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- ] You receive a message that was encrypted using the RSA system with public key (43, 143), where 43 is the public exponent. A message encrypted using this cryptosystem was captured by an enemy and consisted of the following numbers: 6 82 132 115. Can you decipher what was the original message? (Hint: As a first step, you need to find p and q.)The affine cypher with key "9x+3" was used to encrypt the following text. (To encode, multiply the letter by 9 and then add 3 (mod 26).) The encryption result of the letter "A" is, for example, "D." Decrypt the ciphertext below. GNVBAXSL XG QWBUse the two prime numbers p 5 and q =13 in the first step to give ONE integrated example to show how the five steps in the basic process of RSA Cryptography works. Based on your example, demonstrate how the number 60 as an original message is encrypted into ciphertext and decrypted back correctly to the original message. If you need to choose a number in any step, you must choose it from the set {r e R: 8 Sx< 12}. You must show all detailed steps involved.
- The answer above is NOT correct. 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 11100010 — ХоXјX2XҙX4XsX6X7 when encrypted by the LFSR produced the ciphertext 11010110 — 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). 0010Encrypt the word WEST using p=53 and q=71. Choose e=3. (Map A - Z to 126 and perform the encryption one letter at a time.)Encrypt the 12-bit plaintext 101010100111 using Simplified B-DES and the 9-bit key K = 010001111. This is a non-computer problem and you need to show all the steps of the encryption process
- Analyze the public key cryptography named RSA in which you have to choose p=3 and q=7 and encode the word “CSEDAY”. Finally, you have to apply the decryption algorithm to the encrypted version to recover the original plain text message.I need help with the following problem: A ciphertext was generated using an affine cipher. The most frequent letter in the ciphertext is “C” and the second most frequent letter is “V”. Break the code. Assume that the most frequent letter in English is “E” and the second most frequent letter is “T”.Use the two prime numbers p = 5 and q =13 in the first step to give ONE integrated example to show how the five steps in the basic process of RSA Cryptography works. Based on your example, demonstrate how the number 60 as an original message is encrypted into ciphertext and decrypted back correctly to the original message. If you need to choose a number in any step, you must choose it from the set {x e R: 8 s x< 12}. You must show all detailed steps involved.
- 5. Your opponent uses RSA with n = pq and encryption exponent e and encrypts a message m. This yields the ciphertext c = m mod n. A spy tells you that, for this message, 12345 =1 mod n. Describe how to determine m. Note that you don't know p, q, 6(n), or a decryption exponent d. However, you should find a de- cryption exponent that works for this particular ciphertext. Moreover,explain carefully why your decryption works (Your explanation must include how the spy's information is used.)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. NoNote: 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).