   Chapter 8.4, Problem 36ES ### 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
1 views

# In 36,37,39 and 40, use the RSA cipher with public key n = 713 = 23.31 and e = 43. 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. HELP

To determine

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

Explanation

Given information:

The numeric equivalents of H,E,LandP are 08,05,12and16.

Concept used:

To encrypt these letters, we have to compute 843and543and1243and 1643 modulo 713 since,

It is given that n=713ande=43

Calculation:

Write 43 as a sum of factors of two

43=32+8+2+1=25+23+21+1

Calculate 82kfor k=1,2,3,4,5

For H:88(mod713)by the definition of mod

82=64(mod713)84mod713=( 8 2mod713)2mod713=(64)2mod713=4096mod713since642=4096=531since 4096mod713=531

88( 8 4)2=(531)2mod713=281961(mod713)=326

816( 8 8)2=(326)2mod713=106276mod713since3262=106276=39mod713by definition of mod

832( 8 16)2(39)2mod713=1521mod713since392=1521=95mod713

Now,

843mod713=(8 32+8+2+1)mod713=(8 3288828)mod713=(95326648)mod713=158556640mod713using a calculator=233by the definition of mod

For E:

55(mod713)by definition of mod52=25(mod713)since 52=2554=625(mod713)since54=625

58=( 5 4)2=(625)2(mod713)=390625(mod713)=532(mod713)by the definition of mod

516=( 5 8)2=(614)2mod713=376996(mod713)using a calculator=532(mod713)by definition of mod

532=( 5 16)2=(532)2(mod713)=283024mod713using a calculator=676(mod713)

Thus,

543mod713=532+8+2+1mod713=(5 32585251)mod713=(676614255)mod713by algebra=(51883000)mod713=129using a calculator and by the definition of mod

For L:

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 