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 4.4 - Page 189: 24

Answer

See below.

Work Step by Step

Given m mod 5=2 and n mod 5=3, we can write: m=5k+2 and n=5p+3, where k, p are integers. We have $mn=(5k+2)(5p+3)=25kp+15k+10p+6\\ =5(5kp+3k+2p+1)+1$ Thus mn mod 5 = 1 Note: some textbooks may have a printing error for n mod 5 = 3.
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