Math Problem Statement

In a Public key system using RSA, the ciphertext C = 20 is sent to a user whose public key is e = 13, n = 77. What is the plaintext M?

Solution

The problem provided relates to the RSA cryptosystem. Let's break it down step by step to determine the plaintext MM:

Problem Breakdown

  1. RSA Encryption Formula: C=MemodnC = M^e \mod n where:

    • CC is the ciphertext (given as 20),
    • MM is the plaintext to be found,
    • ee is the public key exponent (given as 13),
    • nn is the modulus (given as 77).
  2. Goal: To find MM, we need to compute the modular inverse operation: M=CdmodnM = C^d \mod n where dd is the private key, calculated as: d=e1modϕ(n)d = e^{-1} \mod \phi(n) Here, ϕ(n)\phi(n) (Euler's totient function) is calculated as: ϕ(n)=(p1)(q1)\phi(n) = (p - 1)(q - 1) where n=pqn = p \cdot q, and pp, qq are the prime factors of nn.


Step-by-Step Solution

Step 1: Factorize n=77n = 77

The prime factors of n=77n = 77 are p=7p = 7 and q=11q = 11.

Step 2: Compute ϕ(n)\phi(n)

ϕ(n)=(p1)(q1)=(71)(111)=610=60\phi(n) = (p - 1)(q - 1) = (7 - 1)(11 - 1) = 6 \cdot 10 = 60

Step 3: Find dd (the private key)

We need to compute dd such that: de1modϕ(n)d \cdot e \equiv 1 \mod \phi(n) or equivalently: d131mod60d \cdot 13 \equiv 1 \mod 60

Using the Extended Euclidean Algorithm, we find: d=37d = 37

Step 4: Decrypt to find MM

The plaintext MM is given by: M=CdmodnM = C^d \mod n Substituting C=20C = 20, d=37d = 37, and n=77n = 77, we calculate: M=2037mod77M = 20^{37} \mod 77

Simplify with Modular Arithmetic

To compute 2037mod7720^{37} \mod 77, we use repeated squaring:

  1. Compute powers of 20 modulo 77:

    • 202mod77=400mod77=1520^2 \mod 77 = 400 \mod 77 = 15,
    • 204mod77=152mod77=225mod77=7120^4 \mod 77 = 15^2 \mod 77 = 225 \mod 77 = 71,
    • 208mod77=712mod77=5041mod77=2220^8 \mod 77 = 71^2 \mod 77 = 5041 \mod 77 = 22,
    • 2016mod77=222mod77=484mod77=2220^{16} \mod 77 = 22^2 \mod 77 = 484 \mod 77 = 22.
  2. Use the exponents to calculate 2037mod7720^{37} \mod 77: 2037=203220420120^{37} = 20^{32} \cdot 20^4 \cdot 20^1 Substituting:

    • 2032mod77=(2016)2mod77=222mod77=2220^{32} \mod 77 = (20^{16})^2 \mod 77 = 22^2 \mod 77 = 22,
    • 204mod77=7120^4 \mod 77 = 71,
    • 201mod77=2020^1 \mod 77 = 20.

    Combine: 2037mod77=(227120)mod77=15620mod77=4420^{37} \mod 77 = (22 \cdot 71 \cdot 20) \mod 77 = 15620 \mod 77 = 44

Thus, the plaintext M=44M = 44.


Final Answer:

The plaintext MM is: 44\boxed{44}

Would you like further clarification or a deeper explanation of any step?


Relative Questions:

  1. How does the Extended Euclidean Algorithm find the modular inverse?
  2. Why is modular arithmetic crucial in cryptographic algorithms like RSA?
  3. Can nn be any number, or must it have specific properties in RSA encryption?
  4. What happens if the primes pp and qq are too small?
  5. How is the security of RSA affected by advances in computational power?

Tip:

When performing modular exponentiation, always use the method of repeated squaring to simplify calculations efficiently.

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Math Problem Analysis

Mathematical Concepts

RSA Cryptography
Modular Arithmetic
Euler's Totient Function
Extended Euclidean Algorithm
Prime Factorization

Formulas

C = M^e mod n
M = C^d mod n
d = e^(-1) mod φ(n)
φ(n) = (p-1)(q-1)

Theorems

RSA Encryption and Decryption

Suitable Grade Level

Undergraduate or Advanced High School