   Chapter 8.4, Problem 37ES ### 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. COME

To determine

To encode the given message into their numeric equivalents and encrypt them.

Explanation

Given information:

The numeric equivalents of C,O,Mand E are 03,15,13,05

Concept used:

To encrypt the letter,compute the values of 343 mod 713,1543mod713,1343mod713,543mod713.

Calculation:

We use the fact that 43=32+8+2+1

Calculate 32kfor k=1,2,3,4,5

For C:

33(mod713)since aamod713

329mod713by the definition of mod and 32=93481mod713since34=8138( 3 4)2mod713=(81)2mod713since 34mod713=81=6561mod713since 812=6561=144mod713since by the definition of mod

316( 3 8)2mod713=(144)2mod713since38mod713=144=20736mod713since 1442=20736=59mod713by the definition of mod

332( 3 16)2mod713(59)2mod713since 316mod713=59

=3481mod713since 592=3481=629(mod713)by the definition of mod

Now,

343mod713=(3 32+8+2+1)mod713=(3 3238323)mod713by algebra=(62914493)mod713=2445552mod713using calculator=675

For O: Calculate first 152kfork=1,2,3,4,5

1515(mod713)since aa(modn)152225(mod713)since 152=225154( 15 2)2mod713=(225)2mod713since 152mod713=225=50625(mod713)using calculator=2mod713by the definition of mod

158( 15 4)2mod713=22mod713since154mod713=2=4mod713by 22=4

1516( 15 8)2mod713=42mod713since158mod713=4=16mod713since42=16

1532( 15 16)2mod713=162mod713since 1516mod713=16=256mod713since 162=256

Now,

1543mod713=( 15 32+8+2+1)mod713=( 15 32 158 152 151)mod713by algebra=(256422515)mod713=(3456000)mod713using calculator=89

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