Divisibility By 3 Or 4: Numbers In The Set {1, 2, ..., 100}

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Hey guys! Today, we're diving into a fun math problem: figuring out how many numbers within the set A = {1, 2, ..., 100} are divisible by either 3 or 4. This is a classic problem that combines basic divisibility rules with a bit of set theory. So, let's break it down step by step and make sure we understand the solution clearly. We'll make it super easy, even if math isn't your favorite thing. Let's get started!

Understanding the Problem

Before we jump into calculations, let's make sure we really understand what the problem is asking. We have a set of numbers, A, which includes all the integers from 1 to 100. Our mission is to find out how many of these numbers can be divided evenly by 3, by 4, or by both. Think of it like this: we're looking for the multiples of 3 and the multiples of 4 within this set. But, and this is important, we need to be careful not to count any numbers twice! If a number is divisible by both 3 and 4, we only want to count it once. This is where the concept of the least common multiple and the principle of inclusion-exclusion will come in handy. So, stay with me, and we'll tackle this together.

Why This Matters

You might be wondering, "Why are we even doing this?" Well, these types of problems aren't just about getting the right answer; they help us develop important problem-solving skills. Understanding divisibility and how numbers relate to each other is crucial in many areas, from computer science to everyday financial calculations. Plus, it's a great way to sharpen your logical thinking! So, even though it might seem like a simple counting problem, it's actually a building block for more advanced math and real-world applications. Let's see how we can approach this systematically.

Finding Multiples of 3

Okay, first up, let’s figure out how many numbers in our set are divisible by 3. To do this, we need to find the largest multiple of 3 that is less than or equal to 100. Think of it like this: we're asking, "How many times does 3 fit into 100?" We can easily find this by dividing 100 by 3. So, 100 ÷ 3 = 33.333.... This means that 3 fits into 100 thirty-three whole times. Therefore, the largest multiple of 3 in our set is 3 × 33 = 99. This tells us that there are 33 numbers in the set A that are divisible by 3. These numbers are 3, 6, 9, all the way up to 99. See? Not so scary!

The Arithmetic Behind It

To make it super clear, we're essentially using the concept of integer division here. When we divide 100 by 3, we're interested in the whole number part of the result, which is 33. This is because each of these 33 numbers (3 × 1, 3 × 2, 3 × 3, ..., 3 × 33) is a multiple of 3 and falls within our set of 1 to 100. Understanding this simple division trick makes counting multiples much easier. Now, let's move on to the multiples of 4!

Finding Multiples of 4

Now, let's tackle the multiples of 4 within our set A. We'll use the same method we used for multiples of 3. We need to find the largest multiple of 4 that's less than or equal to 100. So, let's divide 100 by 4. Guess what? 100 ÷ 4 = 25. This is a whole number, which makes our job even easier! It means that there are 25 numbers in the set A that are divisible by 4. These numbers are 4, 8, 12, and so on, up to 100. We're making good progress, guys!

Why Simple Division Works

The beauty of this approach lies in its simplicity. By dividing the upper limit of our set (100) by the number we're checking divisibility for (4 in this case), we directly get the count of multiples. Each whole number from 1 to 25, when multiplied by 4, gives us a number that's both a multiple of 4 and within our set. This consistent method helps us avoid missing any numbers and keeps our calculations straightforward. But, here comes the tricky part: we need to consider the numbers that are multiples of both 3 and 4. This is where we avoid double-counting!

Avoiding Double Counting: Multiples of Both 3 and 4

Here's where things get a little more interesting. We've counted the multiples of 3 and the multiples of 4 separately. But some numbers are multiples of both 3 and 4! If we simply add the number of multiples of 3 and 4, we'll be counting these numbers twice. We don't want that! So, how do we find these numbers? Well, a number that is divisible by both 3 and 4 must be divisible by their least common multiple (LCM). The LCM of 3 and 4 is 12. So, we need to find out how many multiples of 12 there are in our set.

Finding the Least Common Multiple (LCM)

The LCM is the smallest number that is a multiple of both numbers. In this case, the LCM of 3 and 4 is 12. You can find this by listing the multiples of each number until you find a common one: Multiples of 3: 3, 6, 9, 12, 15,... Multiples of 4: 4, 8, 12, 16,... The smallest number that appears in both lists is 12. Now that we know the LCM, we can proceed to find out how many multiples of 12 are within our set.

Counting Multiples of 12

Let’s find out how many multiples of 12 are in our set A. We'll use the same division method again. We divide 100 by 12: 100 ÷ 12 = 8.333.... This means that there are 8 whole multiples of 12 within our set. So, the numbers 12, 24, 36, 48, 60, 72, 84, and 96 are divisible by both 3 and 4. We have 8 numbers that we've counted twice, and we need to subtract these from our initial count to get the correct answer. We’re almost there!

The Final Calculation: Inclusion-Exclusion Principle

Okay, guys, we're in the home stretch! We've done all the hard work, and now it's time to put it all together. We know:

  • There are 33 numbers divisible by 3.
  • There are 25 numbers divisible by 4.
  • There are 8 numbers divisible by both 3 and 4.

To find the total number of elements divisible by 3 or 4, we use the principle of inclusion-exclusion. This principle helps us avoid double-counting when we're dealing with overlapping sets. The formula is:

Total = (Multiples of 3) + (Multiples of 4) - (Multiples of both 3 and 4)

So, let's plug in our numbers: Total = 33 + 25 - 8 = 50

Breaking Down the Formula

The inclusion-exclusion principle is super useful in situations like this. Here’s why it works:

  1. We start by adding the number of multiples of 3 and the number of multiples of 4. This gives us an initial count, but it includes the numbers that are multiples of both 3 and 4 twice.
  2. To correct for this double-counting, we subtract the number of multiples of both 3 and 4. This removes the extra count, giving us the accurate total.

And there you have it! There are 50 numbers in the set A that are divisible by either 3 or 4. High five!

Conclusion

Awesome job, everyone! We've successfully navigated this divisibility problem using a combination of basic arithmetic and a clever principle to avoid double-counting. Remember, the key to solving these types of problems is to break them down into smaller, manageable steps. We found the multiples of 3, the multiples of 4, and then we cleverly handled the overlap by using the least common multiple and the inclusion-exclusion principle. Math problems like this might seem a bit daunting at first, but with a systematic approach and a little bit of logical thinking, you can conquer them all!

Why This Matters Beyond the Problem

The skills we used today aren't just for math class; they're valuable in many areas of life. The ability to break down problems, identify overlaps, and think logically are crucial skills in everything from planning a project to making financial decisions. So, keep practicing these concepts, and you'll find that they come in handy more often than you might think. Keep up the great work, and I'll catch you in the next math adventure!