (ii) is true and the other is false | |
Both are true |
Question 6 Explanation:
i) Hash function is many to one function. It is not one-one (or) injective.
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
Question 7 |
The total number of keys required for a set of n individuals to be able to communicate with each other using secret key and public key crypto-systems, respectively are:
n(n-1) and 2n | |
2n and n(n-1)/2 | |
n(n-1)/2 and 2n | |
n(n-1)/2 and n |
Question 7 Explanation:
For private key crypto, a key used for encryption as well as decryption. So, no. of keys required for n individuals is same as no. of communication link between any two individuals which is
nC2 = n(n-1)/2
In case of public key, each sender has its own public key as well as private key. So, no. of keys are 2n.
nC2 = n(n-1)/2
In case of public key, each sender has its own public key as well as private key. So, no. of keys are 2n.
Question 8 |
Suppose that everyone in a group of N people wants to communicate secretly with the N–1 others using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is
2N | |
N(N – 1) | |
N(N – 1)/2 | |
(N – 1)2 |
Question 8 Explanation:
For N people, the number of unique pairs can be calculated using the combination formula (N 2), which is given by:
(N 2) = N(N – 1)/2
The number of keys required in the system as a whole to satisfy the confidentiality requirement is:
N(N – 1)/2
(N 2) = N(N – 1)/2
The number of keys required in the system as a whole to satisfy the confidentiality requirement is:
N(N – 1)/2
Question 9 |
A layer-4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT
block entire HTTP traffic during 9:00PM and 5:00AM | |
block all ICMP traffic | |
stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address | |
block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00AM |
Question 9 Explanation:
(A) It is possible to block entire HTTP traffic by blocking port no.80.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
Question 10 |
Using public key cryptography, X adds a digital signature to message M, encrypts <M, σ>, and sends it to Y, where it is decrypted. Which one of the following sequences of keys is used for the operations?
Encryption: X’s private key followed by Y’s private key; Decryption: X’s public key followed by Y’s public key | |
Encryption: X’s private key followed by Y’s public key; Decryption: X’s public key followed by Y’s private key | |
Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public key followed by X’s private key | |
Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private key followed by X’s public key |
Question 10 Explanation:
Encryption: Source has to encrypt with its private key for forming Digital signature for Authentication. Source has to encrypt the (M, σ) with Y’s public key to send it confidentially.
Decryption: Destination Y has to decrypt first with its private key, then decrypt using source public key.
Question 11 |
In the RSA public key cryptosystem, the private and public keys are (e, n) and (d, n) respectively, where n = p*q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0 < M < n and f(n) = (p- 1)(q-1). Now consider the following equations.
I. M’= Me mod n
M = (M’)d mod n
II. ed ≡ 1 mod n
III. ed ≡ 1 mod f(n)
IV. M’= Me mod f(n)
M = (M’)d mod f(n)
Which of the above equations correctly represent RSA cryptosystem?
I and II | |
I and III | |
II and IV | |
III and IV |
Question 11 Explanation:
To generate the encryption and decryption keys, we can proceed as follows.
1. Generate randomly two “large” primes p and q.
2. Compute n=pq and ∅=(p-1)(q-1).
3. Choose a number e so that
gcd(e,∅)=1
4. Find the multiplicative inverse of e modulo ∅, i.e., find d so that
ed≡1 (mod ∅)
This can be done efficiently using Euclid’s Extended Algorithm.
The encryption public key is KE=(n,e) and the decryption private key is KD=(n,d).
The encryption function is
E(M)=Me mod n
The decryption function is
D(M)=Md mod n
1. Generate randomly two “large” primes p and q.
2. Compute n=pq and ∅=(p-1)(q-1).
3. Choose a number e so that
gcd(e,∅)=1
4. Find the multiplicative inverse of e modulo ∅, i.e., find d so that
ed≡1 (mod ∅)
This can be done efficiently using Euclid’s Extended Algorithm.
The encryption public key is KE=(n,e) and the decryption private key is KD=(n,d).
The encryption function is
E(M)=Me mod n
The decryption function is
D(M)=Md mod n
Question 12 |
Avalanche effect in cryptography refers
Large changes in cipher text when the keyword is changed minimally | |
Large changes in cipher text when the plain text is changed | |
Large impact of keyword change to the length of the cipher text | |
None of the above |
Question 12 Explanation:
Avalanche effect describes a concept in cryptography, where a small change in the input value (keyword) causes a significant change in the output (hash value/ cipher text)
Question 13 |
In a columnar transposition cipher, the plain text is “the tomato is a plant in the night shade family”, keyword is “TOMATO”. The ciphertext is
“TINESAX / EOAHTFX / HTLTHEY / MAIIAIX / TAPNGDL / OSTNHMX” | |
“TINESAX / EOAHTFX / MAIIAIX / HTLTHEY / TAPNGDL / OS TN HMX” | |
“TINESAX / EOAHTFX / HTLTHEY / MAIIAIX / OSTNHMX / TAPNGDL” | |
“EOAHTFX / TINESAX / HTLTHEY / MIIAIX / TAPNGDL / OSTNHMX” |
Question 13 Explanation:
We first pick a keyword for our encryption. We write the plaintext out in a grid where the number of columns is the number of letters in the keyword. We then title each column with the respective letter from the keyword. We take the letters in the keyword in alphabetical order, and read down the columns in this order. If a letter is repeated, we do the one that appears first, then the next and so on.
encrypt the message "The tomato is a plant in the nightshade family" using the keyword tomato. We get the grid given below.
