Answered: Let P = 3, Q = 11, N = 33, ? =20, and e…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter2: The Integers

Section2.8: Introduction To Cryptography (optional)

Problem 19E: Suppose that in an RSA Public Key Cryptosystem. Encrypt the message "algebra" using the -letter...

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Suppose that in an RSA Public Key Cryptosystem. Encrypt the message "algebra" using the -letter alphabet from Example 4. Use two-digit blocks. Use three-digit blocks. What is the secret key?
Suppose that in an RSA Public Key Cryptosystem, the public key is. Encrypt the message "pay me later” using two-digit blocks and the -letter alphabet from Example 2. What is the secret key? Example 2 Translation Cipher Associate the letters of the "alphabet" with the integers. Let and define the mapping by where is the key, the number of positions from the plaintext to the ciphertext. If our alphabet consists of through, in natural order, followed by a blank, then we have "letters" that we associate with the integers as follows:
Rework Example 5 by breaking the message into two-digit blocks instead of three-digit blocks. What is the enciphered message using the two-digit blocks? Example 5: RSA Public Key Cryptosystem We first choose two primes (which are to be kept secret): p=17, and q=43. Then we compute m (which is to be made public): m=pq=1743=731. Next we choose e (to be made public), where e must be relatively prime to (p1)(q1)=1642=672. Suppose we take e=205. The Euclidean Algorithm can be used to verify that (205,672)=1. Then d is determined by the equation 1=205dmod672 Using the Euclidean Algorithm, we find d=613 (which is kept secret). The mapping f:AA, where A=0,1,2,...,730, defined by f(x)=x205mod731 is used to encrypt a message. Then the inverse mapping g:AA, defined by g(x)=x613mod731 can be used to recover the original message. Using the 27-letter alphabet as in Examples 2 and 3, the plaintext message no problem is translated into the message as follows: plaintext:noproblemmessage:13142615171401110412 The message becomes 13142615171401110412. This message must be broken into blocks mi, each of which is contained in A. If we choose three-digit blocks, each block mim=731. mi:13142615171401110412f(mi)=mi205mod731=ci:082715376459551593320 The enciphered message becomes 082715376459551593320 where we choose to report each ci with three digits by appending any leading zeros as necessary. To decipher the message, one must know the secret key d=613 and apply the inverse mapping g to each enciphered message block ci=f(mi): ci:082715376459551593320g(ci)=ci613mod731:13142615171401110412 Finally, by re-breaking the message back into two-digit blocks, one can translate it back into plaintext. Three-digitblockmessage:13142615171401110412Two-digitblockmessage:13142615171401110412Plaintext:noproblem The RSA Public Key Cipher is an example of an exponentiation cipher.

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Let P = 3, Q = 11, N = 33, ? =20, and e = 3 Public Key is: (33, 3) Private Key is : (33, 7) 1 8 27 31 26 18 13 is the message needing to be decrypted.