Let p = 29 and q = 31, and let e == 101.%3D(a) Find n =pq and o(n).(b) Find d.(c) Then let m = 555 be the message and encode by computing s me mod n.(d) Decode by computing z = sd mod n.

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Answered: Let p= 29 and q = 31, and let e = = 101…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.2: Arithmetic Sequences

Problem 54E

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Suppose that in a long ciphertext message the letter occurred most frequently, followed in frequency by. Using the fact that in the -letter alphabet, described in Example, "blank" occurs most frequently, followed in frequency by, read the portion of the message enciphered using an affine mapping on. Write out the affine mapping and its inverse. 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.
a. Prove that 10n(1)n(mod11) for every positive integer n. b. Prove that a positive integer z is divisible by 11 if and only if 11 divides a0-a1+a2-+(1)nan, when z is written in the form as described in the previous problem. a. Prove that 10n1(mod9) for every positive integer n. b. Prove that a positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. (Hint: Any integer can be expressed in the form an10n+an110n1++a110+a0 where each ai is one of the digits 0,1,...,9.)

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Let p= 29 and q = 31, and let e = = 101 %3D (a) Find n = pq and o(n). (b) Find d.