   Chapter 8.4, Problem 40ES ### 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. 028 018 675 129

To determine

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

028018675129.

Explanation

Given information:

Cipher text: 028018675129

Concept used:

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

Public key is n=713=23.31 and e=43.

Calculation:

Now, decrypt the message to compute

28307mod713,18307mod713,675307mod713,129307mod713

Convert 307 to a sum of powers of 2 that is,

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

Perform 282k for k=1,2,3,4.

28=¯28(mod713)282=¯784(mod713) since 282=784=71mod713 since 784mod713=71284=¯( 28 2)2mod713=¯712mod713 since 282mod713=71=5041mod713 since 712=5041=50mod713 since 5041mod713=50

288=¯( 28 4)2mod713=¯502mod713 since 284mod713=50=2500mod713 since 502=2500=361mod713 since 2500mod713=3612816=¯( 28 8)2mod713=3612mod713 since 282mod713=361=¯130321mod713 since 3612=130321=¯555mod713 since 130321mod713=555

Perform 282k for k=5,6,7,8

2832=¯( 28 16)2mod713=¯(555)2mod713 since 2816mod713=555=¯308025mod713 since 5552=308025=¯9mod713 since 308025mod713=92864=¯( 28 32)2mod713=¯92mod713 since 2832mod713=9=¯81mod713 since 92=81

28128=¯( 28 64)2mod713=¯(81)2mod713 since 2864mod713=81=¯6561mod713 since 812=6561=¯144mod713 since 6561mod713=14428256=¯( 28 128)2mod713=¯(144)2mod713 since 28128mod713=144=¯20736mod713 since 1442=20736=¯59mod713 since 20736mod713=59

Use the fact that,

28307mod713=( 28 256+32+16+2+1)mod713=( 28 256 28 32 28 16 282 281)mod713=(5995557128)mod713 By algebra=585873540mod713 Use a calculator=14

Thus, the decryption for 28 is 14, which corresponds to the letter N.

Calculate the 282k for k=1,2,3,4

18=¯18mod713 since a=¯a(modn)182=¯324mod713 since182=324184=¯( 18 2)2mod713=¯3242mod713 since 3242=104976=104976mod713 and 182mod713=324=165mod713104976mod713=165

188=¯( 18 4)2mod713=1652mod713 since 184mod713=165=27225mod713 since 1652=27225=131mod71327225mod713=131

1816=¯( 18 8)2mod713=1312mod713 since 188mod713=131=17161mod713 since 1312=17161=49mod713 since 17161mod713=49

Perform the 282k for k=5,6,7,8

1832=¯( 18 16)2mod713=(49)2mod713 since 1816mod713=49=2401mod713 since 492=2401=262mod713 since 2401mod713=2621864=¯( 18 32)2mod713=(262)2mod713 since 1832mod713=262=68644mod713 since 2622=68644=196mod713 since 68644mod713=196

18128=¯( 18 64)2mod713=(196)2713 since 1864mod713=196=38416mod713 since 1962=38416=627mod713 since 38416mod713=62718256=¯( 18 128)2mod713=(627)2mod713 since 18128mod713=627=393129mod713 since 6272=393129=266mod713 since 393129mod713=266

Thus, the decryption for 18 we have to calculate is 18307mod713.

18307mod713=( 18 256+32+16+2+1)mod713=( 18 256 18 32 18 16 182 181)mod713=(2662624932418)mod713=(19915743456)mod713 Using a calculator=9mod713 by the definiton of mod.=9

Thus, the decryption for 18 is 9, which corresponds to the letter l.

Perform 6752k for k=1,2,3,4

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