Finding Numbers Divisible by 3 or 7 Between 1 and 200 Using Inclusion-Exclusion Principle
In this article, we will explore how to determine the count of numbers between 1 and 200 that are evenly divisible by 3 or 7. We will use the principle of inclusion-exclusion, a fundamental concept in combinatorics, to accurately calculate this count. This method will provide a clear and concise solution to the problem.
Understanding the Problem
We start by breaking down the problem into simpler parts. We need to find the number of integers between 1 and 200 that are divisible by either 3 or 7. To solve this, we will use the principle of inclusion-exclusion, which is a powerful tool for dealing with problems that involve counting overlapping sets.
Counting Multiples of 3
The first step is to count the multiples of 3 within the range 1 to 200. We can use the formula for the n-th term of an arithmetic sequence to find how many multiples of 3 there are.
The formula for the n-th term a_n of an arithmetic sequence is given by:
a_n a (n-1)d, where a is the first term and d is the common difference.
For multiples of 3:
a 3 d 3 a_n ≤ 200Substituting the values, we get:
3 (n-1)3 ≤ 200
(n-1)3 ≤ 197
3n - 3 ≤ 197
3n ≤ 200
n ≤ 200/3
n ≤ 66.67
Since n must be a whole number, n 66. Therefore, there are 66 multiples of 3 between 1 and 200.
Counting Multiples of 7
Next, we count the multiples of 7 within the same range.
Using the same formula for the n-th term of an arithmetic sequence:
a 7 d 7 a_n ≤ 200Substituting the values, we get:
7 (n-1)7 ≤ 200
(n-1)7 ≤ 193
7n - 7 ≤ 193
7n ≤ 200
n ≤ 200/7
n ≤ 28.57
Since n must be a whole number, n 28. Therefore, there are 28 multiples of 7 between 1 and 200.
Counting Multiples of 21
Now, we count the multiples of 21 (i.e., numbers that are divisible by both 3 and 7) to avoid double-counting.
a 21 d 21 a_n ≤ 200Substituting the values, we get:
21 (n-1)21 ≤ 200
(n-1)21 ≤ 179
21n - 21 ≤ 179
21n ≤ 200
n ≤ 200/21
n ≤ 9.52
Since n must be a whole number, n 9. Therefore, there are 9 multiples of 21 between 1 and 200.
Applying the Principle of Inclusion-Exclusion
The principle of inclusion-exclusion allows us to find the total count by subtracting the overlap counted twice. The total count of numbers divisible by 3 or 7 is given by:
Total Multiples of 3 Multiples of 7 - Multiples of both 3 and 7
Total 66 28 - 9
Total 85
Therefore, there are 85 numbers between 1 and 200 that are evenly divisible by 3 or 7.
Conclusion
Using the principle of inclusion-exclusion, we accurately determined that there are 85 numbers between 1 and 200 that are divisible by either 3 or 7. Understanding and applying this principle can be very useful in solving similar problems involving overlapping sets in arithmetic sequences.
Related Keywords
divisibility rules arithmetic sequences inclusion-exclusion principleAdditional Resources:
Learn more about arithmetic sequences Explore the inclusion-exclusion principle