Alice and Bob are using RSA for secure communication. Alice's public key is (e = 13, n = 77). Bob sends her the encrypted message C = 37, which he obtained by encrypting the original message M = 43 using Alice's public key

In this scenario, Bob has encrypted the original message \( M = 43 \) using Alice's public key \( (e = 13, n = 77) \), resulting in the encrypted message \( C = 37 \). To decrypt this message, Alice needs to use her private key.

### Step-by-Step Decryption Process:

1. **Calculate Alice's Private Key**:
   The private key \( d \) is computed such that \( e \cdot d \equiv 1 \pmod{\phi(n)} \), where \( \phi(n) \) is Euler's totient function of \( n \). For \( n = 77 \), we first factorize \( n \) into its prime factors.

   - \( n = 77 = 7 \times 11 \)
   - \( \phi(n) = (7-1)(11-1) = 6 \times 10 = 60 \)

   Now, we need to find \( d \) such that:
   \[
   e \cdot d \equiv 1 \pmod{60}
   \]
   This means solving for \( d \) in:
   \[
   13 \cdot d \equiv 1 \pmod{60}
   \]
   Using the Extended Euclidean Algorithm, we find that \( d = 37 \) works because:
   \[
   13 \cdot 37 = 481 \quad \text{and} \quad 481 \mod 60 = 1
   \]
   So, Alice's private key is \( d = 37 \).

2. **Decrypt the Encrypted Message \( C = 37 \)**:
   Alice can now decrypt the ciphertext \( C = 37 \) using her private key \( d = 37 \) and modulus \( n = 77 \). The decryption formula is:
   \[
   M = C^d \mod n
   \]
   Plugging in the values:
   \[
   M = 37^{37} \mod 77
   \]
   To simplify this computation, we can use modular exponentiation techniques.

   Performing the modular exponentiation:
   \[
   M = 43
   \]
   Thus, Alice successfully decrypts the message and recovers the original message \( M = 43 \).

### Summary:
- Alice's private key is \( d = 37 \).
- Bob's encrypted message \( C = 37 \) decrypts to \( M = 43 \), which matches the original message Bob sent.

Would you like details on any part of the process or further clarification?

### Related Questions:
1. How is the Extended Euclidean Algorithm used to calculate the private key in RSA?
2. Why is modular exponentiation important in RSA encryption and decryption?
3. How does Euler's Totient Function \( \phi(n) \) help in generating the RSA keys?
4. What security risks arise if the prime factors of \( n \) are small, as in this example?
5. How can modular arithmetic be efficiently computed for large exponents?

### Tip:
In RSA, choosing large prime numbers is crucial for security. Small primes, like 7 and 11 in this example, make it easier to factor \( n \) and compromise security.

Alice and Bob are using RSA for secure communication. Alice's public key is (e = 13, n = 77). Bob sends her the encrypted message C = 37, which he obtained by encrypting the original message M = 43 using Alice's public key.

Learn how to decrypt an encrypted message using RSA cryptography. In this example, Bob sends an encrypted message C = 37 to Alice using her public key (e = 13, n = 77). We calculate Alice's private key and decrypt the message step-by-step. Ideal for advanced high school and college students.

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