when P is equals to 11 Q equals to 13 e equals to 11 and M is equals to 7 derive the encryption and decryption using RSA algorithm

The RSA algorithm involves key generation, encryption, and decryption. Given the values:

- \( P = 11 \)
- \( Q = 13 \)
- \( e = 11 \)
- \( M = 7 \) (the plaintext message)

We will go through each step of the RSA process: key generation, encryption, and decryption.

### 1. **Key Generation:**

#### Step 1: Calculate \( N \)
\( N = P \times Q = 11 \times 13 = 143 \)

#### Step 2: Calculate \( \phi(N) \) (Euler's Totient Function)
\( \phi(N) = (P - 1) \times (Q - 1) = (11 - 1) \times (13 - 1) = 10 \times 12 = 120 \)

#### Step 3: Public Key \( (N, e) \)
We are given \( e = 11 \). Ensure that \( e \) is coprime to \( \phi(N) \), i.e., \( \gcd(11, 120) = 1 \). Since \( e = 11 \) is coprime with \( 120 \), we can proceed.

- The public key is \( (N = 143, e = 11) \).

#### Step 4: Calculate the Private Key \( d \)
We need to find \( d \), the modular inverse of \( e \) modulo \( \phi(N) \). This means solving for \( d \) in the equation:

\[
d \times e \equiv 1 \, (\text{mod} \, \phi(N))
\]

This can be solved using the Extended Euclidean Algorithm to find \( d \) such that:

\[
11d \equiv 1 \, (\text{mod} \, 120)
\]

Using the Extended Euclidean Algorithm:

- \( 120 = 10 \times 11 + 10 \)
- \( 11 = 1 \times 10 + 1 \)
- \( 10 = 10 \times 1 + 0 \)

Reversing the steps:
- \( 1 = 11 - 1 \times 10 \)
- \( 1 = 11 - 1 \times (120 - 10 \times 11) \)
- \( 1 = 11 \times 11 - 1 \times 120 \)

Thus, \( d = 11 \).

- The private key is \( (N = 143, d = 11) \).

### 2. **Encryption:**

The encryption formula in RSA is:

\[
C = M^e \, (\text{mod} \, N)
\]

Given \( M = 7 \), \( e = 11 \), and \( N = 143 \):

\[
C = 7^{11} \, (\text{mod} \, 143)
\]

Let's calculate \( 7^{11} \, (\text{mod} \, 143) \):

\[
7^2 = 49
\]
\[
7^4 = 49^2 = 2401 \quad \Rightarrow \quad 2401 \, (\text{mod} \, 143) = 115
\]
\[
7^8 = 115^2 = 13225 \quad \Rightarrow \quad 13225 \, (\text{mod} \, 143) = 4
\]
\[
7^{11} = 7^8 \times 7^2 \times 7 = 4 \times 49 \times 7 = 1372 \quad \Rightarrow \quad 1372 \, (\text{mod} \, 143) = 84
\]

Thus, the ciphertext \( C = 84 \).

### 3. **Decryption:**

The decryption formula in RSA is:

\[
M = C^d \, (\text{mod} \, N)
\]

Given \( C = 84 \), \( d = 11 \), and \( N = 143 \):

\[
M = 84^{11} \, (\text{mod} \, 143)
\]

Let's calculate \( 84^{11} \, (\text{mod} \, 143) \):

\[
84^2 = 7056 \quad \Rightarrow \quad 7056 \, (\text{mod} \, 143) = 61
\]
\[
84^4 = 61^2 = 3721 \quad \Rightarrow \quad 3721 \, (\text{mod} \, 143) = 7
\]
\[
84^8 = 7^2 = 49
\]
\[
84^{11} = 84^8 \times 84^2 \times 84 = 49 \times 61 \times 84 = 250284 \quad \Rightarrow \quad 250284 \, (\text{mod} \, 143) = 7
\]

Thus, the decrypted message \( M = 7 \), which matches the original plaintext.

### Summary:

- **Public Key**: \( (N = 143, e = 11) \)
- **Private Key**: \( (N = 143, d = 11) \)
- **Encrypted Message (Ciphertext)**: \( C = 84 \)
- **Decrypted Message**: \( M = 7 \)

---

Would you like any further details or clarification? Here are five related questions to consider:

1. How is the modular inverse calculated in more complex cases?
2. Why is it important for \( e \) and \( \phi(N) \) to be coprime?
3. What is the significance of the totient function \( \phi(N) \)?
4. How is RSA security impacted by large prime numbers?
5. What optimizations can be made to RSA for faster encryption?

**Tip:** In real RSA applications, much larger prime numbers are used to ensure security, typically over 1024 or 2048 bits in size.

Explore how to derive RSA encryption and decryption using P=11, Q=13, e=11, and M=7. This detailed walkthrough covers key cryptography concepts like number theory, modular arithmetic, and the Extended Euclidean Algorithm, suitable for advanced high school or undergraduate students.

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