The encoding matrix Awas used to encrypt a six-letter word, resulting in the ciphertext: WSKUNWThe word that was encrypted is:(Enter your answer with no spaces.)You may find the following table of multiplicative inverses mod(26) helpful:a 13579 11 15 17 19 21 23 25a-119 21 15319723 11 5 17 25

Given that, The ciphertext is W S K U N WHere, W = 22, S = 18, K = 10, U =…

Answered: The encoding matrix A was used to…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter2: The Integers

Section2.7: Introduction To Coding Theory (optional)

Problem 26E

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Decode the number using the given private key. Decode the number M = 38 using the private key d = 17 and n = 77. Assume you are decoding the number using the RSA cryptosystem. Note: You can use the Modular Exponentiation calculator to help with the calculation.
Convert (7206)8 and (A0EB)16 to a binary representation.
Encode the number using the given public key. Encode the number M = 14 using the public key n = 77 and e = 13. Assume you are encoding the number using the RSA cryptosystem. Note: You can use the Modular Exponentiation calculator to help with the calculation. b 46 (mod 77), encoded number = 46 47 (mod 77), encoded number = 47 48 (mod 77), encoded number = 48 49 (mod 77), encoded number = 49
Suppose the RSA cryptosystem is used with public key (35, 7). What is the corresponding cipher text when the plain text "2" is sent? O 23 15 12 O 8
Suppose that the ciphertext ERC WYJJMGMIRXPC EHZERGIH XIGLRSPSKC MW MRHMWXMRKYMWLEFPI JVSQ QEKMG was produced by encrypting a plaintext message using a shift cipher. What is the original plaintext?
A edfinity.com/assessments/6024929991fbb7005dbf9ca9#modal-6 x Practice problem Encrypt the message " MATH " by translating the letters into numbers and then applying the encryption function given, and then translating the numbers back into letters. (a) f(p) = (19p+2) mod 26 b) f(p) = (5p + 10) mod 26 (e) f(p) = (7p+7) mod 26 Use A = 0.B= LC – 2.D = 3.E – 4F = 5.G 6.H= 7.I = 8, J = 9.K = 10 L= 11 M = 12.N 13.0 = 14.P 15.Q = 16 R = 17. S – 18 T = 19,U = 2 Submit answer Answers in progress) Po Answer Correct Answer WIZA SKBT NHKE tiit O Type here to search
The ciphertext OIKYWVHBX was produced by encrypting a plaintext message using the Vigen`ere cipher with key HOT. What is the plaintext message?
Find the Order and the inverse of 23 in Z25.
Suppose we decide to use Data Encryption Standard (DES) for data encryption. Given the initial 64-bit key is: 5B5A5767 6A56676E Given the 64-bit block of plaintext to be encrypted is: 675A6967 5E5A6B5 A
Suppose a message is encrypted using the RSA system with n = 53*61 and e = 19. Suppose the ciphertext message is 2701 7352 6410. Show how it is decrypted, i.e., find the secret key d and show how it is used to do the decryption. Do not calculate the modular exponentiations, only show what they should be. Thank you for your help!
Use the RSA cipher with public key (n, e) = (713, 43) to encrypt the word "TEE." Start by encoding the letters of the word "TEE" into their numeric equivalents. Assume the letters of the alphabet are encoded as follows: A = 01, 8 = 02, C = 03, ..., Z = 26. Since the code for T is 20 and since e = 43 = 32 + 8 + 2 + 1, the first letter of the encrypted message is found by computing 2043 mod 713. 20¹a (mod 713) 20² Eb (mod 713) 204 c (mod 713) 208 d (mod 713) 2032 = f (mod 713) 2016 e (mod 713) b = The result is that a = C = d = e = and f = Thus, 2043 mod 713 = (a · b. d. f) mod 713 = So the first number in the encrypted message is Repeat these computations for each letter to find the complete encrypted message and enter your answer below. (Enter the message as a sequence of integer triples separated by a single space, where each triple is written using a fixed number of digits: 001 for 1, 002 for 2, ..., 099 for 99.)
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