Math Problem Statement

For the given RSA algorithm. Given p = 3, q = 11 in RSA algorithm, e = 3 and the ciphertext C = 27, find d and the plaintext M.

Solution

The text in the image reads:

"5. (a) For the given RSA algorithm. Given p=3p = 3, q=11q = 11 in RSA algorithm, e=3e = 3 and the ciphertext C=27C = 27, find dd and the plaintext MM."

To solve this, let's break it down step-by-step.

Step 1: Understanding RSA Parameters

In the RSA algorithm:

  1. pp and qq are prime numbers used to compute n=p×qn = p \times q.
  2. ϕ(n)=(p1)(q1)\phi(n) = (p-1)(q-1) is the Euler's totient function.
  3. The public key consists of (e,n)(e, n), where ee is the public exponent.
  4. The private key consists of (d,n)(d, n), where dd is the private exponent.
  5. dd is calculated as the modular multiplicative inverse of ee modulo ϕ(n)\phi(n), meaning dd satisfies the equation e×d1(modϕ(n))e \times d \equiv 1 \pmod{\phi(n)}.

Step 2: Calculate nn and ϕ(n)\phi(n)

Given:

  • p=3p = 3
  • q=11q = 11

We compute: n=p×q=3×11=33n = p \times q = 3 \times 11 = 33 ϕ(n)=(p1)(q1)=(31)(111)=2×10=20\phi(n) = (p-1)(q-1) = (3-1)(11-1) = 2 \times 10 = 20

Step 3: Find dd

We need to find dd such that: e×d1(modϕ(n))e \times d \equiv 1 \pmod{\phi(n)} Given e=3e = 3, we want: 3d1(mod20)3d \equiv 1 \pmod{20}

We need the multiplicative inverse of 3 modulo 20. The value of dd that satisfies this can be found using the Extended Euclidean Algorithm.

Step 4: Calculate the Multiplicative Inverse of ee

Using the Extended Euclidean Algorithm, we find that: d=7(since 3×7=211(mod20))d = 7 \quad \text{(since } 3 \times 7 = 21 \equiv 1 \pmod{20}\text{)}

Step 5: Decrypt the Ciphertext

Now, to find the plaintext MM: M=Cd(modn)M = C^d \pmod{n} Given C=27C = 27, d=7d = 7, and n=33n = 33, we calculate: M=277(mod33)M = 27^7 \pmod{33}

This is a large calculation, so let's simplify by computing 277mod3327^7 \mod 33.

Step 6: Simplify Using Modular Exponentiation

Let's compute this step-by-step: 276(mod33)27 \equiv -6 \pmod{33} 272(6)2=363(mod33)27^2 \equiv (-6)^2 = 36 \equiv 3 \pmod{33} 27432=9(mod33)27^4 \equiv 3^2 = 9 \pmod{33} 277(27427227)(mod33)27^7 \equiv (27^4 \cdot 27^2 \cdot 27) \pmod{33} (9327)(mod33)\equiv (9 \cdot 3 \cdot 27) \pmod{33} (27)(mod33)\equiv (27) \pmod{33} M=6M = 6

Conclusion

  • The private key d=7d = 7
  • The plaintext M=6M = 6

Would you like more details on any step, or do you have any further questions?

Related Questions

  1. How do you compute the modular multiplicative inverse using the Extended Euclidean Algorithm?
  2. What is the significance of the public and private keys in RSA encryption?
  3. Can you provide more examples of decrypting ciphertext using the RSA algorithm?
  4. How does RSA encryption ensure security in communication?
  5. What are the practical applications of RSA in today's technology?

Tip

To perform modular exponentiation efficiently, use the "square and multiply" method.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Cryptography
Modular Arithmetic

Formulas

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

Theorems

Euler's Totient Function
Modular Multiplicative Inverse

Suitable Grade Level

Undergraduate (Cryptography/Computer Science)