## Network-Security

Question 1 |

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 |

__public key__and receiver decrypts it using his own

__private key__.

Question 2 |

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 |

(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 3 |

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 |

ii) It uses the P-Box permutation.

Statement-I is false, II is true.

Question 4 |

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 |

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 5 |

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 |

^{n}C

_{2}= 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 6 |

The value of 3^{51} mod 5 is ______.

3 | |

5 | |

2 | |

1 |

^{51}mod 5

⇒ 3

^{1}= 3 ⇒ 3 mod 5 = 3

3

^{2}⇒ 9 mod 5 = 4

3

^{3}⇒ 27 mod 5 = 2

3

^{4}⇒ 81 mod 5 = 1

3

^{5}⇒ 243 mod 5 = 3

For every four numbers sequence is repeating.

So, (51 % 4) = 3

⇒ 3

^{3}= 27

⇒ 27 mod 5 = 2

Question 7 |

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 |

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,

x

^{2}− (p+q)x + pq = (x-p)(x-q)

Substituting for p+q and pq in the above equation

x

^{2}- (n+1-ϕ(n))x + n

Question 8 |

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’= M^{e}mod n M = (M’)^{d}mod n II. ed ≡ 1 mod n III. ed ≡ 1 mod f(n) IV. M’= M^{e}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 |

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 K

_{E}=(n,e) and the decryption private key is K

_{D}=(n,d).

The encryption function is

E(M)=M

^{e}mod n

The decryption function is

D(M)=M

^{d}mod n

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 |

(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 |

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 |

Symmetric key cryptography is computationally infeasible. | |

Symmetric key cryptography involves key distribution. | |

Symmetric key cryptography is unreliable. | |

Digital signature requires a pair of private – public keys. |

Question 12 |

+ 3 | |

+ 2 | |

– 3 | |

– 2 |

O-2 = M

N - 2 = L

G-2 = E

R-2 = P

A-2 = Y

T-2 = R

S-2 = Q

Question 13 |

(a) Masquerading (i) Session is intercepted (b) Phishing (ii) One pretends to be someone else (c) Hijacking (iii) A email misleads a user into entering confidential information Codes: (a) (b) (c)

(i) (ii) (iii)
| |

(i) (iii) (ii)
| |

(iii) (ii) (i) | |

(ii) (iii) (i) |

Phishing-Phishing is a cybercrime in which a target or targets are contacted by email, telephone or text message by someone posing as a legitimate institution to lure individuals into providing sensitive data such as personally identifiable information, banking and credit card details, and passwords.

Hijacking-Hijacking is a type of network security attack in which the attacker takes control of a communication - just as an airplane hijacker takes control of a flight - between two entities and masquerades as one of them. So interception between two entities is done.

Question 14 |

packet filter | |

deep inspection firewall | |

reverse proxy | |

stateful packet inspection firewall |

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 |

→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 |

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 |

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 17 |

Blowfish | |

IDEA | |

AES | |

All the above |

Question 18 |

Confidentiality and Integrity | |

Authentication
| |

Non-repudiation
| |

All the above |

Integrity-No one can change the information. Authentication-It guarantees the identity of the sender. Non-repudiation-

**Nonrepudiation**refers to the ability to ensure that a party to a contract or a communication cannot deny the authenticity of their signature on a document or the sending of a message that they originated. To

**repudiate**means to deny.

Hence all are the security requirement.

Question 19 |

John Richradson, John Smith and Len Adleman | |

Ron Rivest, John Smith and L Hospital | |

Ron Rivest, Adi Shamir and Len Adleman | |

None of the above |

**RSA algorithm is named after**Ron Rivest, Adi Shamir and Len Adleman, who invented it in 1977. The RSA algorithm can be used for both public key encryption and digital signatures. Its security is based on the difficulty of factoring large integers.

Question 20 |

Digital Encryption Standard
| |

Digital Encryption Specification
| |

Data Encryption Standard | |

Data Encryption Specification |

Question 21 |

Data Security | |

Data Integrity | |

Data Redundancy | |

(1) and (2) |

Question 22 |

Public-key cryptography is also known as asymmetric cryptography | |

Asymmetric cryptography uses a pair of cryptographic keys
| |

A message encrypted with the private key can be decrypted only with the corresponding public key | |

The private key is kept secret, while the public key is widely distributed |

Question 23 |

CTTOEWMROPNRUEK | |

MROUEKCTTPNROEW | |

OEWPNRCTTUEKMRO | |

UEKPNRMROOEWCTT |

Question 24 |

“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” |

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 25 |

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 26 |

substitution | |

transposition | |

substitution and transposition | |

normalization |