Four-Digit Number Divisible By 2, 5, & 9: How To Find It?
Hey guys! Let's dive into a cool math problem: figuring out a four-digit number that plays nice with 2, 5, and 9 – meaning it can be divided by all of them without leaving any leftovers. This isn't just some random number game; it's a chance to flex our math muscles and understand how divisibility rules work. So, buckle up, and let's crack this numerical puzzle together!
Understanding Divisibility Rules
Before we jump into finding the number, let's quickly refresh our memory on divisibility rules. These rules are super handy shortcuts that tell us if a number can be divided evenly by another number without actually doing the long division. Knowing these rules makes our task way easier and faster.
- Divisibility by 2: A number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, or 8). Think of it like this: if you can split the number into pairs, it’s divisible by 2.
- Divisibility by 5: A number is divisible by 5 if its last digit is either 0 or 5. Easy peasy, right? Just check the last digit!
- Divisibility by 9: This one's a bit trickier but still manageable. A number is divisible by 9 if the sum of its digits is divisible by 9. So, if you add up all the digits and the total can be divided by 9, you’ve got a number that's divisible by 9.
These rules are our secret weapons in solving this problem. By keeping them in mind, we can narrow down our options and find the four-digit number we're looking for.
Applying the Divisibility Rules to Find the Number
Okay, now that we've got our divisibility rules locked and loaded, let's put them to work! Our mission is to find a four-digit number that ticks all three boxes: divisible by 2, 5, and 9. This might sound like a tall order, but by tackling each rule one by one, we can simplify the challenge.
First off, let's consider the rules for 2 and 5. For a number to be divisible by both 2 and 5, it must end in 0. Why? Because that's the only digit that satisfies both rules – it's even (divisible by 2) and it's either 0 or 5 (divisible by 5). So, we know our mystery number looks something like this: _ _ _ 0.
Now, let's bring in the divisibility rule for 9. This rule tells us that the sum of the digits has to be divisible by 9. So, if our number is ABCO, then A + B + C + 0 should be a multiple of 9. Since 0 doesn't change the sum, we really just need A + B + C to be divisible by 9.
Here's where a little bit of trial and error – combined with some smart thinking – comes in. We need to find three digits that add up to a multiple of 9. Remember, we're looking for a four-digit number, so the first digit (A) can't be 0. Let's start thinking about multiples of 9: 9, 18, 27… We don't need to go too high, because the maximum sum we can get from three single-digit numbers is 9 + 9 + 9 = 27.
Let's try aiming for the smallest four-digit number first. If we want a small number, we should start with a small digit for A. Let's try 1. Now we need two more digits (B and C) that, when added to 1, give us a multiple of 9. If we aim for the sum to be 9, then B + C would need to be 8. There are a few combinations that work here, like 1 + 7, 2 + 6, 3 + 5, and 4 + 4.
So, let's try the smallest of these, making B = 1 and C = 7. That gives us the number 1170. Let's check if it works: 1 + 1 + 7 + 0 = 9, which is divisible by 9! And it ends in 0, so it's divisible by 2 and 5. Bingo! We found one!
But hold on, is this the only number? Let's explore a bit more. What if we aimed for the next multiple of 9, which is 18? If A + B + C = 18, and we still start with A = 1, then B + C would need to be 17. The only single-digit combination that adds up to 17 is 8 + 9. So, we could have the number 1890. Let's check: 1 + 8 + 9 + 0 = 18, which is divisible by 9. It ends in 0, so it’s also divisible by 2 and 5. Another one bites the dust!
We could keep going, trying different combinations and aiming for the sum of the digits to be 27, but you get the idea. By methodically applying the divisibility rules and using a bit of trial and error, we can find several four-digit numbers that meet our criteria.
Examples of Four-Digit Numbers Divisible by 2, 5, and 9
To solidify our understanding, let's take a look at a few more examples of four-digit numbers that are divisible by 2, 5, and 9. We've already discovered 1170 and 1890. Now, let's see if we can find a couple more without too much hassle.
Remember, the key is that the number must end in 0 (to be divisible by 2 and 5), and the sum of its digits must be divisible by 9. Let's try working backwards this time. Instead of starting with the first digit, let's think about what sums we can make using different digits that are divisible by 9. We know 9 and 18 work; let's try for 27.
If the sum of the digits is 27, and the last digit is 0, then the first three digits must add up to 27. The highest digits we can use are 9s, so 9 + 9 + 9 = 27. That gives us the number 9990. Cool! Let's check: 9 + 9 + 9 + 0 = 27, which is divisible by 9, and it ends in 0, so it's also divisible by 2 and 5. That was an easy one!
Now, let's aim for something in the middle. Let's say we want the first digit to be a 5. If we want the sum of the digits to be 18 (a multiple of 9), then the remaining two digits (excluding the 0 at the end) need to add up to 13 (since 5 + 13 = 18). What combinations can we use for 13? We could use 4 + 9 or 5 + 8 or 6 + 7. Let's pick 5 and 8. That gives us the number 5580.
Let's make sure it works: 5 + 5 + 8 + 0 = 18, which is divisible by 9. And, of course, it ends in 0, so it's divisible by 2 and 5. Awesome! We've got another one.
So, just to recap, we've found these four-digit numbers that are divisible by 2, 5, and 9:
- 1170
- 1890
- 9990
- 5580
And there are plenty more out there! The fun part is that you can keep playing with different digits and combinations to discover even more numbers that fit the bill. It's like a mathematical treasure hunt!
Conclusion
Alright, guys, we've successfully navigated the world of divisibility and uncovered the secrets of four-digit numbers that play nicely with 2, 5, and 9. By understanding and applying the divisibility rules, we've turned what might have seemed like a daunting task into a fun and engaging puzzle.
We've learned that a number divisible by 2, 5, and 9 must end in 0 and have its digits add up to a multiple of 9. Armed with this knowledge, we've discovered several examples, like 1170, 1890, 9990, and 5580. But remember, this is just the tip of the iceberg! There are countless other numbers out there waiting to be found.
So, the next time you're faced with a math problem that seems tricky, remember the power of breaking it down into smaller steps and using the tools at your disposal – in this case, the divisibility rules. And who knows? You might just discover a whole new world of mathematical wonders!
Keep exploring, keep questioning, and most importantly, keep having fun with numbers! Math isn't just about formulas and equations; it's about problem-solving, logical thinking, and the joy of discovery. Until next time, happy calculating!