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 6.1 - The Basics of Counting - Exercises - Page 396: 20

Answer

a) 9 b) 6 c) 2

Work Step by Step

Note that neither 5 nor 31 is divisible by either 3 or 4. a) There are $\left \lfloor{\frac{31}{3}}\right \rfloor$=10 integers less than 31 that are divisible by 3, and $\left \lfloor{\frac{5}{3}}\right \rfloor$=1 of them are less than 5 as well. This leaves 10-1=9 numbers between 5 and 31 that are divisible by 3. They are 6, 9, 12, 15, 18, 21, 24, 27 and 30. b) There are $\left \lfloor{\frac{31}{4}}\right \rfloor$=7 integers less than 31 that are divisible by 4, and $\left \lfloor{\frac{5}{4}}\right \rfloor$=1 of them are less than 5 as well. This leaves 7-1=6 numbers between 5 and 31 that are divisible by 3. They are 8, 12, 16, 20, 24 and 28. c) A number is divisible by both 3 and 4 if and only if it is divisible by their least common multiple, which is 12. Obviously there are only two such number between 5 and 31, namely 12 and 24. This could also be done like previous parts : $\left \lfloor{\frac{31}{12}}\right \rfloor$-$\left \lfloor{\frac{5}{12}}\right \rfloor$=2-0=2. This can also be seen as intersection of integers obtained in part a and b.
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