Question

How many integers between 1 and 250 are divisible by 3 or \(5 ?\)

Short answer

There are 117 integers between 1 and 250 divisible by 3 or 5.

Step-by-step solution

Determine numbers divisible by 3

First, find the integers between 1 and 250 that are divisible by 3. The sequence of numbers that are divisible by 3 between 1 and 250 is an arithmetic sequence where the first term is 3 and the last term is the largest multiple of 3 less than or equal to 250. Calculate this last term: We divide 250 by 3 and take the integer part, which gives \(\lfloor 250/3 \rfloor = 83 \). Thus, the integers divisible by 3 are 3, 6, ..., 249. To find the number of terms, divide 249 by 3: \(n = \lfloor 249/3 \rfloor = 83\). There are 83 numbers divisible by 3.

Determine numbers divisible by 5

Next, find the integers between 1 and 250 that are divisible by 5. These numbers form a sequence with the first term as 5 and the last term as the largest multiple of 5 less than or equal to 250, which is 250. Calculate the number of terms: Divide 250 by 5, \( \lfloor 250/5 \rfloor = 50 \). So, there are 50 numbers divisible by 5.

Determine numbers divisible by both 3 and 5 (i.e., 15)

Find integers between 1 and 250 that are divisible by both 3 and 5, which means they are divisible by 15. The first term in this sequence is 15, and the largest term is 240. Determine the number of terms: Divide 240 by 15, \( \lfloor 240/15 \rfloor = 16 \). So, there are 16 numbers divisible by 15.

Apply the inclusion-exclusion principle

Use the inclusion-exclusion principle to find the count of numbers divisible by either 3 or 5. The formula is: \( |A \cup B| = |A| + |B| - |A \cap B| \). From previous steps: \(|A| = 83\), \(|B| = 50\), and \(|A \cap B| = 16\). Compute \(83 + 50 - 16 = 117\).

Conclusion

The total number of integers between 1 and 250 that are divisible by either 3 or 5 is 117. These calculations confirm the solution by considering the principles of divisibility and inclusion-exclusion.

Key concepts

Arithmetic Sequence

An arithmetic sequence is a simple yet powerful concept in mathematics. It is a sequence of numbers in which each term after the first is formed by adding a constant difference to the previous term. For example, in the exercise, you'd find numbers divisible by 3, 5, and 15.

• The sequence of numbers divisible by 3 starts at 3, like 3, 6, 9, ..., up to the largest number under or equal to 250. This sequence has a common difference of 3.
• Similarly, numbers divisible by 5 start at 5, such as 5, 10, 15, ..., up to 250.
• The arithmetic sequence for numbers divisible by both 3 and 5 is numbers divisible by 15. Start from 15, like 15, 30, 45, ..., up to 240.

In solving such problems, identifying the sequence's common difference and the largest relevant term helps ensure that all the numbers fitting the criteria are accurately counted.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is essential when you want to calculate the union of two sets. This principle ensures that numbers not counted twice by subtracting those in the intersection from the total.
Imagine you have two groups: numbers divisible by 3 and numbers divisible by 5. Calculating separately, we need to find those appearing in both groups, i.e., divisible by 15.
The formula for inclusion-exclusion is:

• \[|A \cup B| = |A| + |B| - |A \cap B|\]

This formula tells us how to account for overlap. In the problem, the result is:

• 83 numbers divisible by 3
• 50 numbers divisible by 5
• Subtract 16 that appear in both groups (15)

This subtraction gives the correct total of 117 unique numbers. This is how the overlap or duplicate counting is managed by the principle.

Divisibility

Divisibility in mathematics refers to one number being able to divide another number evenly with no remainder. This concept helps identify specific sequences or sets of numbers. For example, to solve the exercise, numbers divisible by 3 are those that can be expressed as 3 times an integer, such as 3, 6, 9, etc.

• A number is divisible by 3 if the sum of its digits is a multiple of 3.
• Divisibility by 5 is easily checked if the last digit is 0 or 5.
• For complex calculations, divisibility rules help quickly identify qualifying numbers without recounting.

Understanding these rules streamlines finding integers within a certain range that meet particular divisibility criteria.

Integer Count

Counting integers meeting certain conditions is a foundational skill in discrete mathematics. The exercise illustrated this by finding integers divisible by either 3 or 5 between 1 and 250.
We can express the process in a few simple steps:

• Identify the sequences using divisibility rules.
• Count the number of integers in each sequence by dividing the largest term by the common difference.
• Apply any necessary principles (like inclusion-exclusion) to handle overlaps.

These steps ensure an accurate count, avoiding common pitfalls like missing or double-counting numbers. Ultimately, you become adept at finding numbers in defined ranges that meet specific conditions efficiently.