What is the sum of all natural numbers from 1 to 100 which are divisible by 2 and 5?

Many of us encounter math problems that seem a bit tricky at first glance. One such question that might pop up is: "What is the sum of all natural numbers from 1 to 100 which are divisible by 2 and 5?" This might sound complicated, but by breaking it down, we can find a clear and straightforward answer.

Understanding the Requirements

First, let's clarify what "natural numbers" are. In mathematics, natural numbers are the positive whole numbers starting from 1. So, we're looking at the numbers 1, 2, 3, and so on, all the way up to 100.

Next, we need to understand what "divisible by 2 and 5" means. A number is divisible by another number if it can be divided by that number with no remainder. If a number is divisible by both 2 and 5, it means it's a multiple of both 2 and 5.

The Magic of Multiples

When a number is divisible by both 2 and 5, it's also divisible by their least common multiple. The least common multiple (LCM) of 2 and 5 is 10.

Therefore, any number divisible by both 2 and 5 is essentially a multiple of 10.

Identifying the Numbers

Now that we know we're looking for multiples of 10 between 1 and 100, let's list them out:

These are all the natural numbers from 1 to 100 that are perfectly divisible by both 2 and 5.

Calculating the Sum

The final step is to add these numbers together. We can do this by simple addition:

10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 = ?

Let's calculate this sum:

10 + 20 = 30
30 + 30 = 60
60 + 40 = 100
100 + 50 = 150
150 + 60 = 210
210 + 70 = 280
280 + 80 = 360
360 + 90 = 450
450 + 100 = 550

So, the sum of all natural numbers from 1 to 100 which are divisible by 2 and 5 is 550.

An Alternative Approach: Arithmetic Series

For those who enjoy a bit more mathematical elegance, this problem can also be solved using the concept of an arithmetic series. An arithmetic series is a sequence of numbers such that the difference between the consecutive terms is constant. In our case, the sequence of multiples of 10 (10, 20, 30, ..., 100) forms an arithmetic series with:

The first term (a₁) = 10
The last term (a_n) = 100
The common difference (d) = 10
The number of terms (n) = 10 (as we listed them out earlier)

The formula for the sum of an arithmetic series is:

S_n = n/2 * (a₁ + a_n)

Plugging in our values:

S₁₀ = 10/2 * (10 + 100)

S₁₀ = 5 * (110)

S₁₀ = 550

This confirms our earlier calculation. Both methods lead to the same correct answer.

The key to solving this problem is understanding that being divisible by both 2 and 5 is the same as being divisible by 10.

Frequently Asked Questions (FAQ)

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

To find numbers divisible by both 2 and 5, you need to find their least common multiple (LCM). The LCM of 2 and 5 is 10. Therefore, any number divisible by both 2 and 5 will be a multiple of 10.

Why is the LCM important in this problem?

The LCM simplifies the problem. Instead of checking each number from 1 to 100 for divisibility by 2 and then by 5, we can simply look for multiples of 10, which directly satisfies both conditions simultaneously.

Are there any other ways to calculate the sum of these numbers?

Yes, as shown above, you can use the formula for the sum of an arithmetic series if you identify the numbers as a sequence with a common difference. This can be particularly useful for larger ranges of numbers.

What if the range was different, like 1 to 200?

If the range was 1 to 200, you would again find the multiples of 10 (10, 20, ..., 200). There would be 20 such numbers. Using the arithmetic series formula: S₂₀ = 20/2 * (10 + 200) = 10 * 210 = 2100.