We have written the keyword above the grid of the plaintext, and also the numbers telling us which order to read the columns in. Notice that the first "O" is 3 and the second "O" is 4, and the same thing for the two "T"s.
The plaintext is written in a grid beneath the keyword. The numbers represent the alphabetical order of the keyword, and so the order in which the columns will be read.
Starting with the column headed by "A", our ciphertext begins "TINESAX" from this column. We now move to the column headed by "M", and so on through the letters of the keyword in alphabetical order to get the ciphertext "TINESAX / EOAHTFX / HTLTHEY / MAIIAIX / TAPNGDL / OSTNHMX" (where the / tells you where a new column starts). The final ciphertext is thus "TINESAX EOAHTFX HTLTHEY MAIIAIX TAPNGDL OSTNHMX".
encrypt the message "The tomato is a plant in the nightshade family" using the keyword tomato. We get the grid given below.
We have written the keyword above the grid of the plaintext, and also the numbers telling us which order to read the columns in. Notice that the first "O" is 3 and the second "O" is 4, and the same thing for the two "T"s.
The plaintext is written in a grid beneath the keyword. The numbers represent the alphabetical order of the keyword, and so the order in which the columns will be read.
Starting with the column headed by "A", our ciphertext begins "TINESAX" from this column. We now move to the column headed by "M", and so on through the letters of the keyword in alphabetical order to get the ciphertext "TINESAX / EOAHTFX / HTLTHEY / MAIIAIX / TAPNGDL / OSTNHMX" (where the / tells you where a new column starts). The final ciphertext is thus "TINESAX EOAHTFX HTLTHEY MAIIAIX TAPNGDL OSTNHMX".
Question 14 |
Consider an information exchange scenario where Anthony is the sender and Bond is the intended recipient of the data.
Match the following appropriately.
I-C, II-B, III-D, IV-A | |
I-C, II-D, III-A, IV-B | |
I-C, II-D, III-B, IV-A | |
I-A, II-D, III-B, IV-C |
Question 14 Explanation:
Message Authentication: Bond needs to be sure of anthony’s identity and that an imposter has not sent the message.
Message confidentiality: The transmitted message must make sense to only bond and to all others it must be garbage.
Message Integrity: The message must arrive at the bond’s side exactly as it was sent. Message Non Repudiation: Anthony must not be able to deny sending a message that he or she in fact, did send
Options C is correct.
Message confidentiality: The transmitted message must make sense to only bond and to all others it must be garbage.
Message Integrity: The message must arrive at the bond’s side exactly as it was sent. Message Non Repudiation: Anthony must not be able to deny sending a message that he or she in fact, did send
Options C is correct.
Question 15 |
Considering the following key using a block of five characters, encryption of the message "NETWORKING" using the Transposition Cipher is:
Plaintext : 5 4 3 2 1
Ciphertext 1 2 3 4 5
GNIKROWTEN | |
OGWNTIEKNR | |
OWTENGNIKR | |
NREKTIWNOG |
Question 15 Explanation:
→A transposition cipher is a method of encryption by which the positions held by units of plaintext (which are commonly characters or groups of characters) are shifted according to a regular system, so that the ciphertext constitutes a permutation of the plaintext. That is, the order of the units is changed (the plaintext is reordered).
→According to given question ,
→Plaintext : 5 4 3 2 1
→Ciphertext 1 2 3 4 5
→Given message "NETWO RKING".
→The message is divided into two equal messages of 5 characters length.
→According to cipher text, The reverse of 5 characters of message is OWTEN GNIKR
→According to given question ,
→Plaintext : 5 4 3 2 1
→Ciphertext 1 2 3 4 5
→Given message "NETWO RKING".
→The message is divided into two equal messages of 5 characters length.
→According to cipher text, The reverse of 5 characters of message is OWTEN GNIKR
Question 16 |
Which of the following is NOT a type of firewall for network security?
Circuit level gateways | |
Application level gateways | |
Packet filters | |
Digital Immune System |
Question 16 Explanation:
A circuit-level gateway is a type of firewall. Circuit-level gateways work at the session layer of the OSI model, or as a "shim-layer" between the application layer and the transport layer of the TCP/IP stack. They monitor TCP handshaking between packets to determine whether a requested session is legitimate.
