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Find the number of integers between 1 and 1000 , including both 1 and 1000 , that are not divisible by any of \(5,6,\) or 8.

Short Answer

Expert verified
There are 600 integers between 1 and 1000 that are not divisible by 5, 6, or 8.

Step by step solution

01

Define the Total Set

Determine the total number of integers between 1 and 1000, inclusive. There are 1000 integers from 1 to 1000.
02

Calculate Multiples of 5

Find the number of integers between 1 and 1000 that are divisible by 5.Using the formula for the number of multiples, we have:\[ \left\lfloor \frac{1000}{5} \right\rfloor = 200 \] Thus, there are 200 integers divisible by 5.
03

Calculate Multiples of 6

Find the number of integers between 1 and 1000 that are divisible by 6.Using the formula for the number of multiples, we have:\[ \left\lfloor \frac{1000}{6} \right\rfloor = 166 \]Thus, there are 166 integers divisible by 6.
04

Calculate Multiples of 8

Find the number of integers between 1 and 1000 that are divisible by 8.Using the formula for the number of multiples, we have:\[ \left\lfloor \frac{1000}{8} \right\rfloor = 125 \]Thus, there are 125 integers divisible by 8.
05

Apply Inclusion-Exclusion for Two-Sets

Apply the inclusion-exclusion principle to find multiples of combinations of two divisors.For multiples of both 5 and 6 (i.e., 30):\[ \left\lfloor \frac{1000}{30} \right\rfloor = 33 \]For multiples of both 5 and 8 (i.e., 40):\[ \left\lfloor \frac{1000}{40} \right\rfloor = 25 \]For multiples of both 6 and 8 (i.e., 24):\[ \left\lfloor \frac{1000}{24} \right\rfloor = 41 \]
06

Apply Inclusion-Exclusion for Three-Sets

Apply the inclusion-exclusion principle to find multiples of all three divisors simultaneously.For multiples of 5, 6, and 8 (i.e., 120):\[ \left\lfloor \frac{1000}{120} \right\rfloor = 8 \]
07

Use Inclusion-Exclusion Principle

Use the inclusion-exclusion principle to calculate the total number of integers divisible by 5, 6, or 8.\[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \]Substitute the values:\[ 200 + 166 + 125 - 33 - 25 - 41 + 8 = 400 \]Thus, there are 400 integers between 1 and 1000 that are divisible by 5, 6, or 8.
08

Find Non-Divisible Integers

Subtract the number of integers divisible by either 5, 6, or 8 from the total number of integers.\[ 1000 - 400 = 600 \]Thus, there are 600 integers between 1 and 1000 that are not divisible by 5, 6, or 8.

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Key Concepts

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

Divisibility
Divisibility is a mathematical concept that deals with whether one integer can be divided by another without leaving a remainder. In our exercise, we are interested in numbers between 1 and 1000 that are not divisible by the integers 5, 6, and 8. To determine this, the first step is to count how many numbers are divisible by each of these integers. For example, to find the quantity of numbers divisible by 5, divide 1000 by 5, and consider the integer portion of the result: \[ \left\lfloor \frac{1000}{5} \right\rfloor = 200 \] This indicates there are 200 numbers between 1 and 1000 that can be cleanly divided by 5. We repeat this process for other divisors as well, like 6 and 8, to understand which numbers fit each category. Comprehending divisibility allows us to sort the integers by their divisibility properties, which sets the stage for more complex combinatorial counting methods.
Combinatorics
Combinatorics is an area of mathematics concerned with counting, arranging, and finding patterns in numbers. It becomes powerful when combined with concepts like the Inclusion-Exclusion Principle. In this exercise, we use combinatorics to count how many integers from 1 to 1000 are not divisible by 5, 6, or 8. Instead of simply adding up integers, combinatorics helps us to adjust for any overlaps—integers that are divisible by multiple numbers in our case (like both 5 and 6). By applying combinatorial techniques, we prevent overcounting. This involves subtracting counts of combinations (e.g., numbers divisible by both 5 and 6) and adding back those divisible by all three divisors, ensuring accurate totals. The Inclusion-Exclusion Principle is integral as it refines our counting by considering intersections of sets (e.g., integers divisible by a pair like 5 and 8), hence ensuring no double-counting occurs.
Integer Counting
Integer counting is the process of determining the number of whole numbers within a certain range that meet specified criteria. In this exercise, integer counting helps us determine how many numbers from 1 to 1000 are not divisible by the numbers 5, 6, or 8. To achieve this, we consider the total amount of integers in the range first.
  • Total integers from 1 to 1000: 1000
  • Integers that are divisible by 5, 6 or 8 (using divisibility and combinatorics): 400
Subtracting the divisible count from the total gives us the count of integers that do not meet those divisibility criteria:\[ 1000 - 400 = 600 \]This approach illustrates how integer counting leverages both direct counting strategies and corrective adjustments using combinatorial principles.

