How many numbers are divisible by 2 and 5 both lying between 101 and 999?

This is a great question that involves a bit of number theory, but we can break it down easily for anyone to understand. We're looking for numbers within a specific range – strictly between 101 and 999 – that have a special property: they must be divisible by *both* 2 and 5.

Understanding Divisibility by 2 and 5

Let's start with what it means for a number to be divisible by 2 and 5.

A number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, or 8). In simpler terms, it's an even number.
A number is divisible by 5 if its last digit is either a 0 or a 5.

Now, for a number to be divisible by *both* 2 and 5, it needs to satisfy both conditions. If a number must end in an even digit (for divisibility by 2) *and* end in a 0 or 5 (for divisibility by 5), the only digit that works for both is 0.

This means any number divisible by both 2 and 5 must end in a 0. Numbers ending in 0 are commonly known as multiples of 10.

Finding Multiples of 10 Between 101 and 999

So, our problem has now been simplified: we need to find how many multiples of 10 are strictly between 101 and 999.

The numbers we are considering are:

102, 103, 104, ..., 997, 998, 999.

We are looking for numbers in this list that end in 0. These will be:

Method 1: Listing and Counting (Illustrative)

While we won't list every single number, we can see a pattern. The first multiple of 10 after 101 is 110. The last multiple of 10 before 999 is 990.

To find the count, we can think of this as an arithmetic sequence where the first term is 110, the last term is 990, and the common difference is 10. The formula for the number of terms in an arithmetic sequence is:

Number of terms = ((Last Term - First Term) / Common Difference) + 1

Applying this:

Number of multiples of 10 = ((990 - 110) / 10) + 1

Number of multiples of 10 = (880 / 10) + 1

Number of multiples of 10 = 88 + 1

Number of multiples of 10 = 89

Method 2: Using Integer Division

Another way to approach this is by using integer division (which essentially means dividing and taking the whole number part of the answer).

We want numbers less than 999 that are divisible by 10. The largest multiple of 10 less than or equal to 999 is 990. The number of multiples of 10 up to 999 is 999 // 10 = 99.

We also want numbers greater than 101 that are divisible by 10. The largest multiple of 10 less than or equal to 101 is 100. The number of multiples of 10 up to 101 is 101 // 10 = 10.

Since we want numbers *between* 101 and 999, we need to subtract the multiples of 10 that are less than or equal to 101 from the multiples of 10 that are less than or equal to 999.

Total numbers divisible by 10 and less than or equal to 999 = 99

Total numbers divisible by 10 and less than or equal to 101 = 10

Therefore, the numbers divisible by 10 and strictly between 101 and 999 are:

99 - 10 = 89

Conclusion

Both methods confirm that there are 89 numbers that are divisible by both 2 and 5 (which means they are divisible by 10) lying strictly between 101 and 999.

Frequently Asked Questions (FAQ)

How do you find numbers divisible by both 2 and 5?

A number divisible by both 2 and 5 must be a multiple of their least common multiple. The least common multiple of 2 and 5 is 10. So, numbers divisible by both 2 and 5 are simply multiples of 10. This means they must end in the digit 0.

Why does a number divisible by 2 and 5 have to end in 0?

For a number to be divisible by 2, its last digit must be even (0, 2, 4, 6, 8). For a number to be divisible by 5, its last digit must be 0 or 5. The only digit that satisfies both conditions simultaneously is 0. Therefore, any number divisible by both 2 and 5 must end in 0.

What does "lying between 101 and 999" mean in terms of the numbers included?

"Lying between 101 and 999" means we are considering numbers that are strictly greater than 101 and strictly less than 999. So, 101 itself and 999 itself are not included in this range. The smallest number we consider is 102, and the largest is 998.