   Chapter 8.4, Problem 38ES ### 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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# Find the least positive inverse for 43 modulo 660.

To determine

To find:

The least positive inverse for 43modulo 660.

Explanation

Given information:

43 mod 660

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:

First check if gcd(660,43)=1.

Use Euclidean algorithm to find gcd(660,43).

Divide 660 by 43 to get quotient=15 and remainder=15.

Then, 660=4315+15 and hence gcd(660,43)=gcd(43,15)

Divide 43 by 15 to get quotient=2 and remainder=13

Then, 43=152+13 and hence gcd(43,15)=gcd(15,13)

Divide 15 by 13 to get quotient=1 and remainder=2

Then, 15=131+2 and hence gcd(15,13)=gcd(13,2)

Divide 13 by 2 to get quotient=6 and remainder=1

Then, 13=26+1 and hence gcd(13,2)=gcd(2,1)

Divide 2 by 1 to get quotient=2 and remainder=0

Then, 2=12+0 and hence gcd(2,1)=gcd(1,0)

Now, gcd(1,0)=1.

Hence,

gcd(660,43)=gcd(43,15)=gcd(15,13)=gcd(13,2)=gcd(2,1)=gcd(1,0)=1

Since gcd(660,43)=1, the inverse for 43modulo 660 exists.

Find the inverse for 43modulo 660.

To do so, find a linear combination of 43and660 that equals 1.

Now,

660=4315+1515=6604315

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