The following parameters are chosen to generate the public and private keys for the RSA cryptosystem.

p=7
q=17
e=13
d=37

If the plaintext is m=6, then what is the ciphertext c?

c=

To find the ciphertext \( c \) using the RSA encryption method, follow these steps:

### Step 1: Calculate \( n \)
The modulus \( n \) is calculated by multiplying the primes \( p \) and \( q \):
\[
n = p \times q = 7 \times 17 = 119
\]

### Step 2: Encrypt the plaintext
The ciphertext \( c \) is calculated using the formula:
\[
c = m^e \mod n
\]
Substitute the given values:
\[
c = 6^{13} \mod 119
\]

### Step 3: Compute \( 6^{13} \mod 119 \)
To compute \( 6^{13} \mod 119 \) efficiently, we can use modular exponentiation:

\[
6^2 = 36
\]
\[
6^4 = 36^2 = 1296 \equiv 103 \mod 119
\]
\[
6^8 = 103^2 = 10609 \equiv 25 \mod 119
\]
\[
6^{13} = 6^8 \times 6^4 \times 6 = 25 \times 103 \times 6 = 15450 \equiv 121 \equiv 2 \mod 119
\]

So, the ciphertext \( c \) is \( \boxed{2} \).

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

Here are five related questions:
1. How does the size of \( n \) affect the security of RSA encryption?
2. What would happen if \( e \) and \( \phi(n) \) are not coprime in RSA?
3. How can one verify the correctness of the decryption process?
4. Why is it necessary to use large primes in RSA?
5. How does the choice of \( e \) impact the efficiency of encryption?

**Tip:** To quickly compute large exponents in modular arithmetic, break the exponentiation down using the method of successive squaring (modular exponentiation).

Learn how to calculate the RSA encryption ciphertext c using given parameters p, q, e, d, and plaintext m = 6. Follow a detailed step-by-step solution involving modular arithmetic and Euler's theorem. Suitable for advanced high school students and cryptography enthusiasts.

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