/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 An integer is called squarefree ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 6

An integer is called squarefree if it is not divisible by the square of a positive integer greater than \(1 .\) Find the number of squarefree positive integers less than 100 .

Short Answer

Expert verified
There are 71 square-free positive integers less than 100.

Step by step solution

01

- Identify squares of integers greater than 1

List the squares of positive integers greater than 1 that are less than 100. These squares are 4, 9, 16, 25, 36, 49, 64, and 81.
02

- Count multiples of each square

Count the multiples of each square number that are less than 100. The multiples are:- 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 (25 multiples)- 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99 (11 multiples)- 16: 16, 32, 48, 64, 80, 96 (6 multiples)- 25: 25, 50, 75 (3 multiples)- 36: 36, 72 (2 multiples)- 49: 49, 98 (2 multiples)- 64: 64 (1 multiple)- 81: 81 (1 multiple)
03

- Use the inclusion-exclusion principle

Since some numbers are counted multiple times, use the inclusion-exclusion principle to avoid double-counting.- Multiples of 4 and 9: 36, 72 (2 multiples)- Multiples of 4 and 16: 64, 96 (2 multiples)- Adjust for these overlaps.
04

- Calculate and adjust for overlaps

Adjust the counts by subtracting the overlaps to find the unique multiples of squares less than 100. Total overlaps: 4 (there are 132 - 4 = 28 unique multiples).
05

- Find the number of squarefree integers

Subtract the number of unique multiples from 99 (since we are considering numbers from 1 to 99):99 - 28 = 71

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

square divisors
A squarefree integer is a number that isn鈥檛 divisible by any perfect square greater than 1. To understand this, let's break it down into simple terms. A perfect square is a number that results from squaring an integer, such as 4 (which is 2 squared) or 9 (which is 3 squared). The concept of 'divisors' means you are looking for numbers that can be divided without a remainder. In this case, we are interested in perfect squares as divisors. For example, 12 is not squarefree since it can be divided by 4. However, 10 is squarefree because it cannot be divided by any perfect squares other than 1.
inclusion-exclusion principle
The inclusion-exclusion principle helps us count the number of elements in the union of overlapping sets more accurately. Imagine you have two overlapping sets: Set A and Set B. If you simply add the sizes of Set A and Set B, you count the overlap twice. Therefore, you need to subtract the overlap once to correct the count. In our context, this principle is crucial because the multiples of squares such as 4 and 9 overlap (e.g., 36 and 72). By using the inclusion-exclusion principle, we adjust our count to avoid this double-counting. Subtracting these overlaps ensures we get the exact number of unique multiples of the squares.
counting multiples
Counting multiples is key to determining which numbers are not squarefree. To count multiples of a given perfect square under 100, list all the multiples incrementally. For instance, to find multiples of 4 under 100: 4, 8, 12, 16, and so on, up to 96. Similarly, do this for other perfect squares like 9, 16, 25, etc. By counting these, you identify the numbers that undermine the squarefree status. After counting the multiples of each square and then correctly adjusting for overlaps (using the inclusion-exclusion principle), you subtract this from the total numbers under 100 (i.e., 99) to find the squarefree integers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use generating functions to solve the recurrence relation \(a_{k}=7 a_{k-1}\) with the initial condition \(a_{0}=5\)

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that if \(n\) is a positive integer, then the number of partitions of \(n\) into distinct parts equals the number of partitions of \(n\) into odd parts with repetitions allowed; that is, \(p_{o}(n)=p_{d}(n) .[\text { Hint: Show that the generating }\) functions for \(p_{o}(n)\) and \(p_{d}(n)\) are equal. \(]\)

a) Find a recurrence relation for the number of ternary strings of length \(n\) that do not contain consecutive symbols that are the same. b) What are the initial conditions? c) How many ternary strings of length six do not contain consecutive symbols that are the same?

Exercises 33鈥�37 deal with a variation of the Josephus problem described by Graham, Knuth, and Patashnik in [GrKnPa94]. This problem is based on an account by the historian Flavius Josephus, who was part of a band of 41 Jewish rebels trapped in a cave by the Romans during the Jewish Roman war of the first century. The rebels preferred suicide to capture; they decided to form a circle and to repeatedly count off around the circle, killing every third rebel left alive. However, Josephus and another rebel did not want to be killed this way; they determined the positions where they should stand to be the last two rebels remaining alive. The variation we consider begins with n people, numbered 1 to n, standing around a circle. In each stage, every second person still left alive is eliminated until only one survives. We denote the number of the survivor by J(n). Determine \(J(100), J(1000),\) and \(J(10,000)\) from your formula for \(J(n) .\)

Suppose someone picks a number \(x\) from a set of \(n\) numbers. A second person tries to guess the number by successively selecting subsets of the \(n\) numbers and asking the first person whether \(x\) is in each set. The first person answers either "yes" or "no." When the first person answers each query truthfully, we can find x Links using log n queries by successively splitting the sets used in each query in half. Ulam鈥檚 problem, proposed by Stanislaw Ulam in 1976, asks for the number of queries required to find x, supposing that the first person is allowed to lie exactly once. a) Show that by asking each question twice, given a number \(x\) and a set with \(n\) elements, and asking one more question when we find the lie, Ulam's problem can be solved using \(2 \log n+1\) queries. b) Show that by dividing the initial set of \(n\) elements into four parts, each with \(n / 4\) elements, 1\(/ 4\) of the elements can be eliminated using two queries. [Hint: Use two queries, where each of the queries asks whether the element is in the union of two of the subsets with \(n / 4\) elements and where one of the subsets of \(n / 4\) elements is used in both queries.] c) Show from part (b) that if \(f(n)\) equals the number of queries used to solve Ulam's problem using the method from part (b) and \(n\) is divisible by \(4,\) then \(f(n)=f(3 n / 4)+2\) . d) Solve the recurrence relation in part (c) for \(f(n)\) e) Is the naive way to solve Ulam鈥檚 problem by asking each question twice or the divide-and-conquer method based on part (b) more efficient? The most efficient way to solve Ulam鈥檚 problem has been determined by A. Pelc [Pe87].
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.