Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.8 - Page 225: 16

Answer

gcd(4131,2431) 4131 = 2431*1 + 1700 2431 =1700*1 + 731 1700 = 731*2 + 238 731 = 238*3+17 (last non zero remainder = gcd) 238 = 17*14+0 Therefore gcd (4131,2431) = 17

Work Step by Step

Steps: 1. Break down the larger number in terms of a smaller number. For eg: if a,b such that a>b then a = bm+x (x is remainder) 2. Now break b in terms of terms of x such that b = xm' + c (c is remainder) 3. Repeat step 2 until the term c becomes 0. The last non-zero remainder term is the gcd.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.