30 + Interview Questions in Public Key Cryptography And Rsa in Cryptograph & Network Security Page 1 InterviewSolution

1.	In Merkle-Hellman Cryptosystem, the public key can be used to decrypt messages, but cannot be used to decrypt messages. The private key encrypts the messages.(a) True(b) FalseI have been asked this question in an online quiz.Asked question is from Knapsack/ Merkle topic in division Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT ANSWER is (b) False Best explanation: The public key can be used to ENCRYPT messages, but cannot be used to decrypt messages. The private key decrypts the messages.

Discussion

2.	Consider knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, and 38}. Find the ‘n’.(a) 011111(b) 010011(c) 010111(d) 010010I had been asked this question in semester exam.This intriguing question originated from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct option is (b) 010011 Easy explanation: v0=1, V1=2, v2=4, v3=9, v4=20, v5=38 K=6, V=23 Starting from LARGEST NUMBER: v5 > V thenϵ_5=0 v4 < V then V = V – v4 = 23 – 20 = 3 ϵ_4=1 v3 > V thenϵ_3=0 v2> V thenϵ_2=0 v1 < V then V = V – v1= 3 – 2 = 1 ϵ_1=1 v0 =1 then V = V – v0= 1 – 1 = 0ϵ_0=1 n= ϵ_5 ϵ_4 ϵ_3 ϵ_2 ϵ_1 ϵ_0 = 010011.

Discussion

3.	p = 7; q = 11; M = 8(a) 19(b) 57(c) 76(d) 59This question was posed to me in exam.I would like to ask this question from Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT CHOICE is (b) 57 For explanation I would SAY: n = 77; f(n) = 60; d = 53; C = 57.

Discussion

4.	For p = 11 and q = 19 and choose d=17. Apply RSA algorithm where Cipher message=80 and thus find the plain text.(a) 54(b) 43(c) 5(d) 24This question was addressed to me in an interview.This interesting question is from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» RIGHT option is (C) 5 Explanation: N = pq = 11 × 19 = 209. C=M^e mod n ;C=5^17 mod 209 ; C = 80 mod 209.

Discussion

5.	In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.(a) True(b) FalseThe question was posed to me in class test.Question is taken from Knapsack/ Merkle in division Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» CORRECT answer is (b) False Best explanation: The hard knapsack becomes the public KEY and the EASY knapsack becomes the private key.

Discussion

6.	USENET falls under which category of public key sharing?(a) Public announcement(b) Publicly available directory(c) Public-key authority(d) Public-key certificatesThis question was addressed to me by my college director while I was bunking the class.I want to ask this question from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The correct ANSWER is (a) Public announcement Best explanation: MANY users have adopted the practice of appending their public key to MESSAGES that they send to public forums, such as USENET newsgroups and Internet mailing lists.

Discussion

7.	RSA is also a stream cipher like Merkel-Hellman.(a) True(b) FalseThe question was asked in examination.I want to ask this question from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The correct ANSWER is (a) True For EXPLANATION: RSA is a block CIPHER system.

Discussion

8.	Set {1, 2, 3, 9, 10, and 24} is superincreasing.(a) True(b) FalseI had been asked this question in semester exam.My enquiry is from Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» RIGHT ANSWER is (B) False Explanation: It is not because 10 < 1+2+3+9.

Discussion

9.	Another name for Merkle-Hellman Cryptosystem is(a) RC4(b) Knapsack(c) Rijndael(d) Diffie-HellmanI have been asked this question during an interview.My question is taken from Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT option is (B) Knapsack The BEST explanation: Knapsack is another name for Merkel-Hellman Cryptosystem.

Discussion

10.	Which Cryptographic system uses C1 = (e1^r) mod p and C1 = (e2^r x P) mod p at the encryption side?(a) Elgamal(b) RSA(c) Rabin(d) WhirlpoolThis question was posed to me in an international level competition.I would like to ask this question from Rabin/ Elgamal Algorithm in division Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Right option is (a) Elgamal To EXPLAIN I would say: TheElgamal cryptographic system USES the above FORMULAE to COMPUTE the CT.

