Finding ABC: Division By 5 Problem Solved!

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Finding ABC: Division by 5 Problem Solved!

Hey guys! Let's dive into a cool math problem today. We're going to figure out how to find a three-digit number, which we'll call ABC, that has a special property. When you divide this number by 5, you get a quotient (the result of the division) that looks like BC (another number), and a remainder that's equal to A. Sounds intriguing, right? Let's break it down step by step and make sure we understand every little detail. Get ready to put on your thinking caps, because we're about to embark on a mathematical adventure! This problem might seem a bit tricky at first, but don't worry, we'll tackle it together and make sure everyone understands the solution. Remember, the key to solving math problems is to break them down into smaller, more manageable parts. So, let's get started and unravel the mystery of ABC!

Understanding the Problem

Okay, before we jump into solving, let’s make sure we really understand what the question is asking. The core of this problem lies in understanding how division works with remainders. We're told that our mystery number, ABC, when divided by 5, results in a quotient BC and a remainder A. To visualize this, think about it like this: if you have a bunch of objects (ABC of them!) and you divide them into groups of 5, you'll have BC complete groups, and A objects left over. This remainder, A, is super important because it tells us something crucial about the number ABC. Specifically, since it's a remainder after dividing by 5, A can only be a few values: 0, 1, 2, 3, or 4. Got it? This limitation on A is going to be a key piece of the puzzle later on. It's like finding a secret passage in a maze – recognizing this restriction narrows down the possibilities and guides us closer to the solution. So, let's keep this in mind as we move forward. This foundational understanding of remainders is essential for cracking the problem.

Breaking Down ABC and BC

Now, let's dig deeper into what ABC and BC actually mean in mathematical terms. Remember, these aren't just letters; they represent digits in a number. ABC is a three-digit number, which means we can express it as 100A + 10B + C. Think of it like expanding the number based on place values: A is in the hundreds place, B is in the tens place, and C is in the ones place. Similarly, BC is a two-digit number, so we can write it as 10*B + C. Seeing these numbers in their expanded form is super helpful because it allows us to work with the individual digits and their values. For instance, we can clearly see how the value of A in ABC is actually 100 times its digit value. This is a crucial step because it transforms the problem from abstract symbols to concrete mathematical expressions. By understanding the place values, we can manipulate the equation in a way that reveals the relationships between A, B, and C. It's like translating a secret code – once you understand the language, you can unlock the message. So, let's use these expanded forms to rewrite the original division problem as an equation.

Setting Up the Equation

Time to get those algebraic muscles working! We know that when ABC is divided by 5, we get BC as the quotient and A as the remainder. We can write this relationship as an equation: ABC = 5 * BC + A. Remember that this equation is just a mathematical way of stating the division problem. It captures the essence of what's happening when we divide ABC by 5. Now, let's substitute the expanded forms of ABC and BC that we figured out earlier. This gives us: 100A + 10B + C = 5 * (10*B + C) + A. This equation might look a bit intimidating, but don't worry! It's just a combination of the place values and the division relationship. The next step is to simplify this equation. This is where our algebra skills come in handy. We'll need to distribute, combine like terms, and rearrange things to get a clearer picture of the connection between A, B, and C. Think of it like tidying up a messy room – once we organize the equation, the solution will be much easier to find. So, let's get ready to simplify and see what secrets this equation holds.

Simplifying the Equation

Alright, let's roll up our sleeves and simplify the equation we've got: 100A + 10B + C = 5 * (10B + C) + A. The first step is to distribute the 5 on the right side: 100A + 10B + C = 50B + 5C + A. Now, let's gather similar terms together. We want to get all the A's, B's, and C's on the same sides of the equation. Subtract 'A' from both sides: 99A + 10B + C = 50B + 5C. Next, subtract 10B and C from both sides: 99A = 40B + 4C. Look at that! The equation is much simpler now. We've managed to isolate 99A on one side and an expression involving B and C on the other. This simplified equation is a huge step forward because it highlights the relationship between A and the combination of B and C. It's like zooming in on a map – we're getting a clearer view of the key features. Now, we can start using this simplified equation to figure out possible values for A, B, and C. Remember that these are digits, so they can only be numbers from 0 to 9. This limitation is going to be crucial in the next stage of our solution.

