Find the following using a technique discussed in Section 8.4.162 mod 45 =164 mod 45 =168 mod 45 =1616 mod 45 =Need Help?Watch ItRead It Example 8.4.4Computing a mod n When k Is a Power of 2Find 144 mod 713.Solution Use property (8.4.1) to write 144 = (1442)?. Then1444 mod 713 = (1442)² mod 713= (1442 mod 713)? mod 713= (20736 mod 713) mod 713because 1442 = 20736= 592 mod 713because 20736 mod 713 = 59= 3481 mod 713because 592 = 3481= 629because 3481 mod 713 = 629.Example 8.4.5Computing a" mod n When k Is Not a Power of 2Find 123 mod 713.anon2 JedsSolution First write the exponent as a sum of powers of 2:43 = 2° + 23 + 2+1 = 32+8+2+1.Next compute 12 for k = 0, 1, 2, 3, 4, and 5.12 mod 713 = 12122 mod 713 = 14412 mod 713 = 144?mod 713 = 59by Example 8.4.4128 mod 713 = 592 mod 713= 629 by Example 8.4.41216 mod 713 = 6292 mod T713 = 639 by the method of Example 8.4.41232 mod 713 = 6392 mod 713 = 485 by the method of Example 8.4.4.By property (8.4.2),1243 = 1232+8 +2+1 = 1232.128. 12²- 12'.Thus, by Corollary 8.4.4,1243 mod 713= {(1232 mod 713)·(12° mod 713)·(122 mod 713)·(12 mod 713)} mod 713.By substitution,1245 mod 713 = (485·629·144·12) mod 713= 527152320 mod 713= 48.You should know how to do the computations in Example 8.4.5 by hand using only asimple electronic calculator, but if you are computing a lot of residues, especially onesinvolving large numbers, you may want to write a short computer or calculator program todo the computations for you.Extending the Euclidean AlgorithmAn extended version of the Euclidean algorithm can be used to find a concrete expressionfor the greatest common divisor of integers a and b.

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Answered: Find the following using a technique…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter2: The Integers

Section2.5: Congruence Of Integers

Problem 57E

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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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Find the following using a technique discussed in Section 8.4. 162 mod 45 = 164 mod 45 = 168 mod 45 = 1616 mod 45% = Need Help? Watch It Read It