Discussion

11.	For p = 11 and q = 19 and choose e=17. Apply RSA algorithm where message=5 and find the cipher text.(a) C=80(b) C=92(c) C=56(d) C=23The question was posed to me in examination.Origin of the question is Knapsack/ Merkle topic in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» RIGHT choice is (a) C=80 The EXPLANATION: N = pq = 11 × 19 = 209.

Discussion

12.	In an RSA system the public key of a given user is e = 31, n = 3599. Whatis the private key of this user?(a) 3031(b) 2412(c) 2432(d) 1023This question was posed to me in a national level competition.Question is from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct option is (a) 3031 To explain: By trail and error, we DETERMINE that p = 59 and Q = 61. HENCE f(n) = 58 x 60 = 3480. Then, using the extended Euclidean algorithm, we find that the multiplicative inverse of 31 modulo f(n) is 3031.

Discussion

13.	Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153).(a) 35212(b) 12543(c) 19367(d) 32432I had been asked this question in final exam.Origin of the question is Knapsack/ Merkle in division Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Right answer is (c) 19367 The explanation: Since N=233 ´ 241=56,153, P=233 and q=241 f(n) = (p – 1)(q – 1) = 55,680 Using Extended EUCLIDEAN algorithm, we obtain d = 23–1 MOD 55680 = 19,367.

Discussion

14.	p = 3; q = 11; M = 5(a) 28(b) 26(c) 18(d) 12I have been asked this question in an online interview.My enquiry is from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct option is (B) 26 To elaborate: n = 33; f(n) = 20; d = 3; C = 26.

Discussion

15.	In RSA, Ф(n) = _______ in terms of p and q.(a) (p)/(q)(b) (p)(q)(c) (p-1)(q-1)(d) (p+1)(q+1)I have been asked this question during an interview.Origin of the question is Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT OPTION is (c) (p-1)(q-1) Easiest EXPLANATION: Ф(N) = (p-1)(q-1).

Discussion

16.	Imagine you had a set of weights {62, 93, 26, 52, 166, 48, 91, and 141}. Find subset that sums to V = 302.(a) {62, 48, 166, 52}(b) {141, 26, 52, 48}(c) {93, 26, 91, 48}(d) {62, 26, 166, 48}I have been asked this question during a job interview.I want to ask this question from Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» RIGHT CHOICE is (d) {62, 26, 166, 48} To explain: {62, 26, 166, 48} =302.

Discussion

17.	For the Knapsack: {1 6 8 15 24}, find the plain text code if the ciphertext is 38.(a) 10010(b) 01101(c) 01001(d) 01110The question was posed to me in an internship interview.My question is based upon Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Right CHOICE is (b) 01101 The EXPLANATION is: If someone sends you the code 38 this can only have come from the PLAIN TEXT 01101.

Discussion

18.	For the Knapsack: {1 6 8 15 24}, Find the cipher text value for the plain text 10011.(a) 40(b) 22(c) 31(d) 47I had been asked this question during an interview for a job.This key question is from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» CORRECT CHOICE is (a) 40 Easiest EXPLANATION: 1+15+24 = 40.

Discussion

19.	In the RSA algorithm, we select 2 random large values ‘p’ and ‘q’. Which of the following is the property of ‘p’ and ‘q’?(a) p and q should be divisible by Ф(n)(b) p and q should be co-prime(c) p and q should be prime(d) p/q should give no remainderThe question was asked in an online interview.The above asked question is from Knapsack/ Merkle topic in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT choice is (C) p and q should be prime To explain I WOULD SAY: ‘p’ and ‘q’ should have large random values which are both prime numbers.

