Finding Three 3-Digit Numbers Summing To 1998

Hey guys! Let's dive into a cool math problem today. We're going to figure out how to find three 3-digit numbers – specifically, numbers in the forms aab, aba, and baa – that add up to 1998. Sounds like fun, right? This is a classic type of problem that blends a bit of algebra with number sense, and it's a great way to sharpen those mathematical skills. So, let’s break it down and find out how we can crack this puzzle!

Understanding the Problem

Okay, so first things first, let's make sure we totally get what the problem is asking. We've got three numbers: aab, aba, and baa. Notice how they're structured? The letters a and b represent digits, and they're shuffled around in each number. The main keyword here is understanding place value. Remember that in a three-digit number, the position of a digit matters a lot! For instance, in the number aab, the first a is in the hundreds place, the second a is in the ones place, and b is in the tens place. This means we can express aab as 100a + 10b + a.

Similarly, aba can be written as 100a + 10b + a, and baa is 100b + 10a + a. The problem tells us that when we add these three numbers together, we should get 1998. So, our goal is to figure out what digits a and b need to be to make this equation true. This involves some algebraic manipulation and a bit of logical thinking to narrow down the possibilities. We're not just blindly guessing numbers; we're using math to guide us to the solution. Keep in mind that a and b must be single-digit numbers (0-9), and a cannot be zero because that would make aab not a three-digit number. So, with a solid grasp of the problem, we're ready to move on to the next step: setting up the equation!

Setting Up the Equation

Alright, let's get our hands dirty with some algebra! We've already talked about how to break down the numbers aab, aba, and baa using their place values. Now, we're going to use that knowledge to set up an equation that represents the problem. Remember, we said that aab + aba + baa = 1998. So, let’s rewrite each of these three-digit numbers in terms of a and b based on their place values:

• aab = 100a + 10b + a
• aba = 100a + 10a + b
• baa = 100b + 10a + a

Now, let's add these expressions together. We combine like terms – the a terms with a terms and the b terms with b terms. This gives us:

(100a + 10b + a) + (100a + 10a + b) + (100b + 10a + a) = 1998

Let's simplify this by adding up all the a and b coefficients:

(100a + 100a + 10a + a + 10a + a) + (10b + b + 100b) = 1998

This simplifies to:

222a + 111b = 1998

Now we have a nice, clean equation that we can work with! This equation is the key word to solving the problem. It tells us the relationship between a and b that must be true for the sum of the three numbers to be 1998. We can even simplify this equation further by noticing that both 222 and 111 are divisible by 111. So, let’s divide both sides of the equation by 111 to make the numbers smaller and easier to handle:

(222a + 111b) / 111 = 1998 / 111

This gives us:

2a + b = 18

Great! Now we have a super simple equation: 2a + b = 18. This is where the fun really begins, as we can now start using this equation to figure out the possible values for a and b. Onward to the next step!

Finding Possible Values for a and b

Okay, now we've got the simplified equation 2a + b = 18. This is a much easier equation to work with, and it's going to help us nail down the possible values for a and b. Remember, a and b are digits, which means they can only be whole numbers from 0 to 9. Also, a quick reminder: a can't be 0 because that would make aab a two-digit number, not a three-digit number. So, a must be between 1 and 9, and b can be between 0 and 9.

So, how do we find the right values? Well, we can try plugging in different values for a and see what value of b we get. Let's start by thinking about the range of possible values for a. If a is too big, then 2a will be a large number, and b would have to be negative to make the equation true, which isn't possible since b must be a digit. So, we need to find the maximum value a can be.

If we rearrange our equation, we get b = 18 - 2a. Since the smallest b can be is 0, we can set 18 - 2a greater than or equal to 0: 18 - 2a >= 0. Solving for a, we get 2a <= 18, which means a <= 9. So, the largest a can be is 9.

Now, let's consider the smallest a can be. We already know a can't be 0, so let's start testing values from 1 upwards. A systematic way to find the solutions is to create a little table. We'll put in a value for a, calculate 2a, and then find b by subtracting 2a from 18. Here’s how it looks:

Notice that we’ve added a “Valid?” column. This is because b also has to be a digit between 0 and 9. From our table, we can see that when a is 1, 2, 3, or 4, b is greater than 9, which is not allowed. But when a is 5, 6, 7, 8, or 9, b is a valid digit. So, we have five possible pairs of values for a and b. This keyword analysis is super important. Let's write these pairs down:

• a = 5, b = 8
• a = 6, b = 6
• a = 7, b = 4
• a = 8, b = 2
• a = 9, b = 0

Now that we’ve found these pairs, the final step is to use them to construct our three-digit numbers and check if they add up to 1998. Let's do that next!

Constructing the Numbers and Checking the Solutions

Alright, we're in the home stretch now! We’ve identified five possible pairs of values for a and b: (5, 8), (6, 6), (7, 4), (8, 2), and (9, 0). Now, we need to plug these pairs into our three-digit numbers (aab, aba, and baa) and see if they actually add up to 1998. This is a crucial step to make sure we haven't made any mistakes along the way.

Let's take each pair one by one and construct the numbers:

1. a = 5, b = 8
   • aab = 558
   • aba = 585
   • baa = 855
   • Sum = 558 + 585 + 855 = 1998 (This works!)
2. a = 6, b = 6
   • aab = 666
   • aba = 666
   • baa = 666
   • Sum = 666 + 666 + 666 = 1998 (This works too!)
3. a = 7, b = 4
   • aab = 774
   • aba = 747
   • baa = 477
   • Sum = 774 + 747 + 477 = 1998 (Another one!)
4. a = 8, b = 2
   • aab = 882
   • aba = 828
   • baa = 288
   • Sum = 882 + 828 + 288 = 1998 (Yes!)
5. a = 9, b = 0
   • aab = 990
   • aba = 909
   • baa = 099 (which is just 99)
   • Sum = 990 + 909 + 99 = 1998 (And this one works as well!)

Wow, all five pairs work! That's awesome! So, we have found all the sets of three-digit numbers aab, aba, and baa that add up to 1998. The keyword here is confirmation. We didn't just stop at finding the possible values for a and b; we actually plugged them back into the original problem to make sure they work. This is a great habit to get into when solving math problems – always check your solutions!

Conclusion

So, there you have it, guys! We've successfully found all the sets of three-digit numbers in the forms aab, aba, and baa that sum up to 1998. It was a bit of a journey, but we tackled it step by step. First, we made sure we understood the problem and what it was asking. Then, we set up an equation using place values to represent the numbers in terms of a and b. We simplified the equation to make it easier to work with, and then we found the possible values for a and b. Finally, we plugged those values back into the original problem to confirm that they worked.

This type of problem is a fantastic way to practice your algebra skills, your number sense, and your logical thinking. Remember, math isn't just about memorizing formulas; it's about understanding how numbers and equations work and using that knowledge to solve puzzles. I hope you had fun working through this problem with me! Keep practicing, and you'll become a math whiz in no time!