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Most popular questions from this chapter

Let \(X=\\{2,4 \mid, Y=\\{1,4 \mid,\) and \(Z=\\{0,4,8 \mid\). Construct the following sets: (a) \(X \times Y\) (b) \(X \times Y \times Z\) (c) \(Y \times Z\) (d) \(Z \times Y \times X\) (c) \(Z \times X \times Y\)

Prove by induction that the following identities are true for the Fibonacci numbers: (a) \(\sum_{i=0}^{n} F_{2 j+1}=F_{2 n+2}-1\) for \(n \geq 0\) (b) \(\sum_{i=1}^{n} F_{i}^{2}=F_{n} \cdot F_{n+1}-1\) for \(n \geq 1\) (c) \(\sum_{i=0}^{n} F_{i}=F_{n+2}-1\) for \(n \geq 0\)

A fixed dose of a given drug increases the concentration of that drug above normal levels in the bloodstream by an amount \(C_{0}\) (measured in percent). The effect of the drug wears off over time such that the concentration at some time \(t\) is \(C_{0} e^{-k t}\) where \(k\) is the known rate at which the concentration of the drug in the bloodstream declines. (a) Find the residual concentration \(R\), the accumulated amount of the drug above normal levels in the bloodstream, at time \(t\) after \(n\) doses given at intervals of \(t_{0}\) hours starting with the first dose at \(t=0\). (b) If the drug is alcohol and 1 oz. of alcohol has \(C_{0}=0.05 \%\), how often can a "dose" be taken so that the residual concentration is never more than \(0.15 \%\) ? Assume \(k=(1 / 3) \ln (2)\)

(a) Suppose you take out a mortgage for \(A\) dollars at a monthly interest rate \(I\) and a monthly payment \(P\). (To calculate \(I\) : if the annual interest rate is \(12 \%\), divide by 12 to get a monthly rate of \(1 \%,\) then replace the percentage with the decimal fraction 0.01.) Let \(A_{n}\) denote the amount you have left to pay off after \(n\) months. So, \(A_{0}=A\) by definition. At the end of each month, you are first charged interest on all the money you owed during the month, and then your payment is subtracted. So. $$A_{n+1}=A_{n}(1+I)-P$$ Prove by induction that $$A_{n}=\left(A-\frac{P}{I}\right)(1+I)^{n}+\frac{P}{I}$$ (b) Use this to calculate the monthly payment on a 30 -year loan of \(\$ 100,000\) at \(12 \%\) interest per year. (Note that the formula is inexact, since moncy is always rounded off to a whole number of cents. The derivation here does not do that. We use \(12 \%\) to make the arithmetic easier. You should consult a local bank to find a current value.)

Find the expression tree for the formula $$((((\neg(\neg p)) \wedge(\neg q)) \wedge r) \vee(((\neg(\neg q)) \wedge(\neg r)) \wedge s)) \leftrightarrow(s \rightarrow p)$$

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