Discussion

20.	In RSA, we select a value ‘e’ such that it lies between 0 and Ф(n) and it is relatively prime to Ф(n).(a) True(b) FalseI had been asked this question in examination.My question is from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» CORRECT answer is (B) False To explain I WOULD say: gcd(E, Ф(n))=1;and1 < e < Ф(n).

Discussion

21.	For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.(a) 23(b) 64(c) 11(d) 54The question was posed to me in an interview for internship.The question is from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct choice is (C) 11 Easiest EXPLANATION: N = pq = 11 × 19 = 187. C=M^e mod n ;C=88^7 mod 187 ; C = 11 mod 187.

Discussion

22.	“Rabin Cryptosystem is a variant of the Elgamal Cryptosystem”(a) True(b) FalseI had been asked this question in an internship interview.The above asked question is from Rabin/ Elgamal Algorithm in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct ANSWER is (b) False Easy explanation: RABIN CRYPTOSYSTEM is a VARIANT of the RSA Cryptosystem.

Discussion

23.	p = 5; q = 11; M = 9(a) 43(b) 14(c) 26(d) 37This question was posed to me in exam.My question is taken from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT option is (B) 14 Explanation: n = 55; F(n) = 40; d = 27; C = 14.

Discussion

24.	Find the ciphertext for the message {100110101011011} using superincreasing sequence { 1, 3, 5, 11, 35 } and private keys a = 5 and m=37.(a) C = ( 33, 47, 65 )(b) C = ( 65, 33, 47 )(c) C = ( 47, 33, 65 )(d) C = ( 47, 65, 33 )This question was posed to me in an interview for job.My doubt stems from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Right option is (c) C = ( 47, 33, 65 ) To explain: {vi} = { 1, 3, 5, 11, 35 } a = 5and m = 37 Public key GENERATION: {wi} = avi MOD m wi = {5, 15, 25, 18, 27} Break the message into k-bit tuple i.e. 5-bit tuple 1001101010 11011 Encoding of M as follows: MCi 1001147 0101033 1101165 Ciphertext SENT will be: C = (47, 33, and 65).

Discussion

25.	For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.(a) 88(b) 122(c) 143(d) 111I had been asked this question in examination.The doubt is from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct OPTION is (a) 88 The explanation is: n = pq = 11 × 19 = 187. C=M^e MOD n ;C=11^23 mod 187 ; C = 88 mod 187.

Discussion

26.	n = 35; e = 5; C = 10. What is the plaintext (use RSA) ?(a) 3(b) 7(c) 8(d) 5I have been asked this question in semester exam.This question is from Knapsack/ Merkle topic in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Right choice is (d) 5 For explanation: Use RSA SYSTEM to decrypt and get PT = 5.

Discussion

27.	p = 17; q = 31; M = 2(a) 254(b) 423(c) 128(d) 523This question was posed to me in exam.This interesting question is from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» Correct option is (c) 128 Explanation: N = 527; F(n) = 480; d = 343; C = 128.

Discussion

28.	A superincreasing knapsack problem is ____ to solve than a jumbled knapsack.(a) Easier(b) Tougher(c) Shorter(d) LengthierI had been asked this question in an interview for internship.This question is from Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The correct OPTION is (a) Easier Easy EXPLANATION: A superincreasing knapsack is chosen to make computations easier while manual CALCULATIONS of knapsack problems.

Discussion

29.	Sender chooses p = 107, e1 = 2, d = 67, and the random integer is r=45. Find the plaintext to be transmitted if the ciphertext is (28,9).(a) 45(b) 76(c) 66(d) 13The question was posed to me in semester exam.My query is from Rabin/ Elgamal Algorithm topic in portion Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» The CORRECT OPTION is (c) 66 The best explanation: P = [C2 (C1d)-1] MOD p = 66.

Discussion

30.	p = 11; q = 13; M = 7(a) 84(b) 124(c) 106(d) 76The question was posed to me in semester exam.The above asked question is from Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security
Answer» CORRECT ANSWER is (c) 106 Best explanation: N = 143; F(n) = 120; d = 11; C = 106.

