How many natural numbers between 101 and 999 which are divisible by 2 or 5? A Detailed Guide

This article will dive deep into answering a common mathematical question: "How many natural numbers between 101 and 999 are divisible by 2 or 5?" We'll break down the problem step-by-step, using clear language and focusing on practical ways to arrive at the correct answer. This isn't just about a number; it's about understanding the principles behind it.

Understanding the Question

First, let's clarify what we're looking for. We need to count all the whole, positive numbers (natural numbers) that fall within the range of 101 up to, but not including, 999. The key condition is that these numbers must be divisible by either 2 or 5. Divisible means that when you divide the number by 2 or 5, there's no remainder.

What are Natural Numbers?

Natural numbers are the positive counting numbers: 1, 2, 3, 4, and so on. In this problem, our range starts after 100 and ends before 1000, so we're considering numbers like 101, 102, ..., 998.

What does "Divisible by 2 or 5" Mean?

A number is divisible by 2 if it's an even number (ends in 0, 2, 4, 6, or 8). A number is divisible by 5 if it ends in 0 or 5. When we say "divisible by 2 *or* 5," we mean numbers that satisfy at least one of these conditions. This includes numbers divisible by both 2 and 5 (which are numbers divisible by 10).

Strategy: The Principle of Inclusion-Exclusion

To solve this, we'll use a powerful mathematical concept called the Principle of Inclusion-Exclusion. This principle helps us count items in the union of two sets without double-counting. In our case, the two sets are:

Set A: Numbers between 101 and 999 divisible by 2.
Set B: Numbers between 101 and 999 divisible by 5.

The principle states:

Total (A or B) = |A| + |B| - |A and B|

Where:

|A| is the number of elements in Set A.
|B| is the number of elements in Set B.
|A and B| is the number of elements that are in *both* Set A and Set B (i.e., divisible by both 2 and 5).

Step 1: Count Numbers Divisible by 2

We need to find how many numbers between 101 and 999 are divisible by 2. These are the even numbers.

Finding the First and Last Even Number in the Range

The first natural number greater than 101 that is divisible by 2 is 102.

The last natural number less than 999 that is divisible by 2 is 998.

Calculating the Count

We can use a formula to find the count of an arithmetic sequence: Count = (Last Term - First Term) / Common Difference + 1.

In this case, the common difference is 2.

Count (divisible by 2) = (998 - 102) / 2 + 1

Count (divisible by 2) = 896 / 2 + 1

Count (divisible by 2) = 448 + 1

So, there are 449 numbers between 101 and 999 that are divisible by 2.

Step 2: Count Numbers Divisible by 5

Now, let's find how many numbers between 101 and 999 are divisible by 5.

Finding the First and Last Number Divisible by 5

The first natural number greater than 101 that is divisible by 5 is 105 (since 101, 102, 103, 104 are not).

The last natural number less than 999 that is divisible by 5 is 995 (since 996, 997, 998 are not).

Calculating the Count

Using the same formula with a common difference of 5:

Count (divisible by 5) = (995 - 105) / 5 + 1

Count (divisible by 5) = 890 / 5 + 1

Count (divisible by 5) = 178 + 1

So, there are 179 numbers between 101 and 999 that are divisible by 5.

Step 3: Count Numbers Divisible by Both 2 and 5 (i.e., by 10)

Numbers divisible by both 2 and 5 are numbers divisible by their least common multiple, which is 10. These are numbers ending in 0.

Finding the First and Last Number Divisible by 10

The first natural number greater than 101 that is divisible by 10 is 110.

The last natural number less than 999 that is divisible by 10 is 990.

Calculating the Count

Using the formula with a common difference of 10:

Count (divisible by 10) = (990 - 110) / 10 + 1

Count (divisible by 10) = 880 / 10 + 1

Count (divisible by 10) = 88 + 1

So, there are 89 numbers between 101 and 999 that are divisible by both 2 and 5.

Step 4: Apply the Principle of Inclusion-Exclusion

Now we put it all together:

Total (divisible by 2 or 5) = (Numbers divisible by 2) + (Numbers divisible by 5) - (Numbers divisible by both 2 and 5)

Total = 449 + 179 - 89

Total = 628 - 89

Total = 539

Therefore, there are 539 natural numbers between 101 and 999 that are divisible by 2 or 5.

Alternative Approach: Counting Numbers NOT Divisible

Another way to think about this is to count the numbers that are *not* divisible by 2 and *not* divisible by 5, and then subtract that from the total number of integers in the range.

Total Numbers in the Range

The total number of natural numbers from 101 to 998 (inclusive) is 998 - 101 + 1 = 898.

Numbers NOT Divisible by 2 or 5

A number is NOT divisible by 2 if it's odd. A number is NOT divisible by 5 if it doesn't end in 0 or 5.

This means we are looking for odd numbers that do not end in 5. These are numbers ending in 1, 3, 7, or 9.

In every block of 10 consecutive numbers (e.g., 101-110), there are 4 numbers that fit this criteria (101, 103, 107, 109). Let's verify our range.

Numbers from 101 to 999. Total 898 numbers.

Let's consider the numbers from 1 to 999. Total 999 numbers.

Numbers not divisible by 2 or 5 from 1 to 999: In every 10 numbers, 4 are not divisible by 2 or 5 (e.g., 1, 3, 7, 9). So, (999 / 10) * 4 = 99 * 4 = 396. We need to be careful with the remainder.

A more direct way is to consider the count of numbers divisible by 2 or 5 within a full range and then adjust for our specific range.

Let's stick to the Inclusion-Exclusion Principle for clarity, as it directly answers the question.

Summary of the Answer

We have systematically determined that:

There are 449 numbers between 101 and 999 divisible by 2.
There are 179 numbers between 101 and 999 divisible by 5.
There are 89 numbers between 101 and 999 divisible by both 2 and 5 (i.e., by 10).

Using the Principle of Inclusion-Exclusion (Total = Set A + Set B - Both), we get:

449 + 179 - 89 = 539

So, the final answer to "How many natural numbers between 101 and 999 which are divisible by 2 or 5?" is 539.

This problem highlights the importance of systematic counting and avoiding double-counting, a fundamental concept in mathematics and data analysis.

Frequently Asked Questions (FAQ)

How do I find the first and last number in a range that is divisible by a certain number?

To find the first number greater than or equal to a starting number (X) that is divisible by a divisor (D), you can divide X by D, take the ceiling of the result (round up to the nearest whole number), and then multiply by D. For the last number less than or equal to an ending number (Y) that is divisible by D, divide Y by D, take the floor of the result (round down to the nearest whole number), and then multiply by D.

Why do we subtract the numbers divisible by both 2 and 5?

We subtract them because the Principle of Inclusion-Exclusion states that when you add the counts of two sets, you've counted the elements that belong to *both* sets twice. To correct this overcounting, you must subtract the count of those overlapping elements (the intersection of the sets) once.

What if the range included 100 or 1000?

If the range included 100 or 1000, we would adjust the "First" and "Last" numbers accordingly. For instance, if the range was "between 100 and 1000 inclusive," then 100 would be the first number divisible by 2, 5, and 10, and 1000 would be the last.

Is there a simpler way to think about divisibility by 2 or 5?

Yes, numbers divisible by 2 end in 0, 2, 4, 6, or 8. Numbers divisible by 5 end in 0 or 5. So, numbers divisible by 2 *or* 5 are simply numbers that end in 0, 2, 4, 5, 6, or 8. This means any number ending in 1, 3, 7, or 9 is *not* divisible by 2 or 5.