Sum Of Integers 1-100 Divisible By 2 Or 5: Explained!

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Sum of Integers from 1 to 100 Divisible by 2 or 5: Explained!

Hey guys! Ever get those math problems that seem a bit tricky at first? Let's break down one of those today: finding the sum of all the integers from 1 to 100 that are divisible by either 2 or 5. It might sound daunting, but trust me, we'll make it super clear and even a little fun! So, grab your thinking caps, and let’s dive in!

Understanding the Problem

Before we jump into calculations, let's really understand what we're trying to find. We need to identify all the numbers between 1 and 100 that can be divided evenly by 2 or 5 (or both!). Think about it: numbers like 2, 4, 5, 6, 8, 10... all the way up to 100. Then, we need to add all those numbers together. Sounds like a task, right? But we'll tackle it step-by-step.

Divisibility by 2: These are all the even numbers. They’re easy to spot – 2, 4, 6, 8, and so on. In our range of 1 to 100, there are exactly 50 of these even numbers. Why? Because every other number is divisible by 2. So, half of 100 is 50. Keep this number in mind; it’s important!

Divisibility by 5: Numbers divisible by 5 end in either a 0 or a 5. Think 5, 10, 15, 20, and so forth. Between 1 and 100, we have 20 numbers divisible by 5. You can find this by simply dividing 100 by 5. Simple, right?

The Tricky Part: Now, here's where it gets a little interesting. Some numbers are divisible by both 2 and 5. These are the multiples of 10 (like 10, 20, 30...). If we just added the sum of numbers divisible by 2 and the sum of numbers divisible by 5, we'd be counting these multiples of 10 twice. We don’t want that! It's like inviting the same guest to a party twice – we’d have too many folks and not enough cake!

So, our strategy is going to be:

  1. Find the sum of numbers divisible by 2.
  2. Find the sum of numbers divisible by 5.
  3. Find the sum of numbers divisible by 10 (the ones we counted twice).
  4. Subtract the sum of numbers divisible by 10 from the combined sums of numbers divisible by 2 and 5. This gets rid of the duplicates.

Clear as mud? Don’t worry, it will be! Let’s get calculating!

Calculating the Sums

Okay, let's put on our calculation hats! We're going to use a handy little formula for finding the sum of an arithmetic series. An arithmetic series is just a sequence of numbers that increase by a constant amount (like 2, 4, 6, 8... or 5, 10, 15, 20...).

The formula is:

Sum = (n / 2) * (first term + last term)

Where:

  • n = the number of terms in the series
  • first term = the first number in the series
  • last term = the last number in the series

Let's break it down for each case.

Sum of Numbers Divisible by 2

  • First term: 2
  • Last term: 100
  • Number of terms (n): 50 (as we figured out earlier)

Plugging these values into our formula:

Sum (divisible by 2) = (50 / 2) * (2 + 100) = 25 * 102 = 2550

So, the sum of all the even numbers from 1 to 100 is 2550. Awesome!

Sum of Numbers Divisible by 5

  • First term: 5
  • Last term: 100
  • Number of terms (n): 20 (we calculated this earlier)

Using the formula again:

Sum (divisible by 5) = (20 / 2) * (5 + 100) = 10 * 105 = 1050

Alright! The sum of all numbers divisible by 5 is 1050. We're on a roll!

Sum of Numbers Divisible by 10

These are the numbers we've counted twice, so we need to subtract their sum. Multiples of 10 are 10, 20, 30... 100.

  • First term: 10
  • Last term: 100
  • Number of terms (n): 10 (100 / 10 = 10)

Let’s use that formula one more time:

Sum (divisible by 10) = (10 / 2) * (10 + 100) = 5 * 110 = 550

Got it! The sum of numbers divisible by 10 is 550.

Putting It All Together

Now for the grand finale! We have all the pieces of the puzzle. Remember our strategy?

  1. Sum (divisible by 2): 2550
  2. Sum (divisible by 5): 1050
  3. Sum (divisible by 10): 550

We need to add the sums from steps 1 and 2, and then subtract the sum from step 3:

Total Sum = Sum (divisible by 2) + Sum (divisible by 5) - Sum (divisible by 10)

Total Sum = 2550 + 1050 - 550

Total Sum = 3600 - 550

Total Sum = 3050

The Answer!

Drumroll, please! The sum of the integers from 1 to 100 that are divisible by 2 or 5 is 3050! How cool is that?

Why This Works: The Principle of Inclusion-Exclusion

You might be wondering why we had to subtract the sum of the multiples of 10. This is a classic concept in mathematics called the Principle of Inclusion-Exclusion. It's a fancy name, but the idea is pretty straightforward.

Imagine you have two overlapping circles. One circle represents numbers divisible by 2, and the other represents numbers divisible by 5. The overlapping area represents numbers divisible by both 2 and 5 (which are the multiples of 10).

If we just added the numbers in each circle, we'd be counting the numbers in the overlap twice. To get the correct total, we need to:

  1. Include everything in the first circle.
  2. Include everything in the second circle.
  3. Exclude the overlap (because we counted it twice).

That's exactly what we did with our sums! We included the multiples of 2, included the multiples of 5, and then excluded the multiples of 10. The Principle of Inclusion-Exclusion helps us avoid double-counting in many different situations.

Real-World Applications

Okay, so you might be thinking,