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Advanced Cryptography Problem Set with Solutions, Exams of Cryptography and System Security

Queen Mary, University of London (QMUL)Cryptography and System Security

A comprehensive problem set on advanced cryptography, focusing on mathematical concepts and their applications. It covers modular exponentiation, rsa key generation, multiplicative inverses, and diffie-hellman key exchange. Each question includes a detailed solution, demonstrating the mathematical reasoning behind the answer. The problem set is designed to challenge and enhance understanding of core cryptographic principles, making it a valuable resource for students and professionals in the field. It provides practical exercises and step-by-step solutions to reinforce learning and problem-solving skills in cryptography. Useful for university students.

Typology: Exams

2024/2025

Available from 05/30/2025

Emma_Johnson
Emma_Johnson 🇬🇧

2.1K documents

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Advanced Cryptography Problem Set
Questions with Solutions
This comprehensive cryptography problem set contains challenging, math-
based multiple-choice questions focusing on core concepts such as modular
exponentiation, RSA key generation, multiplicative inverses, and Diffie-
Hellman key exchange. Each question is accompanied by a detailed
solution, demonstrating the mathematical reasoning and methods (like
repeated squaring, Fermat’s Little Theorem, and the Extended Euclidean
Algorithm) used to arrive at the correct answer. What is 7^13 mod 55?
✅ A. 47
B. 17
C. 28
D. 4
Solution:
Convert 13 to binary: 1101 = 2³ + 2² + 2⁰
Compute powers:
- 7² = 49
- 7⁴ = 49² = 2401 → 2401 mod 55 = 36
- 7 = 36² = 1296 → 1296 mod 55 = 31⁸
Now, 7 × 7⁴ × 7 = 31 × 36 × 7 = 7836 → 7836 mod 55 = 47⁸
Answer: ✅ A. 47
1. 2. Find the inverse of 11 modulo 26.
✅ A. 19
B. 17
C. 7
D. 21
Solution:
pf3
pf4
pf5

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Advanced Cryptography Problem Set

Questions with Solutions

This comprehensive cryptography problem set contains challenging, math- based multiple-choice questions focusing on core concepts such as modular exponentiation, RSA key generation, multiplicative inverses, and Diffie- Hellman key exchange. Each question is accompanied by a detailed solution, demonstrating the mathematical reasoning and methods (like repeated squaring, Fermat’s Little Theorem, and the Extended Euclidean Algorithm) used to arrive at the correct answer. What is 7^13 mod 55?  ✅ A. 47  B. 17  C. 28  D. 4 Solution: Convert 13 to binary: 1101 = 2³ + 2² + 2⁰ Compute powers:

  • 7² = 49
  • 7⁴ = 49² = 2401 → 2401 mod 55 = 36
  • 7 ⁸= 36² = 1296 → 1296 mod 55 = 31 Now, 7 ⁸× 7⁴ × 7 = 31 × 36 × 7 = 7836 → 7836 mod 55 = 47 Answer: ✅ A. 47
    1. Find the inverse of 11 modulo 26.  ✅ A. 19  B. 17  C. 7  D. 21 Solution:

Use the Extended Euclidean Algorithm: 26 = 211 + 4 11 = 24 + 3 4 = 13 + 1 Back-substitute: 1 = 4 - 13 = 211 - 1 So, inverse of 11 mod 26 is 19. Answer: ✅ A. 19

    1. What is the RSA encryption of message m = 7 using e = 3 and n = 33?  A. 10  ✅ B. 13  C. 28  D. 31 Solution: RSA Encryption: c = m^e mod n = 7^3 mod 33 = 343 mod 33 343 ÷ 33 = 10 with remainder 13 → Answer: ✅ B. 13
    1. If p = 23 and g = 5, what is the public key A given private key a = 6 in Diffie-Hellman?  A. 8  B. 20  C. 2  ✅ D. 21 Solution: A = g^a mod p = 5^6 mod 23 5^2 = 25 → 2, 5^4 = 2² = 4, 5^6 = 4 * 2 = 8 → 8 mod 23 = 8 Wait, this contradicts. Actually: 5² = 25 mod 23 = 2 5⁴ = 2² = 4, then 5 ⁶= 4*2 = 8

 D. 14

Solution: Use exponentiation by squaring: 14² = 196 mod 33 = 31 14⁴ = 31² = 961 mod 33 = 4 14 ⁸= 4² = 16 Combine: 14 14²14 = 163114 = 6944 mod 33 = 7⁸ Answer: ✅ B. 7

    1. Given RSA primes p=7, q=11, compute n and phi(n).  φA. n=77, ✅=  B. n=72, φ=  C. n=77, φ=  D. n=66, φ= Solution: n = pq = 711 = 77 φ (n) = (p-1)(q-1) = 610 = 60 Answer: A. n=77,φ ✅=
    1. If ciphertext c = 10, e = 3, n = 33, find m (RSA decryption) if d = 7.  A. 7  B. 10  ✅ C. 4  D. 16 Solution: m = c^d mod n = 10^7 mod 33 10² = 100 → 1, 10⁴ = 1² = 1, 10 ⁷= 1*10³ = 1000 → 1000 mod 33 = 4 Answer: ✅ C. 4
    1. Compute 9^23 mod 77.

Solution:

  • → Correct is A.
  • Oops! Correction made. ✅ A.
  •  A.
  •  B.
  •  ✅ C.
  •  D.
  • Shared key = B^a mod p = 9^4 mod Solution:
  • 9² = 81 →
  • 12² = 144 → 144 mod 23 =
  • Answer: ✅ A.
  •  A. 5. 6. What is the modular inverse of 17 mod 43?
  •  ✅ B.
  •  C.
  •  D.
  • 43 = 2*17 + Use Extended Euclidean Algorithm:
  • 17 = 1*9 +
  • 9 = 1*8 +
  • Back-substitute gives inverse of 17 mod 43 =
  • Answer: ✅ A.
  •  A. 6. 7. What is 14^11 mod 33?
  •  ✅ B.
  •  C.
  •  A.
  •  B.
  •  ✅ C.
  •  D.
  • 9² = 81 → Use repeated squaring:
  • 9⁴ = 16, 9 ⁸ = 256 → 25, 9¹ ⁶= 625 → 625 mod 77 =
  • Now combine: 9¹ ⁶* 9⁴ * 9² * 9 = 9164*9 = 5184 mod 77 =
  • Answer: ✅ C.