   Chapter 8.4, Problem 32ES ### 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 an inverse for 41 modulo 660. Find the least positive solution for the following congruence: 41x=125(mod 660).

To determine

(a)

To find:

An inverse for 41 modulo 660.

Explanation

Given information:

41mod660

Concept used:

amod b = Remainder, when a is divided by b.

Calculation:

The objective is to find an inverse for 41 modulo 660.

First. Find the greatest common divisor using the quotient remainder theorem as,

41=0660+41660=1641+441=104+14=41+0

The greatest common divisor when we divide 41 by 660 is the last non zero remainder.

Thus, the greatest common divisor is 1.

Now, express the greatest common divisor as a linear combination of 41 and 660.

1=41104=141104=14110⋅<

To determine

(b)

To find:

The least positive solution for the following congruence: 41x=125(mod660)

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