Question

CIPHER THAT LETS LOOK PA RSA AT USES Two PRIMES p=23 AND q=17 PUBLIC KEY e=3 A) PRIVATE DECRYPTING KEY d. FIND IN B) DESCRIBE STEPS HOW TO FIND IS c=165. PLAIN TEXT CIPHERTEXTI IF

Answer 1

RSA algorithm is an assymetric cryptographic algirithm which used to encrypt and decrypt messages.Assymetric means it involves two keys, public key and private key, as one of the keys can be given to anyone and the other key will be private.

Here p and q are two prime numbers and given as p=23, q=17

> First, we need to find n, which is the product of p and q.

n = p x q

n = 23 x 17

n = 391

> After that compute Φ = (p-1)(q-1)

Φ = (23-1)(17-1) = 22 x 16

Φ = 352

> Public key value is given in the question, e = 3.

> Next is to find the private key d, so d is the inverse of e mod Φ = 1

i.e., de mod Φ = 1

d = e^-1 mod Φ

Here Φ is 352, 352k+1 = 353, 705, 1057, 1409 where k=1,2,3,4...

Now check among those values which one is divisible by 3(public key), 705/3 = 235

705 is exactly divisible by 3, and d = 235

> So the two keys are {e,Φ} and {d, n}

key 1 = {3, 352} and key 2 = {235, 391} used for encryption and decryption respectively.

> Here asked to find the plain text from cipher text which is decryption.

To find decryption, t = c^d mod n (where c is cipher text, d is private key)  

Given c = 165 , we found d = 235 and n = 391

Plain text, t = 165²³⁵ mod 391

The same is handwritten and attached that image, can go through which is comfortable to you either typed or handwritten.

Given p=23, 9-17 m=p xq n = 23x17 n = 391 -> 22 X 16 Porublic key 0=(p-1) (9-1) = (23-1) (17-1) ø= 352 is given, e = 3 Private kayld) need to be computed, d=é modo 352k+I 353, 705, 1054,1409 (k = 1, 2, 3, 4, um Check which rumbers are exactly exactly divisible by e=3 705 is divisible by 3. Hence, d = 235 1. {e, &} = {3, 352} 2. Id ny = {235, 3913 → Asked to find the plain text, which is decryption. given C = 165, Plain tent, t= cd mod in Plain tent, t = 165 +05 - 235 3 Two keys 235 mod 391