### RSA Public Key Cryptosystem Example**Problem:**Bob chooses the secret primes $ p = 11 $ and $ q = 23 $. Bob chooses the encryption exponent $ e = 7 $. Bob publishes $ N = 253 $ and $ e = 7 $.Alice chooses the plaintext $ m_1 = 5 $. Compute the ciphertext $ c_1 $ that Alice sends to Bob.**Solution:**To encrypt the plaintext $ m_1 = 5 $ using Bob's public key values $ N = 253 $ and $ e = 7 $, Alice uses the RSA encryption formula:\[ c \equiv m^e \, (\text{mod} \, N) \]Given:- $ m = 5 $- $ e = 7 $- $ N = 253 $1. Compute $ 5^7 $: \[ 5^7 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 78125 \]2. Compute $ 78125 \mod 253 $: \[ 78125 \div 253 \approx 308.8745 \quad \Rightarrow \quad 78125 = 253 \times 308 + 181 \quad \Rightarrow \quad 78125 \mod 253 = 181 \]Therefore, the ciphertext $ c_1 $ that Alice sends to Bob is:\[ c_1 = 181 \]

We are given a plain text m1 and using RSA public key cryptosystem, we will find out the ciphertext…

(7) (RSA public key cryptosystem) Bob chooses the secret primes p = 11 and q = 23. Bob chooses the encryp- tion exponent e = 7. Bob publishes N = 253 and e = 7. Alice chooses the plaintext m₁ = 5. Compute the ciphertext c₁ that Alice sends to Bob.

(7) (RSA public key cryptosystem) Bob chooses the secret primes p = 11 and q = 23. Bob chooses the encryp- tion exponent e = 7. Bob publishes N = 253 and e = 7. Alice chooses the plaintext m₁ = 5. Compute the ciphertext c₁ that Alice sends to Bob.

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Chapter1: Introduction

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Question

### RSA Public Key Cryptosystem Example

**Problem:**

Bob chooses the secret primes $ p = 11 $ and $ q = 23 $. Bob chooses the encryption exponent $ e = 7 $. Bob publishes $ N = 253 $ and $ e = 7 $.

Alice chooses the plaintext $ m_1 = 5 $. Compute the ciphertext $ c_1 $ that Alice sends to Bob.

**Solution:**

To encrypt the plaintext $ m_1 = 5 $ using Bob's public key values $ N = 253 $ and $ e = 7 $, Alice uses the RSA encryption formula:

\[ c \equiv m^e \, (\text{mod} \, N) \]

Given:
- $ m = 5 $
- $ e = 7 $
- $ N = 253 $

1. Compute $ 5^7 $:
\[
5^7 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 78125
\]

2. Compute $ 78125 \mod 253 $:
\[
78125 \div 253 \approx 308.8745 \quad \Rightarrow \quad 78125 = 253 \times 308 + 181 \quad \Rightarrow \quad 78125 \mod 253 = 181
\]

Therefore, the ciphertext $ c_1 $ that Alice sends to Bob is:

\[ c_1 = 181 \]

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