Use the RSA cipher with public key n = 713 = 23 · 31 and e = 43 to encode the messages HELP into their numeric equivalents and encrypt them. Performs all step.
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A: SUMMARY: - Hence, we discussed all the points.
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Use the RSA cipher with public key n = 713 = 23 · 31 and e = 43 to encode the messages HELP into their numeric equivalents and encrypt them. Performs all step.
Use pencil and paper to write the answer.
show your work as well.
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- Explain and Apply Encrypted a plaintext "Meet me at the park" with key "Study" using a keyed columnar transposition cypher. It should be noted that all sides agreed on the number of columns to be five.The following ciphertext is encrypted using a monoalphabetic substitution cipher. Use the frequency analysis technique described in the lecture to decrypt this. Show the frequency of letters in the ciphertext and show your work. You can use different tools (with citing the source) to find the frequency of letters, or write your own code. But you are not allowed to brute force the ciphertext for finding the plaintext. It is important to show your work. If you only submit the final plaintext no points will be given. Explain your thought process as you try to decrypt this. This ciphertext is included in file hw1q1cipher.txt. ZIW VQKD LAFLIOFW YOSZWKWR ZIKGAUI ZIW SWQCWL GY ZIW ZKWWL EQLZOFU RQHHSWR LIQRGVL GF ZIW UKGAFR TWSGV TOKRL EIOKHWR QFR LQFU OF ZIW TKQFEIWL QRROFU ZG ZIW HWQEWYAS QDTOQFEW GY ZIW YGKWLZ Q SOUIZ TKWWMW KALZSWR ZIW SWQCWL EKWQZOFU Q UWFZSW VIOLHWKOFU LGAFR ZIQZ YOSSWR ZIW QOK (please type answer not write by hend)Practice using multiplicative ciphers to encode and decode a message: 1. Find quote that you like. Make sure that the total number of characters is at least 50 and no more than 100. 2. Then pick an acceptable number (not every number between 1 and 26 is acceptable) that is between 3 and 25 as your multiplicative key. Share this key. 3. Create a table of correspondence of 26 alphabets in plaintext and the ciphertext. (4) Use the table of correspondence to encode the quote you found in Part (1). Post the encrypted message in groups of three letters in the discussion forum. For example, here is another quote from John Dewey Education is not preparation for life; education is life itself. We will use multiplicative key 17. The table of correspondence between plaintext and ciphertext is as follows. (image attached Thus the encrypted message is GPS YQB WUD WKD UBL TGL QTQ BWU DXU TVW XGG PSY QBW UDW KVW XGW BKG VX (5) Pick an encrypted message from your classmate and try to decrypt…
- Help with this pleaseSuppose you were given a number of different ciphertexts and you are told that they are each encryptions of plaintexts using a Caesar cipher-type encryption but with different keys and different plaintext languages from around the world. You are asked to try to work out the most likely language of each ciphertext. Explain how you would approach this problem.You are given a string that was encoded by a Caesar cipher with an unknown distance value. The text can contain any of the printable ASCII characters. Suggest an algorithm for cracking this code.
- Let's pretend a Vigen'ere cipher was used to encrypt a long string of text, and that the encrypted text included many instances of the same string. Describe the methods you'll use to calculate the length of the key based on this information.Try it yourself to make sure you've got the concept. 1. Use a Caesar cipher with key 6 (AàG) to encrypt the following message: ANCIENT OF DAYS4. Decipher the following ciphertext, which was enciphered with the Caesar cipher: TEBKFKQEBZLROPBLCERJXKBSBKQP, using the statistical cryptoanalysis and the 1-gram model of English (see the table below). Write a program to calculate the correlation (i) for 0 ≤ i ≤ 25 to help identify the key and the plaintext. a b с d e f 61) 0.080 0.015 0.030 0.040 0.130 0.020 0.015 h i k 0.060 j 0.005 1 0.065 0.005 0.035 m 0.030 n 0 9 Р 0.020 r 0.070 S 0.080 0.002 0.065 0.060 t u V X y N 0.090 0.030 0.010 0.015 0.005 0.020 0.002
- . Using the RSA public key cryptosystem, with a = 1, b = 2, etc., a. If p = 7 and q = 11, list five legal values for d. b. If p = 13, q = 31, and d = 7, find e. c. Using p = 5, q= 11, and d = 27, find e and encrypt "abcdefahii".8. Also, in the Caesar cipher you could not encrypt two different letters to be the same letter (meaning if “p" → "S" then nothing else can go to “S"). Is this true of the Vigen'ere cipher too?Give me answer please.