Application level gateways is also a type of firewall. Packet filtering is a firewall technique used to control network access by monitoring outgoing and incoming packets and allowing them to pass or halt based on the source and destination Internet Protocol (IP) addresses, protocols and ports.
The Digital Immune system is a comprehensive approach to virus protection, and is not a type of firewall.
Application level gateways is also a type of firewall. Packet filtering is a firewall technique used to control network access by monitoring outgoing and incoming packets and allowing them to pass or halt based on the source and destination Internet Protocol (IP) addresses, protocols and ports.
The Digital Immune system is a comprehensive approach to virus protection, and is not a type of firewall.
Question 17 |
Consider the sequence of steps involved in the process of using digital signatures for a message in PGP; which of the following steps is WRONGLY presented?
SHA-1 is used to generate hash code of message | |
The hash code is prepended to the message and sent to the receiver | |
The receiver uses RSA with senders public key to decrypt and recovered code | |
Receiver generates a new hash code for the message and compares with the recovered code and accept the message as authentic, if only, they match |
Question 17 Explanation:
Point 2 is wrongly presented.
Point 2 should be the hash code is first encrypted and then preoended to the message and sent to the receiver.
Question 18 |
Which of the following is a secret – key encryption algorithm?
RSA | |
Diffie-Hellman key exchange | |
Advanced Encryption Standard (AES) | |
Elliptic Curve Cryptography (ECC) |
Question 18 Explanation:
RSA is a public key encryption algorithm.
Diffie-Hellman key exchange algorithm is not at all a encryption algorithm ,instead it is a key exchange algorithm.
AES is a secret key encryption algorithm.
Diffie-Hellman key exchange algorithm is not at all a encryption algorithm ,instead it is a key exchange algorithm.
AES is a secret key encryption algorithm.
Question 19 |
Which of the following operations are generally used for transforming plain text to cipher text?
substitution | |
transposition | |
substitution and transposition | |
normalization |
Question 20 |
With reference to network security across a packet switching network, which of the following provide the most effective solution?
End-to-end encryption | |
Link encryption | |
Combination of both link and end-to-end encryption | |
Either link encryption or end-to-end encryption but not both |
Question 20 Explanation:
The implementation of encryption in packet-switched networks must ensure that essential addressing information can be accessed by the relevant network devices such as switches, bridges and routers. Encryption is broadly termed link layer encryption or end-to-end encryption depending on whether it is applied and re-applied at each end of each link in a communication path, or whether it is applied over the whole path between end systems.
Question 21 |
The private key in public key encryption is used for
encryption | |
hashing
| |
decryption | |
decryption and hashing
|
Question 21 Explanation:
The private key in public key encryption is used for hashing.
Question 22 |
Public key encryption makes use of
one key | |
two keys | |
hash function
| |
All the given options |
Question 22 Explanation:
Public key encryption makes use of two keys, public key and private key.
Question 23 |
A firewall is
an established network performance reference point
| |
software or hardware used to isolate a private network from a public network
| |
a virus that infects macros | |
a predefined encryption key used to encrypt and decrypt data transmissions |
Question 23 Explanation:
A firewall is a software used to isolate a private network from a public network.
Question 24 |
A digital signature is
scanned signature
| |
signature in binary form
| |
encrypting information | |
handwritten signature
|
Question 24 Explanation:
Digital Signature is a process that guarantees that the contents of a message have not been altered in transit. When you, the server, digitally sign a document, you add a one-way hash (encryption) of the message content using your public and private key pair.
Question 25 |
In____authentication, the claimant proves that she knows the secret without sending it to the verifier.
Asymmetric | |
Zero knowledge
| |
Symmetric | |
Challenge response
|
Question 25 Explanation:
In password authentication ,the claimant proves her identity by demonstrating that she knows the secret, the password.However, because the claimant reveals this secret , it is susceptible to interception by the adversary.In challenge response authentication , the claimant proves that she knows the secret without sending it. In other words the claimant does not send the secret to the verifier, the verifier either has it or finds it.
In challenge response authentication, the claimant proves that she knows the secret without sending it to the verifier.
In challenge response authentication, the claimant proves that she knows the secret without sending it to the verifier.
Question 26 |
In RSA algorithm if p=7, q=11 and e=13 then what will be the value of d?
23 | |
40 | |
37 | |
13 |
Question 26 Explanation:
We know that in RSA algorithm,
e * d ≡ 1 mod φ(n)
Lets first find φ(n),
φ(n) = (p-1)(q-1)
= (7-1)(11-1)
= 60
Now,
e * d ≡ 1 mod φ(n)
13 * d ≡ 1 mod 60
For d = 37 the above equation satisfies.
e * d ≡ 1 mod φ(n)
Lets first find φ(n),
φ(n) = (p-1)(q-1)
= (7-1)(11-1)
= 60
Now,
e * d ≡ 1 mod φ(n)
13 * d ≡ 1 mod 60
For d = 37 the above equation satisfies.
There are 26 questions to complete.