   Chapter 9.3, Problem 37ES ### 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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# For each of exercises 37-39, the number of elements in a certain set can be found by computing the number in a larger universe that are not in the set and subtracting this from the total in the larger universe. In each of these, as was the case for the solution to Example 93.6(b). De Morgan's laws and the inclusion/ exclusion rule can be used. How many positive integer less than 1.000 have no common factors with 1.000?

To determine

To find the number of positive integers less than 1000 have no comon factors with 1000.

Explanation

Given information:

Principle of inclusion/exclusion

Concept used:

n(AB)=n(A)+n(B)n(AB)

Calculation:

Use the fundamental theorem of arithmetic to write the number 1000 as product of prime powers.

1000=10×10×10=23×53

Let A be the set of all positive integers less than 1000 that are multiples of 2.

Let B be the set of all positive integers less than 1000 that are multiples of 5.

Since the only prime factors of 1000 are 2 and 5, the number of positive integers less than 1000 that have a common factor with 1000 is n(AB).

The set AB represents the set of positive integers less than 1000 that are multiples of 2 and 5, that is, multiple of 10.

The positive integers that are multiples of 2 and less than 1000 are 2,4,6,......,998.

The positive integers that are multiples of 5 and less than 1000 are 5,10,15,

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