Finding Two-Digit Numbers Equal To 4 Times The Sum Of Their Digits

Hey guys! Today, we're diving into a fun math problem that involves finding two-digit numbers with a cool property. We want to figure out how many two-digit natural numbers exist where the number itself is exactly four times the sum of its digits. Sounds intriguing, right? Let's break it down step by step and get to the bottom of this! We will explore the concept of two-digit numbers, understand how to represent them algebraically, and then apply this knowledge to solve our specific problem. Let’s get started and see how many such numbers we can find!

Understanding the Problem

So, the core of the problem revolves around two-digit numbers. To really nail this, let's get a solid handle on what makes a number a two-digit number and how we can play with their digits. A two-digit number is any whole number between 10 and 99, inclusive. Each of these numbers has two digits: a tens digit and a ones digit. For example, in the number 42, 4 is the tens digit, and 2 is the ones digit. We can express any two-digit number using a bit of algebra, which will come in super handy later on. If we call the tens digit 'a' and the ones digit 'b', then the number can be written as 10a + b. Think about it: 42 is the same as (10 * 4) + 2. This representation is key because it lets us turn our word problem into a mathematical equation. The problem states that we're looking for numbers where the number itself is four times the sum of its digits. So, if our number is 10a + b, and the sum of its digits is a + b, we need to find situations where 10a + b = 4(a + b). This equation is our ticket to solving the puzzle. We can manipulate it, try out different values, and see what fits the bill. It's like detective work with numbers, and I'm excited to see what we uncover!

Setting Up the Equation

Now, let’s transform our word problem into a clear and workable equation. This is where the magic of algebra really shines! As we discussed, a two-digit number can be represented as 10a + b, where 'a' is the tens digit and 'b' is the ones digit. The problem tells us that the number is equal to four times the sum of its digits. So, the sum of the digits is simply a + b. Now, we can put it all together. The equation we need to solve is: 10a + b = 4(a + b). This equation is the heart of our solution. It captures the relationship described in the problem in a concise mathematical form. To solve it, we'll need to use some algebraic techniques to simplify and rearrange the equation. Our goal is to isolate the variables and see if we can find any patterns or restrictions on the values of 'a' and 'b'. Once we have a simplified equation, it will be much easier to test different possibilities and find the two-digit numbers that fit the criteria. So, let's roll up our sleeves and get ready to manipulate this equation into something more manageable!

Solving the Equation

Alright, let's dive into solving the equation we set up: 10a + b = 4(a + b). This is where we put on our algebraic hats and get to work! First, we need to simplify the equation by distributing the 4 on the right side: 10a + b = 4a + 4b. Now, our goal is to get all the 'a' terms on one side and all the 'b' terms on the other. Let's subtract 4a from both sides: 10a - 4a + b = 4a - 4a + 4b, which simplifies to 6a + b = 4b. Next, let's subtract 'b' from both sides: 6a + b - b = 4b - b, which gives us 6a = 3b. We're getting closer! Now, we can simplify further by dividing both sides by 3: (6a) / 3 = (3b) / 3, which simplifies to 2a = b. This is a fantastic result! It tells us that the ones digit ('b') is exactly twice the tens digit ('a'). This gives us a clear relationship between the digits, making it much easier to find the solutions. Now, all we need to do is think about what possible values 'a' can take and then calculate 'b' accordingly. Remember, 'a' and 'b' are digits, so they can only be whole numbers from 0 to 9. However, since 'a' is the tens digit of a two-digit number, it can't be 0. So, let's explore the possibilities!

Finding Possible Values

Okay, we've arrived at the crucial equation: 2a = b. This equation is our golden ticket because it tells us exactly how the tens digit ('a') and the ones digit ('b') are related. Remember, 'a' and 'b' are digits, meaning they must be whole numbers between 0 and 9. But since 'a' is the tens digit of a two-digit number, it can't be 0. So, let's systematically explore the possible values for 'a' and see what 'b' turns out to be. If a = 1, then b = 2 * 1 = 2. This gives us the number 12. If a = 2, then b = 2 * 2 = 4. This gives us the number 24. If a = 3, then b = 2 * 3 = 6. This gives us the number 36. If a = 4, then b = 2 * 4 = 8. This gives us the number 48. Now, let's think about what happens if we go higher. If a = 5, then b = 2 * 5 = 10. But wait a minute! 'b' has to be a single digit, so it can't be 10. This means we've hit our limit for possible values of 'a'. So, we've found four potential numbers: 12, 24, 36, and 48. But before we declare victory, we need to make sure these numbers actually satisfy the original condition of the problem.

Verifying the Solutions

We've got a list of potential solutions: 12, 24, 36, and 48. But it's super important to double-check that these numbers actually fit the problem's condition. Remember, the condition is that the two-digit number must be equal to four times the sum of its digits. Let's test each number one by one: For 12: The sum of the digits is 1 + 2 = 3. Four times the sum is 4 * 3 = 12. Bingo! 12 works. For 24: The sum of the digits is 2 + 4 = 6. Four times the sum is 4 * 6 = 24. Another winner! For 36: The sum of the digits is 3 + 6 = 9. Four times the sum is 4 * 9 = 36. We're on a roll! For 48: The sum of the digits is 4 + 8 = 12. Four times the sum is 4 * 12 = 48. Awesome! All four numbers satisfy the condition. This means we've successfully found all the two-digit numbers that are equal to four times the sum of their digits. It's always a good idea to verify your solutions, especially in math problems. This helps prevent errors and gives you confidence that your answer is correct.

Final Answer

Alright guys, we've cracked the code! We started with a tricky problem about two-digit numbers and used a combination of algebraic representation, equation solving, and systematic testing to find the answer. The question was: How many two-digit natural numbers are there where the number is equal to four times the sum of its digits? We went through the process step-by-step, setting up an equation, simplifying it, finding possible values, and then verifying our solutions. And after all that hard work, we found that there are exactly four such numbers: 12, 24, 36, and 48. So, the final answer is 4. This was a fantastic journey through the world of numbers and equations! We used our mathematical skills to solve a real problem, and that's something to be proud of. Remember, math isn't just about formulas and calculations; it's about problem-solving and critical thinking. And we nailed it! Keep practicing, keep exploring, and keep having fun with math!