Discussion

Explore topic-wise InterviewSolutions in .

p = 7; q = 11; M = 8(a) 19(b) 57(c) 76(d) 59This question was posed to me in exam.I would like to ask this question from Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security

In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.(a) True(b) FalseThe question was posed to me in class test.Question is taken from Knapsack/ Merkle in division Public Key Cryptography and RSA of Cryptograph & Network Security

RSA is also a stream cipher like Merkel-Hellman.(a) True(b) FalseThe question was asked in examination.I want to ask this question from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security

Set {1, 2, 3, 9, 10, and 24} is superincreasing.(a) True(b) FalseI had been asked this question in semester exam.My enquiry is from Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security

Another name for Merkle-Hellman Cryptosystem is(a) RC4(b) Knapsack(c) Rijndael(d) Diffie-HellmanI have been asked this question during an interview.My question is taken from Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security

Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153).(a) 35212(b) 12543(c) 19367(d) 32432I had been asked this question in final exam.Origin of the question is Knapsack/ Merkle in division Public Key Cryptography and RSA of Cryptograph & Network Security

p = 3; q = 11; M = 5(a) 28(b) 26(c) 18(d) 12I have been asked this question in an online interview.My enquiry is from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security

In RSA, Ф(n) = _______ in terms of p and q.(a) (p)/(q)(b) (p)(q)(c) (p-1)(q-1)(d) (p+1)(q+1)I have been asked this question during an interview.Origin of the question is Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security

For the Knapsack: {1 6 8 15 24}, find the plain text code if the ciphertext is 38.(a) 10010(b) 01101(c) 01001(d) 01110The question was posed to me in an internship interview.My question is based upon Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security

For the Knapsack: {1 6 8 15 24}, Find the cipher text value for the plain text 10011.(a) 40(b) 22(c) 31(d) 47I had been asked this question during an interview for a job.This key question is from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security

In RSA, we select a value ‘e’ such that it lies between 0 and Ф(n) and it is relatively prime to Ф(n).(a) True(b) FalseI had been asked this question in examination.My question is from Knapsack/ Merkle topic in section Public Key Cryptography and RSA of Cryptograph & Network Security

For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.(a) 23(b) 64(c) 11(d) 54The question was posed to me in an interview for internship.The question is from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security

“Rabin Cryptosystem is a variant of the Elgamal Cryptosystem”(a) True(b) FalseI had been asked this question in an internship interview.The above asked question is from Rabin/ Elgamal Algorithm in chapter Public Key Cryptography and RSA of Cryptograph & Network Security

p = 5; q = 11; M = 9(a) 43(b) 14(c) 26(d) 37This question was posed to me in exam.My question is taken from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security

For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.(a) 88(b) 122(c) 143(d) 111I had been asked this question in examination.The doubt is from Knapsack/ Merkle in portion Public Key Cryptography and RSA of Cryptograph & Network Security

n = 35; e = 5; C = 10. What is the plaintext (use RSA) ?(a) 3(b) 7(c) 8(d) 5I have been asked this question in semester exam.This question is from Knapsack/ Merkle topic in portion Public Key Cryptography and RSA of Cryptograph & Network Security

p = 17; q = 31; M = 2(a) 254(b) 423(c) 128(d) 523This question was posed to me in exam.This interesting question is from Knapsack/ Merkle in chapter Public Key Cryptography and RSA of Cryptograph & Network Security

A superincreasing knapsack problem is ____ to solve than a jumbled knapsack.(a) Easier(b) Tougher(c) Shorter(d) LengthierI had been asked this question in an interview for internship.This question is from Knapsack/ Merkle in section Public Key Cryptography and RSA of Cryptograph & Network Security

p = 11; q = 13; M = 7(a) 84(b) 124(c) 106(d) 76The question was posed to me in semester exam.The above asked question is from Knapsack/ Merkle topic in chapter Public Key Cryptography and RSA of Cryptograph & Network Security