   Chapter 8.4, Problem 39ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# In 36,37,39 and 40, use the RSA cipher with public key and In 36 and 37, encode the messages into their numeric equivalents and encrypt them. In 39-40, decrypt the given cipher text and find the original messages. 675 089 089 048

To determine

To decrypt the given cipher text and calculate the original message:

675089089048.

Explanation

Given information:

The given cipher text 675089089048.

Concept used:

Inverse of amodulon: For all integers aandn, if gcd(a,n)=1, then there exists an integer s

such that as1(modn). The integer s is called the inverse of amodulon.

Calculation:

Consider the cipher text,

675089089048

The objective is to decrypt the given cipher text and find the original message.

The decryption key is d=307, since 307 is the inverse of 43.

Public key is n=713=2331 and e=43.

Thus, to decrypt the message, compute the following:

675307mod713,89307mod713,89307mod713,48307mod713

Convert 307 to a sum of powers of 2.

307=256+32+16+2+1=28+25+24+21+1

Perform 6752kfork=1,2,3,4.

675675(mod713)6752455625(mod713)since 6752=455625=18mod713since455625mod713=18

6754( 675 2)2mod713182mod713since6752mod713=18=324mod713since182=324

6758( 675 4)2mod713(324)2mod713since 6754mod713=324104976mod713since 3242=104976165mod713since 104976mod713=165

67516( 675 8)2mod713(165)2mod713since6758mod713=16527225mod713since1652=27225131mod713since27225mod713=131

Perform 6752kfork=5,6,7,8

67532( 675 16)2mod713(131)2mod713since67516mod713=13117161mod713since1312=1716149mod713since17161mod713=49

67564( 675 32)2mod713(49)2mod713since67532mod713=492401mod713since492=2401262mod713since2401mod713=262

675128( 675 64)2mod713(262)2mod713since67564mod713=26268644mod713since2622=68644196mod713since68644mod713=196

675256( 675 128)2mod713(196)2mod713since675128=196mod17338416mod713since1962=38416627mod713since38416mod713=627

Use the fact that

675307mod173=( 675 256+32+16+8+2+1)mod173=( 675 256 675 32 675 16 6752 6751)mod173=(6274913118675)mod173=48900262950mod173=3mod173using calculator=3

Thus, 675 is decrypted as 3, which corresponds to letter C.

Calculate 892kfor k=1,2,3,4.

8989mod713since by aa(modn)892892mod713=7921mod713since 892=7921=78mod713since 7921mod713=78

894( 89 2)2mod713(78)2mod713since892mod713=786084mod713since782=6084=380mod713since380=6084mod713

898( 89 4)2mod713(380)2mod713since894mod713=380144400mod713since3802=144400=374mod713

8916( 89 8)2mod713(374)2mod713since898mod713=374139876mod713since3742=139876=128mod713since139876mod713=126

Perform 892kfor k=5,6,7,8

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