In RSA, suppose p = 3, q = 11, e = 7, and d = 3. Given the message m = 5, show the encryption process to obtain the ciphertext c from m and the decryption process to derive m back from c.

Let M be the message to be sent by using RSA cryptosystem. If (n, e) is the public key and (n, d) is the private key, which of the following is the encryption function that will yield the cipher text that corresponds to M? О Е(M) — Ме mod (p - 1) (q — 1) О Е[M) — ма mod (p - 1)(q — 1) O E(M) = Me mod n О ЕМ) — Ма mod n

Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m = 4. The plaintext given by 11000101 — ХоXјX2XҙХ4X5X6X7 when encrypted by the LFSR produced the ciphertext 10010000 Уo У1 У2 Уз Уз У5 У6 Ут . What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent P3 = 0, p2 = 1, p1 = 0, po = 1).

4. Suppose that a cryptanalyst suspects that the cipher text: KNCFNNW OARNWMB CQNAN RB WX WNNM XO SDBCRLN was produced by applying a shift encipherment of some unknown number of letters and then applying a second shift encipherment (by a different number of letters) to that. How will the work to obtain the plaintext in this case compare with the work to find it if the cryptanalyst suspected a single shift encipherment? Decipher the message.

For this problem you will perform RSA encryption and decryption with p = 13, q = 17, and e = 5 (not 3!). Show your work. Compute n and d. Encrypt the message m = 14 and give the ciphertext. Perform decryption on the ciphertext obtained in part (b) to get the plaintext message.

Messages are to be encoded using the RSA method, and the primes chosen are p “ 17 and q “ 19, so that n “ pq “ 323, and e “ 19. Thus, the public key is p323, 19q.   (a) Show that the decryption exponent d (your private key) is 91.

Messages are to be encoded using the RSA method, and the primes chosen arep = 13 and q = 23, so that n = pq = 299. The encryption exponent is e = 13.Thus, the public key is (299, 13).(i) Show that the decryption exponent d (the private key) is 61.(ii) Use the repeated squaring algorithm to find the encrypted form c of themessage m = 84.  (iii)

In an RSA system, the public key (n,e) of a given user is (323, 11). 1. What is the value of the exponent in the private key (n, d), of this user? 2. Suppose you want to send this user the message m = 45, write down the expression to generate the ciphertext for this message. 3. Suppose that your ciphertext, c, is 5, write down the expression to generate the plaintext matching this ciphertext. 4. Can you encrypt the message m=322 with this public key? O a. Yes O b. No

Let (M, C, e, d, Ke, Kd) denote the RSA cryptosystem with the public key (s, n) = (49, 8909).Suppose that Malice has obtained the prime factorization 8909 = 59 · 151. Find thevalue of the private key t.

The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m=4The plaintext given by 10001001=x0x1x2x3x4x5x6x7 when encrypted by the LFSR produced the ciphertext 10110011=y0y1y2y3y3y5y6y7 What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3=0,p2=1,p1=0,p0=1