Finding Possible Values

Okay, now for the exciting part – let's find out what numbers A, B, and C could be! We've got the simplified equation: 99A = 40B + 4C. The key here is to remember that A, B, and C are single digits (0-9). Let's think about A first. Since 99A is on one side of the equation, the result must be a multiple of 99. Also, 40B + 4C must also be a multiple of 99. Since B and C are single digits, the maximum value of 40B + 4C occurs when B=9 and C=9, giving us 409 + 49 = 360 + 36 = 396. So, 99*A must be less than or equal to 396. This means A can only be 1, 2, 3, or 4. But remember what we learned at the beginning? A is also the remainder when ABC is divided by 5, so A can only be 0, 1, 2, 3, or 4. Combining these two restrictions, we now have a limited set of possibilities for A. This is like narrowing down suspects in a detective story – we're getting closer to the culprit! Now, let's test each possible value of A in our equation and see what we can find out about B and C.

Solving for B and C

Let's put on our detective hats and try out the possible values of A one by one. Remember our equation: 99A = 40B + 4*C.

  • If A = 1: 99 * 1 = 40B + 4C, so 99 = 40B + 4C. Let's divide the entire equation by the greatest common divisor, which is 1. No simplification here, it stays as 99 = 40B + 4C. Now we have to think, what values of B and C (between 0 and 9) would make this equation true? Notice that 40B must end in 0. Thus, 4C would give us the final digit which is 9. However, no single-digit value of C when multiplied by 4 yields a number ending in 9. Therefore, A cannot be 1.
  • If A = 2: 99 * 2 = 40B + 4C, so 198 = 40B + 4C. Let's divide everything by the greatest common divisor, which is 2, giving 99 = 20B + 2C. Now, if we divided further by 2, we'd get a non-integer value on one side, so let's keep it as 99 = 20B + 2C. Similar to the previous case, 20B ends with 0. For the number on the right to end with 9, 2C must end in 9. Again, no digit C multiplied by 2 can result in a number that ends with 9. So, A cannot be 2.
  • If A = 3: 99 * 3 = 40B + 4C, so 297 = 40B + 4C. If we divide by the GCD, which is 1, then the equation remains 297 = 40B + 4C. Let's consider the final digits. For the final digits to match, 4C must end in 7 since 40B always ends in 0. But 4 times a single-digit never ends in 7, therefore A cannot be 3.
  • If A = 4: 99 * 4 = 40B + 4C, so 396 = 40B + 4C. Now let's find the GCD, which is 4. Dividing everything by 4, we get 99 = 10*B + C. Aha! This looks promising. Since B and C are single digits, we can easily see that B = 9 and C = 9 will satisfy this equation. So, A = 4, B = 9, and C = 9 is a solution!

The Solution

We've cracked the code, guys! We found that A = 4, B = 9, and C = 9. This means our mystery number ABC is 499. Let's double-check if this works with the original problem. If we divide 499 by 5, we get a quotient of 99 (BC) and a remainder of 4 (A). It checks out! This feels like that amazing moment in a movie when the puzzle pieces finally fall into place. The whole journey, from understanding the problem to simplifying the equation and finding the values, has led us to this satisfying solution. So, remember the next time you're faced with a challenging math problem, break it down, stay organized, and don't be afraid to explore different possibilities. You've got this!

Conclusion

So, there you have it! We successfully determined the number ABC, which is 499, that satisfies the condition of giving a quotient BC (99) and a remainder A (4) when divided by 5. This problem highlights the beauty of mathematics – how seemingly complex puzzles can be solved with a methodical approach and a bit of algebraic manipulation. Remember, math isn't just about numbers and equations; it's about problem-solving, logical thinking, and the thrill of discovering solutions. I hope you enjoyed this mathematical journey as much as I did! Keep practicing, keep exploring, and most importantly, keep having fun with math! Who knows what other amazing mathematical mysteries we'll uncover together? Until next time, happy problem-solving! This is a great example of how breaking down a problem into smaller, manageable steps can lead to the solution. Remember to always double-check your work and celebrate your successes!