Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 6 - Section 6.2 - The Pigeonhole Principle - Exercises - Page 405: 12

Answer

26 ordered pairs

Work Step by Step

a $mod$ 5 can take 5 values i.e {0,1,2,3,4} So for the ordered pair (a,b) we have 5$\times$5 = 25 combinations {(0,1),(0,2)..(1,0)......(4,4)} Hence by pigeonhole principle we can see that we need at least 26 ordered pairs ($a,b$).
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