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"5. (a) 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 \( M \)."

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

### Step 1: Understanding RSA Parameters

In the RSA algorithm:
1. \( p \) and \( q \) are prime numbers used to compute \( n = p \times q \).
2. \( \phi(n) = (p-1)(q-1) \) is the Euler's totient function.
3. The public key consists of \( (e, n) \), where \( e \) is the public exponent.
4. The private key consists of \( (d, n) \), where \( d \) is the private exponent.
5. \( d \) is calculated as the modular multiplicative inverse of \( e \) modulo \( \phi(n) \), meaning \( d \) satisfies the equation \( e \times d \equiv 1 \pmod{\phi(n)} \).

### Step 2: Calculate \( n \) and \( \phi(n) \)

Given:
- \( p = 3 \)
- \( q = 11 \)

We compute:
\[
n = p \times q = 3 \times 11 = 33
\]
\[
\phi(n) = (p-1)(q-1) = (3-1)(11-1) = 2 \times 10 = 20
\]

### Step 3: Find \( d \)

We need to find \( d \) such that:
\[
e \times d \equiv 1 \pmod{\phi(n)}
\]
Given \( e = 3 \), we want:
\[
3d \equiv 1 \pmod{20}
\]

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

### Step 4: Calculate the Multiplicative Inverse of \( e \)

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

### Step 5: Decrypt the Ciphertext

Now, to find the plaintext \( M \):
\[
M = C^d \pmod{n}
\]
Given \( C = 27 \), \( d = 7 \), and \( n = 33 \), we calculate:
\[
M = 27^7 \pmod{33}
\]

This is a large calculation, so let's simplify by computing \( 27^7 \mod 33 \).

### Step 6: Simplify Using Modular Exponentiation

Let's compute this step-by-step:
\[
27 \equiv -6 \pmod{33}
\]
\[
27^2 \equiv (-6)^2 = 36 \equiv 3 \pmod{33}
\]
\[
27^4 \equiv 3^2 = 9 \pmod{33}
\]
\[
27^7 \equiv (27^4 \cdot 27^2 \cdot 27) \pmod{33}
\]
\[
\equiv (9 \cdot 3 \cdot 27) \pmod{33}
\]
\[
\equiv (27) \pmod{33}
\]
\[
M = 6
\]

### Conclusion

- The private key \( d = 7 \)
- The plaintext \( M = 6 \)

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.

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.

Learn how to solve an RSA algorithm problem by finding the private key d and decrypting the ciphertext C = 27. This detailed solution includes key RSA formulas such as n = p × q, φ(n), and the modular inverse calculation. Suitable for undergraduate cryptography students.

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