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How many numbers between 1 and 1000 are not divisible by \(3,7,\) or \(9 ?\)

Short Answer

Expert verified
There are 572 numbers between 1 and 1000 that are not divisible by 3, 7, or 9.

Step by step solution

01

Calculate Total Numbers

Identify the total number of integers between 1 and 1000. Since the numbers start from 1 and end at 1000, the total count of numbers is 1000.
02

Use Inclusion-Exclusion Principle

To find the count of numbers divisible by 3 or 7 or 9, we use the inclusion-exclusion principle: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \] where:- \( A \) is the set of numbers divisible by 3- \( B \) is the set of numbers divisible by 7- \( C \) is the set of numbers divisible by 9
03

Count Numbers Divisible by Each Individual Number

- \(|A|\) (divisible by 3): To get the count, divide 1000 by 3, floor the result: \( \left\lfloor \frac{1000}{3} \right\rfloor = 333 \)- \(|B|\) (divisible by 7): \( \left\lfloor \frac{1000}{7} \right\rfloor = 142 \)- \(|C|\) (divisible by 9): \( \left\lfloor \frac{1000}{9} \right\rfloor = 111 \)
04

Count Numbers Divisible by Pairs Using LCM

- \(|A \cap B|\) (divisible by LCM of 3 and 7 = 21): \( \left\lfloor \frac{1000}{21} \right\rfloor = 47 \)- \(|A \cap C|\) (divisible by LCM of 3 and 9 = 9): This is the same as \(|C|\), which is 111, because every 9 is a 3.- \(|B \cap C|\) (divisible by LCM of 7 and 9 = 63): \( \left\lfloor \frac{1000}{63} \right\rfloor = 15 \)
05

Count Numbers Divisible by All Three Using LCM

- \(|A \cap B \cap C|\) (divisible by LCM of 3, 7, and 9 = 63): Since it is already calculated in the intersection of two sets \(|B \cap C|\), the value is 15.
06

Apply Inclusion-Exclusion Formula

\[|A \cup B \cup C| = 333 + 142 + 111 - 47 - 111 - 15 + 15 = 428\]This is the count of numbers between 1 and 1000 that are divisible by either 3, 7, or 9.
07

Find Numbers Not Divisible by 3, 7, or 9

Subtract the count of numbers divisible by 3, 7, or 9 from the total numbers:\[1000 - 428 = 572\]So there are 572 numbers that are not divisible by 3, 7, or 9.

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Key Concepts

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

Divisibility Rules
Understanding divisibility rules is essential when solving problems involving finding the number of integers within a specific range that meet particular conditions. Divisibility rules help us quickly determine whether one number can be divided by another without leaving a remainder. For example, a number is divisible by 3 if the sum of its digits is divisible by 3. Similarly, a number is divisible by 7 if doubling the last digit and subtracting it from the rest of the number results in a number that is divisible by 7. For 9, the rule is similar to that of 3: a number is divisible by 9 if the sum of its digits is divisible by 9.

Learning these rules can significantly ease calculations, especially when dealing with larger sets of numbers. By applying divisibility rules, we can quickly assess how many numbers in a range, such as from 1 to 1000, meet certain criteria. These rules become even more powerful when combined with concepts like the Least Common Multiple (LCM) and integer counting to solve extensive counting problems.
Least Common Multiple (LCM)
The Least Common Multiple, often abbreviated as LCM, is the smallest multiple that is shared by two or more numbers. It plays a crucial role in solving problems involving divisibility, especially when we need to count numbers that are divisible by the least common multiples of several numbers. The LCM helps us in accounting for overlaps where numbers are divisible by two or more factors, ensuring we do not double count them.

To find the LCM of two numbers, you list the multiples of each number until you find the smallest common one. Alternatively, you can use prime factorization, where you multiply the highest powers of all prime factors that appear in any of the numbers. For instance, the LCM of 3 and 7 is 21, while the LCM of 3 and 9 is 9, as every multiple of 9 is also a multiple of 3 due to the factors involved.

