Consider an RSA system with n=44916803. Let public exponent e = 6146419. What is the message x if the encrypted word is 1767607? How did you arrive at your answer? Show and explain all your steps.
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Consider an RSA system with n=44916803. Let public exponent e = 6146419. What is the message x if the encrypted word is 1767607? How did you arrive at your answer? Show and explain all your steps.
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- ONLY THE TAP BITS. 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=4m=4.The plaintext given by 01001011=x0x1x2x3x4x5x6x701001011=x0x1x2x3x4x5x6x7 when encrypted by the LFSR produced the ciphertext 11010110=y0y1y2y3y3y5y6y711010110=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=1p3=0,p2=1,p1=0,p0=1).An animal name has been encoded using the letter to number correspondence abcde t 8hij k 1 m nop qrstuVH x y z e 1 2 3 4 5 6 7 8 9 18 11 12 13 14 15 16 17 18 19 28 21 22 23 24 25 and the encryption function e :n + (n + 2) mod 26 The result is the secret identification kiwcpc . What is the animal? The answer will be a nonsense word (no numbers).We analyzed a program which was supposedly used for cracking some cryptographic algorithms. During the investigations we determined that the program input size can be varied in a wide range and for N-bit input the program result is also always N bit long. Additionally we found that the program working time depends significantly on input length N, especially when N is greater than 10-15. Our test also revealed, that the program working time depends only on input length, not the input itself. During our tests, we fixed the following working times (with an accuracy of one hundredth of a second): N=2-16.38 seconds N = 5-16.38 seconds N = 10-16.44 seconds N=15-18.39 seconds N=20-1 minute 4.22 seconds We also planned to test the program for N=25 and N = 30, but for both cases the program didn't finish within half an hour and was forced to terminate it. Finally, we decided for N= 30 to not terminate the program, but to wait a little bit longer. The result was 18 hours 16 minutes 14.62…
- Please explain the problem in detail so that I can understand.An animal name has been encoded using the letter to number correspondence abc de f 8hijk 1 m nopP arstu vH x y z e 1 2 3 4 5 6 7 8 9 1e 11 12 13 14 15 16 17 18 19 28 21 22 23 24 25 and the encryption function e:n + (n + 2) mod 26 The result is the secret identification kiwcpc . What is the animal? The answer will be a nonsense word (no numbers).Someone sends you a message of “4”. They also send you a copy of their message encrypted with their private key. The “signed”, or encrypted copy is “49”. Their public key is (exponent 23, clock 55). Show the math you can do to authenticate the 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.In a public-key system using RSA, you intercept the ciphertext C = 10 sent to a user whose public key is e = 5,n = 35. What is the plaintext M?Perform encryption and decryption using the RSA algorithm for the letter "S". Show all steps in details when p=5 , q=7 , e=5.
- The operator of a Vigenere encryption machine is bored and encrypts a plaintext consisting of the same letter of the alphabet repeated several hundred times. The key is a seven-letter English word. Eve knows that the key is a word but does not yet know its length. What property of the ciphertext will make Eve suspect that the plaintext is one repeated letter and will allow her to guess that the key length is seven? What will the number of matches be for the different displacements?Alice and Bod have decided to use a symmetric encryption algorithm. They have some assumptions about their messages:- Messages only contain capital letters (i.e. A to Z)- The length of their shared key must be greater than or equal to the length of the plaintext- They assign each letter a number as follows: (A,0), (B,1), (C,2), (D,3),…, (Z,25)Their algorithm combines the key and the message using modular addition. The numerical values of corresponding message and key letters are added together, modulo 26. For example, if the plain text is “HELLO” and the key is “SECRET” then the encrypted message is calculated as following:Since the length of the plaintext is 5, we just need the first 5 letters of the key (i.e. “SECRE”), then for each letter, we should add corresponding letters in both the plaintext and the key modulo 26.Plaintext: H (7) E (4) L (11) L (11) O (14)Key: S (18) E (4) C (2) R (17) E(4)Cipher: Z (25) I (8) N(13) C(2) S (18) Write a program in Python, C/C++ or JavaScript to…B