This concept is utilized in the Inclusion-Exclusion Principle, where the LCM helps find how many numbers between 1 and 1000 are divisible by pairs or combinations of numbers like 3, 7, and 9. Knowing the LCM allows us to correctly identify and count such numbers without missing or overcounting.
Integer Counting
Integer counting is a fundamental concept in mathematics that deals with counting the number of integers in a particular set. This concept is especially useful when you're tasked with finding how many numbers within a certain range meet specific criteria. For example, to count how many integers are between 1 and 1000 that are not divisible by 3, 7, or 9, you start by counting all integers present in the given range, which totals to 1000.

From there, you apply methods like the Inclusion-Exclusion Principle to eliminate overcounts when dealing with multiple divisibility conditions. Using integer counting, you systematically determine how many numbers are divisible by one or more criteria and then subtract this from the total to find those that do not meet the given conditions.

Integer counting, combined with other mathematical principles and rules, allows us to tackle complex problems in a structured way, making it possible to reach accurate conclusions in problems involving large sets of numbers.

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Most popular questions from this chapter

Which of the following statements are correct? Prove each correct statement. Disprove each incorrect statement by finding a counterexample. (a) \(A\) and \(B\) are disjoint if and only if \(B\) and \(A\) are disjoint. (Read the statement carefully - the order in which the sets are listed might matter') (b) \(A \cup B\) and \(C\) are disjoint if and only if both the following are true: (i) \(A\) and \(C\) are disjoint and (ii) \(B\) and \(C\) are disjoint. (c) \(A \cap B\) and \(C\) are disjoint if and only if both the following are true: (i) \(A\) and \(C\) are disjoint and (ii) \(B\) and \(C\) are disjoint. (d) \(A \cup B\) and \(C\) are disjoint if and only if one of the following is true: (i) \(A\) and \(C\) are disjoint or (ii) \(B\) and \(C\) are disjoint. (e) \(A \cap B\) and \(C\) are disjoint if and only if one of the following is true: (i) \(A\) and \(C\) are disjoint or (ii) \(B\) and \(C\) are disjoint. (f) Let \(U\) be a universal set with \(A, B \subseteq U, A\) and \(B\) are disjoint if and only if \(\bar{A}\) and \(\bar{B}\) are disjoint.

Prove by induction: (a) \(0 \cdot 2^{0}+1 \cdot 2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\cdots+n \cdot 2^{n}=(n-1) 2^{n+1}+2\) for \(n \geq 0\) (b) \(1^{2}+3^{2}+5^{2}+\cdots+(2 n+1)^{2}=(n+1)(2 n+1)(2 n+3) / 3\) for \(n \geq 0\) (c) \(1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n-1} n^{2}=(-1)^{n-1} n(n+1) / 2\) for \(n \geq 0\) (d) \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n \cdot(n+1)=n(n+1)(n+2) / 3\) for \(n \geq 0\) (e) \(1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\cdots+n \cdot(n+1) \cdot(n+2)=n(n+1)(n+2)$$(n+3) / 4\) for \(n \geq 0\)

The terms of a sequence are given recursively as \(a_{0}=2, a_{1}=6,\) and \(a_{n}=2 a_{n-1}+\) \(3 a_{n-2}\) for \(n \geq 2\). Prove by induction that \(b_{n}=2 \cdot 3^{n}\) is a closed form for the sequence.

For (a) and (b), prove the stated result. For (c) and (d), find a counterexample to show that these conjcctures are false. (a) \(A \oplus B=(A \cup B)-(A \cap B)\) (b) \(A \cap(B \oplus C)=(A \cap B) \oplus(A \cap C)\) (c) \((A \cap B) \oplus(C \cap D) \subseteq(A \oplus C) \cap(B \oplus D)\) (d) \((A \cup B) \oplus(C \cup D) \subseteq(A \cup C) \oplus(B \cup D)\)

Prove by induction for all \(n \in \mathbb{N}\) : (a) 7 divides \(n^{7}-n\) (b) 11 divides \(n^{11}-n\) (c) 13 divides \(n^{13}-n\) (d) 120 divides \(n^{5}-5 n^{3}+4 n\)

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